Bismillah hir Rehman nir Raheem Assalat o Wasalam o Allika Ya RasoolALLAH
Calculus (11
th
Ed. Text Book)
Thomas, Weir, Hass, Giordano
Ch Ch
Published By: Muhammad Hassan Riaz Yousufi
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
11
INFINITE SEQUENCES AND SERIES
You
Chapter
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OVERVIEW While everyone knows how to add together two numbers, or even several, how to add together infinitely many numbers is not so clear. In this chapter we study such questions, the subject of the theory of infinite series. Infinite series sometimes have a finite sum, as in 1 1 1 1 + + + + Á = 1. 2 4 8 16
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This sum is represented geometrically by the areas of the repeatedly halved unit square shown here. The areas of the small rectangles add together to give the area of the unit square, which they fill. Adding together more and more terms gets us closer and closer to the total.
1/8 1/16
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1/2
1/4
Mu
ham
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Other infinite series do not have a finite sum, as with 1 + 2 + 3 + 4 + 5 + Á.
The sum of the first few terms gets larger and larger as we add more and more terms. Taking enough terms makes these sums larger than any prechosen constant. With some infinite series, such as the harmonic series 1 +
1 1 1 1 1 + + + + + Á 5 2 3 4 6
it is not obvious whether a finite sum exists. It is unclear whether adding more and more terms gets us closer to some sum, or gives sums that grow without bound. As we develop the theory of infinite sequences and series, an important application gives a method of representing a differentiable function ƒ(x) as an infinite sum of powers of x. With this method we can extend our knowledge of how to evaluate, differentiate, and integrate polynomials to a class of functions much more general than polynomials. We also investigate a method of representing a function as an infinite sum of sine and cosine functions. This method will yield a powerful tool to study functions.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
i
Sequences A sequence is a list of numbers
HISTORICAL ESSAY
a1, a2 , a3 , Á , an , Á
Sequences and Series
747
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11.1
Sequences
You
in a given order. Each of a1, a2, a3 and so on represents a number. These are the terms of the sequence. For example the sequence 2, 4, 6, 8, 10, 12, Á , 2n, Á
has first term a1 = 2, second term a2 = 4 and nth term an = 2n. The integer n is called the index of an , and indicates where an occurs in the list. We can think of the sequence
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a1, a2 , a3 , Á , an , Á
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as a function that sends 1 to a1 , 2 to a2 , 3 to a3 , and in general sends the positive integer n to the nth term an . This leads to the formal definition of a sequence.
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DEFINITION Infinite Sequence An infinite sequence of numbers is a function whose domain is the set of positive integers.
The function associated to the sequence 2, 4, 6, 8, 10, 12, Á , 2n, Á
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sends 1 to a1 = 2, 2 to a2 = 4, and so on. The general behavior of this sequence is described by the formula an = 2n.
12, 14, 16, 18, 20, 22 Á
is described by the formula an = 10 + 2n. It can also be described by the simpler formula bn = 2n, where the index n starts at 6 and increases. To allow such simpler formulas, we let the first index of the sequence be any integer. In the sequence above, 5an6 starts with a1 while 5bn6 starts with b6 . Order is important. The sequence 1, 2, 3, 4 Á is not the same as the sequence 2, 1, 3, 4 Á . Sequences can be described by writing rules that specify their terms, such as
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We can equally well make the domain the integers larger than a given number n0 , and we allow sequences of this type also. The sequence
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an = 2n, 1 bn = s 1dn + 1 n , cn =
n  1 n ,
dn = s 1dn + 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
5an6 =
E 21, 22, 23, Á , 2n, Á F
i
or by listing terms,
1 1 1 1 5bn6 = e 1,  , ,  , Á , s 1dn + 1 n , Á f 2 3 4
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n  1 1 2 3 4 5cn6 = e 0, , , , , Á , n , Á f 2 3 4 5
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748
5dn6 = 51, 1, 1, 1, 1, 1, Á , s 1dn + 1, Á 6. We also sometimes write 5an6 =
E 2n F n = 1. . q
nR
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Figure 11.1 shows two ways to represent sequences graphically. The first marks the first few points from a1, a2 , a3 , Á , an , Á on the real axis. The second method shows the graph of the function defining the sequence. The function is defined only on integer inputs, and the graph consists of some points in the xyplane, located at s1, a1 d, s2, a2 d, Á , sn, an d, Á . an
a2 a3 a4 a5
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a1 1
0
2
an 兹n
a3 a 2
a1
0
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1
Mu
ham
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a2 a4
a5 a3
1 an n
Diverges
3 2 1
0 an
Converges to 0
1 0
1
2
an a1
0
3
4
5
n
Converges to 0
1
1 an (1) n1 1n
n
1 2 3 4 5
0
n
FIGURE 11.1 Sequences can be represented as points on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis an is its value.
Convergence and Divergence Sometimes the numbers in a sequence approach a single value as the index n increases. This happens in the sequence 1 1 1 1 e 1, , , , Á , n , Á f 2 3 4 whose terms approach 0 as n gets large, and in the sequence
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1 1 2 3 4 e 0, , , , , Á , 1  n , Á f 2 3 4 5
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
Sequences
749
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E 21, 22, 23, Á , 2n, Á F
i
whose terms approach 1. On the other hand, sequences like
have terms that get larger than any number as n increases, and sequences like 51, 1, 1, 1, 1, 1, Á , s 1dn + 1, Á 6
0
a2 a3
aN
a1
L L
an
an
bounce back and forth between 1 and 1, never converging to a single value. The following definition captures the meaning of having a sequence converge to a limiting value. It says that if we go far enough out in the sequence, by taking the index n to be larger then some value N, the difference between an and the limit of the sequence becomes less than any preselected number P 7 0.
You
L
N
n
n
n 7 N
FIGURE 11.2 an : L if y = L is a horizontal asymptote of the sequence of points 5sn, an d6 . In this figure, all the an’s after aN lie within P of L.
Nicole Oresme (ca. 1320–1382)
ƒ an  L ƒ 6 P.
If no such number L exists, we say that 5an6 diverges. If 5an6 converges to L, we write limn: q an = L, or simply an : L, and call L the limit of the sequence (Figure 11.2).
The definition is very similar to the definition of the limit of a function ƒ(x) as x tends to q (limx: q ƒsxd in Section 2.4). We will exploit this connection to calculate limits of sequences.
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HISTORICAL BIOGRAPHY
Q
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1 2 3
DEFINITIONS Converges, Diverges, Limit The sequence 5an6 converges to the number L if to every positive number P there corresponds an integer N such that for all n,
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L (N, aN)
0
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L (n, an)
L
EXAMPLE 1
Applying the Definition
Show that
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1 (a) lim n = 0 n: q
(b) lim k = k n: q
sany constant kd
Mu
ham
Solution
(a) Let P 7 0 be given. We must show that there exists an integer N such that for all n, n 7 N
Q
1 ` n  0 ` 6 P.
This implication will hold if s1>nd 6 P or n 7 1>P. If N is any integer greater than 1>P, the implication will hold for all n 7 N. This proves that limn: q s1>nd = 0. (b) Let P 7 0 be given. We must show that there exists an integer N such that for all n, n 7 N
Q
ƒ k  k ƒ 6 P.
Since k  k = 0, we can use any positive integer for N and the implication will hold. This proves that limn: q k = k for any constant k.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
EXAMPLE 2
A Divergent Sequence
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Show that the sequence 51, 1, 1, 1, 1, 1, Á , s 1dn + 1, Á 6 diverges.
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Suppose the sequence converges to some number L. By choosing P = 1>2 in the definition of the limit, all terms an of the sequence with index n larger than some N must lie within P = 1>2 of L. Since the number 1 appears repeatedly as every other term of the sequence, we must have that the number 1 lies within the distance P = 1>2 of L. It follows that ƒ L  1 ƒ 6 1>2, or equivalently, 1>2 6 L 6 3>2. Likewise, the number 1 appears repeatedly in the sequence with arbitrarily high index. So we must also have that ƒ L  s 1d ƒ 6 1>2, or equivalently, 3>2 6 L 6 1>2. But the number L cannot lie in both of the intervals (1> 2, 3> 2) and s 3>2, 1>2d because they have no overlap. Therefore, no such limit L exists and so the sequence diverges. Note that the same argument works for any positive number P smaller than 1, not just 1> 2.
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Solution
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The sequence {1n} also diverges, but for a different reason. As n increases, its terms become larger than any fixed number. We describe the behavior of this sequence by writing lim 2n = q . n: q
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In writing infinity as the limit of a sequence, we are not saying that the differences between the terms an and q become small as n increases. Nor are we asserting that there is some number infinity that the sequence approaches. We are merely using a notation that captures the idea that an eventually gets and stays larger than any fixed number as n gets large.
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DEFINITION Diverges to Infinity The sequence 5an6 diverges to infinity if for every number M there is an integer N such that for all n larger than N, an 7 M. If this condition holds we write lim an = q
n: q
or
an : q .
Mu
ham
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Similarly if for every number m there is an integer N such that for all n 7 N we have an 6 m, then we say 5an6 diverges to negative infinity and write lim an =  q
n: q
or
an :  q .
A sequence may diverge without diverging to infinity or negative infinity. We saw this in Example 2, and the sequences 51, 2, 3, 4, 5, 6, 7, 8, Á 6 and 51, 0, 2, 0, 3, 0, Á 6 are also examples of such divergence.
Calculating Limits of Sequences If we always had to use the formal definition of the limit of a sequence, calculating with P’s and N’s, then computing limits of sequences would be a formidable task. Fortunately we can derive a few basic examples, and then use these to quickly analyze the limits of many more sequences. We will need to understand how to combine and compare sequences. Since sequences are functions with domain restricted to the positive integers, it is not too surprising that the theorems on limits of functions given in Chapter 2 have versions for sequences.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
Sequences
751
5.
Quotient Rule:
limn: q san + bn d = A + B limn: q san  bn d = A  B limn: q san # bn d = A # B limn: q sk # bn d = k # B sAny number kd an A limn: q = if B Z 0 B bn
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Sum Rule: Difference Rule: Product Rule: Constant Multiple Rule:
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1. 2. 3. 4.
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THEOREM 1 Let 5an6 and 5bn6 be sequences of real numbers and let A and B be real numbers. The following rules hold if limn: q an = A and limn: q bn = B.
The proof is similar to that of Theorem 1 of Section 2.2, and is omitted.
EXAMPLE 3
Applying Theorem 1
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By combining Theorem 1 with the limits of Example 1, we have: 1 1 (a) lim a n b = 1 # lim n = 1 # 0 = 0 n: q n: q n: q
(c)
n  1 1 1 lim a1  n b = lim 1  lim n = 1  0 = 1 n b = n: q n: q n: q
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(b) lim a
Constant Multiple Rule and Example 1a
lim
n: q
5 1 = 5 # lim n n: q n2
#
1 lim n = 5 # 0 # 0 = 0
n: q
s4>n 6 d  7 0  7 4  7n 6 = lim = = 7. 1 + 0 n: q n 6 + 3 n: q 1 + s3>n 6 d
Product Rule
Sum and Quotient Rules
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(d) lim
Difference Rule and Example 1a
Mu
ham
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Be cautious in applying Theorem 1. It does not say, for example, that each of the sequences 5an6 and 5bn6 have limits if their sum 5an + bn6 has a limit. For instance, 5an6 = 51, 2, 3, Á 6 and 5bn6 = 51, 2, 3, Á 6 both diverge, but their sum 5an + bn6 = 50, 0, 0, Á 6 clearly converges to 0. One consequence of Theorem 1 is that every nonzero multiple of a divergent sequence 5an6 diverges. For suppose, to the contrary, that 5can6 converges for some number c Z 0. Then, by taking k = 1>c in the Constant Multiple Rule in Theorem 1, we see that the sequence 1 e c # can f = 5an6
converges. Thus, 5can6 cannot converge unless 5an6 also converges. If 5an6 does not converge, then 5can6 does not converge. The next theorem is the sequence version of the Sandwich Theorem in Section 2.2. You are asked to prove the theorem in Exercise 95.
THEOREM 2 The Sandwich Theorem for Sequences Let 5an6, 5bn6, and 5cn6 be sequences of real numbers. If an … bn … cn holds for all n beyond some index N, and if limn: q an = limn: q cn = L, then limn: q bn = L also.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 752
Chapter 11: Infinite Sequences and Series
EXAMPLE 4
Applying the Sandwich Theorem
Since 1>n : 0, we know that cos n n :0
1 :0 2n 1 (c) s 1dn n : 0
cos n 1 1  n … n … n;
because
0 …
because
1 1 1  n … s 1dn n … n .
1 1 … n; 2n
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(b)
because
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(a)
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An immediate consequence of Theorem 2 is that, if ƒ bn ƒ … cn and cn : 0, then bn : 0 because cn … bn … cn . We use this fact in the next example.
The application of Theorems 1 and 2 is broadened by a theorem stating that applying a continuous function to a convergent sequence produces a convergent sequence. We state the theorem without proof (Exercise 96). y
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y 2x (1, 2)
THEOREM 3 The Continuous Function Theorem for Sequences Let 5an6 be a sequence of real numbers. If an : L and if ƒ is a function that is continuous at L and defined at all an , then ƒsan d : ƒsLd.
1 , 21/2 2 1 , 21/3 3
1
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2
EXAMPLE 5
Applying Theorem 3
Show that 2sn + 1d>n : 1.
We know that sn + 1d>n : 1. Taking ƒsxd = 1x and L = 1 in Theorem 3 gives 1sn + 1d>n : 11 = 1.
0
1 3
1 2
1
x
EXAMPLE 6
The Sequence 521>n6
The sequence 51>n6 converges to 0. By taking an = 1>n, ƒsxd = 2x , and L = 0 in Theorem 3, we see that 21>n = ƒs1>nd : ƒsLd = 20 = 1. The sequence 521>n6 converges to 1 (Figure 11.3).
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FIGURE 11.3 As n : q , 1>n : 0 and 21>n : 20 (Example 6).
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Solution
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Using l’Hôpital’s Rule The next theorem enables us to use l’Hôpital’s Rule to find the limits of some sequences. It formalizes the connection between limn: q an and limx: q ƒsxd.
THEOREM 4 Suppose that ƒ(x) is a function defined for all x Ú n0 and that 5an6 is a sequence of real numbers such that an = ƒsnd for n Ú n0 . Then lim ƒsxd = L
x: q
Q
lim an = L.
n: q
Proof Suppose that limx: q ƒsxd = L. Then for each positive number P there is a number M such that for all x, x 7 M
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Q
ƒ ƒsxd  L ƒ 6 P.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
Sequences
753
EXAMPLE 7
an = ƒsnd
and
ƒ an  L ƒ = ƒ ƒsnd  L ƒ 6 P.
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Q
n 7 N
i
Let N be an integer greater than M and greater than or equal to n0 . Then
Applying L’Hôpital’s Rule
lim
n: q
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Show that ln n n = 0.
The function sln xd>x is defined for all x Ú 1 and agrees with the given sequence at positive integers. Therefore, by Theorem 5, limn: q sln nd>n will equal limx: q sln xd>x if the latter exists. A single application of l’Hôpital’s Rule shows that
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Solution
1>x 0 ln x = lim = = 0. x 1 x: q x: q 1
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lim
We conclude that limn: q sln nd>n = 0.
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When we use l’Hôpital’s Rule to find the limit of a sequence, we often treat n as a continuous real variable and differentiate directly with respect to n. This saves us from having to rewrite the formula for an as we did in Example 7.
EXAMPLE 8
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Find
Applying L’Hôpital’s Rule
By l’Hôpital’s Rule (differentiating with respect to n),
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Solution
ham
EXAMPLE 9
Mu
2n . 5n n: q lim
2n 2n # ln 2 = lim 5 n: q 5n n: q q = . lim
Applying L’Hôpital’s Rule to Determine Convergence
Does the sequence whose nth term is an = a
n + 1 b n  1
n
converge? If so, find limn: q an . q The limit leads to the indeterminate form 1 . We can apply l’Hôpital’s Rule if # we first change the form to q 0 by taking the natural logarithm of an :
Solution
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ln an = ln a
n + 1 b n  1
= n ln a
n
n + 1 b. n  1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 754
Chapter 11: Infinite Sequences and Series
= lim
ln a
n: q
= lim
n + 1 b n  1
n + 1 b n  1 1>n
2>sn 2  1d
1>n 2 2n 2 = 2. = lim 2 n: q n  1 n: q
q #0
0 0
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n: q
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lim ln an = lim n ln a
n: q
i
Then,
l’Hôpital’s Rule
The sequence 5an6 converges to e 2 .
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Since ln an : 2 and ƒsxd = e x is continuous, Theorem 4 tells us that an = e ln an : e 2 .
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Commonly Occurring Limits
The next theorem gives some limits that arise frequently.
1.
3. 4.
n: q
ln n n = 0 n
lim 2n = 1
n: q
lim x 1>n = 1
5.
ham
e n (rounded)
n!
Mu
n 1 5 10 20
3 148 22,026 4.9 * 10 8
1 120 3,628,800 2.4 * 10 18
6.
sx 7 0d
n: q
lim x n = 0
s ƒ x ƒ 6 1d
n: q
n
x lim a1 + n b = e x n: q
ma
Factorial Notation The notation n! (“n factorial”) means the product 1 # 2 # 3 Á n of the integers from 1 to n. Notice that sn + 1d! = sn + 1d # n! . Thus, 4! = 1 # 2 # 3 # 4 = 24 and 5! = 1 # 2 # 3 # 4 # 5 = 5 # 4! = 120 . We define 0! to be 1. Factorials grow even faster than exponentials, as the table suggests.
lim
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2.
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THEOREM 5 The following six sequences converge to the limits listed below:
xn = 0 n: q n! lim
sany xd
sany xd
In Formulas (3) through (6), x remains fixed as n : q .
Proof The first limit was computed in Example 7. The next two can be proved by taking logarithms and applying Theorem 4 (Exercises 93 and 94). The remaining proofs are given in Appendix 3.
EXAMPLE 10 (a)
Applying Theorem 5
ln sn 2 d 2 ln n = n :2#0 = 0 n n
(b) 2n 2 = n 2>n = sn 1/n d2 : s1d2 = 1 n
(c) 23n = 3 sn d : 1 # 1 = 1 1>n
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1/n
Formula 1 Formula 2 Formula 3 with x = 3 and Formula 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
n
1 (d) a b : 0 2 n
100 n :0 n!
Formula 5 with x = 2
Formula 6 with x = 100
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(f)
n
n  2 2 2 n b = a1 + n b : e
1 2
755
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(e) a
Formula 4 with x = 
Sequences
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11.1
Recursive Definitions
So far, we have calculated each an directly from the value of n. But sequences are often defined recursively by giving
EXAMPLE 11
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The value(s) of the initial term or terms, and A rule, called a recursion formula, for calculating any later term from terms that precede it.
Sequences Constructed Recursively
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1. 2.
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(a) The statements a1 = 1 and an = an  1 + 1 define the sequence 1, 2, 3, Á , n, Á of positive integers. With a1 = 1, we have a2 = a1 + 1 = 2, a3 = a2 + 1 = 3, and so on. (b) The statements a1 = 1 and an = n # an  1 define the sequence 1, 2, 6, 24, Á , n!, Á of factorials. With a1 = 1, we have a2 = 2 # a1 = 2, a3 = 3 # a2 = 6, a4 = 4 # a3 = 24, and so on. (c) The statements a1 = 1, a2 = 1, and an + 1 = an + an  1 define the sequence 1, 1, 2, 3, 5, Á of Fibonacci numbers. With a1 = 1 and a2 = 1, we have a3 = 1 + 1 = 2, a4 = 2 + 1 = 3, a5 = 3 + 2 = 5, and so on. (d) As we can see by applying Newton’s method, the statements x0 = 1 and xn + 1 = xn  [ssin xn  x n2 d>scos xn  2xn d] define a sequence that converges to a solution of the equation sin x  x 2 = 0.
Bounded Nondecreasing Sequences
Mu
ham
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The terms of a general sequence can bounce around, sometimes getting larger, sometimes smaller. An important special kind of sequence is one for which each term is at least as large as its predecessor.
DEFINITION Nondecreasing Sequence A sequence 5an6 with the property that an … an + 1 for all n is called a nondecreasing sequence.
EXAMPLE 12
Nondecreasing Sequences
(a) The sequence 1, 2, 3, Á , n, Á of natural numbers n 1 2 3 ,Á (b) The sequence , , , Á , 2 3 4 n + 1 (c) The constant sequence 536
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 756
Chapter 11: Infinite Sequences and Series
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There are two kinds of nondecreasing sequences—those whose terms increase beyond any finite bound and those whose terms do not.
EXAMPLE 13
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DEFINITIONS Bounded, Upper Bound, Least Upper Bound A sequence 5an6 is bounded from above if there exists a number M such that an … M for all n. The number M is an upper bound for 5an6. If M is an upper bound for 5an6 but no number less than M is an upper bound for 5an6, then M is the least upper bound for 5an6.
Applying the Definition for Boundedness
(a) The sequence 1, 2, 3, Á , n, Á has no upper bound.
n 1 2 3 (b) The sequence , , , Á , , Á is bounded above by M = 1. 2 3 4 n + 1
nR
y
No number less than 1 is an upper bound for the sequence, so 1 is the least upper bound (Exercise 113).
yL
L
(8, a 8 )
(5, a 5) (1, a 1)
0
1
2
3
4
5
6
7
8
x
(a) an … L for all values of n and (b) given any P 7 0, there exists at least one integer N for which aN 7 L  P. The fact that 5an6 is nondecreasing tells us further that
Mu
ham
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FIGURE 11.4 If the terms of a nondecreasing sequence have an upper bound M, they have a limit L … M .
A nondecreasing sequence that is bounded from above always has a least upper bound. This is the completeness property of the real numbers, discussed in Appendix 4. We will prove that if L is the least upper bound then the sequence converges to L. Suppose we plot the points s1, a1 d, s2, a2 d, Á , sn, an d, Á in the xyplane. If M is an upper bound of the sequence, all these points will lie on or below the line y = M (Figure 11.4). The line y = L is the lowest such line. None of the points sn, an d lies above y = L, but some do lie above any lower line y = L  P, if P is a positive number. The sequence converges to L because
ass a
M
dH
yM
an Ú aN 7 L  P
for all n Ú N.
Thus, all the numbers an beyond the Nth number lie within P of L. This is precisely the condition for L to be the limit of the sequence {an}. The facts for nondecreasing sequences are summarized in the following theorem. A similar result holds for nonincreasing sequences (Exercise 107).
THEOREM 6 The Nondecreasing Sequence Theorem A nondecreasing sequence of real numbers converges if and only if it is bounded from above. If a nondecreasing sequence converges, it converges to its least upper bound.
Theorem 6 implies that a nondecreasing sequence converges when it is bounded from above. It diverges to infinity if it is not bounded from above.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1 Sequences
Finding Terms of a Sequence
23. an = 2 + s0.1dn
1  n n2 s 1dn + 1 3. an = 2n  1
2. an =
an + 1 = an + s1>2n d an + 1 = an>sn + 1d
an + 1 = s 1dn + 1an>2
an + 1 = nan>sn + 1d
11. a1 = a2 = 1, 12. a1 = 2,
29. an =
n 2  2n + 1 n  1
an + 2 = an + 1 + an
a2 = 1,
an + 2 = an + 1>an
Finding a Sequence’s Formula
15. The sequence 1, 4, 9, 16, 25, Á
1’s with alternating signs
Squares of the positive integers; with alternating signs
Reciprocals of squares of the positive integers, with alternating signs
ma
1 1 1 1 16. The sequence 1,  , ,  , , Á 4 9 16 25
1’s with alternating signs
Squares of the positive integers diminished by 1
18. The sequence 3, 2, 1, 0, 1, Á
Integers beginning with 3
ham
17. The sequence 0, 3, 8, 15, 24, Á
19. The sequence 1, 5, 9, 13, 17, Á
Every other odd positive integer Every other even positive integer
21. The sequence 1, 0, 1, 0, 1, Á
Alternating 1’s and 0’s
Mu
20. The sequence 2, 6, 10, 14, 18, Á
22. The sequence 0, 1, 1, 2, 2, 3, 3, 4, Á
Each positive integer repeated
Finding Limits
Which of the sequences 5an6 in Exercises 23–84 converge, and which diverge? Find the limit of each convergent sequence.
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28. an =
n + 3 n 2 + 5n + 6
30. an =
1  n3 70  4n 2
1 32. an = s 1dn a1  n b
33. an = a
34. an = a2 
1 1 b a3 + n b 2n 2 n
n + 1 1 b a1  n b 2n
35. an =
s 1dn + 1 2n  1
1 36. an = a b 2
37. an =
2n An + 1
38. an =
39. an = sin a
dH
14. The sequence 1, 1, 1, 1, 1, Á
2n + 1
1  32n
31. an = 1 + s 1dn
In Exercises 13–22, find a formula for the nth term of the sequence. 13. The sequence 1, 1, 1, 1, 1, Á
n + s 1dn n
iaz
2n  1 6. an = 2n
26. an =
nR
10. a1 = 2,
1  5n 4 n 4 + 8n 3
ass a
9. a1 = 2,
27. an =
1 n!
Each of Exercises 7–12 gives the first term or two of a sequence along with a recursion formula for the remaining terms. Write out the first ten terms of the sequence. 8. a1 = 1,
1  2n 1 + 2n
4. an = 2 + s 1dn
2n 5. an = n + 1 2
7. a1 = 1,
25. an =
24. an =
You
Each of Exercises 1–6 gives a formula for the nth term an of a sequence 5an6 . Find the values of a1, a2 , a3 , and a4 . 1. an =
suf
i
EXERCISES 11.1
757
p 1 + nb 2
1 s0.9dn
40. an = np cos snpd
41. an =
sin n n
42. an =
sin2 n 2n
43. an =
n 2n
44. an =
3n n3
46. an =
ln n ln 2n
45. a n =
ln sn + 1d 2n
47. an = 81>n
48. an = s0.03d1>n
7 49. an = a1 + n b n
51. an = 210n 3 53. an = a n b 55. an =
n! nn
n
n
52. an = 2n 2 54. an = sn + 4d1>sn + 4d 56. an = ln n  ln sn + 1d
57. an = 24nn 59. an =
1 50. an = a1  n b
1>n
ln n n 1>n n
n
n
58. an = 232n + 1
(Hint: Compare with 1> n.)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 758
Chapter 11: Infinite Sequences and Series
1 64. an = ln a1 + n b n b 66. an = a n + 1
n
65. an = a
n
xn b 67. an = a 2n + 1 n
68. an = a1 
1 b n2 s10>11dn 70. an = s9/10dn + s11/12dn 72. an = sinh sln nd
i
,
x 7 0
Do the sequences converge? If so, to what value? In each case, begin by identifying the function ƒ that generates the sequence.
3n # 6n 2n # n!
a. x0 = 1,
71. an = tanh n n2 1 sin n 2n  1
1 1 77. an = a b + 3 22n sln nd200 79. an = n
tan1 n
2
sln nd5
81. an = n  2n  n 2
2n 1
2n 2  1  2n 2 + n L1
n
1 x dx
84. an =
Theory and Examples
1 p dx, L1 x
p 7 1
dH
n
ƒsxn d . ƒ¿sxn d
xn + 1 = xn 
x n2  2 xn 1 = + xn 2xn 2 tan xn  1
b. x0 = 1,
xn + 1 = xn 
c. x0 = 1,
xn + 1 = xn  1
sec2 xn
88. a. Suppose that ƒ(x) is differentiable for all x in [0, 1] and that ƒs0d = 0 . Define the sequence 5an6 by the rule an = nƒs1>nd . Show that lim n: q an = ƒ¿s0d . Use the result in part (a) to find the limits of the following sequences 5an6.
1 b. an = n tan1 n
c. an = nse 1>n  1d
ass a
2n
1 83. an = n
xn + 1 = xn 
1>n
75. an = tan1 n
78. an = 2n + n
82. an =
87. Newton’s method The following sequences come from the recursion formula for Newton’s method,
n
n
1 n
80. an =
69. an =
73. an =
1 74. an = n a1  cos n b 76. an =
3n + 1 b 3n  1
suf
n! 2n # 3n
You
62. an =
b. The fractions rn = xn>yn approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that r n2  2 = ;s1>yn d2 and that yn is not less than n.)
n! 10 6n 1>sln nd 1 63. an = a n b 61. an =
iaz
s 4dn n!
nR
60. an =
85. The first term of a sequence is x1 = 1 . Each succeeding term is the sum of all those that come before it:
2 d. an = n ln a1 + n b
89. Pythagorean triples A triple of positive integers a, b, and c is called a Pythagorean triple if a 2 + b 2 = c 2 . Let a be an odd positive integer and let b = j
a2 k 2
and
c = l
a2 m 2
be, respectively, the integer floor and ceiling for a 2>2 .
ma
xn + 1 = x1 + x2 + Á + xn .
Write out enough early terms of the sequence to deduce a general formula for xn that holds for n Ú 2 . 86. A sequence of rational numbers is described as follows:
ham
a a + 2b 1 3 7 17 , , , ,Á, , ,Á. 1 2 5 12 b a + b
Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let xn and yn be, respectively, the numerator and the denominator of the nth fraction rn = xn>yn .
Mu
a. Verify that x 12  2y 12 = 1, x 22  2y 22 = +1 and, more generally, that if a 2  2b 2 = 1 or +1 , then sa + 2bd2  2sa + bd2 = +1
or
a 2 2
a 2 2
a
a. Show that a 2 + b 2 = c 2 . (Hint: Let a = 2n + 1 and express b and c in terms of n.)
1 ,
respectively.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1 Sequences
106. The first term of a sequence is x1 = cos s1d . The next terms are x2 = x1 or cos (2), whichever is larger; and x3 = x2 or cos (3), whichever is larger (farther to the right). In general,
a2 m 2
90. The nth root of n! a. Show that limn: q s2npd1>s2nd = 1 and hence, using Stirling’s approximation (Chapter 8, Additional Exercise 50a), that n n 2n! L e T
for large values of n .
b. Test the approximation in part (a) for n = 40, 50, 60, Á , as far as your calculator will allow.
91. a. Assuming that limn: q s1>n c d = 0 if c is any positive constant, show that ln n = 0 nc
b. Prove that limn: q s1>n c d = 0 if c is any positive constant. (Hint: If P = 0.001 and c = 0.04 , how large should N be to ensure that ƒ 1>n c  0 ƒ 6 P if n 7 N ?)
dH
92. The zipper theorem Prove the “zipper theorem” for sequences: If 5an6 and 5bn6 both converge to L, then the sequence a1, b1, a2 , b2 , Á , an , bn , Á n
93. Prove that limn: q 2n = 1 .
ma
94. Prove that limn: q x 1>n = 1, sx 7 0d . 95. Prove Theorem 2.
96. Prove Theorem 3.
In Exercises 97–100, determine if the sequence is nondecreasing and if it is bounded from above.
99. an =
3n + 1 n + 1
s2n + 3d! sn + 1d! 2 1 100. an = 2  n  n 2
ham
97. an =
2n3n n!
98. an =
Which of the sequences in Exercises 101–106 converge, and which diverge? Give reasons for your answers.
Mu
1 101. an = 1  n 103. an =
2n  1 2n
108. an =
n + 1 n
109. an =
1 + 22n
2n 4n + 1 + 3n 111. an = 4n
1  4n 110. an = 2n 112. a1 = 1,
if c is any positive constant.
converges to L.
(Continuation of Exercise 107.) Using the conclusion of Exercise 107, determine which of the sequences in Exercises 108–112 converge and which diverge.
an + 1 = 2an  3
113. The sequence 5n>sn + 1d6 has a least upper bound of 1 Show that if M is a number less than 1, then the terms of 5n>sn + 1d6 eventually exceed M. That is, if M 6 1 there is an integer N such that n>sn + 1d 7 M whenever n 7 N . Since n>sn + 1d 6 1 for every n, this proves that 1 is a least upper bound for 5n>sn + 1d6 .
ass a
lim
n: q
107. Nonincreasing sequences A sequence of numbers 5an6 in which an Ú an + 1 for every n is called a nonincreasing sequence. A sequence 5an6 is bounded from below if there is a number M with M … an for every n. Such a number M is called a lower bound for the sequence. Deduce from Theorem 6 that a nonincreasing sequence that is bounded from below converges and that a nonincreasing sequence that is not bounded from below diverges.
You
l
xn + 1 = max 5xn , cos sn + 1d6 .
.
iaz
a: q
a2 k 2
nR
lim
j
suf
i
b. By direct calculation, or by appealing to the figure here, find
759
1 102. an = n  n 104. an =
2n  1 3n
n + 1 105. an = ss 1d + 1d a n b n
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114. Uniqueness of least upper bounds Show that if M1 and M2 are least upper bounds for the sequence 5an6 , then M1 = M2 . That is, a sequence cannot have two different least upper bounds. 115. Is it true that a sequence 5an6 of positive numbers must converge if it is bounded from above? Give reasons for your answer.
116. Prove that if 5an6 is a convergent sequence, then to every positive number P there corresponds an integer N such that for all m and n, m 7 N
and
n 7 N
Q
ƒ am  an ƒ 6 P .
117. Uniqueness of limits Prove that limits of sequences are unique. That is, show that if L1 and L2 are numbers such that an : L1 and an : L2 , then L1 = L2 . 118. Limits and subsequences If the terms of one sequence appear in another sequence in their given order, we call the first sequence a subsequence of the second. Prove that if two subsequences of a sequence 5an6 have different limits L1 Z L2 , then 5an6 diverges. 119. For a sequence 5an6 the terms of even index are denoted by a2k and the terms of odd index by a2k + 1 . Prove that if a2k : L and a2k + 1 : L , then an : L . 120. Prove that a sequence 5an6 converges to 0 if and only if the sequence of absolute values 5ƒ an ƒ6 converges to 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 760
Calculator Explorations of Limits
i
b. Graph ƒsxd = 7.25s0.94dx and use Trace to find where the graph crosses the line y = 3.5 .
122. ƒ 2n  1 ƒ 6 10 3 124. 2n>n! 6 10 7
125. Sequences generated by Newton’s method Newton’s method, applied to a differentiable function ƒ(x), begins with a starting value x0 and constructs from it a sequence of numbers 5xn6 that under favorable circumstances converges to a zero of ƒ. The recursion formula for the sequence is xn + 1 = xn 
ƒsxn d . ƒ¿sxn d
a. Show that the recursion formula for ƒsxd = x 2  a, a 7 0 , can be written as xn + 1 = sxn + a>xn d>2 . b. Starting with x0 = 1 and a = 3 , calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.
COMPUTER EXPLORATIONS
Use a CAS to perform the following steps for the sequences in Exercises 129–140. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit L? b. If the sequence converges, find an integer N such that ƒ an  L ƒ … 0.01 for n Ú N . How far in the sequence do you have to get for the terms to lie within 0.0001 of L?
dH
127. A recursive definition of P/ 2 If you start with x1 = 1 and define the subsequent terms of 5xn6 by the rule xn = xn  1 + cos xn  1 , you generate a sequence that converges rapidly to p>2 . a. Try it. b. Use the accompanying figure to explain why the convergence is so rapid. y
cos xn 1
ma
1
xn 1
xn 1
x
1
ham
0
Mu
128. According to a frontpage article in the December 15, 1992, issue of the Wall Street Journal, Ford Motor Company used about 7 14 hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in 1980. The Japanese needed only about 3 12 hours. Ford’s improvement since 1980 represents an average decrease of 6% per year. If that rate continues, then n years from 1992 Ford will use about n
Sn = 7.25s0.94d
hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3 12 hours per vehicle,
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0.5 130. an = a1 + n b
n
129. an = 2n
an + 1 = an +
132. a1 = 1,
an + 1 = an + s 2dn 1 134. an = n sin n
133. an = sin n 135. an =
sin n n
ln n 136. an = n
137. an = s0.9999dn 139. an =
8n n!
n
1 5n
131. a1 = 1,
ass a
126. (Continuation of Exercise 125.) Repeat part (b) of Exercise 125 with a = 2 in place of a = 3 .
You
n
121. ƒ 20.5  1 ƒ 6 10 3 123. s0.9dn 6 10 3
iaz
n
T
suf
In Exercises 121–124, experiment with a calculator to find a value of N that will make the inequality hold for all n 7 N . Assuming that the inequality is the one from the formal definition of the limit of a sequence, what sequence is being considered in each case and what is its limit?
how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence 5Sn6 that is less than or equal to 3.5.
nR
T
Chapter 11: Infinite Sequences and Series
138. an = 1234561>n 140. an =
n 41 19n
141. Compound interest, deposits, and withdrawals If you invest an amount of money A0 at a fixed annual interest rate r compounded m times per year, and if the constant amount b is added to the account at the end of each compounding period (or taken from the account if b 6 0), then the amount you have after n + 1 compounding periods is r An + 1 = a1 + m bAn + b .
(1)
a. If A0 = 1000, r = 0.02015, m = 12 , and b = 50 , calculate and plot the first 100 points sn, An d . How much money is in your account at the end of 5 years? Does 5An6 converge? Is 5An6 bounded?
b. Repeat part (a) with A0 = 5000, r = 0.0589, m = 12 , and b = 50 . c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5% annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have 20,000 dollars? What if the CD earns 6.25%?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1 Sequences
142. Logistic difference equation The recursive relation an + 1 = rans1  an d is called the logistic difference equation, and when the initial value a0 is given the equation defines the logistic sequence 5an6 . Throughout this exercise we choose a0 in the interval 0 6 a0 6 1 , say a0 = 0.3 .
i
e. The situation gets even more interesting. There is actually an increasing sequence of bifurcation values 3 6 3.45 6 3.54 6 Á 6 cn 6 cn + 1 Á such that for cn 6 r 6 cn + 1 the logistic sequence 5an6 eventually oscillates steadily among 2n values, called an attracting 2ncycle. Moreover, the bifurcation sequence 5cn6 is bounded above by 3.57 (so it converges). If you choose a value of r 6 3.57 you will observe a 2ncycle of some sort. Choose r = 3.5695 and plot 300 points. f. Let us see what happens when r 7 3.57 . Choose r = 3.65 and calculate and plot the first 300 terms of 5an6 . Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of an + 1 from previous values of the sequence.
ass a
a. Choose r = 3>4 . Calculate and plot the points sn, an d for the first 100 terms in the sequence. Does it appear to converge? What do you guess is the limit? Does the limit seem to depend on your choice of a0 ?
suf
For the values of the constants A0 , r, m, and b given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation (2) satisfy the recursion formula in Equation (1).
You
(2)
iaz
k
mb mb r Ak = a1 + m b aA0 + r b  r .
d. Next explore the behavior for r values near the endpoints of each of the intervals 3.45 6 r 6 3.54 and 3.54 6 r 6 3.55 . Plot the first 200 terms of the sequences. Describe in your own words the behavior observed in your plots for each interval. Among how many values does the sequence appear to oscillate for each interval? The values r = 3.45 and r = 3.54 (rounded to two decimal places) are also called bifurcation values because the behavior of the sequence changes as r crosses over those values.
nR
d. It can be shown that for any k Ú 0 , the sequence defined recursively by Equation (1) satisfies the relation
b. Choose several values of r in the interval 1 6 r 6 3 and repeat the procedures in part (a). Be sure to choose some points near the endpoints of the interval. Describe the behavior of the sequences you observe in your plots.
g. For r = 3.65 choose two starting values of a0 that are close together, say, a0 = 0.3 and a0 = 0.301 . Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for r = 3.75 . Can you see how the plots look different depending on your choice of a0 ? We say that the logistic sequence is sensitive to the initial condition a0 .
Mu
ham
ma
dH
c. Now examine the behavior of the sequence for values of r near the endpoints of the interval 3 6 r 6 3.45 . The transition value r = 3 is called a bifurcation value and the new behavior of the sequence in the interval is called an attracting 2cycle. Explain why this reasonably describes the behavior.
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761
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o Y iaz 11.2 Infinite Series
11.2
Infinite Series
R n ssa
761
An infinite series is the sum of an infinite sequence of numbers
a H d a m m a h u M
a1 + a2 + a3 + Á + an + Á
The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infinitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at what we get by summing the first n terms of the sequence and stopping. The sum of the first n terms
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sn = a1 + a2 + a3 + Á + an
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 762
Chapter 11: Infinite Sequences and Series
1 1 1 1 + + + + Á 2 4 8 16
You
1 +
suf
i
is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n gets larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in Section 11.1. For example, to assign meaning to an expression like
iaz
We add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.
Suggestive expression for partial sum
s1 = 1
First:
nR
Partial sum
2  1 1 2 2 1 2 4 o
1 s2 = 1 + 2 1 1 s3 = 1 + + 2 4 o 1 1 1 sn = 1 + + + Á + n  1 2 4 2
Second:
ass a
Third: o
dH
nth:
2 
1 2n  1
Value 1 3 2 7 4 o n 2  1 2n  1
Indeed there is a pattern. The partial sums form a sequence whose nth term is sn = 2 
1 . 2n  1
ma
This sequence of partial sums converges to 2 because limn: q s1>2n d = 0. We say 1 1 1 + + Á + n  1 + Á is 2.” 2 4 2
Is the sum of any finite number of terms in this series equal to 2? No. Can we actually add an infinite number of terms one by one? No. But we can still define their sum by defining it to be the limit of the sequence of partial sums as n : q , in this case 2 (Figure 11.5). Our knowledge of sequences and limits enables us to break away from the confines of finite sums.
1
1/4
0
1
1/2
Mu
ham
“the sum of the infinite series 1 +
1/8
2
FIGURE 11.5 As the lengths 1, 12 , 14 , 18 , Á are added one by one, the sum approaches 2.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2
763
i
DEFINITIONS Infinite Series, nth Term, Partial Sum, Converges, Sum Given a sequence of numbers 5an6, an expression of the form a1 + a2 + a3 + Á + an + Á
suf
HISTORICAL BIOGRAPHY
Infinite Series
Blaise Pascal (1623–1662)
s1 = a1 s2 = a1 + a2 o
You
is an infinite series. The number an is the nth term of the series. The sequence 5sn6 defined by
n
iaz
sn = a1 + a2 + Á + an = a ak k=1
o
nR
is the sequence of partial sums of the series, the number sn being the nth partial sum. If the sequence of partial sums converges to a limit L, we say that the series converges and that its sum is L. In this case, we also write q
a1 + a2 + Á + an + Á = a an = L. n=1
ass a
If the sequence of partial sums of the series does not converge, we say that the series diverges.
dH
When we begin to study a given series a1 + a2 + Á + an + Á , we might not know whether it converges or diverges. In either case, it is convenient to use sigma notation to write the series as q
a an ,
n=1
A useful shorthand when summation from 1 to q is understood
q
a ak ,
or
k=1
a an
ma
Geometric Series
Mu
ham
Geometric series are series of the form q
a + ar + ar 2 + Á + ar n  1 + Á = a ar n  1 n=1
in which a and r are fixed real numbers and a Z 0. The series can also be written as q g n = 0 ar n . The ratio r can be positive, as in 1 +
1 1 1 + + Á + a b 2 4 2
n1
+ Á,
or negative, as in 1 
1 1 1 +  Á + a b 3 9 3
n1
+ Á.
If r = 1, the nth partial sum of the geometric series is sn = a + as1d + as1d2 + Á + as1dn  1 = na,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 764
Chapter 11: Infinite Sequences and Series
as1  r n d , 1  r
sn =
sr Z 1d.
Multiply sn by r. Subtract rsn from sn . Most of the terms on the right cancel. Factor.
You
sn = a + ar + ar 2 + Á + ar n  1 rsn = ar + ar 2 + Á + ar n  1 + ar n sn  rsn = a  ar n sns1  rd = as1  r n d
suf
i
and the series diverges because limn: q sn = ; q , depending on the sign of a. If r = 1, the series diverges because the nth partial sums alternate between a and 0. If ƒ r ƒ Z 1, we can determine the convergence or divergence of the series in the following way:
We can solve for sn if r Z 1 .
iaz
If ƒ r ƒ 6 1, then r n : 0 as n : q (as in Section 11.1) and sn : a>s1  rd. If ƒ r ƒ 7 1, then ƒ r n ƒ : q and the series diverges.
nR
If ƒ r ƒ 6 1, the geometric series a + ar + ar 2 + Á + ar n  1 + Á converges to a>s1  rd: q
a n1 , = a ar 1  r n=1
ƒ r ƒ 6 1.
ass a
If ƒ r ƒ Ú 1, the series diverges.
dH
We have determined when a geometric series converges or diverges, and to what value. Often we can determine that a series converges without knowing the value to which it converges, as we will see in the next several sections. The formula a>s1  rd for the sum of a geometric series applies only when the summation index begins with n = 1 in the exq q pression g n = 1 ar n  1 (or with the index n = 0 if we write the series as g n = 0 ar n).
EXAMPLE 1
Index Starts with n = 1
ma
The geometric series with a = 1>9 and r = 1>3 is
Mu
ham
EXAMPLE 2
1 1 1 1 1 + + + Á = a a b 9 27 81 n=1 9 3 q
n1
=
1>9 1 = . 6 1  s1>3d
Index Starts with n = 0
The series
s 1dn5 5 5 5 = 5  + + Á a 4n 4 16 64 n=0 q
is a geometric series with a = 5 and r = 1>4. It converges to 5 a = = 4. 1  r 1 + s1>4d
EXAMPLE 3
A Bouncing Ball
You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find the total distance the ball travels up and down (Figure 11.6).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2 Solution
765
The total distance is
i
a
Infinite Series
This sum is 2ar>s1  rd. ar
s = 6 ar 3
1 + s2>3d 5>3 = 6a b = 30 m. 1>3 1  s2>3d
Repeating Decimals
iaz
EXAMPLE 4
You
If a = 6 m and r = 2>3, for instance, the distance is ar 2
suf
2ar 1 + r s = a + 2ar + 2ar 2 + 2ar 3 + Á = a + = a . 1 r 1  r (''''''')'''''''*
Express the repeating decimal 5.232323 Á as the ratio of two integers. (a)
23 23 23 + + + Á 100 s100d2 s100d3
nR
Solution
5.232323 Á = 5 +
ass a
= 5 +
2
23 1 1 + a b + Áb a1 + 100 100 100
a = 1, r = 1>100
('''''''')''''''''* 1>s1  0.01d
= 5 +
23 518 23 1 b = 5 + = a 100 0.99 99 99
dH
Unfortunately, formulas like the one for the sum of a convergent geometric series are rare and we usually have to settle for an estimate of a series’ sum (more about this later). The next example, however, is another case in which we can find the sum exactly.
EXAMPLE 5
A Nongeometric but Telescoping Series q
(b)
ma
1 Find the sum of the series a . n = 1 nsn + 1d
Mu
ham
FIGURE 11.6 (a) Example 3 shows how to use a geometric series to calculate the total vertical distance traveled by a bouncing ball if the height of each rebound is reduced by the factor r. (b) A stroboscopic photo of a bouncing ball.
We look for a pattern in the sequence of partial sums that might lead to a formula for sk . The key observation is the partial fraction decomposition
Solution
1 1 1 = n , n + 1 nsn + 1d
so k
k
1 1 1 a nsn + 1d = a a n  n + 1 b n=1 n=1 and sk = a
1 1 1 1 1 1 1 1  b + a  b + a  b + Á + a b. 1 2 2 3 3 4 k k + 1
Removing parentheses and canceling adjacent terms of opposite sign collapses the sum to sk = 1 
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1 . k + 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 766
Chapter 11: Infinite Sequences and Series
i
We now see that sk : 1 as k : q . The series converges, and its sum is 1: q
suf
1 a nsn + 1d = 1. n=1
You
Divergent Series
One reason that a series may fail to converge is that its terms don’t become small.
EXAMPLE 6
Partial Sums Outgrow Any Number
(a) The series q
iaz
2 Á + n2 + Á an = 1 + 4 + 9 +
n=1
q
a
3 n + 1 2 4 Á + n + 1 + Á n = 1 + 2 + 3 + n
ass a
n=1
nR
diverges because the partial sums grow beyond every number L. After n = 1, the partial sum sn = 1 + 4 + 9 + Á + n 2 is greater than n 2 . (b) The series
diverges because the partial sums eventually outgrow every preassigned number. Each term is greater than 1, so the sum of n terms is greater than n.
The nthTerm Test for Divergence
dH
Observe that limn: q an must equal zero if the series g n = 1 an converges. To see why, let S represent the series’ sum and sn = a1 + a2 + Á + an the nth partial sum. When n is large, both sn and sn  1 are close to S, so their difference, an , is close to zero. More formally, an = sn  sn  1
q
:
S  S = 0.
Difference Rule for sequences
ma
This establishes the following theorem.
Mu
ham
Caution q Theorem 7 does not say that g n = 1 an converges if an : 0 . It is possible for a series to diverge when an : 0 .
THEOREM 7 q
If a an converges, then an : 0. n=1
Theorem 7 leads to a test for detecting the kind of divergence that occurred in Example 6.
The nthTerm Test for Divergence q
lim an fails to exist or is different from zero. a an diverges if n: q
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2
Applying the nthTerm Test
q
n=1 q
n + 1 n + 1 n diverges because n : 1
You
(b) a
suf
(a) a n 2 diverges because n 2 : q n=1 q
767
i
EXAMPLE 7
Infinite Series
(c) a s 1dn + 1 diverges because limn: q s 1dn + 1 does not exist n=1 q
n n 1 =  Z 0. (d) a diverges because limn: q 2n + 5 2 n = 1 2n + 5
an : 0 but the Series Diverges
iaz
EXAMPLE 8 The series
1 1 1 1 1 1 1 1 1 + + + + + + Á + n + n + Á + n + Á 2 2 4 4 4 4 2 2 2
nR
1 +
(')'* 2 terms
('''')''''* 4 terms
(''''')'''''* 2n terms
ass a
diverges because the terms are grouped into clusters that add to 1, so the partial sums increase without bound. However, the terms of the series form a sequence that converges to 0. Example 1 of Section 11.3 shows that the harmonic series also behaves in this manner.
Combining Series
dH
Whenever we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series.
ma
THEOREM 8 If gan = A and gbn = B are convergent series, then 1.
Mu
ham
2. 3.
Sum Rule: Difference Rule: Constant Multiple Rule:
gsan + bn d = gan + gbn = A + B gsan  bn d = gan  gbn = A  B sAny number kd. gkan = kgan = kA
Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section 11.1. To prove the Sum Rule for series, let An = a1 + a2 + Á + an ,
Bn = b1 + b2 + Á + bn .
Then the partial sums of gsan + bn d are
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sn = sa1 + b1 d + sa2 + b2 d + Á + san + bn d = sa1 + Á + an d + sb1 + Á + bn d = An + Bn .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 768
Chapter 11: Infinite Sequences and Series
suf
i
Since An : A and Bn : B, we have sn : A + B by the Sum Rule for sequences. The proof of the Difference Rule is similar. To prove the Constant Multiple Rule for series, observe that the partial sums of gkan form the sequence sn = ka1 + ka2 + Á + kan = ksa1 + a2 + Á + an d = kAn ,
1. 2.
You
which converges to kA by the Constant Multiple Rule for sequences. As corollaries of Theorem 8, we have
Every nonzero constant multiple of a divergent series diverges. If gan converges and gbn diverges, then gsan + bn d and gsan  bn d both diverge.
We omit the proofs.
iaz
CAUTION Remember that gsan + bn d can converge when gan and gbn both diverge. For example, gan = 1 + 1 + 1 + Á and gbn = s 1d + s 1d + s 1d + Á diverge, whereas gsan + bn d = 0 + 0 + 0 + Á converges to 0. EXAMPLE 9
nR
Find the sums of the following series.
3n  1  1 1 1 = a a n1  n1 b n1 6 6 n=1 n=1 2 q
q
(a) a
q
q
ass a
1 1 = a n1  a n1 n=1 2 n=1 6 1 1 = 1  s1>2d 1  s1>6d
dH
= 2 =
6 5
q
4 1 a n = 4 a 2n n=0 2 n=0
ma ham Mu
Geometric series with a = 1 and r = 1>2, 1>6
4 5
q
(b)
Difference Rule
= 4a
1 b 1  s1>2d
Constant Multiple Rule
Geometric series with a = 1, r = 1>2
= 8
Adding or Deleting Terms We can add a finite number of terms to a series or delete a finite number of terms without altering the series’ convergence or divergence, although in the case of convergence this will q q usually change the sum. If g n = 1 an converges, then g n = k an converges for any k 7 1 and q
q
Á + ak  1 + a an = a1 + a2 + a an .
n=1
Conversely, if
q g n = k an
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n=k
converges for any k 7 1, then q
q g n = 1 an
converges. Thus,
q
1 1 1 1 1 a 5n = 5 + 25 + 125 + a 5n n=4
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2 Infinite Series
769
1 1 1 1 1 a 5n = a a 5n b  5  25  125 . n=4 n=1 q
suf
q
Richard Dedekind (1831–1916)
You
Reindexing HISTORICAL BIOGRAPHY
i
and
As long as we preserve the order of its terms, we can reindex any series without altering its convergence. To raise the starting value of the index h units, replace the n in the formula for an by n  h: q
q
n=1
Á. a an  h = a1 + a2 + a3 +
n=1+h
iaz
a an =
To lower the starting value of the index h units, replace the n in the formula for an by n + h: q
q
Á. a an + h = a1 + a2 + a3 +
nR
a an =
n=1
n=1h
ass a
It works like a horizontal shift. We saw this in starting a geometric series with the index n = 0 instead of the index n = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions.
EXAMPLE 10
Reindexing a Geometric Series
We can write the geometric series q
dH
1 1 1 Á a 2n  1 = 1 + 2 + 4 + n=1
as
q
1 a n, n=0 2
q
1
a n  5, n=5 2
q
or even
1 a 2n + 4 .
n = 4
Mu
ham
ma
The partial sums remain the same no matter what indexing we choose.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
11.2 Infinite Series
EXERCISES 11.2 Finding nth Partial Sums
1. 2 +
H d a
2 2 2 2 + + + Á + n1 + Á 3 9 27 3
9 9 9 9 Á + + Á + + 2. n + 2 3 100 100 100 100
m m a
1 1 1 1 +  + Á + s 1dn  1 n  1 + Á 2 4 8 2 4. 1  2 + 4  8 + Á + s 1dn  1 2n  1 + Á 3. 1 
h u M
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Y z a i R
5.
1 1 1 1 + # + # + Á + + Á 2#3 3 4 4 5 sn + 1dsn + 2d
6.
5 5 5 5 + # + # + Á + + Á 1#2 2 3 3 4 nsn + 1d
n a s as
In Exercises 1–6, find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.
769
Series with Geometric Terms In Exercises 7–14, write out the first few terms of each series to show how the series starts. Then find the sum of the series. s 1dn 4n n=0 q
7. a
q
1 8. a n n=2 4
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 770
Chapter 11: Infinite Sequences and Series q
5 1 12. a a n  n b 3 n=0 2
41. a s 1dnx n
n
s 1d 1 b 13. a a n + 5n n=0 2
q n=0 q
43. a 3 a
2 14. a a n b 5 n=0 q
n+1
n=0
Telescoping Series Use partial fractions to find the sum of each series in Exercises 15–22. q
q
6 16. a n = 1 s2n  1ds2n + 1d
4 15. a n = 1 s4n  3ds4n + 1d
q
x  1 b 2
suf
q
i
Then express the inequality ƒ r ƒ 6 1 in terms of x and find the values of x for which the inequality holds and the series converges.
n=0 q
5 1 11. a a n + n b 3 n=0 2 q
5 10. a s 1dn n 4
42. a s 1dnx 2n n=0 q
n s 1dn 1 a b 2 3 + sin x n=0
n
44. a
You
q
7 9. a n n=1 4
In Exercises 45–50, find the values of x for which the given geometric series converges. Also, find the sum of the series (as a function of x) for those values of x. q
q
45. a 2nx n
46. a s 1dnx 2n
n=0 q
n=0 q
n
2n + 1 18. a 2 2 n = 1 n sn + 1d
47. a s 1dnsx + 1dn
1 48. a a b sx  3dn 2 n=0
19. a a
1 1 20. a a 1>n  1>sn + 1d b 2 n=1 2
49. a sinn x
50. a sln xdn
q
1
n=1
2n

1 2n + 1
b
q
n=0 q
q
22. a stan1 snd  tan1 sn + 1dd
51. 0.23 = 0.23 23 23 Á
n=1
52. 0.234 = 0.234 234 234 Á
ass a
Convergence or Divergence
53. 0.7 = 0.7777 Á
Which series in Exercises 23–40 converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. 22
q
25. a s 1d
n+1
n=1 q
n
3 2n
26. a s 1d
n+1
n=1 q n=0 q
q
57. 1.24123 = 1.24 123 123 123 Á
n
cos np 5n
ma
n=0 q
1 32. a n , n=0 x
q
ƒxƒ 7 1
1 34. a a1  n b q
n
ham
2n  1 3n n=0
nn 36. a n = 1 n!
n b 37. a ln a n + 1 n=1
n b 38. a ln a 2n + 1 n=1
n
Mu
q
Theory and Examples 59. The series in Exercise 5 can also be written as q
1 a sn + 1dsn + 2d n=1
n=0
q
1 a sn + 3dsn + 4d . n = 1
and
Write it as a sum beginning with (a) n = 2 , (b) n = 0 , (c) n = 5 . 60. The series in Exercise 6 can also be written as
n=1 q
n! 35. a n n = 0 1000
e 39. a a p b
58. 3.142857 = 3.142857 142857 Á
n=1 q
2 31. a n n = 1 10
where d is a digit
55. 0.06 = 0.06666 Á
1 30. a ln n
29. a e 2n
q
54. 0.d = 0.dddd Á ,
56. 1.414 = 1.414 414 414 Á
q
28. a
n=0
q
q
n=0
27. a cos np
33. a
24. a A 22 B n
dH
n=0
b
n=0
Express each of the numbers in Exercises 51–58 as the ratio of two integers.
q
1
n=0
nR
q
q
q
Repeating Decimals
1 1 b 21. a a ln sn + 2d ln sn + 1d n=1
23. a a
iaz
40n 17. a 2 2 n = 1 s2n  1d s2n + 1d
q
q
5 a n = 1 nsn + 1d
q
and
5 . a n = 0 sn + 1dsn + 2d
q
Write it as a sum beginning with (a) n = 1 , (b) n = 3 , (c) n = 20 .
q
61. Make up an infinite series of nonzero terms whose sum is
e np 40. a ne n=0 p
Geometric Series
In each of the geometric series in Exercises 41–44, write out the first few terms of the series to find a and r, and find the sum of the series.
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a. 1
b. 3
c. 0.
62. (Continuation of Exercise 61.) Can you make an infinite series of nonzero terms that converges to any number you want? Explain.
63. Show by example that gsan>bn d may diverge even though gan and gbn converge and no bn equals 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 771
11.2 Infinite Series
1/8
65. Show by example that gsan>bn d may converge to something other than A> B even when A = gan, B = gbn Z 0 , and no bn equals 0.
67. What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer.
a. a = 2
that converges to the number
1/2
77. Helga von Koch’s snowflake curve Helga von Koch’s snowflake is a curve of infinite length that encloses a region of finite area. To see why this is so, suppose the curve is generated by starting with an equilateral triangle whose sides have length 1.
iaz
69. Make up a geometric series gar 5 if
n1
You
1/4
66. If gan converges and an 7 0 for all n, can anything be said about gs1>an d ? Give reasons for your answer.
68. If gan converges and gbn diverges, can anything be said about their termbyterm sum gsan + bn d ? Give reasons for your answer.
suf
i
64. Find convergent geometric series A = gan and B = gbn that illustrate the fact that gan bn may converge without being equal to AB.
b. a = 13>2 .
a. Find the length Ln of the nth curve Cn and show that limn: q Ln = q .
1 + e b + e 2b + e 3b + Á = 9 .
nR
70. Find the value of b for which
b. Find the area An of the region enclosed by Cn and calculate limn: q An .
71. For what values of r does the infinite series 1 + 2r + r 2 + 2r 3 + r 4 + 2r 5 + r 6 + Á
ass a
converge? Find the sum of the series when it converges.
72. Show that the error sL  sn d obtained by replacing a convergent geometric series with one of its partial sums sn is ar n>s1  rd .
Curve 1 Curve 2
dH
73. A ball is dropped from a height of 4 m. Each time it strikes the pavement after falling from a height of h meters it rebounds to a height of 0.75h meters. Find the total distance the ball travels up and down.
74. (Continuation of Exercise 73.) Find the total number of seconds the ball in Exercise 73 is traveling. (Hint: The formula s = 4.9t 2 gives t = 2s>4.9 .)
Mu
ham
ma
75. The accompanying figure shows the first five of a sequence of squares. The outermost square has an area of 4 m2 . Each of the other squares is obtained by joining the midpoints of the sides of the squares before it. Find the sum of the areas of all the squares.
76. The accompanying figure shows the first three rows and part of the fourth row of a sequence of rows of semicircles. There are 2n semicircles in the nth row, each of radius 1>2n . Find the sum of the areas of all the semicircles.
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Curve 3
Curve 4
78. The accompanying figure provides an informal proof that q g n = 1 s1>n 2 d is less than 2. Explain what is going on. (Source: “Convergence with Pictures” by P. J. Rippon, American Mathematical Monthly, Vol. 93, No. 6, 1986, pp. 476–478.)
1 72
1 32 1
1 1 22
1
1 62 1 52
…
1 42
1 2
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1 4
…
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
The Integral Test
i
11.3
suf
772
Given a series gan , we have two questions: Does the series converge? If it converges, what is its sum?
You
1. 2.
nR
iaz
Much of the rest of this chapter is devoted to the first question, and in this section we answer that q question by making a connection to the convergence of the improper integral 11 ƒsxd dx. However, as a practical matter the second question is also important, and we will return to it later. In this section and the next two, we study series that do not have negative terms. The reason for this restriction is that the partial sums of these series form nondecreasing sequences, and nondecreasing sequences that are bounded from above always converge (Theorem 6, Section 11.1). To show that a series of nonnegative terms converges, we need only show that its partial sums are bounded from above. It may at first seem to be a drawback that this approach establishes the fact of convergence without producing the sum of the series in question. Surely it would be better to compute sums of series directly from formulas for their partial sums. But in most cases such formulas are not available, and in their absence we have to turn instead to the twostep procedure of first establishing convergence and then approximating the sum.
ass a
Nondecreasing Partial Sums
Suppose that g n = 1 an is an infinite series with an Ú 0 for all n. Then each partial sum is greater than or equal to its predecessor because sn + 1 = sn + an : s1 … s2 … s3 … Á … sn … sn + 1 … Á . q
dH
Since the partial sums form a nondecreasing sequence, the Nondecreasing Sequence Theorem (Theorem 6, Section 11.1) tells us that the series will converge if and only if the partial sums are bounded from above.
ma
Corollary of Theorem 6 q A series g n = 1 an of nonnegative terms converges if and only if its partial sums are bounded from above.
EXAMPLE 1
ham
HISTORICAL BIOGRAPHY
The Harmonic Series
The series
Mu
Nicole Oresme (1320–1382)
q
1 1 1 Á + 1 + Á a n = 1 + 2 + 3 + n n=1 is called the harmonic series. The harmonic series is divergent, but this doesn’t follow from the nthTerm Test. The nth term 1> n does go to zero, but the series still diverges. The reason it diverges is because there is no upper bound for its partial sums. To see why, group the terms of the series in the following way: 1 +
1 1 1 1 1 1 1 1 1 1 + a + b + a + + + b + a + + Á + b + Á. 5 7 2 3 4 6 8 9 10 16 ('')''* 7
2 4
To Read it Online & Download:
=
1 2
(''''')'''''* 7
4 8
=
1 2
('''''')''''''* 7
8 16
=
1 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.3
The Integral Test
773
You
suf
i
The sum of the first two terms is 1.5. The sum of the next two terms is 1>3 + 1>4, which is greater than 1>4 + 1>4 = 1>2. The sum of the next four terms is 1>5 + 1>6 + 1>7 + 1>8, which is greater than 1>8 + 1>8 + 1>8 + 1>8 = 1>2. The sum of the next eight terms is 1>9 + 1>10 + 1>11 + 1>12 + 1>13 + 1>14 + 1>15 + 1>16, which is greater than 8>16 = 1>2. The sum of the next 16 terms is greater than 16>32 = 1>2, and so on. In general, the sum of 2n terms ending with 1>2n + 1 is greater than 2n>2n + 1 = 1>2. The sequence of partial sums is not bounded from above: If n = 2k , the partial sum sn is greater than k >2. The harmonic series diverges.
The Integral Test
EXAMPLE 2
y
iaz
We introduce the Integral Test with a series that is related to the harmonic series, but whose nth term is 1>n 2 instead of 1> n. Does the following series converge? q
nR
1 1 1 1 Á + 1 + Á a n 2 = 1 + 4 + 9 + 16 + n2
(1, f(1))
n=1
Graph of f(x) 12 x (2, f(2)) 1 32 (3, f(3))
1 42
1 22 0
q 2 11 s1>x d
1
2
3
4 …
1 n2
(n, f(n))
n1 n…
x
ma ham
Mu
Caution The series and integral need not have the same value in the convergent case. As we q noted in Example 2, g n = 1s1>n 2 d = q p2>6 while 11 s1>x 2 d dx = 1 .
1 1 1 1 + 2 + 2 + Á + 2 12 2 3 n = ƒs1d + ƒs2d + ƒs3d + Á + ƒsnd
sn =
dH
FIGURE 11.7 The sum of the areas of the rectangles under the graph of f (x) = 1>x 2 is less than the area under the graph (Example 2).
q
ass a
1 12
We determine the convergence of g n = 1s1>n 2 d by comparing it with dx. To carry out the comparison, we think of the terms of the series as values of the function ƒsxd = 1>x 2 and interpret these values as the areas of rectangles under the curve y = 1>x 2 . As Figure 11.7 shows, Solution
n
6 ƒs1d +
1 dx 2 x L1 q
1 dx x2 6 1 + 1 = 2. 6 1 +
L1
As in Section 8.8, Example 3, q 2 11 s1>x d dx = 1 .
Thus the partial sums of g n = 11>n 2 are bounded from above (by 2) and the series converges. The sum of the series is known to be p2>6 L 1.64493. (See Exercise 16 in Section 11.11.) q
THEOREM 9 The Integral Test Let 5an6 be a sequence of positive terms. Suppose that an = ƒsnd, where ƒ is a continuous, positive, decreasing function of x for all x Ú N (N a positive inteq q ger). Then the series g n = N an and the integral 1N ƒsxd dx both converge or both diverge. Proof We establish the test for the case N = 1. The proof for general N is similar. We start with the assumption that ƒ is a decreasing function with ƒsnd = an for every n. This leads us to observe that the rectangles in Figure 11.8a, which have areas
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 774
Chapter 11: Infinite Sequences and Series
a1, a2 , Á , an , collectively enclose more area than that under the curve y = ƒsxd from x = 1 to x = n + 1. That is,
suf
i
y
n+1
L1
y f (x) a2 an
0
1
n
3
2
n1
x
In Figure 11.8b the rectangles have been faced to the left instead of to the right. If we momentarily disregard the first rectangle, of area a1 , we see that n
(a)
a2 + a3 + Á + an …
y
L1
If we include a1 , we have
2
an n1 n
3
x
Combining these results gives n+1
(b)
L1
n
ƒsxd dx. L1 These inequalities hold for each n, and continue to hold as n : q . q If 11 ƒsxd dx is finite, the righthand inequality shows that gan is finite. If q 11 ƒsxd dx is infinite, the lefthand inequality shows that gan is infinite. Hence the series and the integral are both finite or both infinite. L1
FIGURE 11.8 Subject to the conditions of q the Integral Test, the series g n = 1 an and q the integral 11 ƒsxd dx both converge or both diverge.
ƒsxd dx.
iaz
a3
ƒsxd dx … a1 + a2 + Á + an … a1 +
nR
1
a1 + a2 + Á + an … a1 +
ass a
0
a2
ƒsxd dx.
n
y f (x)
a1
You
a1
ƒsxd dx … a1 + a2 + Á + an .
EXAMPLE 3
The pSeries
Show that the pseries
q
dH
1 1 1 1 Á + 1p + Á a n p = 1p + 2p + 3p + n n=1
( p a real constant) converges if p 7 1, and diverges if p … 1. If p 7 1, then ƒsxd = 1>x p is a positive decreasing function of x. Since
Mu
ham
ma
Solution
q
L1
b
x p + 1 1 x p dx = lim c d p dx = x b: q p + 1 1 L1 q
=
1 1 lim a  1b 1  p b: q b p  1
=
1 1 s0  1d = , 1  p p  1
b p  1 : q as b : q because p  1 7 0.
the series converges by the Integral Test. We emphasize that the sum of the pseries is not 1>s p  1d. The series converges, but we don’t know the value it converges to. If p 6 1, then 1  p 7 0 and q
L1
1 1 dx = lim sb 1  p  1d = q . 1  p b: q xp
The series diverges by the Integral Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.3 The Integral Test
775
1 1 1 + + Á + n + Á. 2 3
suf
1 +
i
If p = 1, we have the (divergent) harmonic series
We have convergence for p 7 1 but divergence for every other value of p.
EXAMPLE 4
iaz
You
The pseries with p = 1 is the harmonic series (Example 1). The pSeries Test shows that the harmonic series is just barely divergent; if we increase p to 1.000000001, for instance, the series converges! The slowness with which the partial sums of the harmonic series approaches infinity is impressive. For instance, it takes about 178,482,301 terms of the harmonic series to move the partial sums beyond 20. It would take your calculator several weeks to compute a sum with this many terms. (See also Exercise 33b.)
A Convergent Series
The series
q
nR
1 a n2 + 1
n=1
converges by the Integral Test. The function ƒsxd = 1>sx 2 + 1d is positive, continuous, and decreasing for x Ú 1, and b 1 dx = lim C arctan x D 1 q b: x + 1 = lim [arctan b  arctan 1]
ass a
q
b: q
=
p p p = . 2 4 4
dH
L1
2
Again we emphasize that p>4 is not the sum of the series. The series converges, but we do not know the value of its sum.
Mu
ham
ma
Convergence of the series in Example 4 can also be verified by comparison with the series g1>n 2 . Comparison tests are studied in the next section.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
u o zY
11.3 The Integral Test
EXERCISES 11.3 Determining Convergence or Divergence
a s as
Which of the series in Exercises 1–30 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) q
1 1. a n n = 1 10 q
q
2. a e n n=1 q
m m a h u M
5 4. a n=1 n + 1 q
H ad
q
1 7. a  n 8 n=1
5. a
3
2n 8 8. a n n=1 q n=1
n 3. a n=1 n + 1 q
6. a
a i R n
q
2
n2n ln n 9. a n n=1 q
10. a
2n 2 13. a n=0 n + 1 n=2 q
q
5n 12. a n n=1 4 + 3
q
2n 15. a n + 1 n=1
2n 11. a n n=1 3
ln n
1 14. a n = 1 2n  1
q
q
n 2n 1 17. a 18. a a1 + n b n = 1 2n A 2n + 1 B n = 2 ln n n=1 q q 1 1 19. a 20. a n n n = 1 sln 2d n = 1 sln 3d q q s1>nd 1 21. a 22. a 2 n = 3 sln nd2ln2 n  1 n = 1 ns1 + ln nd q
16. a
1
q
q
n=2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 776
Chapter 11: Infinite Sequences and Series q
e 25. a 2n n=1 1 + e q
2 26. a n n=1 1 + e
i
a. Show that
n=1 q
n
q
L2
dx xsln xd p
s p a positive constantd
q
1
8 tan n 27. a 2 n=1 1 + n
n 28. a 2 n=1 n + 1
q
converges if and only if p 7 1 .
b. What implications does the fact in part (a) have for the convergence of the series
q
29. a sech n
30. a sech2 n
n=1
n=1
You
n=1 q
39. Logarithmic pseries
1 24. a n tan n
suf
q
1 23. a n sin n
q
1 p ? a n = 2 nsln nd
Theory and Examples 2a 1 b 32. a a n + 1 n=3 n  1
q
q
33. a. Draw illustrations like those in Figures 11.7 and 11.8 to show that the partial sums of the harmonic series satisfy the inequalities L1
1 b. a 1.01 n = 2 nsln nd
q
1 1 Á + 1 x dx … 1 + 2 + n n
… 1 +
q
1 a. a n = 2 nsln nd
L1
1 x dx = 1 + ln n.
dH
T b. There is absolutely no empirical evidence for the divergence of the harmonic series even though we know it diverges. The partial sums just grow too slowly. To see what we mean, suppose you had started with s1 = 1 the day the universe was formed, 13 billion years ago, and added a new term every second. About how large would the partial sum sn be today, assuming a 365day year?
34. Are there any values of x for which g n = 1s1>snxdd converges? Give reasons for your answer. q
q g n = 1 an
q
1 1 c. a d. a 3 n ln sn d nsln nd3 n=2 n=2 41. Euler’s constant Graphs like those in Figure 11.8 suggest that as n increases there is little change in the difference between the sum
ass a
n+1
ln sn + 1d =
q
iaz
a 1 b 31. a a n + 4 n=1 n + 2
Give reasons for your answer.
40. (Continuation of Exercise 39.) Use the result in Exercise 39 to determine which of the following series converge and which diverge. Support your answer in each case.
nR
For what values of a, if any, do the series in Exercises 31 and 32 converge?
1 1 + Á + n 2
1 +
and the integral n
ln n =
L1
1 x dx .
To explore this idea, carry out the following steps. a. By taking ƒsxd = 1>x in the proof of Theorem 9, show that
or
36. (Continuation of Exercise 35.) Is there a “largest” convergent series of positive numbers? Explain.
0 6 ln sn + 1d  ln n … 1 +
ham
ma
35. Is it true that if is a divergent series of positive numbers q then there is also a divergent series g n = 1 bn of positive numbers with bn 6 an for every n? Is there a “smallest” divergent series of positive numbers? Give reasons for your answers.
37. The Cauchy condensation test The Cauchy condensation test says: Let 5an6 be a nonincreasing sequence (an Ú an + 1 for all n) of positive terms that converges to 0. Then gan converges if and only if g2na2n converges. For example, gs1>nd diverges because g2n # s1>2n d = g1 diverges. Show why the test works. 38. Use the Cauchy condensation test from Exercise 37 to show that q
Mu
1 a. a diverges; n = 2 n ln n q
1 b. a p converges if p 7 1 and diverges if p … 1 . n=1 n
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ln sn + 1d … 1 +
1 1 + Á + n … 1 + ln n 2
1 1 + Á + n  ln n … 1 . 2
Thus, the sequence an = 1 +
1 1 + Á + n  ln n 2
is bounded from below and from above. b. Show that 1 6 n + 1 Ln
n+1
1 x dx = ln sn + 1d  ln n ,
and use this result to show that the sequence 5an6 in part (a) is decreasing.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.3 The Integral Test
42. Use the integral test to show that
1 1 1 + + Á + n  ln n : g . 2
q
ae
You
converges.
Mu
ham
ma
dH
ass a
nR
iaz
The number g , whose value is 0.5772 Á , is called Euler’s constant. In contrast to other special numbers like p and e, no other
n2
n=0
i
expression with a simple law of formulation has ever been found for g .
suf
Since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1), the numbers an defined in part (a) converge:
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Comparison Tests
usu
11.4
777
fi
11.4 Comparison Tests
Yo
We have seen how to determine the convergence of geometric series, pseries, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is known.
iaz
THEOREM 10 The Comparison Test Let gan be a series with no negative terms.
(a) gan converges if there is a convergent series gcn with an … cn for all n 7 N, for some integer N.
nR
(b) gan diverges if there is a divergent series of nonnegative terms gdn with an Ú dn for all n 7 N, for some integer N.
Ha ssa
Proof In Part (a), the partial sums of gan are bounded above by HISTORICAL BIOGRAPHY
q
M = a1 + a2 + Á + aN +
Albert of Saxony (ca. 1316–1390)
a cn .
n=N+1
ad
They therefore form a nondecreasing sequence with a limit L … M. In Part (b), the partial sums of gan are not bounded from above. If they were, the partial sums for gdn would be bounded by q
M * = d1 + d2 + Á + dN +
a an
n=N+1
Mu
ha
mm
and gdn would have to converge instead of diverge.
EXAMPLE 1
Applying the Comparison Test
(a) The series q
5 a 5n  1 n=1 diverges because its nth term 5 = 5n  1
1 7 n 1 n 5 is greater than the nth term of the divergent harmonic series.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 778
Chapter 11: Infinite Sequences and Series
i
(b) The series 1 1 1 1 Á a n! = 1 + 1! + 2! + 3! + n=0
suf
q
You
converges because its terms are all positive and less than or equal to the corresponding terms of q
1 1 1 1 + a n = 1 + 1 + + 2 + Á. 2 2 n=0 2 The geometric series on the left converges and we have q
iaz
1 1 1 + a n = 1 + = 3. 1  s1>2d n=0 2
The fact that 3 is an upper bound for the partial sums of g n = 0 s1>n!d does not mean that the series converges to 3. As we will see in Section 11.9, the series converges to e. (c) The series
nR
2 1 1 1 1 1 + + 1 + + + + Á + + Á n 7 3 2 + 21 4 + 22 8 + 23 2 + 2n
ass a
5 +
q
converges. To see this, we ignore the first three terms and compare the remaining terms q with those of the convergent geometric series g n = 0 s1>2n d. The term 1>s2n + 2nd of the truncated sequence is less than the corresponding term 1>2n of the geometric series. We see that term by term we have the comparison, 1
dH 1 +
2 + 21
+
1
4 + 22
+
1
1 1 1 + Á … 1 + + + + Á 2 4 8 8 + 23
So the truncated series and the original series converge by an application of the Comparison Test.
ma
The Limit Comparison Test
Mu
ham
We now introduce a comparison test that is particularly useful for series in which an is a rational function of n.
THEOREM 11 Limit Comparison Test Suppose that an 7 0 and bn 7 0 for all n Ú N (N an integer). an = c 7 0, then gan and gbn both converge or both diverge. n: q bn
1.
If lim
2.
If lim
an = 0 and gbn converges, then gan converges. bn
3.
If lim
an = q and gbn diverges, then gan diverges. bn
n: q
n: q
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.4 Comparison Tests
779
n 7 N Q `
suf
i
Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b). Since c>2 7 0, there exists an integer N such that for all n Limit definition with P = c>2, L = c , and an replaced by an>bn
an c  c` 6 . 2 bn

You
Thus, for n 7 N, an c c 6  c 6 , 2 2 bn
iaz
an 3c c 6 6 , 2 2 bn
3c c a bbn 6 an 6 a bbn . 2 2
Using the Limit Comparison Test
ass a
EXAMPLE 2
nR
If gbn converges, then gs3c>2dbn converges and gan converges by the Direct Comparison Test. If gbn diverges, then gsc>2dbn diverges and gan diverges by the Direct Comparison Test.
Which of the following series converge, and which diverge? q
(a)
q
5 7 9 3 2n + 1 2n + 1 + + + + Á = a = a 2 2 4 9 16 25 sn + 1d n + 2n + 1 n=1 n=1 q
1 1 1 1 1 + + + + Á = a n 7 1 3 15 2  1 n=1
(c)
1 + 3 ln 3 1 + n ln n 1 + 4 ln 4 1 + 2 ln 2 + + + Á = a 2 9 14 21 n=2 n + 5
dH (b)
ma
q
Solution
Mu
ham
(a) Let an = s2n + 1d>sn 2 + 2n + 1d. For large n, we expect an to behave like 2n>n 2 = 2>n since the leading terms dominate for large n, so we let bn = 1>n. Since q
q
n=1
n=1
1 a bn = a n diverges
and an 2n 2 + n = lim 2 = 2, n: q bn n: q n + 2n + 1 lim
gan diverges by Part 1 of the Limit Comparison Test. We could just as well have taken bn = 2>n, but 1> n is simpler.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 780
Chapter 11: Infinite Sequences and Series
q
q
1 a bn = a 2n converges n=1 n=1
You
and
suf
i
(b) Let an = 1>s2n  1d. For large n, we expect an to behave like 1>2n , so we let bn = 1>2n . Since
an 2n = lim n n: q bn n: q 2  1 lim
= lim
n: q
1 1  s1>2n d
iaz
= 1,
nR
gan converges by Part 1 of the Limit Comparison Test. (c) Let an = s1 + n ln nd>sn 2 + 5d. For large n, we expect an to behave like sn ln nd>n 2 = sln nd>n, which is greater than 1> n for n Ú 3, so we take bn = 1>n. Since q
q
n=2
n=2
and
ass a
1 a bn = a n diverges
an n + n 2 ln n = lim n: q bn n: q n2 + 5 = q, lim
dH
gan diverges by Part 3 of the Limit Comparison Test. q
EXAMPLE 3
ln n converge? 3>2 n=1 n
Does a
Because ln n grows more slowly than n c for any positive constant c (Section 11.1, Exercise 91), we would expect to have
Mu
ham
ma
Solution
ln n n 1>4 1 6 3>2 = 5>4 3>2 n n n
for n sufficiently large. Indeed, taking an = sln nd>n 3>2 and bn = 1>n 5>4 , we have lim
n: q
an ln n = lim 1>4 bn n: q n 1>n = lim n: q s1>4dn 3>4
l’Hôpital’s Rule
4 = 0. n 1>4 Since gbn = gs1>n 5>4 d (a pseries with p 7 1) converges, gan converges by Part 2 of the Limit Comparison Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.4 Comparison Tests
Determining Convergence or Divergence
n
q
8. a
n=1 q
q
1 10. a 2 n = 2 sln nd q
13. a
1
2n ln n 1 16. a 2 n = 1 s1 + ln nd n=2 q
q
n=1 q
14. a
n 2n 2  1 n + 2n 22. a 2 n n=1 n 2 n=2 q
6. a
n + 1
40. Prove that if gan is a convergent series of nonnegative terms, then ga n2 converges.
n 2 2n 1 9. a n = 3 ln sln nd n=1 q
1 2n + 2 sln nd2 3
n sln nd2
q
12. a
sln nd3 n
n=1 q
n 3>2 ln sn + 1d 17. a n + 1 n=2 n=1 q
2n 20. a 2 n=1 n + 1
3
1 15. a n = 1 1 + ln n q
1 18. a 2 n = 1 s1 + ln nd q
1  n 21. a n n = 1 n2
q
1 23. a n  1 + 1 n=1 3
q
sin n 2n
n=1 q
3
q
1
19. a
11. a
3. a
q
3n  1 + 1 3n n=1
24. a
q
1 25. a sin n
1 26. a tan n
10n + 1 27. a nsn + 1dsn + 2d n=1
5n 3  3n 28. a 2 n sn  2dsn 2 + 5d n=3
29. a
n=1 q
32. a
n=1 q
q
1
tan n n 1.1 tanh n n2
30. a
n=1 q
33. a
n=1
1
sec n n 1.3
q
31. a
n=1 q
coth n n2 n
2n 34. a 2 n=1 n
1 n
n 2n
ma
q
n=1 q
dH
n=1 q
1 35. a 1 + 2 + 3 + Á + n n=1
You
n + 2n 2n 5. a n = 1 3n  1 n=1 q
COMPUTER EXPLORATION
41. It is not yet known whether the series
iaz
n b 7. a a n = 1 3n + 1 q
3
2. a
3 2 2n + 2 n 1 + cos n 4. a n2 n=1
n=1 q
39. Suppose that an 7 0 and bn 7 0 for n Ú N (N an integer). If lim n: q san>bn d = q and gan converges, can anything be said about g bn ? Give reasons for your answer.
q
1 a 3 2 n = 1 n sin n
converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps.
nR
1
1. a
2
q
a. Define the sequence of partial sums k
1 sk = a 3 2 . n = 1 n sin n
ass a
q
38. If g n = 1 an is a convergent series of nonnegative numbers, can q anything be said about g n = 1san>nd ? Explain. q
Which of the series in Exercises 1–36 converge, and which diverge? Give reasons for your answers. q
suf
i
EXERCISES 11.4
781
What happens when you try to find the limit of sk as k : q ? Does your CAS find a closed form answer for this limit?
b. Plot the first 100 points sk, sk d for the sequence of partial sums. Do they appear to converge? What would you estimate the limit to be? c. Next plot the first 200 points sk, sk d . Discuss the behavior in your own words.
d. Plot the first 400 points sk, sk d . What happens when k = 355 ? Calculate the number 355> 113. Explain from your calculation what happened at k = 355 . For what values of k would you guess this behavior might occur again?
q
1 36. a 2 2 1 + 2 + 3 + Á + n2 n=1
Theory and Examples
You will find an interesting discussion of this series in Chapter 72 of Mazes for the Mind by Clifford A. Pickover, St. Martin’s Press, Inc., New York, 1992.
Mu
ham
37. Prove (a) Part 2 and (b) Part 3 of the Limit Comparison Test.
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
11.5 The Ratio and Root Tests
11.5
The Ratio and Root Tests
z a i R n a
s s a dH
781
The Ratio Test measures the rate of growth (or decline) of a series by examining the ratio an + 1>an . For a geometric series gar n , this rate is a constant ssar n + 1 d>sar n d = rd, and the series converges if and only if its ratio is less than 1 in absolute value. The Ratio Test is a powerful rule extending that result. We prove it on the next page using the Comparison Test.
Mu
a m m a h
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
THEOREM 12 The Ratio Test Let gan be a series with positive terms and suppose that an + 1 = r. n: q an Then
iaz
(a) the series converges if r 6 1, (b) the series diverges if r 7 1 or r is infinite, (c) the test is inconclusive if r = 1.
You
lim
i
Chapter 11: Infinite Sequences and Series
suf
782
Proof
nR
(a) R<1. Let r be a number between r and 1. Then the number P = r  r is positive. Since an + 1 an : r,
aN + 1 6 raN , aN + 2 6 raN + 1 6 r 2aN , aN + 3 6 raN + 2 6 r 3aN , o aN + m 6 raN + m  1 6 r maN .
dH
That is,
ass a
an + 1>an must lie within P of r when n is large enough, say for all n Ú N. In particular an + 1 when n Ú N. an 6 r + P = r,
Mu
ham
ma
These inequalities show that the terms of our series, after the Nth term, approach zero more rapidly than the terms in a geometric series with ratio r 6 1. More precisely, consider the series gcn , where cn = an for n = 1, 2, Á , N and cN + 1 = raN , cN + 2 = r 2aN , Á , cN + m = r maN , Á . Now an … cn for all n, and q
Á + aN  1 + aN + raN + r 2aN + Á a cn = a1 + a2 +
n=1
= a1 + a2 + Á + aN  1 + aN s1 + r + r 2 + Á d. The geometric series 1 + r + r 2 + Á converges because ƒ r ƒ 6 1, so gcn converges. Since an … cn , gan also converges. (b) 1<R ◊ ˆ . From some index M on, an + 1 an 7 1
and
aM 6 aM + 1 6 aM + 2 6 Á .
The terms of the series do not approach zero as n becomes infinite, and the series diverges by the nthTerm Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.5
783
The Ratio and Root Tests
1 a n
q
and
n=1
1 a n2 n=1
suf
q
i
(c) R = 1. The two series
show that some other test for convergence must be used when r = 1. q
1>sn + 1d an + 1 n : 1. = an = n + 1 1>n
You
1 For a n : n=1
2 1>sn + 1d2 an + 1 n = = a b : 12 = 1. an n + 1 1>n 2
q
1 For a 2 : n=1 n
iaz
In both cases, r = 1, yet the first series diverges, whereas the second converges. The Ratio Test is often effective when the terms of a series contain factorials of expressions involving n or expressions raised to a power involving n.
Applying the Ratio Test
nR
EXAMPLE 1
Investigate the convergence of the following series. q
2n + 5 3n n=0
q s2nd! (b) a n = 1 n!n!
ass a
(a) a
q
4nn!n! (c) a n = 1 s2nd!
Solution
(a) For the series g n = 0 s2n + 5d>3n , q
s2n + 1 + 5d>3n + 1 an + 1 1 an = s2n + 5d>3n = 3
#
2n + 1 + 5 1 = 3 2n + 5
1#2 # a 2 + 5 ## 2n n b : 1 + 5 2
3 1
=
2 . 3
dH
The series converges because r = 2>3 is less than 1. This does not mean that 2> 3 is the sum of the series. In fact, n
2n + 5 5 5 21 2 1 a 3n = a a 3 b + a 3n = 1  s2>3d + 1  s1>3d = 2 . n=0 n=0 n=0
ma
q
q
s2n + 2d! s2nd! and , then an + 1 = n!n! sn + 1d!sn + 1d! n!n!s2n + 2ds2n + 1ds2nd! an + 1 an = sn + 1d!sn + 1d!s2nd! =
s2n + 2ds2n + 1d 4n + 2 : 4. = n + 1 sn + 1dsn + 1d
The series diverges because r = 4 is greater than 1. (c) If an = 4nn!n!>s2nd!, then
Mu
ham
(b) If an =
q
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4n + 1sn + 1d!sn + 1d! # s2nd! an + 1 an = s2n + 2ds2n + 1ds2nd! 4nn!n! =
2sn + 1d 4sn + 1dsn + 1d : 1. = 2n + 1 s2n + 2ds2n + 1d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 784
Chapter 11: Infinite Sequences and Series
suf
i
Because the limit is r = 1, we cannot decide from the Ratio Test whether the series converges. When we notice that an + 1>an = s2n + 2d>s2n + 1d, we conclude that an + 1 is always greater than an because s2n + 2d>s2n + 1d is always greater than 1. Therefore, all terms are greater than or equal to a1 = 2, and the nth term does not approach zero as n : q . The series diverges.
You
The Root Test
The convergence tests we have so far for gan work best when the formula for an is relatively simple. But consider the following.
Solution
Let an = e
n>2n, 1>2n,
n odd n even.
Does gan converge?
iaz
EXAMPLE 2
We write out several terms of the series: q
=
nR
3 5 7 1 1 1 1 Á a an = 21 + 22 + 23 + 24 + 25 + 26 + 27 + n=1 3 5 7 1 1 1 1 + + + + + + + Á. 2 4 8 16 32 64 128
ass a
Clearly, this is not a geometric series. The nth term approaches zero as n : q , so we do not know if the series diverges. The Integral Test does not look promising. The Ratio Test produces 1 , 2n
n odd
n + 1 , 2
n even.
dH
an + 1 an = d
As n : q , the ratio is alternately small and large and has no limit. A test that will answer the question (the series converges) is the Root Test.
ma
THEOREM 13 The Root Test Let gan be a series with an Ú 0 for n Ú N, and suppose that n
lim 2an = r.
n: q
Mu
ham
Then
(a) the series converges if r 6 1, (b) the series diverges if r 7 1 or r is infinite, (c) the test is inconclusive if r = 1.
Proof n
n
(a) R<1. Choose an P 7 0 so small that r + P 6 1. Since 2an : r, the terms 2an eventually get closer than P to r. In other words, there exists an index M Ú N such that
To Read it Online & Download:
n
2an 6 r + P
when n Ú M.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.5
The Ratio and Root Tests
785
an 6 sr + Pdn
suf
for n Ú M.
i
Then it is also true that Now, g n = M sr + Pdn , a geometric series with ratio sr + Pd 6 1, converges. By q comparison, g n = M an converges, from which it follows that q
q
q
You
Á + aM  1 + a an = a1 + a an
n=1
n=M
iaz
converges. n (b) 1<R ◊ ˆ . For all indices beyond some integer M, we have 2an 7 1, so that an 7 1 for n 7 M. The terms of the series do not converge to zero. The series diverges by the nthTerm Test. (c) R = 1. The series g n = 1 s1>nd and g n = 1 s1>n 2 d show that the test is not conclusive when r = 1. The first series diverges and the second converges, but in both cases n 2an : 1.
EXAMPLE 3
q
nR
q
Applying the Root Test
Which of the following series converges, and which diverges? q
q
n2 (a) a n n=1 2
1 b (c) a a n=1 1 + n q
ass a
2n (b) a 2 n=1 n
Solution
n
2 2n 2 n2 n n (a) a n converges because n = n n = B2 n=1 2 22
dH
q
q
n 2n n 2 (b) a 2 diverges because = A n2 n=1 n
n
n nB2 A2
2
:
1 6 1. 2
2 2 : 7 1. n 2 1 2 n A B n
n
1 n 1 1 b converges because a b = : 0 6 1. (c) a a 1 + n B 1 + n n=1 1 + n
ma
q
Mu
ham
EXAMPLE 2 Let an = e Solution
n>2n, 1>2n,
Revisited n odd n even.
Does gan converge?
We apply the Root Test, finding that n 2n>2, n 2an = e 1>2,
n odd n even.
Therefore, n
2n 1 n … 2an … . 2 2 n
n
Since 2n : 1 (Section 11.1, Theorem 5), we have limn: q 2an = 1>2 by the Sandwich Theorem. The limit is less than 1, so the series converges by the Root Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 786
Chapter 11: Infinite Sequences and Series
Determining Convergence or Divergence
n
q
n 1. a n n=1 2
2. a n 2e n n=1 q
q
3. a n!e n
n! 4. a n n = 1 10
n 10 5. a n n = 1 10
6. a a
2 + s 1dn 7. a 1.25n n=1
8. a
n=1 q
q
n=1 q
n
1 b 10. a a1 3n n=1 q
n=1 q
n
q
q
q
ln n 15. a n
16. a
n=1 q
sn + 1dsn + 2d n! n=1
n=1 q
n ln n 2n
18. a e nsn 3 d
sn + 3d! 19. a n n = 1 3!n!3
20. a
dH
17. a
n=1 q
n
n2nsn + 1d! 3nn! n=1
q
q
q
n! 21. a n = 1 s2n + 1d!
n! 22. a n n=1 n
q
q
q
2
ham
q
an + 1 =
28. a1 = 1,
an + 1 =
1 , 3
an + 1 =
30. a1 = 3,
an + 1 =
31. a1 = 2,
an + 1 =
q
40. a
sn!dn 2
n sn d nn 42. a n 2 n = 1 s2 d n=1 q
2sn d 1 # 3 # Á # s2n  1d 43. a 4n2nn! n=1 q 1 # 3 # Á # s2n  1d 44. a # # Á # s2nd]s3n + 1d n = 1 [2 4 n=1 q
Theory and Examples 45. Neither the Ratio nor the Root Test helps with pseries. Try them on q
1 a np
n=1
q
3n 26. a 3 n n=1 n 2
Which of the series g n = 1 an defined by the formulas in Exercises 27–38 converge, and which diverge? Give reasons for your answers.
and show that both tests fail to provide information about convergence. 46. Show that neither the Ratio Test nor the Root Test provides information about the convergence of
1 + sin n an n
1 + tan1 n an n 3n  1 a 2n + 5 n n a n + 1 n 2 n an
Mu
29. a1 =
nn
24. a
n! ln n 25. a n = 1 nsn + 2d!
27. a1 = 2,
41. a
n sn>2d n = 2 sln nd
ma
n 23. a n n = 2 sln nd
q
ass a
1 1 14. a a n  2 b n n=1
s3nd! n!sn + 1d!sn + 2d!
q sn!dn 39. a n 2 n = 1 sn d
sln ndn nn n=1
q
1 , an + 1 2 1 n 35. a1 = , an + 1 = 2an 3 1 36. a1 = , an + 1 = san dn + 1 2 2nn!n! 37. an = s2nd! 34. a1 =
Which of the series in Exercises 39–44 converge, and which diverge? Give reasons for your answers.
12. a
1 1 13. a a n  2 b n n=1
an + 1
38. an =
q
ln n 11. a 3 n=1 n
33. a1 = 1,
nR
q
n
s 2dn 3n n=1
q
3 9. a a1  n b
n  2 n b
an + 1 =
You
22
2n a 2 n 1 + ln n = an n n + ln n = a n + 10 n
32. a1 = 5,
iaz
Which of the series in Exercises 1–26 converge, and which diverge? Give reasons for your answers. (When checking your answers, remember there may be more than one way to determine a series’ convergence or divergence.) q
suf
i
EXERCISES 11.5
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q
1 a sln nd p
s p constantd .
n=2
n>2n, if n is a prime number 1>2n, otherwise. Does gan converge? Give reasons for your answer.
47. Let an = e
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Alternating Series, Absolute and Conditional Convergence
787
suf
11.6
Alternating Series, Absolute and Conditional Convergence
i
11.6
A series in which the terms are alternately positive and negative is an alternating series. Here are three examples: s 1dn + 1 1 1 1 1 + Á +  +  Á + n 5 2 3 4
2 + 1 
You
1 
s 1dn4 1 1 1 +  + Á + + Á 2 4 8 2n
1  2 + 3  4 + 5  6 + Á + s 1dn + 1n + Á
(1) (2) (3)
THEOREM 14 The series
nR
iaz
Series (1), called the alternating harmonic series, converges, as we will see in a moment. Series (2) a geometric series with ratio r = 1>2, converges to 2>[1 + s1>2d] = 4>3. Series (3) diverges because the nth term does not approach zero. We prove the convergence of the alternating harmonic series by applying the Alternating Series Test.
The Alternating Series Test (Leibniz’s Theorem)
ass a
q
a s 1d
n+1
un = u1  u2 + u3  u4 + Á
n=1
converges if all three of the following conditions are satisfied: The un’s are all positive. un Ú un + 1 for all n Ú N, for some integer N. un : 0.
dH
1. 2. 3.
Mu
ham
ma
Proof If n is an even integer, say n = 2m, then the sum of the first n terms is s2m = su1  u2 d + su3  u4 d + Á + su2m  1  u2m d = u1  su2  u3 d  su4  u5 d  Á  su2m  2  u2m  1 d  u2m .
The first equality shows that s2m is the sum of m nonnegative terms, since each term in parentheses is positive or zero. Hence s2m + 2 Ú s2m , and the sequence 5s2m6 is nondecreasing. The second equality shows that s2m … u1 . Since 5s2m6 is nondecreasing and bounded from above, it has a limit, say lim s2m = L.
m: q
(4)
If n is an odd integer, say n = 2m + 1, then the sum of the first n terms is s2m + 1 = s2m + u2m + 1 . Since un : 0, lim u2m + 1 = 0
m: q
and, as m : q , s2m + 1 = s2m + u2m + 1 : L + 0 = L.
(5)
Combining the results of Equations (4) and (5) gives lim sn = L (Section 11.1, Exern: q cise 119).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
The alternating harmonic series
i
EXAMPLE 1
q
a s 1d
n+1
n=1
1 1 1 1 Á n = 1  2 + 3  4 +
suf
788
satisfies the three requirements of Theorem 14 with N = 1; it therefore converges.
You
u4
s2
s4
L
s3
s1
x
FIGURE 11.9 The partial sums of an alternating series that satisfies the hypotheses of Theorem 14 for N = 1 straddle the limit from the beginning.
iaz
u2 u3
0
A graphical interpretation of the partial sums (Figure 11.9) shows how an alternating series converges to its limit L when the three conditions of Theorem 14 are satisfied with N = 1. (Exercise 63 asks you to picture the case N 7 1.) Starting from the origin of the xaxis, we lay off the positive distance s1 = u1 . To find the point corresponding to s2 = u1  u2 , we back up a distance equal to u2 . Since u2 … u1 , we do not back up any farther than the origin. We continue in this seesaw fashion, backing up or going forward as the signs in the series demand. But for n Ú N, each forward or backward step is shorter than (or at most the same size as) the preceding step, because un + 1 … un . And since the nth term approaches zero as n increases, the size of step we take forward or backward gets smaller and smaller. We oscillate across the limit L, and the amplitude of oscillation approaches zero. The limit L lies between any two successive sums sn and sn + 1 and hence differs from sn by an amount less than un + 1 . Because
nR
u1
ass a
for n Ú N, ƒ L  sn ƒ 6 un + 1 we can make useful estimates of the sums of convergent alternating series.
dH
THEOREM 15 The Alternating Series Estimation Theorem q If the alternating series g n = 1 s 1dn + 1un satisfies the three conditions of Theorem 14, then for n Ú N, sn = u1  u2 + Á + s 1dn + 1un
ma
approximates the sum L of the series with an error whose absolute value is less than un + 1 , the numerical value of the first unused term. Furthermore, the remainder, L  sn , has the same sign as the first unused term.
Mu
EXAMPLE 2
We try Theorem 15 on a series whose sum we know:
q
1 1 1 1 1 1 1 n 1 a s 1d 2n = 1  2 + 4  8 + 16  32 + 64  128 n=0

ham
We leave the verification of the sign of the remainder for Exercise 53.
+
1  Á. 256
The theorem says that if we truncate the series after the eighth term, we throw away a total that is positive and less than 1> 256. The sum of the first eight terms is 0.6640625. The sum of the series is 1 2 1 = . = 3 3>2 1  s 1>2d The difference, s2>3d  0.6640625 = 0.0026041666 Á , is positive and less than s1>256d = 0.00390625.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.6
Alternating Series, Absolute and Conditional Convergence
789
suf
i
Absolute and Conditional Convergence
You
DEFINITION Absolutely Convergent A series gan converges absolutely (is absolutely convergent) if the corresponding series of absolute values, g ƒ an ƒ , converges.
The geometric series
1 1 1 +  + Á 2 4 8
iaz
1 
converges absolutely because the corresponding series of absolute values 1 1 1 + + + Á 2 4 8
nR
1 +
ass a
converges. The alternating harmonic series does not converge absolutely. The corresponding series of absolute values is the (divergent) harmonic series.
DEFINITION Conditionally Convergent A series that converges but does not converge absolutely converges conditionally.
ma
dH
The alternating harmonic series converges conditionally. Absolute convergence is important for two reasons. First, we have good tests for convergence of series of positive terms. Second, if a series converges absolutely, then it converges. That is the thrust of the next theorem.
THEOREM 16
The Absolute Convergence Test
q
q
Mu
ham
If a ƒ an ƒ converges, then a an converges. n=1
n=1
Proof For each n,  ƒ an ƒ … an … ƒ an ƒ , q gn=1
so
0 … an + ƒ an ƒ … 2 ƒ an ƒ .
q gn=1
If 2 ƒ an ƒ converges and, by the Direct Comparison Test, ƒ an ƒ converges, then q the nonnegative series g n = 1 san + ƒ an ƒ d converges. The equality an = san + ƒ an ƒ d  ƒ an ƒ q now lets us express g n = 1 an as the difference of two convergent series: q
q
q
q
a an = a san + ƒ an ƒ  ƒ an ƒ d = a san + ƒ an ƒ d  a ƒ an ƒ .
Therefore,
n=1
n=1
q g n = 1 an
converges.
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n=1
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 790
Chapter 11: Infinite Sequences and Series
EXAMPLE 3
Applying the Absolute Convergence Test
q
suf
i
CAUTION We can rephrase Theorem 16 to say that every absolutely convergent series converges. However, the converse statement is false: Many convergent series do not converge absolutely (such as the alternating harmonic series in Example 1).
You
1 1 1 1 (a) For a s 1dn + 1 2 = 1  + + Á , the corresponding series of absolute 4 9 16 n n=1 values is the convergent series q
1 1 1 1 Á. a n 2 = 1 + 4 + 9 + 16 +
n=1
iaz
The original series converges because it converges absolutely. q
nR
sin n sin 1 sin 2 sin 3 (b) For a 2 = + + + Á , the corresponding series of absolute 1 4 9 n=1 n values is q ƒ sin 1 ƒ ƒ sin 2 ƒ sin n + + Á, a ` n2 ` = 1 4 n=1 which converges by comparison with g n = 1 s1>n 2 d because ƒ sin n ƒ … 1 for every n. The original series converges absolutely; therefore it converges.
ass a
q
EXAMPLE 4
Alternating pSeries
If p is a positive constant, the sequence 51>n p6 is a decreasing sequence with limit zero. Therefore the alternating pseries s 1dn  1 1 1 1 = 1  p + p  p + Á, a np 2 3 4 n=1
dH
q
p 7 0
converges. If p 7 1, the series converges absolutely. If 0 6 p … 1, the series converges conditionally.
ma
Conditional convergence: Absolute convergence:
1 1 1 + + Á 22 23 24 1 1 1 1  3>2 + 3>2  3>2 + Á 2 3 4 1 
Mu
ham
Rearranging Series
THEOREM 17
The Rearrangement Theorem for Absolutely Convergent Series q If g n = 1 an converges absolutely, and b1, b2 , Á , bn , Á is any arrangement of the sequence 5an6, then gbn converges absolutely and q
q
a bn = a an .
n=1
n=1
(For an outline of the proof, see Exercise 60.)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.6
791
Applying the Rearrangement Theorem
i
EXAMPLE 5
Alternating Series, Absolute and Conditional Convergence
1 
suf
As we saw in Example 3, the series
1 1 1 1 + + Á + s 1dn  1 2 + Á 4 9 16 n
1 1 1 1 1 1 1 1 1 + + + + Á. 4 16 9 25 49 36 64 100 144
iaz
1 
You
converges absolutely. A possible rearrangement of the terms of the series might start with a positive term, then two negative terms, then three positive terms, then four negative terms, and so on: After k terms of one sign, take k + 1 terms of the opposite sign. The first ten terms of such a series look like this:
(See Exercise 61.)
nR
The Rearrangement Theorem says that both series converge to the same value. In this example, if we had the second series to begin with, we would probably be glad to exchange it for the first, if we knew that we could. We can do even better: The sum of either series is also equal to q q 1 1 . a a 2 2 n = 1 s2n  1d n = 1 s2nd
ass a
If we rearrange infinitely many terms of a conditionally convergent series, we can get results that are far different from the sum of the original series. Here is an example.
EXAMPLE 6
Rearranging the Alternating Harmonic Series
The alternating harmonic series
dH
1 1 1 1 1 1 1 1 1 1 1  +  +  +  + +  Á 5 7 1 2 3 4 6 8 9 10 11
can be rearranged to diverge or to reach any preassigned sum. (a) Rearranging g n = 1 s 1dn + 1>n to diverge. The series of terms g[1>s2n  1d] diverges to + q and the series of terms gs 1>2nd diverges to  q . No matter how far out in the sequence of oddnumbered terms we begin, we can always add enough positive terms to get an arbitrarily large sum. Similarly, with the negative terms, no matter how far out we start, we can add enough consecutive evennumbered terms to get a negative sum of arbitrarily large absolute value. If we wished to do so, we could start adding oddnumbered terms until we had a sum greater than +3, say, and then follow that with enough consecutive negative terms to make the new total less than 4. We could then add enough positive terms to make the total greater than +5 and follow with consecutive unused negative terms to make a new total less than 6, and so on. In this way, we could make the swings arbitrarily large in either direction. q (b) Rearranging g n = 1 s 1dn + 1>n to converge to 1. Another possibility is to focus on a particular limit. Suppose we try to get sums that converge to 1. We start with the first term, 1> 1, and then subtract 1> 2. Next we add 1> 3 and 1> 5, which brings the total back to 1 or above. Then we add consecutive negative terms until the total is less than 1. We continue in this manner: When the sum is less than 1, add positive terms until the total is 1 or more; then subtract (add negative) terms until the total is again less than 1. This process can be continued indefinitely. Because both the oddnumbered
Mu
ham
ma
q
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 792
Chapter 11: Infinite Sequences and Series
suf
i
terms and the evennumbered terms of the original series approach zero as n : q , the amount by which our partial sums exceed 1 or fall below it approaches zero. So the new series converges to 1. The rearranged series starts like this:
You
1 1 1 1 1 1 1 1 1 1 1 1 1 1  + +  + +  + +  + + 5 7 1 2 3 4 9 6 11 13 8 15 17 10 1 1 1 1 1 1 1 1 + + + + + + Á 19 21 12 23 25 14 27 16
5.
Mu
ham
ma
dH
6.
nR
2. 3. 4.
The nthTerm Test: Unless an : 0, the series diverges. Geometric series: gar n converges if ƒ r ƒ 6 1; otherwise it diverges. pseries: g1>n p converges if p 7 1; otherwise it diverges. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test. Series with some negative terms: Does g ƒ an ƒ converge? If yes, so does gan , since absolute convergence implies convergence. Alternating series: gan converges if the series satisfies the conditions of the Alternating Series Test.
ass a
1.
iaz
The kind of behavior illustrated by the series in Example 6 is typical of what can happen with any conditionally convergent series. Therefore we must always add the terms of a conditionally convergent series in the order given. We have now developed several tests for convergence and divergence of series. In summary:
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 792
i f su
Chapter 11: Infinite Sequences and Series
EXERCISES 11.6 Determining Convergence or Divergence
q
Which of the alternating series in Exercises 1â€“10 converge, and which diverge? Give reasons for your answers. q
q
1 1. a s 1dn + 1 2 n n=1 q
3. a s 1d
n+1
n=1 q
2. a s 1dn + 1
n a b 10
n
n=1 q
4. a s 1d
n+1
n=1 q
n
3>2
10 n n 10
H d
ln n 6. a s 1dn + 1 n
1 5. a s 1dn + 1 ln n n=2 q
n=1 q
a m m a h
ln n 7. a s 1dn + 1 ln n 2 n=2 q
2n + 1 9. a s 1dn + 1 n + 1 n=1
1 8. a s 1dn ln a1 + n b n=1 q
10. a s 1dn + 1 n=1
32n + 1 2n + 1
Absolute Convergence
u M
Which of the series in Exercises 11â€“44 converge absolutely, which converge, and which diverge? Give reasons for your answers.
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q
11. a s 1dn + 1s0.1dn n=1 q
13. a s 1dn
n a s s a
1
u o Y z a i R
n=1 q
1
2n
15. a s 1d
n+1
n=1 q
n n3 + 1
1 17. a s 1dn n + 3 n=1 q
3 + n 19. a s 1dn + 1 5 + n n=1 q
21. a s 1dn + 1 n=1 q
1 + n n2
12. a s 1dn + 1 n=1 q
14. a
s0.1dn n
s 1dn
1 + 2n n! 16. a s 1dn + 1 n 2 n=1 n=1 q
q
sin n 18. a s 1dn 2 n n=1 q
1 20. a s 1dn ln sn 3 d n=2 q
s 2dn + 1 22. a n n=1 n + 5
23. a s 1dnn 2s2>3dn
24. a s 1dn + 1 A 210 B
tan1 n 25. a s 1d 2 n + 1 n=1
1 26. a s 1dn + 1 n ln n
q
n=1 q
n
n
n=1 q n=2
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4100 AWL/Thomas_ch11p746847 8/25/04 2:41 PM Page 793
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.6 Alternating Series, Absolute and Conditional Convergence
s 100dn n! n=1 q
q
29. a
30. a s 5dn ln n b 32. a s 1dn a ln n 2 n=2
s 1d 31. a 2 n = 1 n + 2n + 1 33. a
to estimate L. Compute
q
cos np
cos np 34. a n
n 2n s 1dnsn + 1dn 35. a s2ndn n=1 n=1 q
n=1 q
n+1
s20 +
2
s 1d sn!d s2nd! n=1
36. a
sn!d2 3n 38. a s 1dn s2n + 1d! n=1
s2nd! 37. a s 1dn n 2 n!n n=1 q
q
39. a s 1dn A 2n + 1  2n B 40. a s 1dn A 2n 2 + n  n B q
q
n=1 q
n=1
n=1 q
42. a
n=1 q
s 1dn
q
43. a s 1dn sech n
2n + 2n + 1
as an approximation to the sum of the alternating harmonic series. The exact sum is ln 2 = 0.6931. Á 53. The sign of the remainder of an alternating series that satisfies the conditions of Theorem 14 Prove the assertion in Theorem 15 that whenever an alternating series satisfying the conditions of Theorem 14 is approximated with one of its partial sums, then the remainder (sum of the unused terms) has the same sign as the first unused term. (Hint: Group the remainder’s terms in consecutive pairs.) 54. Show that the sum of the first 2n terms of the series
n=1
44. a s 1dn csch n
1 
ass a
n=1
Error Estimation
q
1 45. a s 1dn + 1 n
dH
It can be shown that the sum is ln 2.
n=1 q
1 46. a s 1dn + 1 n 10 n=1 q
s0.01dn 47. a s 1dn + 1 n
ma
q
1 = a s 1dnt n, 48. 1 + t n=0
0 6 t 6 1
ham
T Approximate the sums in Exercises 49 and 50 with an error of magnitude less than 5 * 10 6 . n=0 q
1 50. a s 1dn n! n=0
1 1 1 1 1 + # + # + # + # + Á. 1#2 2 3 3 4 4 5 5 6
Do these series converge? What is the sum of the first 2n + 1 terms of the first series? If the series converge, what is their sum?
55. Show that if g n = 1 an diverges, then g n = 1 ƒ an ƒ diverges. q
56. Show that if
q gn=1
As you will see in Section 11.7, the sum is ln (1.01).
n=1
q
1 1 1 1 1 1 1 1 1 +  +  +  +  + Á 5 5 2 2 3 3 4 4 6
is the same as the sum of the first n terms of the series
In Exercises 45–48, estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series.
1 49. a s 1dn s2nd!
1 21
#
nR
41. a s 1dn A 2n + 1n  2n B
1 2
iaz
q
sn + sn + 1 1 = sn + s 1dn + 2an + 1 2 2
n
You
n=1 q
n1
q
T 52. The limit L of an alternating series that satisfies the conditions of Theorem 14 lies between the values of any two consecutive partial sums. This suggests using the average
i
q
ln n 28. a s 1dn n  ln n n=1
suf
q
n 27. a s 1dn n + 1 n=1
793
As you will see in Section 11.9, the sum is cos 1, the cosine of 1 radian.
As you will see in Section 11.9, the sum is e1 .
Mu
Theory and Examples 51. a. The series
1 1 1 1 1 1 1 1  +  +  + Á + n  n + Á 3 2 9 4 27 8 3 2
does not meet one of the conditions of Theorem 14. Which one?
b. Find the sum of the series in part (a).
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q
an converges absolutely, then
` a an ` … a ƒ an ƒ . q
q
n=1
n=1
57. Show that if g n = 1 an and g n = 1 bn both converge absolutely, then so does q
q
q
q
a. a san + bn d
b. a san  bn d
n=1 q
c. a kan
n=1
(k any number)
n=1
58. Show by example that g n = 1 an bn may diverge even if g n = 1 an q and g n = 1 bn both converge. q
q
T 59. In Example 6, suppose the goal is to arrange the terms to get a new series that converges to 1>2 . Start the new arrangement with the first negative term, which is 1>2 . Whenever you have a sum that is less than or equal to 1>2, start introducing positive terms, taken in order, until the new total is greater than 1>2 . Then add negative terms until the total is less than or equal to 1>2 again. Continue this process until your partial sums have
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series In other words, if a series converges absolutely, its positive terms form a convergent series, and so do its negative terms. Furthermore,
i
been above the target at least three times and finish at or below it. If sn is the sum of the first n terms of your new series, plot the points sn, sn d to illustrate how the sums are behaving. 60. Outline of the proof of the Rearrangement Theorem (Theorem 17) a. Let P be a positive real number, let L = g n = 1 an , and let k sk = g n = 1 an . Show that for some index N1 and for some index N2 Ú N1 ,
q n=1
1 2 1 2 1 2 1 2 1 2  +  +  +  +  + Á. 5 7 5 3 2 3 4 9 11 6
… a ƒ ak ƒ + ƒ sN2  L ƒ 6 P. k = N1
b. The argument in part (a) shows that if g n = 1 an converges q
absolutely then g n = 1 bn converges and g n = 1 bn = g n = 1 an . Now show that because converges to g n = 1 ƒ an ƒ .
ƒ an ƒ converges,
q
61. Unzipping absolutely convergent series ƒ an ƒ converges and
bn = e then
q gn=1
an , 0,
if an Ú 0 if an 6 0,
q gn=1
ƒ bn ƒ
dH
a. Show that if
q gn=1
q
bn converges.
Collect terms with the same denominator, as the arrows indicate, to arrive at
ass a
q
q gn=1
by 2 to get
2S = 2  1 +
q
q
iaz
k=1
` a bk  L ` … ` a bk  sN2 ` + ƒ sN2  L ƒ
1 1 1 1 1 +  +  + 5 2 3 4 6 1 1 1 1 1 1  + + + Á 7 8 9 10 11 12
nR
k=1
You
P ƒ sN2  L ƒ 6 2 .
Since all the terms a1, a2 , Á , aN 2 appear somewhere in the sequence 5bn6 , there is an index N3 Ú N2 such that if n n Ú N3 , then A g k = 1 bk B  sN2 is at most a sum of terms am with m Ú N1 . Therefore, if n Ú N3 , n
n=1
because bn = san + ƒ an ƒ d>2 and cn = san  ƒ an ƒ d>2 . 62. What is wrong here?:
S = 1 
n
n=1
Multiply both sides of the alternating harmonic series
q
and
q
a an = a bn + a cn
q
P a ƒ an ƒ 6 2 n = N1
q
suf
794
2S = 1 
1 1 1 1 1 +  +  + Á. 5 2 3 4 6
The series on the righthand side of this equation is the series we started with. Therefore, 2S = S , and dividing by S gives 2 = 1 . (Source: “Riemann’s Rearrangement Theorem” by Stewart Galanor, Mathematics Teacher, Vol. 80, No. 8, 1987, pp. 675–681.)
63. Draw a figure similar to Figure 11.9 to illustrate the convergence of the series in Theorem 14 when N 7 1 .
b. Use the results in part (a) to show likewise that if g n = 1 ƒ an ƒ converges and
ma
q
cn = e
0, an ,
if an Ú 0 if an 6 0,
then g n = 1 cn converges.
Mu
ham
q
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 794
Chapter 11: Infinite Sequences and Series
11.7
z a i R n a
Power Series
s s a dH
Now that we can test infinite series for convergence we can study the infinite polynomials mentioned at the beginning of this chapter. We call these polynomials power series because they are defined as infinite series of powers of some variable, in our case x. Like polynomials, power series can be added, subtracted, multiplied, differentiated, and integrated to give new power series.
Mu
a m m a h
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
795
Power Series
i
Power Series and Convergence
suf
We begin with the formal definition.
You
DEFINITIONS Power Series, Center, Coefficients A power series about x 0 is a series of the form q
n 2 Á + cn x n + Á . a cn x = c0 + c1 x + c2 x +
n=0
(1)
A power series about x a is a series of the form q
iaz
n 2 Á + cnsx  adn + Á a cnsx  ad = c0 + c1sx  ad + c2sx  ad +
n=0
(2)
nR
in which the center a and the coefficients c0, c1, c2, Á , cn, Á are constants.
Equation (1) is the special case obtained by taking a = 0 in Equation (2).
EXAMPLE 1
A Geometric Series
ass a
Taking all the coefficients to be 1 in Equation (1) gives the geometric power series q
n 2 Á + xn + Á . ax = 1 + x + x +
n=0
dH
This is the geometric series with first term 1 and ratio x. It converges to 1>s1  xd for ƒ x ƒ 6 1. We express this fact by writing 1 = 1 + x + x 2 + Á + x n + Á, 1  x
1 6 x 6 1.
(3)
Mu
ham
ma
Up to now, we have used Equation (3) as a formula for the sum of the series on the right. We now change the focus: We think of the partial sums of the series on the right as polynomials Pnsxd that approximate the function on the left. For values of x near zero, we need take only a few terms of the series to get a good approximation. As we move toward x = 1, or 1, we must take more terms. Figure 11.10 shows the graphs of ƒsxd = 1>s1  xd, and the approximating polynomials yn = Pnsxd for n = 0, 1, 2, and 8. The function ƒsxd = 1>s1  xd is not continuous on intervals containing x = 1, where it has a vertical asymptote. The approximations do not apply when x Ú 1.
EXAMPLE 2
A Geometric Series
The power series n
1 
1 1 1 sx  2d + sx  2d2 + Á + a b sx  2dn + Á 2 4 2
(4)
matches Equation (2) with a = 2, c0 = 1, c1 = 1>2, c2 = 1>4, Á , cn = s 1>2dn . This x  2 is a geometric series with first term 1 and ratio r = . The series converges for 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 796
Chapter 11: Infinite Sequences and Series
1 1x
i
y
suf
y 9 8 7
y8 1 x x 2 x 3 x 4 x 5 x 6 x7 x 8
You
5 4
y2 1 x x2
3
y1 1 x
2 1
y0 1 x
iaz
–1
0
1
`
nR
FIGURE 11.10 The graphs of ƒsxd = 1>s1  xd and four of its polynomial approximations (Example 1).
x  2 ` 6 1 or 0 6 x 6 4. The sum is 2
y
so
y0 1
0
1
Mu
ham
FIGURE 11.11 The graphs of ƒsxd = 2>x and its first three polynomial approximations (Example 2).
0 6 x 6 4.
Series (4) generates useful polynomial approximations of ƒsxd = 2>x for values of x near 2: P0sxd = 1
ma
1
2 y2 3 3x x 4 2 (2, 1) y 2x y1 2 x 2 x 3 2
n sx  2d sx  2d2 1 2 +  Á + a b sx  2dn + Á, x = 1 2 4 2
dH
2
1 2 = x, x  2 1 + 2
ass a
1 = 1  r
P1sxd = 1 
x 1 sx  2d = 2 2 2
P2sxd = 1 
3x x2 1 1 + , sx  2d + sx  2d2 = 3 2 4 2 4
and so on (Figure 11.11).
EXAMPLE 3
Testing for Convergence Using the Ratio Test
For what values of x do the following power series converge? q
xn x2 x3 +  Á (a) a s 1dn  1 n = x 2 3 n=1 q
x3 x5 x 2n  1 = x +  Á (b) a s 1dn  1 5 2n 1 3 n=1 q
xn x2 x3 = 1 + x + + + Á (c) a 2! 3! n = 0 n! q
(d) a n!x n = 1 + x + 2!x 2 + 3!x 3 + Á n=0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
797
Apply the Ratio Test to the series g ƒ un ƒ , where un is the nth term of the series
Solution
suf
i
in question. un + 1 n (a) ` un ` = x : x. n + 1ƒ ƒ ƒ ƒ
–1
0
x
1
iaz
un + 1 2n  1 2 (b) ` un ` = x : x2. 2n + 1
You
The series converges absolutely for ƒ x ƒ 6 1. It diverges if ƒ x ƒ 7 1 because the nth term does not converge to zero. At x = 1, we get the alternating harmonic series 1  1>2 + 1>3  1>4 + Á , which converges. At x = 1 we get 1  1>2 1>3  1>4  Á , the negative of the harmonic series; it diverges. Series (a) converges for 1 6 x … 1 and diverges elsewhere.
ass a
nR
The series converges absolutely for x 2 6 1. It diverges for x 2 7 1 because the nth term does not converge to zero. At x = 1 the series becomes 1  1>3 + 1>5  1>7 + Á , which converges by the Alternating Series Theorem. It also converges at x = 1 because it is again an alternating series that satisfies the conditions for convergence. The value at x = 1 is the negative of the value at x = 1. Series (b) converges for 1 … x … 1 and diverges elsewhere. –1
0
1
x
un + 1 ƒxƒ x n + 1 # n! (c) ` un ` = ` : 0 for every x. n` = n + 1 x sn + 1d!
dH
The series converges absolutely for all x. 0
x
sn + 1d!x n + 1 un + 1 (d) ` un ` = ` ` = sn + 1d ƒ x ƒ : q unless x = 0. n!x n
Mu
ham
ma
The series diverges for all values of x except x = 0. x
0
Example 3 illustrates how we usually test a power series for convergence, and the possible results.
THEOREM 18
The Convergence Theorem for Power Series q
If the power series a an x n = a0 + a1 x + a2 x 2 + Á converges for n=0
x = c Z 0, then it converges absolutely for all x with ƒ x ƒ 6 ƒ c ƒ . If the series diverges for x = d, then it diverges for all x with ƒ x ƒ 7 ƒ d ƒ .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 798
Chapter 11: Infinite Sequences and Series
Proof Suppose the series g n = 0 an c n converges. Then limn: q an c n = 0. Hence, there is an integer N such that ƒ an c n ƒ 6 1 for all n Ú N. That is, 1 ƒ c ƒn
for n Ú N.
Now take any x such that ƒ x ƒ 6 ƒ c ƒ and consider
(5)
You
ƒ an ƒ 6
suf
i
q
ƒ a0 ƒ + ƒ a1 x ƒ + Á + ƒ aN  1x N  1 ƒ + ƒ aN x N ƒ + ƒ aN + 1 x N + 1 ƒ + Á .
There are only a finite number of terms prior to ƒ aN x N ƒ , and their sum is finite. Starting with ƒ aN x N ƒ and beyond, the terms are less than N
x + `c`
N+1
x + `c`
N+2
+ Á
iaz
x `c`
(6)
ass a
nR
because of Inequality (5). But Series (6) is a geometric series with ratio r = ƒ x>c ƒ , which is less than 1, since ƒ x ƒ 6 ƒ c ƒ . Hence Series (6) converges, so the original series converges absolutely. This proves the first half of the theorem. The second half of the theorem follows from the first. If the series diverges at x = d and converges at a value x0 with ƒ x0 ƒ 7 ƒ d ƒ , we may take c = x0 in the first half of the theorem and conclude that the series converges absolutely at d. But the series cannot converge absolutely and diverge at one and the same time. Hence, if it diverges at d, it diverges for all x with ƒ x ƒ 7 ƒ d ƒ .
dH
To simplify the notation, Theorem 18 deals with the convergence of series of the form gan x n . For series of the form gansx  adn we can replace x  a by x¿ and apply the results to the series gansx¿dn .
The Radius of Convergence of a Power Series
Mu
ham
ma
The theorem we have just proved and the examples we have studied lead to the conclusion that a power series gcnsx  adn behaves in one of three possible ways. It might converge only at x = a, or converge everywhere, or converge on some interval of radius R centered at x = a. We prove this as a Corollary to Theorem 18.
COROLLARY TO THEOREM 18 The convergence of the series gcnsx  adn is described by one of the following three possibilities: 1.
2. 3.
There is a positive number R such that the series diverges for x with ƒ x  a ƒ 7 R but converges absolutely for x with ƒ x  a ƒ 6 R. The series may or may not converge at either of the endpoints x = a  R and x = a + R. The series converges absolutely for every x sR = q d. The series converges at x = a and diverges elsewhere sR = 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
799
nR
iaz
You
suf
i
Proof We assume first that a = 0, so that the power series is centered at 0. If the series converges everywhere we are in Case 2. If it converges only at x = 0 we are in Case 3. Otherwise there is a nonzero number d such that gcn d n diverges. The set S of values of x for which the series gcn x n converges is nonempty because it contains 0 and a positive number p as well. By Theorem 18, the series diverges for all x with ƒ x ƒ 7 ƒ d ƒ , so ƒ x ƒ … ƒ d ƒ for all x H S, and S is a bounded set. By the Completeness Property of the real numbers (see Appendix 4) a nonempty, bounded set has a least upper bound R. (The least upper bound is the smallest number with the property that the elements x H S satisfy x … R.) If ƒ x ƒ 7 R Ú p, then x x S so the series gcn x n diverges. If ƒ x ƒ 6 R, then ƒ x ƒ is not an upper bound for S (because it’s smaller than the least upper bound) so there is a number b H S such that b 7 ƒ x ƒ . Since b H S, the series gcn b n converges and therefore the series gcn ƒ x ƒ n converges by Theorem 18. This proves the Corollary for power series centered at a = 0. For a power series centered at a Z 0, we set x¿ = sx  ad and repeat the argument with x¿ . Since x¿ = 0 when x = a, a radius R interval of convergence for gcnsx¿dn centered at x¿ = 0 is the same as a radius R interval of convergence for gcnsx  adn centered at x = a. This establishes the Corollary for the general case.
dH
ass a
R is called the radius of convergence of the power series and the interval of radius R centered at x = a is called the interval of convergence. The interval of convergence may be open, closed, or halfopen, depending on the particular series. At points x with ƒ x  a ƒ 6 R, the series converges absolutely. If the series converges for all values of x, we say its radius of convergence is infinite. If it converges only at x = a, we say its radius of convergence is zero.
How to Test a Power Series for Convergence
ma
1.
Mu
ham
2.
3.
Use the Ratio Test (or nthRoot Test) to find the interval where the series converges absolutely. Ordinarily, this is an open interval ƒx  aƒ 6 R
or
a  R 6 x 6 a + R.
If the interval of absolute convergence is finite, test for convergence or divergence at each endpoint, as in Examples 3a and b. Use a Comparison Test, the Integral Test, or the Alternating Series Test. If the interval of absolute convergence is a  R 6 x 6 a + R, the series diverges for ƒ x  a ƒ 7 R (it does not even converge conditionally), because the nth term does not approach zero for those values of x.
TermbyTerm Differentiation A theorem from advanced calculus says that a power series can be differentiated term by term at each interior point of its interval of convergence.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 800
Chapter 11: Infinite Sequences and Series
suf
i
THEOREM 19 The TermbyTerm Differentiation Theorem If gcnsx  adn converges for a  R 6 x 6 a + R for some R 7 0, it defines a function ƒ: q
ƒsxd = a cnsx  adn,
a  R 6 x 6 a + R.
You
n=0
Such a function ƒ has derivatives of all orders inside the interval of convergence. We can obtain the derivatives by differentiating the original series term by term: q
ƒ¿sxd = a ncnsx  adn  1 n=1 q n=2
iaz
ƒ–sxd = a nsn  1dcnsx  adn  2 ,
EXAMPLE 4
nR
and so on. Each of these derived series converges at every interior point of the interval of convergence of the original series.
Applying TermbyTerm Differentiation
Find series for ƒ¿sxd and ƒ–sxd if
1 = 1 + x + x2 + x3 + x4 + Á + xn + Á 1  x
ass a ƒsxd =
q
= a x n,
1 6 x 6 1
n=0
dH
Solution
ƒ¿sxd =
1 = 1 + 2x + 3x 2 + 4x 3 + Á + nx n  1 + Á s1  xd2 q
= a nx n  1,
1 6 x 6 1
ma
n=1
Mu
ham
ƒ–sxd =
2 = 2 + 6x + 12x 2 + Á + nsn  1dx n  2 + Á s1  xd3 q
= a nsn  1dx n  2,
1 6 x 6 1
n=2
CAUTION
Termbyterm differentiation might not work for other kinds of series. For example, the trigonometric series q sin sn!xd a n2 n=1 converges for all x. But if we differentiate term by term we get the series q n!cos sn!xd , a n2 n=1 which diverges for all x. This is not a power series, since it is not a sum of positive integer powers of x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
801
i
TermbyTerm Integration
THEOREM 20 Suppose that
You
suf
Another advanced calculus theorem states that a power series can be integrated term by term throughout its interval of convergence.
The TermbyTerm Integration Theorem q
ƒsxd = a cnsx  adn n=0
iaz
converges for a  R 6 x 6 a + R sR 7 0d. Then q
a cn
n =0
(x  a) n+1 n + 1
nR
converges for a  R 6 x 6 a + R and q
L
ƒsxd dx = a cn n=0
sx  adn + 1 + C n + 1
ass a
for a  R 6 x 6 a + R.
EXAMPLE 5
A Series for tan1 x, 1 … x … 1
dH
Identify the function
x5 x3 +  Á, 5 3
1 … x … 1.
We differentiate the original series term by term and get
ma
Solution
ƒsxd = x 
ƒ¿sxd = 1  x 2 + x 4  x 6 + Á,
1 6 x 6 1.
Mu
ham
This is a geometric series with first term 1 and ratio x 2 , so ƒ¿sxd =
1 1 = . 1  s x 2 d 1 + x2
We can now integrate ƒ¿sxd = 1>s1 + x 2 d to get L
ƒ¿sxd dx =
dx = tan1 x + C. 2 L1 + x
The series for ƒ(x) is zero when x = 0, so C = 0. Hence ƒsxd = x 
x3 x5 x7 + + Á = tan1 x, 5 7 3
1 6 x 6 1.
In Section 11.10, we will see that the series also converges to tan1 x at x = ;1.
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(7)
4100 AWL/Thomas_ch11p746847 8/25/04 2:41 PM Page 802
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 802
Chapter 11: Infinite Sequences and Series
EXAMPLE 6
suf
i
Notice that the original series in Example 5 converges at both endpoints of the original interval of convergence, but Theorem 20 can guarantee the convergence of the differentiated series only inside the interval.
A Series for ln s1 + xd, 1 6 x … 1
You
The series
1 = 1  t + t2  t3 + Á 1 + t converges on the open interval 1 6 t 6 1. Therefore,
x
x
t2 t3 t4 1 dt = t + + Ád 2 3 4 0 L0 1 + t x2 x3 x4 + + Á, 2 3 4
Theorem 20
1 6 x 6 1.
nR
= x 
iaz
ln s1 + xd =
ass a
It can also be shown that the series converges at x = 1 to the number ln 2, but that was not guaranteed by the theorem.
USING TECHNOLOGY
Study of Series
ma
dH
Series are in many ways analogous to integrals. Just as the number of functions with explicit antiderivatives in terms of elementary functions is small compared to the number of integrable functions, the number of power series in x that agree with explicit elementary functions on xintervals is small compared to the number of power series that converge on some xinterval. Graphing utilities can aid in the study of such series in much the same way that numerical integration aids in the study of definite integrals. The ability to study power series at particular values of x is built into most Computer Algebra Systems. If a series converges rapidly enough, CAS exploration might give us an idea of the q sum. For instance, in calculating the early partial sums of the series g k = 1 [1>s2k  1 d] (Section 11.4, Example 2b), Maple returns Sn = 1.6066 95152 for 31 … n … 200. This suggests that the sum of the series is 1.6066 95152 to 10 digits. Indeed,
Mu
ham
q
q
q
1 1 1 1 = a k1 6 a k  1 = 199 6 1.25 * 10 60 . a k 2 s2  s1>2k  1 dd k = 201 2  1 k = 201 2 k = 201 2
The remainder after 200 terms is negligible. However, CAS and calculator exploration cannot do much for us if the series converges or diverges very slowly, and indeed can be downright misleading. For example, q try calculating the partial sums of the series g k = 1 [1>s10 10kd]. The terms are tiny in comparison to the numbers we normally work with and the partial sums, even for hundreds of terms, are miniscule. We might well be fooled into thinking that the series conq verges. In fact, it diverges, as we can see by writing it as s1>10 10 dg k = 1 s1>kd, a constant times the harmonic series. We will know better how to interpret numerical results after studying error estimates in Section 11.9.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
803
i
Multiplication of Power Series
suf
Another theorem from advanced calculus states that absolutely converging power series can be multiplied the way we multiply polynomials. We omit the proof.
You
THEOREM 21 The Series Multiplication Theorem for Power Series q q If Asxd = g n = 0 an x n and Bsxd = g n = 0 bn x n converge absolutely for ƒ x ƒ 6 R, and n
cn = a0 bn + a1 bn  1 + a2 bn  2 + Á + an  1b1 + an b0 = a ak bn  k , k=0
converges absolutely to A(x)B(x) for ƒ x ƒ 6 R:
iaz
then
q g n = 0 cn x n
a a an x n b # a a bn x n b = a cn x n . q
q
n=0
n=0
nR
n=0
q
EXAMPLE 7
Multiply the geometric series
q
for ƒ x ƒ 6 1,
ass a
n 2 Á + xn + Á = 1 , ax = 1 + x + x + 1  x n=0
by itself to get a power series for 1>s1  xd2 , for ƒ x ƒ 6 1. Solution
Let
q
dH
Asxd = a an x n = 1 + x + x 2 + Á + x n + Á = 1>s1  xd n=0 q
Bsxd = a bn x n = 1 + x + x 2 + Á + x n + Á = 1>s1  xd n=0
Mu
ham
ma
and
Á + ak bn  k + Á + an b0 cn = a(''''''''''')''''''''''''* 0 bn + a1 bn  1 + n + 1 terms
= ('''')''''* 1 + 1 + Á + 1 = n + 1. n + 1 ones
Then, by the Series Multiplication Theorem, q
q
Asxd # Bsxd = a cn x n = a sn + 1dx n n=0
n=0
= 1 + 2x + 3x 2 + 4x 3 + Á + sn + 1dx n + Á is the series for 1>s1  xd2 . The series all converge absolutely for ƒ x ƒ 6 1. Notice that Example 4 gives the same answer because
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d 1 1 a b = . dx 1  x s1  xd2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 804
Chapter 11: Infinite Sequences and Series
In Exercises 1–32, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?
q
4. a
sx  2dn 5. a 10 n n=0
6. a s2xdn n=0 q
q
nx n 7. a n=0 n + 2
8. a
n=1 q
xn
10. a
n 2n 3n s 1dnx n 11. a n! n=0 n=1
q
q
16. a
2n 2 + 3 q nsx + 3dn 17. a 5n n=0 n=0
q
19. a
n=0 q
n=0
s 1dnx n
q
q
n
n=1 q
1 21. a a1 + n b x n
22. a sln ndx n
23. a n nx n
24. a n!sx  4dn
n=1 q
n=1
n=0
s 1dn + 1sx + 2dn n2n n=1 q
Mu 29. a
n=1 q
31. a
n=1
s4x  5d2n + 1 n 3>2
n
sx + pd 2n
s3x + 1dn + 1 30. a 2n + 2 n=1 q
q
32. a
n=0
n
x3 x5 x7 x9 x 11 + + + Á 3! 5! 7! 9! 11!
converges to sin x for all x. a. Find the first six terms of a series for cos x. For what values of x should the series converge? b. By replacing x by 2x in the series for sin x, find a series that converges to sin 2x for all x. c. Using the result in part (a) and series multiplication, calculate the first six terms of a series for 2 sin x cos x. Compare your answer with the answer in part (b). 42. The series ex = 1 + x +
x2 x3 x4 x5 + + + + Á 2! 3! 4! 5!
converges to e x for all x.
Get the information you need about a 1>(n ln n) from Section 11.3, Exercise 38.
n
x 28. a n = 2 n ln n q
n=0
Get the information you need about 2 a 1>(n(ln n) ) from Section 11.3, Exercise 39.
xn 27. a 2 n = 2 nsln nd q
q
26. a s 2dnsn + 1dsx  1dn
ham
25. a
q
ma
n=1 q
n
40. If you integrate the series in Exercise 39 term by term, what new series do you get? For what values of x does the new series converge, and what is another name for its sum?
sin x = x 
20. a 2ns2x + 5dn n
n=0
x2  1 b 2
41. The series
2n 2 + 3
nx n 18. a n 2 n = 0 4 sn + 1d
2nx n 3n
38. a a
converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of x does the new series converge? What is its sum?
14. a
xn
n=0 q
1 1 1 sx  3d + sx  3d2 + Á + a b sx  3dn + Á 2 4 2
ass a
q
2n n n
q
x 2n + 1 n = 0 n!
15. a
sx  1dn
s2x + 3d2n + 1 n! n=0
q
n
39. For what values of x does the series 1 
3x 12. a n = 0 n!
13. a
x + 1 b 3 2
Theory and Examples
s 1dnsx + 2dn n
n=1
q
n=0
dH
q
9. a
n=1 q
q
s3x  2d n
3. a s 1dns4x + 1dn n=0 q
37. a a
n
q
36. a sln xdn
nR
n=0 q
sx + 1d2n 9n n=0 q
34. a
n 2x 35. a a  1b 2 n=0
q
2. a sx + 5dn
n=0 q
sx  1d2n 4n n=0 q
33. a
iaz
q
1. a x n
In Exercises 33–38, find the series’ interval of convergence and, within this interval, the sum of the series as a function of x.
You
Intervals of Convergence
suf
i
EXERCISES 11.7
A x  22 B
2n + 1
2n
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a. Find a series for sd>dxde x . Do you get the series for e x ? Explain your answer. b. Find a series for 1 e x dx . Do you get the series for e x ? Explain your answer. c. Replace x by x in the series for e x to find a series that converges to e x for all x. Then multiply the series for e x and e x to find the first six terms of a series for e x # e x .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7 Power Series
17x 7 62x 9 x3 2x 5 + + + + Á 3 15 315 2835
45. Uniqueness of convergent power series
i
tan x = x +
c. Check your result in part (b) by multiplying the series for sec x by the series given for tan x in Exercise 43.
suf
43. The series
a. Show that if two power series g n = 0 an x n and g n = 0 bn x n are convergent and equal for all values of x in an open interval s c, cd , then an = bn for every n. (Hint: Let q q ƒsxd = g n = 0 an x n = g n = 0 bn x n . Differentiate term by term to show that an and bn both equal f snds0d>sn!d .)
a. Find the first five terms of the series for ln ƒ sec x ƒ . For what values of x should the series converge? b. Find the first five terms of the series for sec2 x . For what values of x should this series converge?
q
You
q
converges to tan x for p>2 6 x 6 p>2 .
805
b. Show that if g n = 0 an x n = 0 for all x in an open interval s c, cd , then an = 0 for every n. q
converges to sec x for p>2 6 x 6 p>2 .
46. The sum of the series g n = 0 sn2>2n d To find the sum of this series, express 1>s1  xd as a geometric series, differentiate both sides of the resulting equation with respect to x, multiply both sides of the result by x, differentiate again, multiply by x again, and set x equal to 1> 2. What do you get? (Source: David E. Dobbs’ letter to the editor, Illinois Mathematics Teacher, Vol. 33, Issue 4, 1982, p. 27.)
a. Find the first five terms of a power series for the function ln ƒ sec x + tan x ƒ . For what values of x should the series converge?
47. Convergence at endpoints Show by examples that the convergence of a power series at an endpoint of its interval of convergence may be either conditional or absolute.
sec x = 1 +
61 6 277 8 Á 5 4 x2 x + x + x + + 2 24 720 8064
48. Make up a power series whose interval of convergence is a. s 3, 3d
b. s 2, 0d
c. (1, 5).
Mu
ham
ma
dH
ass a
b. Find the first four terms of a series for sec x tan x. For what values of x should the series converge?
iaz
44. The series
q
nR
c. Check your result in part (b) by squaring the series given for sec x in Exercise 44.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
11.8 Taylor and Maclaurin Series
11.8
Taylor and Maclaurin Series
805
u o Y
This section shows how functions that are infinitely differentiable generate power series called Taylor series. In many cases, these series can provide useful polynomial approximations of the generating functions.
Series Representations
n a ss
z a i R
We know from Theorem 19 that within its interval of convergence the sum of a power series is a continuous function with derivatives of all orders. But what about the other way around? If a function ƒ(x) has derivatives of all orders on an interval I, can it be expressed as a power series on I? And if it can, what will its coefficients be? We can answer the last question readily if we assume that ƒ(x) is the sum of a power series
a uh
M
m m
ad
a H
q
ƒsxd = a ansx  adn n=0
= a0 + a1sx  ad + a2sx  ad2 + Á + ansx  adn + Á
with a positive radius of convergence. By repeated termbyterm differentiation within the interval of convergence I we obtain ƒ¿sxd = a1 + 2a2sx  ad + 3a3sx  ad2 + Á + nansx  adn  1 + Á ƒ–sxd = 1 # 2a2 + 2 # 3a3sx  ad + 3 # 4a4sx  ad2 + Á ƒ‡sxd = 1 # 2 # 3a3 + 2 # 3 # 4a4sx  ad + 3 # 4 # 5a5sx  ad2 + Á ,
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Chapter 11: Infinite Sequences and Series
i
with the nth derivative, for all n, being
suf
f sndsxd = n!an + a sum of terms with sx  ad as a factor. Since these equations all hold at x = a, we have
and, in general, ƒsndsad = n!an .
You
ƒ¿sad = a1, # ƒ–sad = 1 2a2 , ƒ‡sad = 1 # 2 # 3a3 ,
These formulas reveal a pattern in the coefficients of any power series g n = 0 ansx  adn that converges to the values of ƒ on I (“represents ƒ on I”). If there is such a series (still an open question), then there is only one such series and its nth coefficient is
iaz
q
ƒsndsad . n!
nR
an =
If ƒ has a series representation, then the series must be
ass a
ƒsxd = ƒsad + ƒ¿sadsx  ad + + Á +
ƒ–sad sx  ad2 2!
ƒsndsad sx  adn + Á . n!
(1)
dH
But if we start with an arbitrary function ƒ that is infinitely differentiable on an interval I centered at x = a and use it to generate the series in Equation (1), will the series then converge to ƒ(x) at each x in the interior of I? The answer is maybe—for some functions it will but for other functions it will not, as we will see.
Taylor and Maclaurin Series
HISTORICAL BIOGRAPHIES Brook Taylor (1685–1731)
ma
DEFINITIONS Taylor Series, Maclaurin Series Let ƒ be a function with derivatives of all orders throughout some interval containing a as an interior point. Then the Taylor series generated by ƒ at x = a is
Mu
ham
Colin Maclaurin (1698–1746)
q
a
k=0
ƒ skdsad ƒ–sad sx  adk = ƒsad + ƒ¿sadsx  ad + sx  ad2 2! k! + Á +
ƒ sndsad sx  adn + Á . n!
The Maclaurin series generated by ƒ is ƒ–s0d 2 ƒsnds0d n ƒ skds0d k Á + x = ƒs0d + ƒ¿s0dx + x + x + Á, a k! 2! n! k=0 q
the Taylor series generated by ƒ at x = 0.
The Maclaurin series generated by ƒ is often just called the Taylor series of ƒ.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.8
807
Finding a Taylor Series
i
EXAMPLE 1
Taylor and Maclaurin Series
We need to find ƒs2d, ƒ¿s2d, ƒ–s2d, Á . Taking derivatives we get ƒsxd = x 1,
ƒs2d = 21 =
ƒ¿sxd = x 2,
ƒ¿s2d = 
1 , 22
ƒ–s2d 1 = 23 = 3 , 2! 2
iaz
ƒ–sxd = 2!x 3,
1 , 2
You
Solution
suf
Find the Taylor series generated by ƒsxd = 1>x at a = 2. Where, if anywhere, does the series converge to 1> x?
nR
ƒ‡sxd = 3!x 4,
ƒ‡s2d 1 =  4, 3! 2
o
ass a
ƒ sndsxd = s 1dnn!x sn + 1d,
o ƒ snds2d s 1dn = n+1 . n! 2
The Taylor series is
dH
ƒs2d + ƒ¿s2dsx  2d +
=
ƒ–s2d ƒsnds2d sx  2d2 + Á + sx  2dn + Á 2! n!
n sx  2d2 sx  2d 1 Á + s 1dn sx  2d + Á . + 2 22 23 2n + 1
This is a geometric series with first term 1> 2 and ratio r = sx  2d>2. It converges absolutely for ƒ x  2 ƒ 6 2 and its sum is
ma
1>2 1 1 = = x. 1 + sx  2d>2 2 + sx  2d
Mu
ham
In this example the Taylor series generated by ƒsxd = 1>x at a = 2 converges to 1> x for ƒ x  2 ƒ 6 2 or 0 6 x 6 4.
Taylor Polynomials The linearization of a differentiable function ƒ at a point a is the polynomial of degree one given by P1sxd = ƒsad + ƒ¿sadsx  ad. In Section 3.8 we used this linearization to approximate ƒ(x) at values of x near a. If ƒ has derivatives of higher order at a, then it has higherorder polynomial approximations as well, one for each available derivative. These polynomials are called the Taylor polynomials of ƒ.
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Chapter 11: Infinite Sequences and Series
+
ƒ–sad sx  ad2 + Á 2!
You
Pnsxd = ƒsad + ƒ¿sadsx  ad +
suf
i
DEFINITION Taylor Polynomial of Order n Let ƒ be a function with derivatives of order k for k = 1, 2, Á , N in some interval containing a as an interior point. Then for any integer n from 0 through N, the Taylor polynomial of order n generated by ƒ at x = a is the polynomial
ƒskdsad ƒsndsad sx  adk + Á + sx  adn . n! k!
EXAMPLE 2 y
nR
iaz
We speak of a Taylor polynomial of order n rather than degree n because ƒsndsad may be zero. The first two Taylor polynomials of ƒsxd = cos x at x = 0, for example, are P0sxd = 1 and P1sxd = 1. The firstorder Taylor polynomial has degree zero, not one. Just as the linearization of ƒ at x = a provides the best linear approximation of ƒ in the neighborhood of a, the higherorder Taylor polynomials provide the best polynomial approximations of their respective degrees. (See Exercise 32.)
Finding Taylor Polynomials for e x
3.0
y e x y P3(x)
Solution
we have
ƒ¿sxd = e x,
ƒs0d = e 0 = 1,
dH
2.0
Since
ƒsxd = e x,
y P2(x)
2.5
ass a
Find the Taylor series and the Taylor polynomials generated by ƒsxd = e x at x = 0.
y P1(x)
ƒ¿s0d = 1,
ƒs0d + ƒ¿s0dx +
ma
1.0
0
0.5
1.0
FIGURE 11.12 The graph of ƒsxd = e x and its Taylor polynomials P1sxd = 1 + x P2sxd = 1 + x + sx 2>2!d P3sxd = 1 + x + sx 2>2!d + sx 3>3!d. Notice the very close agreement near the center x = 0 (Example 2).
Mu
Á,
ƒsnds0d = 1,
Á,
.Á
ƒsnds0d n ƒ–s0d 2 x + Á + x + Á 2! n! = 1 + x +
x2 xn + Á + + Á 2 n!
q
xk = a . k = 0 k!
x
ham
–0.5
ƒsndsxd = e x,
The Taylor series generated by ƒ at x = 0 is
1.5
0.5
Á,
This is also the Maclaurin series for e x . In Section 11.9 we will see that the series converges to e x at every x. The Taylor polynomial of order n at x = 0 is Pnsxd = 1 + x +
x2 xn + Á + . 2 n!
See Figure 11.12.
EXAMPLE 3
Finding Taylor Polynomials for cos x
Find the Taylor series and Taylor polynomials generated by ƒsxd = cos x at x = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.8
809
ƒsxd =
i
The cosine and its derivatives are sin x,
ƒ¿sxd =
cos x,
suf
Solution
Taylor and Maclaurin Serie
s3d
ƒ sxd =
ƒ–sxd = cos x, o ƒs2ndsxd = s 1dn cos x,
sin x,
You
o ƒs2n + 1dsxd = s 1dn + 1 sin x.
At x = 0, the cosines are 1 and the sines are 0, so ƒs2nds0d = s 1dn,
ƒs2n + 1ds0d = 0.
The Taylor series generated by ƒ at 0 is
ƒ–s0d 2 ƒ‡s0d 3 ƒsnds0d n x + x + Á + x + Á 2! 3! n!
= 1 + 0#x q
x2 x4 x 2n + 0 # x3 + + Á + s 1dn + Á 2! 4! s2nd!
s 1dkx 2k . s2kd!
nR
= a
iaz
ƒs0d + ƒ¿s0dx +
k=0
ass a
This is also the Maclaurin series for cos x. In Section 11.9, we will see that the series converges to cos x at every x. Because ƒs2n + 1ds0d = 0, the Taylor polynomials of orders 2n and 2n + 1 are identical: P2nsxd = P2n + 1sxd = 1 
x2 x4 x 2n +  Á + s 1dn . 2! 4! s2nd!
dH
Figure 11.13 shows how well these polynomials approximate ƒsxd = cos x near x = 0. Only the righthand portions of the graphs are given because the graphs are symmetric about the yaxis.
y
Mu
ham
ma
2
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1
P4
P0
P8
P12
P16 y cos x
0
1
2
3
4
5
6
7
8
9
x
–1 P2
P6
P10
P14
P18
–2
FIGURE 11.13 The polynomials n
P2nsxd = a
k=0
s 1dkx 2k s2kd!
converge to cos x as n : q . We can deduce the behavior of cos x arbitrarily far away solely from knowing the values of the cosine and its derivatives at x = 0 (Example 3).
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Chapter 11: Infinite Sequences and Series
A Function ƒ Whose Taylor Series Converges at Every x but Converges to ƒ(x) Only at x = 0
suf
i
EXAMPLE 4
It can be shown (though not easily) that 0, 2 e 1>x ,
x = 0 x Z 0
You
ƒsxd = e
(Figure 11.14) has derivatives of all orders at x = 0 and that ƒsnds0d = 0 for all n. This means that the Taylor series generated by ƒ at x = 0 is ƒs0d + ƒ¿s0dx +
ƒ–s0d 2 ƒsnds0d n x + Á + x + Á 2! n!
iaz
= 0 + 0 # x + 0 # x2 + Á + 0 # xn + Á = 0 + 0 + Á + 0 + Á.
The series converges for every x (its sum is 0) but converges to ƒ(x) only at x = 0. 0 , x0 y –1/x2 e , x 0
nR
y
ass a
1
–4
–3
–2
–1
0
1
2
3
4
x
dH
FIGURE 11.14 The graph of the continuous extension of 2 y = e 1>x is so flat at the origin that all of its derivatives there are zero (Example 4).
Two questions still remain. 1.
ma
2.
For what values of x can we normally expect a Taylor series to converge to its generating function? How accurately do a function’s Taylor polynomials approximate the function on a given interval?
Mu
ham
The answers are provided by a theorem of Taylor in the next section.
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Chapter 11: Infinite Sequences and Series
EXERCISES 11.8
R n ssa
Finding Taylor Polynomials
a H d
In Exercises 1–8, find the Taylor polynomials of orders 0, 1, 2, and 3 generated by ƒ at a. 1. ƒsxd = ln x,
a = 1
2. ƒsxd = ln s1 + xd,
a = 0
3. ƒsxd = 1>x,
a = 2
4. ƒsxd = 1>sx + 2d,
a = 0
a m m a h u M
5. ƒsxd = sin x,
a = p>4
6. ƒsxd = cos x,
7. ƒsxd = 2x,
a = 4
8. ƒsxd = 2x + 4,
a = p>4
a = 0
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i f u s u o Y iaz
Finding Taylor Series at x = 0 (Maclaurin Series)
Find the Maclaurin series for the functions in Exercises 9–20. 9. e x
11.
1 1 + x
13. sin 3x
10. e x>2 1 1  x x 14. sin 2 12.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.8 Taylor and Maclaurin Series
e + e 2
18. sinh x =
19. x 4  2x 3  5x + 4
e  e 2
x
20. sx + 1d2
Finding Taylor Series
32. Of all polynomials of degree ◊ n, the Taylor polynomial of order n gives the best approximation Suppose that ƒ(x) is differentiable on an interval centered at x = a and that gsxd = b0 + b1sx  ad + Á + bnsx  adn is a polynomial of degree n with constant coefficients b0, Á , bn . Let Esxd = ƒsxd  gsxd . Show that if we impose on g the conditions
i
17. cosh x =
x
x
2
23. ƒsxd = x + x + 1,
x:a
a = 1
then
a = 2
24. ƒsxd = 3x 5  x 4 + 2x 3 + x 2  2,
gsxd = ƒsad + ƒ¿sadsx  ad +
a = 1
a = 1
26. ƒsxd = x>s1  xd, 27. ƒsxd = e x,
a = 2
28. ƒsxd = 2x,
a = 1
a = 0
ƒ–sad sx  ad2 + Á 2! +
ƒ sndsad sx  adn . n!
Thus, the Taylor polynomial Pnsxd is the only polynomial of degree less than or equal to n whose error is both zero at x = a and negligible when compared with sx  adn .
Theory and Examples 29. Use the Taylor series generated by e x at x = a to show that sx  ad2 e = e c1 + sx  ad + + Á d. 2! a
Quadratic Approximations
The Taylor polynomial of order 2 generated by a twicedifferentiable function ƒ(x) at x = a is called the quadratic approximation of ƒ at x = a . In Exercises 33–38, find the (a) linearization (Taylor polynomial of order 1) and (b) quadratic approximation of ƒ at x = 0 .
ass a
x
The error is negligible when compared to sx  adn .
iaz
25. ƒsxd = 1>x 2,
Esxd = 0, sx  adn
nR
4
b. lim
a = 2
22. ƒsxd = 2x 3 + x 2 + 3x  8,
The approximation error is zero at x = a .
a. Esad = 0
In Exercises 21–28, find the Taylor series generated by ƒ at x = a . 21. ƒsxd = x 3  2x + 4,
suf
16. 5 cos px x
You
15. 7 cos s xd
811
30. (Continuation of Exercise 29.) Find the Taylor series generated by e x at x = 1 . Compare your answer with the formula in Exercise 29.
34. ƒsxd = e sin x
33. ƒsxd = ln scos xd 35. ƒsxd = 1> 21  x 37. ƒsxd = sin x
2
36. ƒsxd = cosh x 38. ƒsxd = tan x
Mu
ham
ma
dH
31. Let ƒ(x) have derivatives through order n at x = a . Show that the Taylor polynomial of order n and its first n derivatives have the same values that ƒ and its first n derivatives have at x = a .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
11.9 Convergence of Taylor Series; Error Estimates
Y z a i R
Convergence of Taylor Series; Error Estimates
11.9
n a s as
811
This section addresses the two questions left unanswered by Section 11.8: 1. 2.
H d a m
m a h
Mu
When does a Taylor series converge to its generating function? How accurately do a function’s Taylor polynomials approximate the function on a given interval?
Taylor’s Theorem We answer these questions with the following theorem.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 812
Chapter 11: Infinite Sequences and Series
ƒsndsad ƒsn + 1dscd sb  adn + sb  adn + 1 . n! sn + 1d!
iaz
+
ƒ–sad sb  ad2 + Á 2!
You
ƒsbd = ƒsad + ƒ¿sadsb  ad +
suf
i
THEOREM 22 Taylor’s Theorem If ƒ and its first n derivatives ƒ¿, ƒ–, Á , ƒsnd are continuous on the closed interval between a and b, and ƒsnd is differentiable on the open interval between a and b, then there exists a number c between a and b such that
nR
Taylor’s Theorem is a generalization of the Mean Value Theorem (Exercise 39). There is a proof of Taylor’s Theorem at the end of this section. When we apply Taylor’s Theorem, we usually want to hold a fixed and treat b as an independent variable. Taylor’s formula is easier to use in circumstances like these if we change b to x. Here is a version of the theorem with this change.
ass a
Taylor’s Formula If ƒ has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, ƒsxd = ƒsad + ƒ¿sadsx  ad +
dH
+
ƒ–sad sx  ad2 + Á 2!
ƒsndsad sx  adn + Rnsxd, n!
(1)
where
ma
Rnsxd =
f sn + 1dscd sx  adn + 1 sn + 1d!
for some c between a and x.
(2)
Mu
ham
When we state Taylor’s theorem this way, it says that for each x H I, ƒsxd = Pnsxd + Rnsxd.
The function Rnsxd is determined by the value of the sn + 1dst derivative ƒsn + 1d at a point c that depends on both a and x, and which lies somewhere between them. For any value of n we want, the equation gives both a polynomial approximation of ƒ of that order and a formula for the error involved in using that approximation over the interval I. Equation (1) is called Taylor’s formula. The function Rnsxd is called the remainder of order n or the error term for the approximation of ƒ by Pnsxd over I. If Rnsxd : 0 as n : q for all x H I, we say that the Taylor series generated by ƒ at x = a converges to ƒ on I, and we write q
ƒsxd = a
k=0
ƒskdsad sx  adk . k!
Often we can estimate Rn without knowing the value of c, as the following example illustrates.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9
813
The Taylor Series for e x Revisited
i
EXAMPLE 1
Convergence of Taylor Series; Error Estimates
suf
Show that the Taylor series generated by ƒsxd = e x at x = 0 converges to ƒ(x) for every real value of x. The function has derivatives of all orders throughout the interval I = s  q , q d. Equations (1) and (2) with ƒsxd = e x and a = 0 give ex = 1 + x +
You
Solution
xn x2 + Á + + Rnsxd 2! n!
and ec xn+1 sn + 1d!
for some c between 0 and x.
iaz
Rnsxd =
Polynomial from Section 11.8, Example 2
nR
Since e x is an increasing function of x, e c lies between e 0 = 1 and e x . When x is negative, so is c, and e c 6 1. When x is zero, e x = 1 and Rnsxd = 0. When x is positive, so is c, and e c 6 e x . Thus, ƒ Rnsxd ƒ …
ass a
and
ƒ x ƒn+1 sn + 1d!
ƒ Rnsxd ƒ 6 e x
xn+1 sn + 1d!
when x … 0,
when x 7 0.
Finally, because
xn+1 = 0 n: q sn + 1d!
dH
lim
for every x,
Section 11.1
lim Rnsxd = 0, and the series converges to e x for every x. Thus,
n: q
q
ma
x2 xk xk ex = a = 1 + x + + Á. + Á + 2! k! k = 0 k!
(3)
Estimating the Remainder
Mu
ham
It is often possible to estimate Rnsxd as we did in Example 1. This method of estimation is so convenient that we state it as a theorem for future reference.
THEOREM 23 The Remainder Estimation Theorem If there is a positive constant M such that ƒ ƒsn + 1dstd ƒ … M for all t between x and a, inclusive, then the remainder term Rnsxd in Taylor’s Theorem satisfies the inequality ƒ Rnsxd ƒ … M
ƒ x  a ƒ n+1 . sn + 1d!
If this condition holds for every n and the other conditions of Taylor’s Theorem are satisfied by ƒ, then the series converges to ƒ(x).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 814
Chapter 11: Infinite Sequences and Series
The Taylor Series for sin x at x = 0
You
EXAMPLE 2
suf
i
We are now ready to look at some examples of how the Remainder Estimation Theorem and Taylor’s Theorem can be used together to settle questions of convergence. As you will see, they can also be used to determine the accuracy with which a function is approximated by one of its Taylor polynomials.
Show that the Taylor series for sin x at x = 0 converges for all x. The function and its derivatives are ƒsxd =
sin x,
ƒ¿sxd =
cos x,
ƒ–sxd =
 sin x,
ƒ‡sxd =
 cos x,
o
iaz
Solution
o
ƒ(2k + 1)sxd = s 1dk cos x,
nR
ƒ(2k)sxd = s 1dk sin x, so
f s2kds0d = 0
and
f s2k + 1ds0d = s 1dk .
ass a
The series has only oddpowered terms and, for n = 2k + 1, Taylor’s Theorem gives sin x = x 
s 1dkx 2k + 1 x3 x5 +  Á + + R2k + 1sxd. 3! 5! s2k + 1d!
dH
All the derivatives of sin x have absolute values less than or equal to 1, so we can apply the Remainder Estimation Theorem with M = 1 to obtain ƒ R2k + 1sxd ƒ … 1 #
ƒ x ƒ 2k + 2 . s2k + 2d!
Mu
ham
ma
Since s ƒ x ƒ 2k + 2>s2k + 2d!d : 0 as k : q , whatever the value of x, R2k + 1sxd : 0, and the Maclaurin series for sin x converges to sin x for every x. Thus,
EXAMPLE 3
q s 1dkx 2k + 1 x3 x5 x7 sin x = a = x + + Á. 3! 5! 7! k = 0 s2k + 1d!
(4)
The Taylor Series for cos x at x = 0 Revisited
Show that the Taylor series for cos x at x = 0 converges to cos x for every value of x. We add the remainder term to the Taylor polynomial for cos x (Section 11.8, Example 3) to obtain Taylor’s formula for cos x with n = 2k:
Solution
cos x = 1 
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x2 x4 x 2k +  Á + s 1dk + R2ksxd. 2! 4! s2kd!
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9
Convergence of Taylor Series; Error Estimates
815
ƒ R2ksxd ƒ … 1 #
ƒ x ƒ 2k + 1 . s2k + 1d!
suf
i
Because the derivatives of the cosine have absolute value less than or equal to 1, the Remainder Estimation Theorem with M = 1 gives
q
cos x = a
k=0
s 1dkx 2k x2 x4 x6 = 1 + + Á. 2! 4! 6! s2kd!
(5)
Finding a Taylor Series by Substitution
iaz
EXAMPLE 4
You
For every value of x, R2k : 0 as k : q . Therefore, the series converges to cos x for every value of x. Thus,
Find the Taylor series for cos 2x at x = 0.
q
cos 2x = a
k=0
nR
Solution We can find the Taylor series for cos 2x by substituting 2x for x in the Taylor series for cos x:
s2xd4 s2xd6 s 1dks2xd2k s2xd2 + + Á = 1 2! 4! 6! s2kd! 22x 2 24x 4 26x 6 + + Á 2! 4! 6!
ass a
= 1 
Equation (5) with 2x for x
q
22kx 2k = a s 1dk . s2kd! k=0
dH
Equation (5) holds for  q 6 x 6 q , implying that it holds for  q 6 2x 6 q , so the newly created series converges for all x. Exercise 45 explains why the series is in fact the Taylor series for cos 2x.
EXAMPLE 5
Finding a Taylor Series by Multiplication
ma
Find the Taylor series for x sin x at x = 0. We can find the Taylor series for x sin x by multiplying the Taylor series for sin x (Equation 4) by x:
Mu
ham
Solution
x sin x = x ax = x2 
x3 x5 x7 + + Áb 3! 5! 7!
x6 x8 x4 + + Á. 3! 5! 7!
The new series converges for all x because the series for sin x converges for all x. Exercise 45 explains why the series is the Taylor series for x sin x.
Truncation Error The Taylor series for e x at x = 0 converges to e x for all x. But we still need to decide how many terms to use to approximate e x to a given degree of accuracy. We get this information from the Remainder Estimation Theorem.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 816
Chapter 11: Infinite Sequences and Series
Calculate e with an error of less than 10 6 .
We can use the result of Example 1 with x = 1 to write e = 1 + 1 +
1 1 + Á + + Rns1d, 2! n!
You
Solution
suf
i
EXAMPLE 6
with Rns1d = e c
1 sn + 1d!
for some c between 0 and 1.
iaz
For the purposes of this example, we assume that we know that e 6 3. Hence, we are certain that
nR
3 1 6 Rns1d 6 sn + 1d! sn + 1d!
because 1 6 e c 6 3 for 0 6 c 6 1. By experiment we find that 1>9! 7 10 6 , while 3>10! 6 10 6 . Thus we should take sn + 1d to be at least 10, or n to be at least 9. With an error of less than 10 6 , 1 1 1 + + Á + L 2.718282. 2 3! 9!
ass a
e = 1 + 1 +
For what values of x can we replace sin x by x  sx 3>3!d with an error of magnitude no greater than 3 * 10 4 ?
EXAMPLE 7
Here we can take advantage of the fact that the Taylor series for sin x is an alternating series for every nonzero value of x. According to the Alternating Series Estimation Theorem (Section 11.6), the error in truncating sin x = x 
Mu
ham
ma
after sx 3>3!d is no greater than
`
x3 x5 +  Á 3! 5! 
dH
Solution
ƒ x ƒ5 x5 ` = . 120 5!
Therefore the error will be less than or equal to 3 * 10 4 if ƒ x ƒ5 6 3 * 10 4 120
or
5 ƒ x ƒ 6 2360 * 10 4 L 0.514.
Rounded down, to be safe
The Alternating Series Estimation Theorem tells us something that the Remainder Estimation Theorem does not: namely, that the estimate x  sx 3>3!d for sin x is an underestimate when x is positive because then x 5>120 is positive. Figure 11.15 shows the graph of sin x, along with the graphs of a number of its approximating Taylor polynomials. The graph of P3sxd = x  sx 3>3!d is almost indistinguishable from the sine curve when 1 … x … 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9
Convergence of Taylor Series; Error Estimates
817
2 P1
P5
P9
P13
P17
5
6
7
y sin x
1
1
2
3
4
–1 P3
P7
8
9
x
You
0
suf
i
y
P11
–2
P19
iaz
FIGURE 11.15 The polynomials
P15
n s 1dkx 2k + 1 P2n + 1sxd = a k = 0 s2k + 1d!
nR
converge to sin x as n : q . Notice how closely P3sxd approximates the sine curve for x 6 1 (Example 7).
ass a
You might wonder how the estimate given by the Remainder Estimation Theorem compares with the one just obtained from the Alternating Series Estimation Theorem. If we write sin x = x 
x3 + R3 , 3!
then the Remainder Estimation Theorem gives ƒ x ƒ4 ƒ x ƒ4 ƒ R3 ƒ … 1 # 4! = 24 ,
dH
which is not as good. But if we recognize that x  sx 3>3!d = 0 + x + 0x 2 sx 3/3!d + 0x 4 is the Taylor polynomial of order 4 as well as of order 3, then
ma
sin x = x 
x3 + 0 + R4 , 3!
Mu
ham
and the Remainder Estimation Theorem with M = 1 gives ƒ x ƒ5 ƒ x ƒ5 ƒ R4 ƒ … 1 # 5! = 120 .
This is what we had from the Alternating Series Estimation Theorem.
Combining Taylor Series On the intersection of their intervals of convergence, Taylor series can be added, subtracted, and multiplied by constants, and the results are once again Taylor series. The Taylor series for ƒsxd + gsxd is the sum of the Taylor series for ƒ(x) and g(x) because the nth derivative of f + g is f snd + g snd , and so on. Thus we obtain the Taylor series for s1 + cos 2xd>2 by adding 1 to the Taylor series for cos 2x and dividing the combined results by 2, and the Taylor series for sin x + cos x is the termbyterm sum of the Taylor series for sin x and cos x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 818
Chapter 11: Infinite Sequences and Series
i
Euler’s Identity
i 3 = i 2i = i,
i 4 = i 2i 2 = 1,
and so on, to simplify the result, we obtain e iu = 1 +
i 5 = i 4i = i,
You
i 2 = 1,
suf
As you may recall, a complex number is a number of the form a + bi, where a and b are real numbers and i = 21. If we substitute x = iu (u real) in the Taylor series for e x and use the relations
i 2u2 i 3u3 i 4u4 i 5u5 i 6u6 iu + + + + + + Á 1! 2! 3! 4! 5! 6!
u4 u6 u3 u5 u2 + + Á b + i au +  Á b = cos u + i sin u. 2! 4! 6! 3! 5!
iaz
= a1 
DEFINITION
nR
This does not prove that e iu = cos u + i sin u because we have not yet defined what it means to raise e to an imaginary power. Rather, it says how to define e iu to be consistent with other things we know.
For any real number u, e iu = cos u + i sin u.
ass a
(6)
Equation (6), called Euler’s identity, enables us to define e a + bi to be e a # e bi for any complex number a + bi. One consequence of the identity is the equation
dH
e ip = 1.
When written in the form e ip + 1 = 0, this equation combines five of the most important constants in mathematics.
A Proof of Taylor’s Theorem
ma
We prove Taylor’s theorem assuming a 6 b. The proof for a 7 b is nearly the same. The Taylor polynomial
Mu
ham
Pnsxd = ƒsad + ƒ¿sadsx  ad +
ƒ–sad f sndsad sx  ad2 + Á + sx  adn 2! n!
and its first n derivatives match the function ƒ and its first n derivatives at x = a. We do not disturb that matching if we add another term of the form Ksx  adn + 1 , where K is any constant, because such a term and its first n derivatives are all equal to zero at x = a. The new function fnsxd = Pnsxd + Ksx  adn + 1 and its first n derivatives still agree with ƒ and its first n derivatives at x = a. We now choose the particular value of K that makes the curve y = fnsxd agree with the original curve y = ƒsxd at x = b. In symbols, ƒsbd = Pnsbd + Ksb  adn + 1,
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or
K =
ƒsbd  Pnsbd sb  adn + 1
.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9 Convergence of Taylor Series; Error Estimates
819
i
With K defined by Equation (7), the function
suf
Fsxd = ƒsxd  fnsxd
F¿sc1 d = 0
You
measures the difference between the original function ƒ and the approximating function fn for each x in [a, b]. We now use Rolle’s Theorem (Section 4.2). First, because Fsad = Fsbd = 0 and both F and F¿ are continuous on [a, b], we know that for some c1 in sa, bd.
Next, because F¿sad = F¿sc1 d = 0 and both F¿ and F– are continuous on [a, c1], we know that for some c2 in sa, c1 d.
iaz
F–sc2 d = 0
Rolle’s Theorem, applied successively to F–, F‡, Á , F sn  1d implies the existence of in sa, c2 d
such that F‡sc3 d = 0,
c4
in sa, c3 d
such that F s4dsc4 d = 0,
nR
c3
o
in sa, cn  1 d
cn
such that F sndscn d = 0.
ass a
Finally, because F snd is continuous on [a, cn] and differentiable on sa, cn d, and F sndsad = F sndscn d = 0, Rolle’s Theorem implies that there is a number cn + 1 in sa, cn d such that F sn + 1dscn + 1 d = 0. n+1
If we differentiate Fsxd = ƒsxd  Pnsxd  Ksx  ad
(8)
a total of n + 1 times, we get
dH
F sn + 1dsxd = ƒsn + 1dsxd  0  sn + 1d!K.
(9)
Equations (8) and (9) together give K =
ƒsn + 1dscd sn + 1d!
for some number c = cn + 1 in sa, bd.
ma
Equations (7) and (10) give ƒsbd = Pnsbd +
ƒsn + 1dscd sb  adn + 1 . sn + 1d!
Mu
ham
This concludes the proof.
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i f u s u o
11.9 Convergence of Taylor Series; Error Estimates
Y z a i R n a s as
EXERCISES 11.9 Taylor Series by Substitution
More Taylor Series
H d a
Use substitution (as in Example 4) to find the Taylor series at x = 0 of the functions in Exercises 1–6. 1. e
m m ha
5x
px 4. sin a b 2
2. e
x>2
5. cos 2x + 1
Mu
819
3. 5 sin s xd
6. cos A x 3>2> 22 B
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Find Taylor series at x = 0 for the functions in Exercises 7–18. 7. xe x
10. sin x  x +
8. x 2 sin x
x3 3!
11. x cos px
9.
x2  1 + cos x 2
12. x 2 cos sx 2 d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 820
Chapter 11: Infinite Sequences and Series
1 s1  xd2
18.
16. x ln s1 + 2xd 2 s1  xd3
i
x2 1  2x
Finding and Identifying Maclaurin Series
Error Estimates 19. For approximately what values of x can you replace sin x by x  sx 3>6d with an error of magnitude no greater than 5 * 10 4 ? Give reasons for your answer. 20. If cos x is replaced by 1  sx 2>2d and ƒ x ƒ 6 0.5 , what estimate can be made of the error? Does 1  sx 2>2d tend to be too large, or too small? Give reasons for your answer.
Recall that the Maclaurin series is just another name for the Taylor series at x = 0 . Each of the series in Exercises 31–34 is the value of the Maclaurin series of a function ƒ(x) at some point. What function and what point? What is the sum of the series? 31. s0.1d 32. 1 
21. How close is the approximation sin x = x when ƒ x ƒ 6 10 3 ? For which of these values of x is x 6 sin x ?
33.
22. The estimate 21 + x = 1 + sx>2d is used when x is small. Estimate the error when ƒ x ƒ 6 0.01 .
34. p 
s 1dkspd2k p4 p2 + 4  Á + 2k + Á 4 # 2! 4 # 4! 4 # s2k!d 2
s 1dkp2k + 1 p5 p p3 + 5  Á + 2k + 1 + Á  3 3 3 #5 3 #3 3 s2k + 1d pk p2 p3 +  Á + s 1dk  1 + Á 2 3 k
35. Multiply the Maclaurin series for e x and sin x together to find the first five nonzero terms of the Maclaurin series for e x sin x . 36. Multiply the Maclaurin series for e x and cos x together to find the first five nonzero terms of the Maclaurin series for e x cos x .
ass a
24. (Continuation of Exercise 23.) When x 6 0 , the series for e x is an alternating series. Use the Alternating Series Estimation Theorem to estimate the error that results from replacing e x by 1 + x + sx 2>2d when 0.1 6 x 6 0 . Compare your estimate with the one you obtained in Exercise 23.
s0.1d3 s0.1d5 s 1dks0.1d2k + 1 +  Á + + Á 3! 5! s2k + 1d!
nR
23. The approximation e x = 1 + x + sx 2>2d is used when x is small. Use the Remainder Estimation Theorem to estimate the error when ƒ x ƒ 6 0.1 .
You
17.
15.
iaz
14. sin2 x
T b. Graph ƒsxd = s1  cos xd>x 2 together with y = s1>2d  sx 2>24d and y = 1>2 for 9 … x … 9 . Comment on the relationships among the graphs.
suf
13. cos2 x (Hint: cos2 x = s1 + cos 2xd>2 .)
dH
25. Estimate the error in the approximation sinh x = x + sx 3>3!d when ƒ x ƒ 6 0.5 . (Hint: Use R4 , not R3 .) 26. When 0 … h … 0.01 , show that e h may be replaced by 1 + h with an error of magnitude no greater than 0.6% of h. Use e 0.01 = 1.01 . 27. For what positive values of x can you replace ln s1 + xd by x with an error of magnitude no greater than 1% of the value of x?
ma
28. You plan to estimate p>4 by evaluating the Maclaurin series for tan1 x at x = 1 . Use the Alternating Series Estimation Theorem to determine how many terms of the series you would have to add to be sure the estimate is good to two decimal places.
ham
29. a. Use the Taylor series for sin x and the Alternating Series Estimation Theorem to show that 1 
sin x x2 6 x 6 1, 6
x Z 0.
Mu
T b. Graph ƒsxd = ssin xd>x together with the functions y = 1  sx 2>6d and y = 1 for 5 … x … 5 . Comment on the relationships among the graphs. 30. a. Use the Taylor series for cos x and the Alternating Series Estimation Theorem to show that 1  cos x x2 1 1 6 , 6 2 24 2 x2
x Z 0.
(This is the inequality in Section 2.2, Exercise 52.)
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37. Use the identity sin2 x = s1  cos 2xd>2 to obtain the Maclaurin series for sin2 x . Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin 2x.
38. (Continuation of Exercise 37.) Use the identity cos2 x = cos 2x + sin2 x to obtain a power series for cos2 x .
Theory and Examples 39. Taylor’s Theorem and the Mean Value Theorem Explain how the Mean Value Theorem (Section 4.2, Theorem 4) is a special case of Taylor’s Theorem. 40. Linearizations at inflection points Show that if the graph of a twicedifferentiable function ƒ(x) has an inflection point at x = a , then the linearization of ƒ at x = a is also the quadratic approximation of ƒ at x = a . This explains why tangent lines fit so well at inflection points. 41. The (second) second derivative test ƒsxd = ƒsad + ƒ¿sadsx  ad +
Use the equation ƒ–sc2 d sx  ad2 2
to establish the following test. Let ƒ have continuous first and second derivatives and suppose that ƒ¿sad = 0 . Then a. ƒ has a local maximum at a if ƒ– … 0 throughout an interval whose interior contains a; b. ƒ has a local minimum at a if ƒ– Ú 0 throughout an interval whose interior contains a.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9 Convergence of Taylor Series; Error Estimates
i
T 48. a. Graph the curves y = s1>3d  sx 2 d>5 and y = sx  tan1 xd>x 3 together with the line y = 1>3 . b. Use a Taylor series to explain what you see. What is
43. a. Use Taylor’s formula with n = 2 to find the quadratic approximation of ƒsxd = s1 + xdk at x = 0 (k a constant).
a. Let P be an approximation of p accurate to n decimals. Show that P + sin P gives an approximation correct to 3n decimals. (Hint: Let P = p + x .)
Euler’s Identity
49. Use Equation (6) to write the following powers of e in the form a + bi .
cos u =
and x4 x5 x e = x + x + + + Á, 2! 3! 2
3
dH
2 x
e iu + e iu 2
obtained by multiplying Taylor series by powers of x, as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent.
ham
ma
46. Taylor series for even functions and odd functions (Continuaq tion of Section 11.7, Exercise 45.) Suppose that ƒsxd = g n = 0 an x n converges for all x in an open interval s c, cd . Show that a. If ƒ is even, then a1 = a3 = a5 = Á = 0 , i.e., the Taylor series for ƒ at x = 0 contains only even powers of x. b. If ƒ is odd, then a0 = a2 = a4 = Á = 0 , i.e., the Taylor series for ƒ at x = 0 contains only odd powers of x. 47. Taylor polynomials of periodic functions a. Show that every continuous periodic function ƒsxd,  q 6 x 6 q , is bounded in magnitude by showing that there exists a positive constant M such that ƒ ƒsxd ƒ … M for all x.
Mu
b. Show that the graph of every Taylor polynomial of positive degree generated by ƒsxd = cos x must eventually move away from the graph of cos x as ƒ x ƒ increases. You can see this in Figure 11.13. The Taylor polynomials of sin x behave in a similar way (Figure 11.15).
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and
sin u =
e iu  e iu . 2i
nR
51. Establish the equations in Exercise 50 by combining the formal Taylor series for e iu and e iu . 52. Show that
a. cosh iu = cos u ,
b. sinh iu = i sin u .
53. By multiplying the Taylor series for e x and sin x, find the terms through x 5 of the Taylor series for e x sin x . This series is the imaginary part of the series for
ass a
x4 x6 x8 + + Á 3! 5! 7!
c. e ip>2
50. Use Equation (6) to show that
q
x sin x = x 2 
b. e ip>4
a. e ip
T b. Try it with a calculator. 45. The Taylor series generated by ƒsxd = g n = 0 an x n is q q g n = 0 an x n A function defined by a power series g n = 0 an x n with a radius of convergence c 7 0 has a Taylor series that converges to the function at every point of s c, cd . Show this by q showing that the Taylor series generated by ƒsxd = g n = 0 an x n is q n the series g n = 0 an x itself. An immediate consequence of this is that series like
You
44. Improving approximations to P
x  tan1 x ? x:0 x3 lim
iaz
b. If k = 3 , for approximately what values of x in the interval [0, 1] will the error in the quadratic approximation be less than 1> 100?
suf
42. A cubic approximation Use Taylor’s formula with a = 0 and n = 3 to find the standard cubic approximation of ƒsxd = 1>s1  xd at x = 0 . Give an upper bound for the magnitude of the error in the approximation when ƒ x ƒ … 0.1 .
821
e x # e ix = e s1 + idx .
Use this fact to check your answer. For what values of x should the series for e x sin x converge?
54. When a and b are real, we define e sa + ibdx with the equation e sa + ibdx = e ax # e ibx = e axscos bx + i sin bxd .
Differentiate the righthand side of this equation to show that d sa + ibdx e = sa + ibde sa + ibdx . dx Thus the familiar rule sd>dxde kx = ke kx holds for k complex as well as real. 55. Use the definition of e iu to show that for any real numbers u, u1 , and u2 , a. e iu1e iu2 = e isu1 + u2d,
b. e iu = 1>e iu .
56. Two complex numbers a + ib and c + id are equal if and only if a = c and b = d . Use this fact to evaluate L
e ax cos bx dx
and
L
e ax sin bx dx
from L
e sa + ibdx dx =
a  ib sa + ibdx e + C, a2 + b2
where C = C1 + iC2 is a complex constant of integration.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
a. For what values of x can the function be replaced by each approximation with an error less than 10 2 ? b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals in Exercises 57–62. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials P1sxd, P2sxd, and P3sxd at x = 0.
i
Step 5: Compare your estimated error with the actual error Ensxd = ƒ ƒsxd  Pnsxd ƒ by plotting Ensxd over the specified interval. This will help answer question (b).
Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. 57. ƒsxd =
1 21 + x
,
3 ƒxƒ … 4
58. ƒsxd = s1 + xd3>2,

1 … x … 2 2
x , ƒxƒ … 2 x2 + 1 60. ƒsxd = scos xdssin 2xd, ƒ x ƒ … 2 61. ƒsxd = e x cos 2x, ƒ x ƒ … 1 59. ƒsxd =
62. ƒsxd = e x>3 sin 2x,
ƒxƒ … 2
Mu
ham
ma
dH
ass a
Step 3: Calculate the sn + 1dst derivative ƒsn + 1dscd associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of c over the specified interval and estimate its maximum absolute value, M.
suf
Taylor’s formula with n = 1 and a = 0 gives the linearization of a function at x = 0 . With n = 2 and n = 3 we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions:
Step 4: Calculate the remainder Rnsxd for each polynomial. Using the estimate M from Step 3 in place of ƒsn + 1dscd, plot Rnsxd over the specified interval. Then estimate the values of x that answer question (a).
You
Linear, Quadratic, and Cubic Approximations
iaz
COMPUTER EXPLORATIONS
nR
822
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 822
i f su
Chapter 11: Infinite Sequences and Series
u o zY
Applications of Power Series
11.10
a i R
This section introduces the binomial series for estimating powers and roots and shows how series are sometimes used to approximate the solution of an initial value problem, to evaluate nonelementary integrals, and to evaluate limits that lead to indeterminate forms. We provide a selfcontained derivation of the Taylor series for tan1 x and conclude with a reference table of frequently used series.
n a s s a
The Binomial Series for Powers and Roots
H d a m am
The Taylor series generated by ƒsxd = s1 + xdm , when m is constant, is 1 + mx +
M
uh
msm  1dsm  2d 3 Á msm  1d 2 x + x + 2! 3! +
msm  1dsm  2d Á sm  k + 1d k Á x + . k!
(1)
This series, called the binomial series, converges absolutely for ƒ x ƒ 6 1. To derive the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
823
ƒsxd = s1 + xdm ƒ¿sxd = ms1 + xdm  1 ƒ–sxd = msm  1ds1 + xdm  2
You
ƒ‡sxd = msm  1dsm  2ds1 + xdm  3
suf
i
series, we first list the function and its derivatives:
o skd
ƒ sxd = msm  1dsm  2d Á sm  k + 1ds1 + xdm  k .
nR
iaz
We then evaluate these at x = 0 and substitute into the Taylor series formula to obtain Series (1). If m is an integer greater than or equal to zero, the series stops after sm + 1d terms because the coefficients from k = m + 1 on are zero. If m is not a positive integer or zero, the series is infinite and converges for ƒ x ƒ 6 1. To see why, let uk be the term involving x k . Then apply the Ratio Test for absolute convergence to see that uk + 1 m  k ` uk ` = ` x` :ƒxƒ k + 1
as k : q .
ass a
Our derivation of the binomial series shows only that it is generated by s1 + xdm and converges for ƒ x ƒ 6 1. The derivation does not show that the series converges to s1 + xdm . It does, but we omit the proof.
dH
The Binomial Series For 1 6 x 6 1,
q m s1 + xdm = 1 + a a b x k , k k=1
ma
where we define m a b = m, 1
msm  1d m a b = , 2! 2
Mu
ham
and msm  1dsm  2d Á sm  k + 1d m a b = k! k
EXAMPLE 1
for k Ú 3.
Using the Binomial Series
If m = 1, a
1 b = 1, 1
a
1s 2d 1 b = = 1, 2! 2
and a
1s 2ds 3d Á s 1  k + 1d 1 k! b = = s 1dk a b = s 1dk . k! k! k
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Chapter 11: Infinite Sequences and Series
suf
i
With these coefficient values and with x replaced by x, the binomial series formula gives the familiar geometric series q
s1 + xd1 = 1 + a s 1dkx k = 1  x + x 2  x 3 + Á + s 1dkx k + Á . k=1
Using the Binomial Series
You
EXAMPLE 2
We know from Section 3.8, Example 1, that 21 + x L 1 + sx>2d for ƒ x ƒ small. With m = 1>2, the binomial series gives quadratic and higherorder approximations as well, along with error estimates that come from the Alternating Series Estimation Theorem:
iaz
s1 + xd1>2
3 1 1 1 1 a b a b a b a b a b 2 2 2 2 2 x = 1 + + x2 + x3 2 2! 3!
= 1 +
nR
3 5 1 1 a b a b a b a b 2 2 2 2 + x4 + Á 4!
5x 4 x x2 x3 + + Á. 2 8 16 128
ass a
Substitution for x gives still other approximations. For example, 21  x 2 L 1 
1 1 1 1  x L 1 2x 8x 2
for ƒ x 2 ƒ small
1 for ` x ` small, that is, ƒ x ƒ large.
dH
A
x2 x4 2 8
Power Series Solutions of Differential Equations and Initial Value Problems
Mu
ham
ma
When we cannot find a relatively simple expression for the solution of an initial value problem or differential equation, we try to get information about the solution in other ways. One way is to try to find a power series representation for the solution. If we can do so, we immediately have a source of polynomial approximations of the solution, which may be all that we really need. The first example (Example 3) deals with a firstorder linear differential equation that could be solved with the methods of Section 9.2. The example shows how, not knowing this, we can solve the equation with power series. The second example (Example 4) deals with an equation that cannot be solved analytically by previous methods.
EXAMPLE 3
Series Solution of an Initial Value Problem
Solve the initial value problem y¿  y = x, Solution
ys0d = 1.
We assume that there is a solution of the form y = a0 + a1 x + a2 x 2 + Á + an  1x n  1 + an x n + Á .
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(2)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
825
(3)
suf
y¿ = a1 + 2a2 x + 3a3 x 2 + Á + nan x n  1 + Á
i
Our goal is to find values for the coefficients ak that make the series and its first derivative
satisfy the given differential equation and initial condition. The series y¿  y is the difference of the series in Equations (2) and (3):
You
y¿  y = sa1  a0 d + s2a2  a1 dx + s3a3  a2 dx 2 + Á + snan  an  1 dx n  1 + Á .
(4)
If y is to satisfy the equation y¿  y = x, the series in Equation (4) must equal x. Since power series representations are unique (Exercise 45 in Section 11.7), the coefficients in Equation (4) must satisfy the equations Constant terms
3a3  a2 = 0
Coefficients of x2
nR
iaz
a1  a0 = 0 2a2  a1 = 1
nan  an  1
Coefficients of x
o
o = 0
Coefficients of xn  1
o
o
ass a
We can also see from Equation (2) that y = a0 when x = 0, so that a0 = 1 (this being the initial condition). Putting it all together, we have a0 = 1,
a3 =
a1 = a0 = 1,
1 + a1 1 + 1 2 = = , 2 2 2 an  1 2 an = n = , Á n! a2 =
a2 2 2 = # = ,Á, 3 3 2 3!
Mu
ham
ma
dH
Substituting these coefficient values into the equation for y (Equation (2)) gives y = 1 + x + 2#
x2 x3 xn + 2# + Á + 2# + Á 2! 3! n!
x2 x3 xn = 1 + x + 2 a + + Á + + Áb 2! 3! n! ('''''')''''''* the Taylor series for ex  1  x
= 1 + x + 2se x  1  xd = 2e x  1  x.
The solution of the initial value problem is y = 2e x  1  x. As a check, we see that ys0d = 2e 0  1  0 = 2  1 = 1 and y¿  y = s2e x  1d  s2e x  1  xd = x.
EXAMPLE 4
Solving a Differential Equation
Find a power series solution for
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y– + x 2y = 0.
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(5)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
We assume that there is a solution of the form y = a0 + a1 x + a2 x 2 + Á + an x n + Á ,
i
Solution
(6)
suf
826
and find what the coefficients ak have to be to make the series and its second derivative y– = 2a2 + 3 # 2a3 x + Á + nsn  1dan x n  2 + Á
(7)
You
satisfy Equation (5). The series for x 2y is x 2 times the righthand side of Equation (6): x 2y = a0 x 2 + a1 x 3 + a2 x 4 + Á + an x n + 2 + Á .
(8)
The series for y– + x 2y is the sum of the series in Equations (7) and (8):
y– + x 2y = 2a2 + 6a3 x + s12a4 + a0 dx 2 + s20a5 + a1 dx 3
iaz
+ Á + snsn  1dan + an  4 dx n  2 + Á .
(9)
2a2 = 0,
nR
Notice that the coefficient of x n  2 in Equation (8) is an  4 . If y and its second derivative y– are to satisfy Equation (5), the coefficients of the individual powers of x on the righthand side of Equation (9) must all be zero: 6a3 = 0,
and for all n Ú 4,
12a4 + a0 = 0,
20a5 + a1 = 0,
ass a
nsn  1dan + an  4 = 0.
(10)
(11)
We can see from Equation (6) that
a0 = ys0d,
a1 = y¿s0d.
dH
In other words, the first two coefficients of the series are the values of y and y¿ at x = 0. Equations in (10) and the recursion formula in Equation (11) enable us to evaluate all the other coefficients in terms of a0 and a1 . The first two of Equations (10) give a2 = 0,
a3 = 0.
ma
Equation (11) shows that if an  4 = 0, then an = 0; so we conclude that a6 = 0,
a7 = 0,
a10 = 0,
a11 = 0,
Mu
ham
and whenever n = 4k + 2 or 4k + 3, an is zero. For the other coefficients we have an  4 an = nsn  1d so that a0 a0 a4 , a8 = # = # # # 4#3 8 7 3 4 7 8 a8 a0 = = # # # # # 11 # 12 3 4 7 8 11 12
a4 = a12 and
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a5 = a13
a5 a1 a1 , a9 = # = # # # # 5 4 9 8 4 5 8 9 a9 a1 = = # # # # # . 12 # 13 4 5 8 9 12 13
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
827
x4 x8 x 12 + + Áb 3#4 3#4#7#8 3 # 4 # 7 # 8 # 11 # 12
+ a1 ax 
x5 x9 x 13 + # # #  # # # # # + Áb . # 4 5 4 5 8 9 4 5 8 9 12 13
You
y = a0 a1 
suf
i
The answer is best expressed as the sum of two separate series—one multiplied by a0 , the other by a1 :
Both series converge absolutely for all x, as is readily seen by the Ratio Test.
Evaluating Nonelementary Integrals
iaz
Taylor series can be used to express nonelementary integrals in terms of series. Integrals like 1 sin x 2 dx arise in the study of the diffraction of light. Express 1 sin x 2 dx as a power series.
EXAMPLE 5
From the series for sin x we obtain
nR
Solution
sin x 2 = x 2 
x3 x 11 x 15 x 10 x7  # + +  Á. # # 3 7 3! 11 5! 15 7! 19 # 9!
ass a
Therefore,
x6 x 10 x 14 x 18 + +  Á. 3! 5! 7! 9!
sin x 2 dx = C +
L
EXAMPLE 6
dx with an error of less than 0.001.
dH
Estimate
Estimating a Definite Integral
1 2 10 sin x
From the indefinite integral in Example 5,
Solution
1
L0
sin x 2 dx =
1 1 1 1 1  # + +  Á. 3 7 3! 11 # 5! 15 # 7! 19 # 9!
Mu
ham
ma
The series alternates, and we find by experiment that 1 L 0.00076 11 # 5!
is the first term to be numerically less than 0.001. The sum of the preceding two terms gives 1
L0
sin x 2 dx L
1 1 L 0.310. 3 42
With two more terms we could estimate 1
L0
sin x 2 dx L 0.310268
with an error of less than 10 6 . With only one term beyond that we have 1
L0
sin x 2 dx L
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 828
Chapter 11: Infinite Sequences and Series
suf
i
with an error of about 1.08 * 10 9 . To guarantee this accuracy with the error formula for the Trapezoidal Rule would require using about 8000 subintervals.
Arctangents
You
In Section 11.7, Example 5, we found a series for tan1 x by differentiating to get d 1 tan1 x = = 1  x2 + x4  x6 + Á dx 1 + x2 and integrating to get
x7 x3 x5 + Á. + 5 7 3
iaz
tan1 x = x 
However, we did not prove the termbyterm integration theorem on which this conclusion depended. We now derive the series again by integrating both sides of the finite formula
nR
n + 1 2n + 2 1 2 4 6 Á + s 1dnt 2n + s 1d t = 1 t + t t + , 1 + t2 1 + t2
(12)
ass a
in which the last term comes from adding the remaining terms as a geometric series with first term a = s 1dn + 1t 2n + 2 and ratio r = t 2 . Integrating both sides of Equation (12) from t = 0 to t = x gives tan1 x = x 
where
x5 x3 x7 x 2n + 1 + + Rnsxd, + Á + s 1dn 5 7 3 2n + 1
s 1dn + 1t 2n + 2 dt. 1 + t2 L0 x
dH
Rnsxd =
The denominator of the integrand is greater than or equal to 1; hence ƒxƒ
ƒ Rnsxd ƒ …
L0
t 2n + 2 dt =
ƒ x ƒ 2n + 3 . 2n + 3
Mu
ham
ma
If ƒ x ƒ … 1, the right side of this inequality approaches zero as n : q . Therefore limn: q Rnsxd = 0 if ƒ x ƒ … 1 and s 1dnx 2n + 1 , 2n + 1 n=0 q
tan1 x = a tan
1
ƒ x ƒ … 1.
x5 x7 x3 + + Á, x = x 5 7 3
(13)
ƒxƒ … 1
We take this route instead of finding the Taylor series directly because the formulas for the higherorder derivatives of tan1 x are unmanageable. When we put x = 1 in Equation (13), we get Leibniz’s formula: s 1dn p 1 1 1 1 = 1  +  +  Á + + Á. 5 7 4 3 9 2n + 1 Because this series converges very slowly, it is not used in approximating p to many decimal places. The series for tan1 x converges most rapidly when x is near zero. For that reason, people who use the series for tan1 x to compute p use various trigonometric identities.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
829
1 b = tan1 , 3
suf
1 2
a = tan1
and
You
then tan sa + bd =
i
For example, if
1 tan a + tan b 2 + = 1  tan a tan b 1 
and
1 3 1 6
= 1 = tan
p 4
iaz
p 1 1 = a + b = tan1 + tan1 . 4 2 3
nR
Now Equation (13) may be used with x = 1>2 to evaluate tan1 (1>2) and with x = 1>3 to give tan1 (1>3). The sum of these results, multiplied by 4, gives p.
Evaluating Indeterminate Forms
ass a
We can sometimes evaluate indeterminate forms by expressing the functions involved as Taylor series.
EXAMPLE 7
Limits Using Power Series
dH
Evaluate
lim
x:1
ln x . x  1
We represent ln x as a Taylor series in powers of x  1. This can be accomplished by calculating the Taylor series generated by ln x at x = 1 directly or by replacing x by x  1 in the series for ln (1 + x) in Section 11.7, Example 6. Either way, we obtain
Mu
ham
ma
Solution
ln x = sx  1d 
1 sx  1d2 + Ă , 2
from which we find that ln x 1 = lim a1  sx  1d + Ă b = 1. x 1 2 x:1 x:1 lim
EXAMPLE 8
Limits Using Power Series
Evaluate
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lim
x:0
sin x  tan x . x3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
The Taylor series for sin x and tan x, to terms in x 5 , are sin x = x 
x3 x5 +  Á, 3! 5!
tan x = x +
x3 2x 5 + + Á. 3 15
sin x  tan x = 
You
Hence,
i
Solution
suf
830
x3 x5 x2 1  Á = x 3 a  Áb 2 8 2 8
and
sin x  tan x x2 1 = lim a Áb 2 8 x:0 x:0 x3
iaz
lim
nR
1 =  . 2
If we apply series to calculate limx:0 ss1>sin xd  s1/xdd, we not only find the limit successfully but also discover an approximation formula for csc x.
Find lim a x:0
1 1  xb. sin x
dH
Solution
Approximation Formula for csc x
ass a
EXAMPLE 9
Mu
ham
ma
x  sin x 1 1  x = = sin x x sin x x3 a
x  ax x # ax 
x3 x5 +  Áb 3! 5! x3 x5 +  Áb 3! 5!
x2 1 + Áb 3! 5!
x2 1 + Á 3! 5! = = x . x2 x2 2 Á Á 1 + + b x a1 3! 3!
Therefore, x2 1 + Á 3! 5! 1 1 lim a  x b = lim § x ¥ = 0. x:0 sin x x:0 x2 1 + Á 3! From the quotient on the right, we can see that if ƒ x ƒ is small, then x 1 1 1  x L x# = 6 3! sin x
To Read it Online & Download:
or
x 1 csc x L x + . 6
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10 Applications of Power Series
suf
i
TABLE 11.1 Frequently used Taylor series q
1 = 1 + x + x 2 + Á + x n + Á = a x n, 1  x n=0
ƒxƒ 6 1
q
1 = 1  x + x 2  Á + s xdn + Á = a s 1dnx n, 1 + x n=0
cos x = 1 
q s 1dnx 2n + 1 x3 x5 x 2n + 1 + Á = a , +  Á + s 1dn 3! 5! s2n + 1d! n = 0 s2n + 1d! q s 1dnx 2n x2 x4 x 2n + Á = a , +  Á + s 1dn 2! 4! s2nd! s2nd! n=0
ƒxƒ 6 q
ƒxƒ 6 q
q s 1dn  1x n x2 x3 xn +  Á + s 1dn  1 n + Á = a , n 2 3 n=1
nR
ln s1 + xd = x 
1 6 x … 1
x 2n + 1 1 + x x3 x5 x 2n + 1 = 2 tanh1 x = 2 ax + + + Á + + Áb = 2 a , 5 1  x 3 2n + 1 n = 0 2n + 1
tan1 x = x 
q s 1dnx 2n + 1 x3 x5 x 2n + 1 +  Á + s 1dn + Á = a , 5 3 2n + 1 2n + 1 n=0
Binomial Series s1 + xdm = 1 + mx +
q
ƒxƒ 6 1
ƒxƒ … 1
ass a
ln
ƒxƒ 6 q
iaz
sin x = x 
x2 xn xn + Á + + Á = a , 2! n! n = 0 n!
You
ƒxƒ 6 1
q
ex = 1 + x +
831
msm  1dsm  2dx 3 msm  1dsm  2d Á sm  k + 1dx k msm  1dx 2 + + Á + + Á 2! 3! k!
where m a b = m, 1
ƒ x ƒ 6 1,
dH
q m = 1 + a a bx k, k k=1
msm  1d m , a b = 2! 2
msm  1d Á sm  k + 1d m a b = k! k
for k Ú 3.
Mu
ham
ma
m Note: To write the binomial series compactly, it is customary to define a b to be 1 and to take x 0 = 1 (even in the usually 0 m q excluded case where x = 0), yielding s1 + xdm = g k = 0 a bx k . If m is a positive integer, the series terminates at x m and the k result converges for all x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Y z a i R n a s as
i f u s u o
11.10 Applications of Power Series
EXERCISES 11.10 Binomial Series
H d ma
Find the first four terms of the binomial series for the functions in Exercises 1–10. 1. s1 + xd1>2
M
3. s1  xd1>2
x 5. a1 + b 2
x 6. a1  b 2
m a uh
1>2
4. s1  2xd
2. s1 + xd1>3
2
2
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7. s1 + x 3 d1>2 1 9. a1 + x b
1>2
8. s1 + x 2 d1>3
2 10. a1  x b
1>3
Find the binomial series for the functions in Exercises 11–14. 11. s1 + xd4
12. s1 + x 2 d3
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831
4100 AWL/Thomas_ch11p746847 8/25/04 2:41 PM Page 832
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 832
Chapter 11: Infinite Sequences and Series x
4
44. Fsxd =
2
t 2e t dt,
L0
[0, 1]
i
x b 2
x
45. Fsxd =
Find series solutions for the initial value problems in Exercises 15–32. 16. y¿  2y = 0,
17. y¿  y = 1,
y s0d = 0
18. y¿ + y = 1,
19. y¿  y = x,
y s0d = 0
20. y¿ + y = 2x,
23. s1  xdy¿  y = 0,
y s0d = 1
22. y¿  x y = 0,
y s0d = 1
y s0d = 2
24. s1 + x 2 dy¿ + 2xy = 0,
49. lim
27. y– + y = x,
y¿s0d = 1 and y s0d = 2
28. y–  y = x,
y¿s0d = 2 and y s0d = 1
51. lim
y¿s0d = b and y s0d = a
31. y– + x 2y = x,
y¿s0d = b and y s0d = a
32. y–  2y¿ + y = 0,
0.2
0.2
L0 0.1
35.
L0
34.
L0
4
dx
36.
3 2 1 + x 2 dx
L0
T Use series to approximate the values of the integrals in Exercises 37–40 with an error of magnitude less than 10 8 . L0
0.1
sin x x dx
38.
0.1
39.
L0
ma
37.
40.
2
e x dx
L0
1
21 + x 4 dx
L0
1  cos x dx x2
ham
41. Estimate the error if cos t 2 is approximated by 1 1
integral 10 cos t 2 dt .
42. Estimate the error if cos 2t is approximated by 1 
t4 t8 + in the 2 4! t t2 t3 + 2 4! 6!
Mu
In Exercises 43–46, find a polynomial that will approximate F(x) throughout the given interval with an error of magnitude less than 10 3 . x
L0
u5
tan1 y  sin y
y:0
y 3 cos y
54. lim sx + 1d sin x: q
1 x + 1
x2  4 x:2 ln sx  1d
56. lim
Theory and Examples 57. Replace x by x in the Taylor series for ln s1 + xd to obtain a series for ln s1  xd . Then subtract this from the Taylor series for ln s1 + xd to show that for ƒ x ƒ 6 1 , ln
1 + x x3 x5 = 2 ax + + + Á b. 5 1  x 3
58. How many terms of the Taylor series for ln s1 + xd should you add to be sure of calculating ln (1.1) with an error of magnitude less than 10 8 ? Give reasons for your answer. 59. According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan1 1 would you have to add to be sure of finding p>4 with an error of magnitude less than 10 3 ? Give reasons for your answer. 60. Show that the Taylor series for ƒsxd = tan1 x diverges for ƒ x ƒ 7 1. T 61. Estimating Pi About how many terms of the Taylor series for tan1 x would you have to use to evaluate each term on the righthand side of the equation
1
in the integral 10 cos 2t dt .
43. Fsxd =
sin u  u + su3>6d
0.25
1 21 + x
0.1
y3
ln s1 + x 2 d x:0 1  cos x
e x  1 dx x
50. lim 52. lim
55. lim
dH
sin x 2 dx
y  tan1 y
2
T In Exercises 33–36, use series to estimate the integrals’ values with an error of magnitude less than 10 3 . (The answer section gives the integrals’ values rounded to five decimal places.)
e x  e x x x:0 u :0
x: q
Approximations and Nonelementary Integrals
33.
t
4
53. lim x 2se 1>x  1d
y¿s0d = 1 and y s0d = 0
(b) [0, 1]
48. lim
nR
30. y–  x y = 0,
2
1  cos t  st 2>2d
y:0
y¿s2d = 2 and y s2d = 0
2
x
ass a
29. y–  y = x,
e x  s1 + xd
t: 0
y¿s0d = 0 and y s0d = 1
ln s1 + td dt, (a) [0, 0.5] t
Use series to evaluate the limits in Exercises 47–56. x:0
y s0d = 3
26. y– + y = 0,
L0
(b) [0, 1]
Indeterminate Forms 47. lim
y¿s0d = 1 and y s0d = 0
25. y–  y = 0,
x
46. Fsxd =
y s0d = 2
2
y s0d = 1
21. y¿  xy = 0,
y s0d = 1
iaz
y s0d = 1
15. y¿ + y = 0,
tan1 t dt, (a) [0, 0.5]
L0
You
Initial Value Problems
suf
14. a1 
13. s1  2xd3
sin t 2 dt,
[0, 1]
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p = 48 tan1
1 1 1 + 32 tan1  20 tan1 57 18 239
with an error of magnitude less than 10 6 ? In contrast, the conq vergence of g n = 1s1>n 2 d to p2>6 is so slow that even 50 terms will not yield twoplace accuracy. 62. Integrate the first three nonzero terms of the Taylor series for tan t from 0 to x to obtain the first three nonzero terms of the Taylor series for ln sec x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10 Applications of Power Series 63. a. Use the binomial series and the fact that
suf
i
69. Series for sin1 x Integrate the binomial series for s1  x 2 d1>2 to show that for ƒ x ƒ 6 1 ,
d sin1 x = s1  x 2 d1>2 dx
1 # 3 # 5 # Á # s2n  1d x 2n + 1 . 2 # 4 # 6 # Á # s2nd 2n + 1 n=1 q
sin1 x = x + a
to generate the first four nonzero terms of the Taylor series for sin1 x . What is the radius of convergence?
70. Series for tan1 x for ƒ x ƒ 7 1 Derive the series
b. Series for cos x Use your result in part (a) to find the first five nonzero terms of the Taylor series for cos1 x . 64. a. Series for sinh x Taylor series for
Find the first four nonzero terms of the
You
1
1
p 1 1 1  x + + Á, x 7 1 2 5x 5 3x 3 p 1 1 1 + Á, x 6 1 , tan1 x =   x + 2 5x 5 3x 3 tan1 x =
x
dt . L0 21 + t 2
T b. Use the first three terms of the series in part (a) to estimate sinh1 0.25 . Give an upper bound for the magnitude of the estimation error. 65. Obtain the Taylor series for 1>s1 + xd2 from the series for 1>s1 + xd . 2
66. Use the Taylor series for 1>s1  x d to obtain a series for 2x>s1  x 2 d2 .
2#4#4#6#6#8# Á p = # # # # # # Á . 4 3 3 5 5 7 7 Find p to two decimal places with this formula.
1 , 9
#
1 1 1 1 1 = 2  4 + 6  8 + Á 1 + s1>t 2 d t t t t
in the first case from x to q and in the second case from  q to x.
71. The value of g n = 1 tan1s2>n2 d q
a. Use the formula for the tangent of the difference of two angles to show that tan stan1 sn + 1d  tan1 sn  1dd =
2 n2
b. Show that N
p 1 2 1 1 a tan n 2 = tan sN + 1d + tan N  4 . n=1
c. Find the value of g n = 1 tan1 q
2 . n2
1 . 13
Mu
ham
ma
1 , 5
dH
T 68. Construct a table of natural logarithms ln n for n = 1, 2, 3, Á , 10 by using the formula in Exercise 57, but taking advantage of the relationships ln 4 = 2 ln 2, ln 6 = ln 2 + ln 3, ln 8 = 3 ln 2, ln 9 = 2 ln 3 , and ln 10 = ln 2 + ln 5 to reduce the job to the calculation of relatively few logarithms by series. Start by using the following values for x in Exercise 57: 1 , 3
1 1 = 2 1 + t2 t
ass a
T 67. Estimating Pi The English mathematician Wallis discovered the formula
iaz
by integrating the series
nR
sinh1 x =
833
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
11.11 Fourier Series
11.11
Fourier Series
HISTORICAL BIOGRAPHY JeanBaptiste Joseph Fourier
We have seen how Taylor series can be used to approximate a function ƒ by polynomials. The Taylor polynomials give a close fit to ƒ near a particular point x = a, but the error in the approximation can be large at points that are far away. There is another method that often gives good approximations on wide intervals, and often works with discontinuous functions for which Taylor polynomials fail. Introduced by Joseph Fourier, this method approximates functions with sums of sine and cosine functions. It is well suited for analyzing periodic functions, such as radio signals and alternating currents, for solving heat transfer problems, and for many other problems in science and engineering.
d a mm
(1766–1830)
a h Mu
n a s s a H
Y z a i R
833
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 834
Chapter 11: Infinite Sequences and Series
suf
i
Suppose we wish to approximate a function ƒ on the interval [0, 2p] by a sum of sine and cosine functions, ƒnsxd = a0 + sa1 cos x + b1 sin xd + sa2 cos 2x + b2 sin 2xd + Á + san cos nx + bn sin nxd
You
or, in sigma notation, n
ƒnsxd = a0 + a sak cos kx + bk sin kxd. k=1
(1)
2. 3.
ƒnsxd and ƒ(x) give the same value when integrated from 0 to 2p. ƒnsxd cos kx and ƒ(x) cos kx give the same value when integrated from 0 to 2p sk = 1, Á , nd. ƒnsxd sin kx and ƒ(x) sin kx give the same value when integrated from 0 to 2p sk = 1, Á , nd.
nR
1.
iaz
We would like to choose values for the constants a0 , a1, a2 , Á an and b1, b2 , Á , bn that make ƒnsxd a “best possible” approximation to ƒ(x). The notion of “best possible” is defined as follows:
Altogether we impose 2n + 1 conditions on ƒn : 2p
ƒsxd dx,
L0
ass a
L0
2p
ƒnsxd dx =
2p
L0
2p
ƒnsxd cos kx dx =
L0
2p
k = 1, Á , n,
ƒsxd sin kx dx,
k = 1, Á , n.
2p
ƒnsxd sin kx dx =
dH
L0
ƒsxd cos kx dx,
L0
It is possible to choose a0 , a1, a2 , Á an and b1, b2 , Á , bn so that all these conditions are satisfied, by proceeding as follows. Integrating both sides of Equation (1) from 0 to 2p gives 2p
ma ham Mu
ƒnsxd dx = 2pa0
L0
since the integral over [0, 2p] of cos kx equals zero when k Ú 1, as does the integral of sin kx. Only the constant term a 0 contributes to the integral of ƒn over [0, 2p]. A similar calculation applies with each of the other terms. If we multiply both sides of Equation (1) by cos x and integrate from 0 to 2p then we obtain 2p
ƒnsxd cos x dx = pa1 .
L0 This follows from the fact that
2p
L0
cos px cos px dx = p
and 2p
L0
2p
cos px cos qx dx =
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L0
2p
cos px sin mx dx =
L0
sin px sin qx dx = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.11
Fourier Series
835
2p
ƒnsxd sin x dx = pb1 .
L0
You
Proceeding in a similar fashion with
suf
i
whenever p, q and m are integers and p is not equal to q (Exercises 9–13). If we multiply Equation (1) by sin x and integrate from 0 to 2p we obtain
cos 2x, sin 2x, Á , cos nx, sin nx
we obtain only one nonzero term each time, the term with a sinesquared or cosinesquared term. To summarize,
L0 2p
ƒnsxd dx = 2pa0
ƒnsxd cos kx dx = pak,
k = 1, Á , n
nR
L0
iaz
2p
2p
ƒnsxd sin kx dx = pbk,
ass a
L0
k = 1, Á , n
We chose ƒn so that the integrals on the left remain the same when ƒn is replaced by ƒ, so we can use these equations to find a0 , a1, a2 , Á an and b1, b2 , Á , bn from ƒ: 2p
Mu
ham
ma
dH
a0 =
1 ak = p
1 ƒsxd dx 2p L0
(2)
2p
L0
1 bk = p
ƒsxd cos kx dx,
k = 1, Á , n
(3)
k = 1, Á , n
(4)
2p
L0
ƒsxd sin kx dx,
The only condition needed to find these coefficients is that the integrals above must exist. If we let n : q and use these rules to get the coefficients of an infinite series, then the resulting sum is called the Fourier series for ƒ(x), q
a0 + a sak cos kx + bk sin kxd.
(5)
k=1
EXAMPLE 1
Finding a Fourier Series Expansion
Fourier series can be used to represent some functions that cannot be represented by Taylor series; for example, the step function ƒ shown in Figure 11.16a.
To Read it Online & Download:
ƒsxd = e
1, 2,
if 0 … x … p if p 6 x … 2p.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 836
Chapter 11: Infinite Sequences and Series
y 2
1
1
0
x
2
–2
–
(a)
You
2
suf
i
y
0
2
3
4
x
(b)
FIGURE 11.16 (a) The step function 1, 2,
0 … x … p p 6 x … 2p
iaz
ƒsxd = e
nR
(b) The graph of the Fourier series for ƒ is periodic and has the value 3>2 at each point of discontinuity (Example 1).
The coefficients of the Fourier series of ƒ are computed using Equations (2), (3), and (4). 2p
1 ƒsxd dx 2p L0
ass a
a0 =
p
=
dH
1 ak = p
2p
ma
2p
p
cos kx dx +
L0
sin kx 1 = p ac d k
1 bk = p
ham Mu
ƒsxd cos kx dx
L0
1 = p a
p 0
+ c
Lp
2 cos kx dxb 2p
2 sin kx d b = 0, k p
k Ú 1
2p
L0
1 = p a
ƒsxd sin kx dx 2p
p
L0
sin kx dx +
cos kx 1 = p a cd k =
2p
3 1 a 1 dx + 2 dxb = 2p L0 2 Lp
p 0
Lp
+ c
2 sin kx dxb 2p
2 cos kx d b k p
s 1dk  1 cos kp  1 = . kp kp
So
To Read it Online & Download:
a0 =
3 , 2
a1 = a2 = Á = 0,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.11
Fourier Series
837
b2 = 0,
b3 = 
2 , 3p
b4 = 0,
b5 = 
2 , 5p
b6 = 0, Á
suf
2 b1 =  p ,
i
and
The Fourier series is
You
3 sin 3x sin 5x 2 + Áb .  p asin x + + 5 2 3
nR
iaz
Notice that at x = p, where the function ƒ(x) jumps from 1 to 2, all the sine terms vanish, leaving 3> 2 as the value of the series. This is not the value of ƒ at p, since ƒspd = 1. The Fourier series also sums to 3> 2 at x = 0 and x = 2p. In fact, all terms in the Fourier series are periodic, of period 2p, and the value of the series at x + 2p is the same as its value at x. The series we obtained represents the periodic function graphed in Figure 11.16b, with domain the entire real line and a pattern that repeats over every interval of width 2p. The function jumps discontinuously at x = np, n = 0, ;1, ;2, Á and at these points has value 3> 2, the average value of the onesided limits from each side. The convergence of the Fourier series of ƒ is indicated in Figure 11.17.
y
ass a
y
f
2 1.5
2
1.5
f1
1
dH
(a)
x
(b)
0
ma
2
1.5
f3
ham
2
1.5
FIGURE 11.17
x
(c)
0
2
x
y
f
f
2 1.5
f9
1
0
f5
1
2
y
f
2
1
0
Mu
f
y
f15
1
(d)
2
x
0
(e)
2
The Fourier approximation functions ƒ1, ƒ3, ƒ5, ƒ9 , and ƒ15 of the function ƒsxd = e
To Read it Online & Download:
x
1, 2,
0 … x … p in Example 1. p 6 x … 2p
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 838
Chapter 11: Infinite Sequences and Series
i
Convergence of Fourier Series
You
suf
Taylor series are computed from the value of a function and its derivatives at a single point x = a, and cannot reflect the behavior of a discontinuous function such as ƒ in Example 1 past a discontinuity. The reason that a Fourier series can be used to represent such functions is that the Fourier series of a function depends on the existence of certain integrals, whereas the Taylor series depends on derivatives of a function near a single point. A function can be fairly “rough,” even discontinuous, and still be integrable. The coefficients used to construct Fourier series are precisely those one should choose to minimize the integral of the square of the error in approximating ƒ by ƒn . That is, 2p
L0
[ƒsxd  ƒnsxd]2 dx
nR
iaz
is minimized by choosing a0 , a1, a2 , Á an and b1, b2 , Á , bn as we did. While Taylor series are useful to approximate a function and its derivatives near a point, Fourier series minimize an error which is distributed over an interval. We state without proof a result concerning the convergence of Fourier series. A function is piecewise continuous over an interval I if it has finitely many discontinuities on the interval, and at these discontinuities onesided limits exist from each side. (See Chapter 5, Additional Exercises 11–18.)
dH
ass a
THEOREM 24 Let ƒ(x) be a function such that ƒ and ƒ¿ are piecewise continuous on the interval [0, 2p]. Then ƒ is equal to its Fourier series at all points where ƒ is continuous. At a point c where ƒ has a discontinuity, the Fourier series converges to ƒsc + d + ƒsc  d 2
Mu
ham
ma
where ƒsc + d and ƒsc  d are the right and lefthand limits of ƒ at c.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 838
Chapter 11: Infinite Sequences and Series
u o zY
EXERCISES 11.11 Finding Fourier Series In Exercises 1–8, find the Fourier series associated with the given functions. Sketch each function. 1. ƒsxd = 1
0 … x … 2p .
1, 2. ƒsxd = e 1,
0 … x … p p 6 x … 2p
3. ƒsxd = e
x, x  2p,
4. ƒsxd = e
x 2, 0,
5. ƒsxd = e x
d a m
0 … x … p p 6 x … 2p
m a h
0 … x … p p 6 x … 2p
u M
0 … x … 2p .
a i R an
6. ƒsxd = e
s s Ha
To Read it Online & Download:
e x, 0,
7. ƒsxd = e
cos x, 0,
8. ƒsxd = e
2, x,
i f su
0 … x … p p 6 x … 2p
0 … x … p p 6 x … 2p
0 … x … p p 6 x … 2p
Theory and Examples Establish the results in Exercises 9–13, where p and q are positive integers. 2p
9.
L0
cos px dx = 0 for all p .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.11 Fourier Series 2p
2p
11.
cos px cos qx dx = e
0, p,
if p Z q . if p = q
L0 sHint: cos A cos B = s1>2d[cossA + Bd + cossA  Bd].d 2p
12.
sin px sin qx dx = e
0, p,
if p Z q . if p = q
L0 sHint: sin A sin B = s1>2d[cos sA  Bd  cos sA + Bd].d 2p
sin px cos qx dx = 0 for all p and q . L0 sHint: sin A cos B = s1>2d[sin sA + Bd + sin sA  Bd].d
a. Use Theorem 24 to verify that the Fourier series for ƒsxd in Exercise 3 converges to ƒsxd for 0 6 x 6 2p .
b. Although ƒ¿sxd = 1 , show that the series obtained by termbyterm differentiation of the Fourier series in part (a) diverges. 16. Use Theorem 24 to find the Value of the Fourier series determined q p2 1 in Exercise 4 and show that = a 2. 6 n=1 n
Mu
ham
ma
dH
ass a
nR
iaz
13.
15. Termbyterm differentiation
suf
L0
14. Fourier series of sums of functions If ƒ and g both satisfy the conditions of Theorem 24, is the Fourier series of ƒ + g on [0, 2p] the sum of the Fourier series of ƒ and the Fourier series of g? Give reasons for your answer.
i
sin px dx = 0 for all p .
You
10.
839
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Chapter 11
Chapter 11
i f u s u o
Questions to Guide Your Review
Questions to Guide Your Review
Y z
839
1. What is an infinite sequence? What does it mean for such a sequence to converge? To diverge? Give examples.
15. When do pseries converge? Diverge? How do you know? Give examples of convergent and divergent pseries.
2. What is a nondecreasing sequence? Under what circumstances does such a sequence have a limit? Give examples.
16. What are the Direct Comparison Test and the Limit Comparison Test? What is the reasoning behind these tests? Give examples of their use.
3. What theorems are available for calculating limits of sequences? Give examples.
iR a
4. What theorem sometimes enables us to use l’Hôpital’s Rule to calculate the limit of a sequence? Give an example.
17. What are the Ratio and Root Tests? Do they always give you the information you need to determine convergence or divergence? Give examples.
5. What six sequence limits are likely to arise when you work with sequences and series?
18. What is an alternating series? What theorem is available for determining the convergence of such a series?
6. What is an infinite series? What does it mean for such a series to converge? To diverge? Give examples.
19. How can you estimate the error involved in approximating the sum of an alternating series with one of the series’ partial sums? What is the reasoning behind the estimate?
n a sa s H d
7. What is a geometric series? When does such a series converge? Diverge? When it does converge, what is its sum? Give examples.
8. Besides geometric series, what other convergent and divergent series do you know? 9. What is the nthTerm Test for Divergence? What is the idea behind the test?
a m m
10. What can be said about termbyterm sums and differences of convergent series? About constant multiples of convergent and divergent series?
20. What is absolute convergence? Conditional convergence? How are the two related? 21. What do you know about rearranging the terms of an absolutely convergent series? Of a conditionally convergent series? Give examples. 22. What is a power series? How do you test a power series for convergence? What are the possible outcomes? 23. What are the basic facts about a. termbyterm differentiation of power series?
11. What happens if you add a finite number of terms to a convergent series? A divergent series? What happens if you delete a finite number of terms from a convergent series? A divergent series?
b. termbyterm integration of power series?
12. How do you reindex a series? Why might you want to do this?
Give examples.
a h u M
c. multiplication of power series?
13. Under what circumstances will an infinite series of nonnegative terms converge? Diverge? Why study series of nonnegative terms?
24. What is the Taylor series generated by a function ƒ(x) at a point x = a ? What information do you need about ƒ to construct the series? Give an example.
14. What is the Integral Test? What is the reasoning behind it? Give an example of its use.
25. What is a Maclaurin series?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 840
Chapter 11: Infinite Sequences and Series 31. How can you sometimes use power series to estimate the values of nonelementary definite integrals?
27. What are Taylor polynomials? Of what use are they?
32. What are the Taylor series for 1>s1  xd, 1>s1 + xd, e x, sin x, cos x, ln s1 + xd, ln [s1 + xd>s1  xd] , and tan1 x ? How do you estimate the errors involved in replacing these series with their partial sums?
suf
28. What is Taylor’s formula? What does it say about the errors involved in using Taylor polynomials to approximate functions? In particular, what does Taylor’s formula say about the error in a linearization? A quadratic approximation?
i
26. Does a Taylor series always converge to its generating function? Explain.
30. How can you sometimes use power series to solve initial value problems?
34. State the theorem on convergence of the Fourier series for ƒ(x) when ƒ and ƒ¿ are piecewise continuous on [0, 2p] .
Mu
ham
ma
dH
ass a
nR
iaz
You
29. What is the binomial series? On what interval does it converge? How is it used?
33. What is a Fourier series? How do you calculate the Fourier coefficients a0 , a1, a2 , Á and b1, b2 , Á for a function ƒ(x) defined on the interval [0, 2p] ?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 840
i f su
Chapter 11: Infinite Sequences and Series
Chapter 11
Practice Exercises
u o Y z
Convergent or Divergent Sequences
Convergent or Divergent Series
Which of the sequences whose nth terms appear in Exercises 1–18 converge, and which diverge? Find the limit of each convergent sequence.
Which of the series in Exercises 25–40 converge absolutely, which converge conditionally, and which diverge? Give reasons for your answers.
1. an = 1 +
s 1dn n
1  2n 3. an = 2n np 5. an = sin 2
2. an =
7. an = 9. an =
ln sn 2 d n
11. an = a 13. an =
n  5 n b
H d
1>n
a m
s 4dn n!
Find the sums of the series in Exercises 19–24. q
h u
1 19. a n = 3 s2n  3ds2n  1d q
9 21. a n = 1 s3n  1ds3n + 2d q
23. a e n=0
n
M
q
n2n 2 + 1
2 20. a n = 2 nsn + 1d
8 22. a n = 3 s4n  3ds4n + 1d 3 24. a s 1d n 4 n=1 n
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1 30. a 2 n = 2 n sln nd
q
ln n 32. a n = 3 ln sln nd q s 1dn 3n 2 34. a 3 n=1 n + 1
s 3dn n! n=1
2n 3n 38. a n n=1 n
q
37. a q
39. a
n=1
1 2nsn + 1dsn + 2d
q
q
40. a
n=2
1 n2n 2  1
Power Series In Exercises 41–50, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally. sx + 4dn n3n n=1 q
41. a 43. a
s 1dn  1s3x  1dn
n=1 q
q
2n
q
q s 1dnsn 2 + 1d 36. a 2 n = 1 2n + n  1
q
q
n=1
n
s 1dn
n + 1 n! n=1
q
35. a
m a
Convergent Series
sa
27. a
s 1d 29. a n = 1 ln sn + 1d q
n sa
n=1
n
16. an = 22n + 1 18. an =
n=1
s 1dn
q
33. a
ln s2n 3 + 1d n
n
sn + 1d! n!
2n
q
ln s2n + 1d n
iR a
q
5 26. a n
q
ln n 31. a 3 n=1 n
3 14. an = a n b
n n 3 An
n=1
6. an = sin np
1 12. an = a1 + n b
15. an = ns21>n  1d 17. an =
2n
q
1
1 28. a 3 n = 1 2n
10. an = n
q
25. a
4. an = 1 + s0.9dn
8. an =
n + ln n n
1  s 1dn
xn 45. a n n n=1
n2
q sx  1d2n  2 42. a n = 1 s2n  1d!
sn + 1ds2x + 1dn s2n + 1d2n n=0 q
44. a q
46. a
n=1
xn 2n
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11 s 1dnsx  1d2n + 1 2n + 1 n=0
Nonelementary Integrals
q
i
48. a
Use series to approximate the values of the integrals in Exercises 77–80 with an error of magnitude less than 10 8 . (The answer section gives the integrals’ values rounded to 10 decimal places.)
q
49. a scsch ndx
50. a scoth ndx
n
n=1
n
n=1
1>2
Maclaurin Series
77.
Each of the series in Exercises 51–56 is the value of the Taylor series at x = 0 of a function ƒ(x) at a particular point. What function and what point? What is the sum of the series?
79.
L0
tan1 x x dx
In Exercises 81–86:
8 4 2 2 +  Á + s 1dn  1 n + Á 3 18 81 n3
81. lim
54. 1 
p2n p2 p4 + Á +  Á + s 1dn 2n # # 9 2! 81 4! 3 s2nd!
83. lim a t: 0
n
sln 2d sln 2d + Á + + Á 2! n!
1

9 23
1
+
45 23
+ s 1dn  1
85. lim
z:0
1
+ Á
Find Taylor series at x = 0 for the functions in Exercises 57–64. 1 58. 1 + x3
59. sin px
60. sin
61. cos sx 5>2 d
62. cos 25x
63. e spx>2d
64. e x
2x 3
ma
at
x = 2
67. ƒsxd = 1>sx + 1d
at
x = 3
68. ƒsxd = 1>x
at
x = a 7 0
Mu
Use power series to solve the initial value problems in Exercises 69–76. 69. y¿ + y = 0,
ys0d = 1
70. y¿  y = 0,
ys0d = 3
71. y¿ + 2y = 0,
ys0d = 3
72. y¿ + y = 1,
ys0d = 0
ys0d = 1 74. y¿ + y = x,
ys0d = 0
75. y¿  y = x,
ys0d = 1
ssin hd>h  cos h h2
h:0
2
86. lim
y:0
y cos y  cosh y
76. y¿  y = x,
lim a
x:0
sin 3x r + 2 + sb = 0 . x3 x
89. a. Show that the series 1 1 a asin 2n  sin 2n + 1 b q
converges.
Initial Value Problems
73. y¿  y = 3x,
1  cos2 z ln s1  zd + sin z
84. lim
n=1
ham
66. ƒsxd = 1>s1  xd
1 1  2b 2  2 cos t t
Theory and Examples
In Exercises 65–68, find the first four nonzero terms of the Taylor series generated by ƒ at x = a . x = 1
e u  e u  2u u  sin u u:0
T b. Compare the accuracies of the approximations sin x L x and sin x L 6x>s6 + x 2 d by comparing the graphs of ƒsxd = sin x  x and gsxd = sin x  s6x>s6 + x 2 dd . Describe what you find.
2
at
dx
88. a. Show that the approximation csc x L 1>x + x>6 in Section 11.10, Example 9, leads to the approximation sin x L 6x>s6 + x 2 d .
dH
1 57. 1  2x
65. ƒsxd = 23 + x 2
2x
87. Use a series representation of sin 3x to find values of r and s for which
 Á
s2n  1d A 23 B 2n  1
Taylor Series
L0
tan1 x
82. lim
nR
1 23
7 sin x x:0 e 2x  1
ass a
56.
1>64
80.
a. Use power series to evaluate the limit.
p3 p2n + 1 p5 +  Á + s 1dn + Á 3! 5! s2n + 1d!
55. 1 + ln 2 +
x sin sx 3 d dx
T b. Then use a grapher to support your calculation.
53. p 
2
L0
Indeterminate Forms
1 1 1 +  Á + s 1dn n + Á 4 16 4 n
52.
1>2
78.
iaz
51. 1 
1
3
e x dx
L0
suf
q
841
You
sn + 1dx 2n  1 3n n=0 q
47. a
Practice Exercises
ys0d = 2
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T b. Estimate the magnitude of the error involved in using the sum of the sines through n = 20 to approximate the sum of the series. Is the approximation too large, or too small? Give reasons for your answer. q 1 1 90. a. Show that the series a atan  tan b converges. 2n 2n + 1 n=1 T b. Estimate the magnitude of the error in using the sum of the tangents through tan s1>41d to approximate the sum of the series. Is the approximation too large, or too small? Give reasons for your answer.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
91. Find the radius of convergence of the series
Show that the series
2 # 5 # 8 # Á # s3n  1d n x . 2 # 4 # 6 # Á # s2nd n=1 q
a3 a1 a2 + + + Á 1 2 3
92. Find the radius of convergence of the series
diverges. q
95. a. Find the interval of convergence of the series 1 3 1 6 Á x + x + 6 180 1 # 4 # 7 # Á # s3n  2d 3n + x + Á. s3nd!
y = 1 +
and find the values of the constants a and b.
96. a. Find the Maclaurin series for the function x 2>s1 + xd . b. Does the series converge at x = 1 ? Explain.
97. If g n = 1 an and g n = 1 bn are convergent series of nonnegative q numbers, can anything be said about g n = 1 an bn ? Give reasons for your answer.
dH
q
98. If g n = 1 an and g n = 1 bn are divergent series of nonnegative q numbers, can anything be said about g n = 1 an bn ? Give reasons for your answer.
ma
99. Prove that the sequence 5xn6 and the series g k= 1 sxk + 1  xk d both converge or both diverge. q
100. Prove that g n = 1 san>s1 + an dd converges if an 7 0 for all n and q g n = 1 an converges.
ham
q
1
2 x 10 x e dx. There are several ways to do this. 1 a. Use the Trapezoidal Rule with n = 2 to estimate 10 x 2e x dx. b. Write out the first three nonzero terms of the Taylor series at x = 0 for x 2e x to obtain the fourth Taylor polynomial P(x) 1 for x 2e x . Use 10 Psxd dx to obtain another estimate for
nR
2 x 10 x e dx. c. The second derivative of ƒsxd = x 2e x is positive for all x 7 0 . Explain why this enables you to conclude that the Trapezoidal Rule estimate obtained in part (a) is too large. (Hint: What does the second derivative tell you about the graph of a function? How does this relate to the trapezoidal approximation of the area under this graph?)
ass a
= x ay + b
q
diverges.
104. Suppose you wish to obtain a quick estimate for the value of
1
b. Show that the function defined by the series satisfies a differential equation of the form
q
1 1 + a n = 2 n ln n
iaz
93. Find a closedform formula for the nth partial sum of the series q g n = 2 ln s1  s1>n 2 dd and use it to determine the convergence or divergence of the series. q 94. Evaluate g k= 2 s1>sk 2  1dd by finding the limits as n : q of the series’ nth partial sum.
q
You
103. Use the result in Exercise 102 to show that
3 # 5 # 7 # Á # s2n + 1d n a 4 # 9 # 14 # Á # s5n  1d sx  1d . n=1 q
dx 2
suf
a
d 2y
i
842
1
e. Use integration by parts to evaluate 10 x 2e x dx.
Fourier Series Find the Fourier series for the functions in Exercises 105–108. Sketch each function. 105. ƒsxd = e
0, 1,
0 … x … p p 6 x … 2p
106. ƒsxd = e
x, 1,
0 … x … p p 6 x … 2p
107. ƒsxd = e
p  x, x  2p,
108. ƒsxd = ƒ sin x ƒ ,
0 … x … p p 6 x … 2p
0 … x … 2p
Mu
101. (Continuation of Section 4.7, Exercise 27.) If you did Exercise 27 in Section 4.7, you saw that in practice Newton’s method stopped too far from the root of ƒsxd = sx  1d40 to give a useful estimate of its value, x = 1 . Prove that nevertheless, for any starting value x0 Z 1 , the sequence x0 , x1, x2 , Á , xn , Á of approximations generated by Newton’s method really does converge to 1.
d. All the derivatives of ƒsxd = x 2e x are positive for x 7 0 . Explain why this enables you to conclude that all Maclaurin polynomial approximations to ƒ(x) for x in [0, 1] will be too small. (Hint: ƒsxd = Pnsxd + Rnsxd .)
102. Suppose that a1, a2 , a3 , Á , an are positive numbers satisfying the following conditions: i. a1 Ú a2 Ú a3 Ú Á ; ii. the series a2 + a4 + a8 + a16 + Á diverges.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11
Chapter 11
q stan1 nd2 2. a 2 n=1 n + 1 q logn sn!d 4. a n3
1 n + s1>2d n = 1 s3n  2d q
3. a s 1dn tanh n
+ 17. Evaluate
6. a1 = a2 = 7,
an + 1 =
n an sn  1dsn + 1d
7. a1 = a2 = 1,
an + 1 =
1 1 + an
ass a
an = n>3n if n is even
a. Use the figure to show that g n = 1 An 6 s1>2dsƒs1d  ƒs2dd .
dH
ƒsn + 1dscd ƒsndsad sx  adn + sx  adn + 1 n! sn + 1d!
x = 0.4
x = 69
12. ln x
near near
14. tan1 x
n
lim c a ƒskd n: q k=1
n
L1
ƒsxd dx d .
If ƒsxd = 1>x , the limit in part (c) is Euler’s constant (Section 11.3, Exercise 41). (Source: “Convergence with Pictures” by P. J. Rippon, American Mathematical Monthly, Vol. 93, No. 6, 1986, pp. 476–478.)
x = 6.3 x = 1.3
near
x = 2
Mu
near
10. sin x
n
c. Then show the existence of
ma
ham near
x = 1
n
1 ƒsxd dx d . lim c a ƒskd  sƒs1d + ƒsndd 2 n: q k=1 L1
expresses the value of ƒ at x in terms of the values of ƒ and its derivatives at x = a . In numerical computations, we therefore need a to be a point where we know the values of ƒ and its derivatives. We also need a to be close enough to the values of ƒ we are interested in to make sx  adn + 1 so small we can neglect the remainder. In Exercises 9–14, what Taylor series would you choose to represent the function near the given value of x? (There may be more than one good answer.) Write out the first four nonzero terms of the series you choose.
13. cos x
q
b. Then show the existence of
ƒ–sad ƒsxd = ƒsad + ƒ¿sadsx  ad + sx  ad2 + Á 2!
11. e
nx n a sn + 1ds2x + 1dn n=1
19. Generalizing Euler’s constant The accompanying figure shows the graph of a positive twicedifferentiable decreasing function ƒ whose second derivative is positive on s0, q d . For each n, the number An is the area of the lunar region between the curve and the line segment joining the points (n, ƒ(n)) and sn + 1, ƒsn + 1dd .
if n Ú 2
Taylor’s formula
near
1 dx . 1 + x2
converges absolutely.
if n Ú 2
Choosing Centers for Taylor Series
x
Ln
q
(Hint: Write out several terms, see which factors cancel, and then generalize.)
9. cos x
n=0
18. Find all values of x for which
nsn + 1d an sn + 2dsn + 3d
if n is odd,
a
iaz
an + 1 =
n+1
q
Which of the series defined by the formulas in Exercises 5–8 converge, and which diverge? Give reasons for your answers.
+
7 3 + + Á. 10 8 10 9
n=2
q g n = 1 an
8. an = 1>3n
7 3 7 3 2 2 2 + + + + + + 10 10 5 10 7 10 2 10 3 10 4 10 6
nR
n=1
1 +
You
Which of the series defined by the formulas in Exercises 1–4 converge, and which diverge? Give reasons for your answers. q
suf
16. Find the sum of the infinite series
q g n = 1 an
1. a
843
i
Additional and Advanced Exercises
Convergence or Divergence
5. a1 = 1,
Additional and Advanced Exercises
Theory and Examples 15. Let a and b be constants with 0 6 a 6 b . Does the sequence 5sa n + b n d1>n6 converge? If it does converge, what is the limit?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 844
Chapter 11: Infinite Sequences and Series
y
suf
i
23. Find a value for the constant b that will make the radius of convergence of the power series q
b nx n a ln n n=2 A1
equal to 5.
A2 A3
You
…
24. How do you know that the functions sin x, ln x, and e x are not polynomials? Give reasons for your answer. 25. Find the value of a for which the limit
f(1)
A2
lim
sin saxd  sin x  x x3
x:0
A3
2
1
3
26. Find values of a and b for which
… 4
iaz
f(4) 0
is finite and evaluate the limit.
y f(x)
f(3)
x
5
lim
2b
ass a
2b
2b
2b
2b
2b • • •
ƒsnd un C un + 1 = 1 + n + n 2 ,
a constant ,
ham
cos sa>nd n lim a1 b , n n: q
appear to depend on the value of a? If so, how? b. Does the value of lim a1 
n: q
cos sa>nd n b , bn
a and b constant, b Z 0 ,
(1)
where ƒ ƒsnd ƒ 6 K for n Ú N , then g n = 1 un converges if C 7 1 and diverges if C … 1 . Show that the results of Raabe’s test agree with what you q q know about the series g n = 1 s1>n 2 d and g n = 1 s1>nd . q
28. (Continuation of Exercise 27.) Suppose that the terms of g n = 1 un are defined recursively by the formulas q
u1 = 1,
2b
ma
2b
21. a. Does the value of
un + 1 =
s2n  1d2 un . s2nds2n + 1d
Apply Raabe’s test to determine whether the series converges. 29. If g n = 1 an converges, and if an Z 1 and an 7 0 for all n, q
a. Show that g n = 1 a n2 converges. q
b. Does
an >s1  an d converge? Explain.
q gn=1
30. (Continuation of Exercise 29.) If g n = 1 an converges, and if q 1 7 an 7 0 for all n, show that g n = 1 ln s1  an d converges. q
(Hint: First show that ƒ ln s1  an d ƒ … an>s1  an d .) 31. Nicole Oresme’s Theorem Prove Nicole Oresme’s Theorem that
appear to depend on the value of b? If so, how?
Mu
T
dH
c. Is every point on the original triangle removed? Explain why or why not.
= 1 .
27. Raabe’s (or Gauss’s) test The following test, which we state without proof, is an extension of the Ratio Test. q Raabe’s test: If g n = 1 un is a series of positive constants and there exist constants C, K, and N such that
b. Find the sum of this infinite series and hence find the total area removed from the original triangle.
2b
2x 2
x:0
20. This exercise refers to the “right side up” equilateral triangle with sides of length 2b in the accompanying figure. “Upside down” equilateral triangles are removed from the original triangle as the sequence of pictures suggests. The sum of the areas removed from the original triangle forms an infinite series. a. Find this infinite series.
cos saxd  b
nR
f(2)
c. Use calculus to confirm your findings in parts (a) and (b).
22. Show that if
q g n = 1 an
converges, then
a a q
n=1
1 + sin san d n b 2
1 +
1 2
#
2 +
1 4
#
3 + Á +
n + Á = 4. 2n  1
(Hint: Differentiate both sides of the equation 1>s1  xd = q 1 + g n = 1 x n .)
converges.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11 32. a. Show that
i
q
nsn + 1d 2x 2 = n x sx  1d3 n=1
suf
for ƒ x ƒ 7 1 by differentiating the identity x2 1  x
c. As an engineer applying statistical control to an industrial operation, you inspect items taken at random from the assembly line. You classify each sampled item as either “good” or “bad.” If the probability of an item’s being good is p and of an item’s being bad is q = 1  p , the probability that the first bad item found is the nth one inspected is p n  1q . The average number inspected up to and including the first q bad item found is g n = 1 np n  1q . Evaluate this sum, assuming 0 6 p 6 1.
You
n=1
twice, multiplying the result by x, and then replacing x by 1> x.
b. Use part (a) to find the real solution greater than 1 of the equation q
nsn + 1d . xn n=1
x = a
33. A fast estimate of P /2 As you saw if you did Exercise 127 in Section 11.1, the sequence generated by starting with x0 = 1 and applying the recursion formula xn + 1 = xn + cos xn converges rapidly to p>2 . To explain the speed of the convergence, let Pn = sp>2d  xn . (See the accompanying figure.) Then p  xn  cos xn 2
= Pn  cos a
a. pk = 2k
p  Pn b 2
c. pk =
= Pn  sin Pn
Use this equality to show that
1 sP d3 . 6 n
0 6 Pn + 1 6
where C0 = the change in concentration achievable by a single dose (mg> mL), k = the elimination constant sh1 d , and t0 = time between doses (h). See the accompanying figure.
⑀n
C
xn
ham
1
cos xn
xn
0
1
x
q gn=1
Mu
34. If an is a convergent series of positive numbers, can anyq thing be said about the convergence of g n = 1 ln s1 + an d ? Give reasons for your answer. 35. Quality control
a. Differentiate the series 1 = 1 + x + x2 + Á + xn + Á 1  x to obtain a series for 1>s1  xd2 .
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1 1 1 = k k + 1 ksk + 1d
5k  1 6k
Rn = C0 e kt0 + C0 e 2k t0 + Á + C0 e nk t0 ,
ma
y
b. pk =
T 37. Safe and effective dosage The concentration in the blood resulting from a single dose of a drug normally decreases with time as the drug is eliminated from the body. Doses may therefore need to be repeated periodically to keep the concentration from dropping below some particular level. One model for the effect of repeated doses gives the residual concentration just before the sn + 1dst dose as
dH
1 1 A P B 3  5! A Pn B 5 + Á . 3! n
=
36. Expected value Suppose that a random variable X may assume the values 1, 2, 3, Á , with probabilities p1, p2, p3, Á , where pk is the probability that X equals k sk = 1, 2, 3, Á d . Suppose also q that pk Ú 0 and that g k = 1 pk = 1 . The expected value of X, deq noted by E(X), is the number g k = 1 k pk , provided the series conq verges. In each of the following cases, show that g k = 1 pk = 1 and find E(X) if it exists. (Hint: See Exercise 35.)
ass a
Pn + 1 =
iaz
=
Concentration (mg/mL)
n+1
nR
q
845
b. In one throw of two dice, the probability of getting a roll of 7 is p = 1>6 . If you throw the dice repeatedly, the probability that a 7 will appear for the first time at the nth throw is q n  1p , where q = 1  p = 5>6 . The expected number of q throws until a 7 first appears is g n = 1 nq n  1p . Find the sum of this series.
a
ax
Additional and Advanced Exercises
C1 C0 C0 e– k t 0
Cn1
C2 Rn C0
0
R2 R1 C0 e– k t 0
R3 t
t0 Time (h)
a. Write Rn in closed form as a single fraction, and find R = limn: q Rn .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series is said to converge if the series q
a ln s1 + an d ,
c. If k = 0.01 h1 and t0 = 10 h , find the smallest n such that Rn 7 s1>2dR .
38. Time between drug doses (Continuation of Exercise 37.) If a drug is known to be ineffective below a concentration CL and harmful above some higher concentration CH , one needs to find values of C0 and t0 that will produce a concentration that is safe (not above CH ) but effective (not below CL). See the accompanying figure. We therefore want to find values for C0 and t0 for which
n=1
obtained by taking the natural logarithm of the product, converges. Prove that the product converges if an 7 1 for every n q and if g n = 1 ƒ an ƒ converges. (Hint: Show that
You
(Source: Prescribing Safe and Effective Dosage, B. Horelick and S. Koont, COMAP, Inc., Lexington, MA.)
i
b. Calculate R1 and R10 for C0 = 1 mg>mL, k = 0.1 h1 , and t0 = 10 h . How good an estimate of R is R10 ?
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846
ƒ ln s1 + an d ƒ …
ƒ an ƒ … 2 ƒ an ƒ 1  ƒ an ƒ
when ƒ an ƒ 6 1>2 .) 40. If p is a constant, show that the series q
and
1 1 + a # p # n = 3 n ln n [ln sln nd]
iaz
R = CL
C0 + R = CH .
a. converges if p 7 1 , b. diverges if p … 1 . In general, if ƒ1sxd = x, ƒn + 1sxd = ln sƒnsxdd , and n takes on the values 1, 2, 3, Á , we find that ƒ2sxd = ln x, ƒ3sxd = ln sln xd , and so on. If ƒnsad 7 1 , then
Highest safe level
CH
nR
Concentration in blood
C
C0
q
CL
Lowest effective level
La
0
t
Time
ass a
t0
1 CH ln . k CL
dH
Thus C0 = CH  CL . When these values are substituted in the equation for R obtained in part (a) of Exercise 37, the resulting equation simplifies to t0 =
dx ƒ1sxdƒ2sxd Á ƒnsxdsƒn + 1sxddp
ma
To reach an effective level rapidly, one might administer a “loading” dose that would produce a concentration of CH mg>mL . This could be followed every t0 hours by a dose that raises the concentration by C0 = CH  CL mg>mL . a. Verify the preceding equation for t0 .
ham
b. If k = 0.05 h1 and the highest safe concentration is e times the lowest effective concentration, find the length of time between doses that will assure safe and effective concentrations. c. Given CH = 2 mg>mL, CL = 0.5 mg/mL , and k = 0.02 h1 , determine a scheme for administering the drug.
Mu
d. Suppose that k = 0.2 h1 and that the smallest effective concentration is 0.03 mg> mL. A single dose that produces a concentration of 0.1 mg> mL is administered. About how long will the drug remain effective?
39. An infinite product The infinite product q
Á q s1 + an d = s1 + a1 ds1 + a2 ds1 + a3 d
converges if p 7 1 and diverges if p … 1 .
41. a. Prove the following theorem: If 5cn6 is a sequence of numbers n such that every sum tn = g k = 1 ck is bounded, then the series q q g n = 1 cn >n converges and is equal to g n = 1 tn>(n(n + 1)) . Outline of proof: Replace c1 by t1 and cn by tn  tn  1 2n + 1 for n Ú 2. If s2n + 1 = g k = 1 ck > k, show that s2n + 1 = t1 a1 
1 1 1 b + t2 a  b 2 2 3
t2n + 1 1 1 b + + Á + t2n a 2n 2n + 1 2n + 1 2n t2n + 1 tk + . = a 2n + 1 k = 1 k sk + 1d
Because ƒ tk ƒ 6 M for some constant M, the series q
tk a k sk + 1d
k=1
converges absolutely and s2n + 1 has a limit as n : q . 2n Finally, if s2n = g k = 1 ck >k, then s2n + 1  s2n = c2n + 1>s2n + 1d approaches zero as n : q because ƒ c2n + 1 ƒ = ƒ t2n + 1  t2n ƒ 6 2M. Hence the sequence of partial sums of the series gck >k converges and the limit is q g k = 1 tk >sk sk + 1dd.
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11 Additional and Advanced Exercises b. Show how the foregoing theorem applies to the alternating harmonic series
L0
xn+2 . n + 2
suf
x
ƒ Rn + 1 ƒ …
t n + 1 dt =
a Hint: As t varies from 0 to x,
1 1 1 1 1 1 1   + +   + Á. 5 7 2 3 4 6
1 + t Ú 1 and
converges. (After the first term, the signs are two negative, two positive, two negative, two positive, and so on in that pattern.)
`
to ln s1 + xd for
t n + 1>s1 + td … t n + 1 ,
You
c. Show that the series
and
x
ƒstd dt ` …
L0
x
L0
ƒ ƒstd ƒ dt. b
iaz
d. If 1 6 x 6 0 , show that
a. Show by long division or otherwise that
ƒ Rn + 1 ƒ
s 1dn + 1t n + 1 1 = 1  t + t 2  t 3 + Á + s 1dnt n + . 1 + t 1 + t b. By integrating the equation of part (a) with respect to t from 0 to x, show that
+ s 1dn
xn+1 + Rn + 1 n + 1
where x
`
n+1
ƒtƒ tn+1 .b ` … 1 + t 1  ƒxƒ
e. Use the foregoing results to prove that the series x 
s 1dnx n + 1 x2 x3 x4 + + Á + + Á 2 3 4 n + 1
converges to ln s1 + xd for 1 6 x … 1 .
Mu
ham
ma
dH
tn+1 dt . Rn + 1 = s 1dn + 1 L0 1 + t
n+2 ƒxƒ tn+1 dt ` = . sn + 2ds1  ƒ x ƒ d L0 1  ƒ x ƒ x
a Hint: If x 6 t … 0, then ƒ 1 + t ƒ Ú 1  ƒ x ƒ and
ass a
x2 x3 x4 ln s1 + xd = x + + Á 2 3 4
… `
nR
42. The convergence of 1 6 x … 1
i
c. If x Ú 0 , show that
1 1 1 1 1 1  +  +  + Á. 5 2 3 4 6
q g n = 1 [s 1dn  1x n]>n
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847
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11
i
Technology Application Projects
Mathematica / Maple Module Bouncing Ball The model predicts the height of a bouncing ball, and the time until it stops bouncing.
You
Mathematica / Maple Module
Mu
ham
ma
dH
ass a
nR
iaz
Taylor Polynomial Approximations of a Function A graphical animation shows the convergence of the Taylor polynomials to functions having derivatives of all orders over an interval in their domains.
To Read it Online & Download:
847
suf
Chapter 11
Technology Application Projects
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suf
i
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
12
VECTORS AND THE GEOMETRY OF SPACE
You
Chapter
ass a
nR
iaz
OVERVIEW To apply calculus in many realworld situations and in higher mathematics, we need a mathematical description of threedimensional space. In this chapter we introduce threedimensional coordinate systems and vectors. Building on what we already know about coordinates in the xyplane, we establish coordinates in space by adding a third axis that measures distance above and below the xyplane. Vectors are used to study the analytic geometry of space, where they give simple ways to describe lines, planes, surfaces, and curves in space. We use these geometric ideas in the rest of the book to study motion in space and the calculus of functions of several variables, with their many important applications in science, engineering, economics, and higher mathematics.
12.1
ThreeDimensional Coordinate Systems
dH ma
z
z = constant
ham
(0, 0, z)
(0, y, z)
(x, 0, z)
0
P(x, y, z)
(0, y, 0)
y
Mu
(x, 0, 0) x
x = constant
To locate a point in space, we use three mutually perpendicular coordinate axes, arranged as in Figure 12.1. The axes shown there make a righthanded coordinate frame. When you hold your right hand so that the fingers curl from the positive xaxis toward the positive yaxis, your thumb points along the positive zaxis. So when you look down on the xyplane from the positive direction of the zaxis, positive angles in the plane are measured counterclockwise from the positive xaxis and around the positive zaxis. (In a lefthanded coordinate frame, the zaxis would point downward in Figure 12.1 and angles in the plane would be positive when measured clockwise from the positive xaxis. This is not the convention we have used for measuring angles in the xyplane. Righthanded and lefthanded coordinate frames are not equivalent.) The Cartesian coordinates (x, y, z) of a point P in space are the numbers at which the planes through P perpendicular to the axes cut the axes. Cartesian coordinates for space are also called rectangular coordinates because the axes that define them meet at right angles. Points on the xaxis have y and zcoordinates equal to zero. That is, they have coordinates of the form (x, 0, 0). Similarly, points on the yaxis have coordinates of the form (0, y, 0), and points on the zaxis have coordinates of the form (0, 0, z). The planes determined by the coordinates axes are the xyplane, whose standard equation is z = 0; the yzplane, whose standard equation is x = 0; and the xzplane, whose standard equation is y = 0. They meet at the origin (0, 0, 0) (Figure 12.2). The origin is also identified by simply 0 or sometimes the letter O. The three coordinate planes x = 0, y = 0, and z = 0 divide space into eight cells called octants. The octant in which the point coordinates are all positive is called the first octant; there is no conventional numbering for the other seven octants.
y = constant
(x, y, 0)
FIGURE 12.1 The Cartesian coordinate system is righthanded.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.1
849
ThreeDimensional Coordinate Systems
You
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i
The points in a plane perpendicular to the xaxis all have the same xcoordinate, this being the number at which that plane cuts the xaxis. The y and zcoordinates can be any numbers. Similarly, the points in a plane perpendicular to the yaxis have a common ycoordinate and the points in a plane perpendicular to the zaxis have a common zcoordinate. To write equations for these planes, we name the common coordinate’s value. The plane x = 2 is the plane perpendicular to the xaxis at x = 2. The plane y = 3 is the plane perpendicular to the yaxis at y = 3. The plane z = 5 is the plane perpendicular to the zaxis at z = 5. Figure 12.3 shows the planes x = 2, y = 3, and z = 5, together with their intersection point (2, 3, 5). z
z
(0, 0, 5)
yzplane: x 0 x yplane: z 0
Line y 3, z 5 Plane z 5 Line x 2, z 5
Plane x 2
nR
Origin
(2, 3, 5)
iaz
xzplane: y 0
Plane y 3 0
(0, 3, 0)
(2, 0, 0)
(0, 0, 0)
ass a
x
y
FIGURE 12.2 The planes x = 0, y = 0 , and z = 0 divide space into eight octants.
y x Line x 2, y 3
FIGURE 12.3 The planes x = 2, y = 3 , and z = 5 determine three lines through the point (2, 3, 5).
ma
dH
The planes x = 2 and y = 3 in Figure 12.3 intersect in a line parallel to the zaxis. This line is described by the pair of equations x = 2, y = 3. A point (x, y, z) lies on the line if and only if x = 2 and y = 3. Similarly, the line of intersection of the planes y = 3 and z = 5 is described by the equation pair y = 3, z = 5. This line runs parallel to the xaxis. The line of intersection of the planes x = 2 and z = 5, parallel to the yaxis, is described by the equation pair x = 2, z = 5. In the following examples, we match coordinate equations and inequalities with the sets of points they define in space.
Mu
ham
EXAMPLE 1
Interpreting Equations and Inequalities Geometrically
(a) z Ú 0 (b) x = 3 (c) z = 0, x … 0, y Ú 0 (d) x Ú 0, y Ú 0, z Ú 0 (e) 1 … y … 1 (f) y = 2, z = 2
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The halfspace consisting of the points on and above the xyplane. The plane perpendicular to the xaxis at x = 3. This plane lies parallel to the yzplane and 3 units behind it. The second quadrant of the xyplane. The first octant. The slab between the planes y = 1 and y = 1 (planes included). The line in which the planes y = 2 and z = 2 intersect. Alternatively, the line through the point s0, 2, 2d parallel to the xaxis.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 850
Chapter 12: Vectors and the Geometry of Space
EXAMPLE 2
The circle x 2 y 2 4, z 3
Graphing Equations
i
z
(0, 2, 3)
x2 + y2 = 4
and
suf
What points P(x, y, z) satisfy the equations
(2, 0, 3)
z = 3?
The points lie in the horizontal plane z = 3 and, in this plane, make up the circle x 2 + y 2 = 4. We call this set of points “the circle x 2 + y 2 = 4 in the plane z = 3” or, more simply, “the circle x 2 + y 2 = 4, z = 3 ” (Figure 12.4).
The plane z3
Distance and Spheres in Space
(0, 2, 0) (2, 0, 0)
You
Solution
y
The formula for the distance between two points in the xyplane extends to points in space.
x
FIGURE 12.4 The circle x 2 + y 2 = 4 in the plane z = 3 (Example 2).
iaz
x 2 y 2 4, z 0
The Distance Between P1sx1 , y1 , z1 d and P2sx2 , y2 , z2 d is
nR
ƒ P1 P2 ƒ = 2sx2  x1 d2 + s y2  y1 d2 + sz2  z1 d2
ass a
Proof We construct a rectangular box with faces parallel to the coordinate planes and the points P1 and P2 at opposite corners of the box (Figure 12.5). If Asx2 , y1 , z1 d and Bsx2 , y2 , z1 d are the vertices of the box indicated in the figure, then the three box edges P1 A, AB, and BP2 have lengths z P1(x1, y1, z 1)
P2(x 2, y 2, z 2 )
ƒ P1 A ƒ = ƒ x2  x1 ƒ ,
ƒ AB ƒ = ƒ y2  y1 ƒ ,
ƒ BP2 ƒ = ƒ z2  z1 ƒ .
Because triangles P1 BP2 and P1 AB are both rightangled, two applications of the Pythagorean theorem give
0 A(x2, y1, z 1)
B(x 2, y 2, z 1) y
Mu
ham
FIGURE 12.5 We find the distance between P1 and P2 by applying the Pythagorean theorem to the right triangles P1 AB and P1 BP2 .
and
ƒ P1 B ƒ 2 = ƒ P1 A ƒ 2 + ƒ AB ƒ 2
(see Figure 12.5). So
ƒ P1 P2 ƒ 2 = ƒ P1 B ƒ 2 + ƒ BP2 ƒ 2
ma
x
dH
ƒ P1 P2 ƒ 2 = ƒ P1 B ƒ 2 + ƒ BP2 ƒ 2
= ƒ P1 A ƒ 2 + ƒ AB ƒ 2 + ƒ BP2 ƒ 2
Substitute 2 2 2 ƒ P1 B ƒ = ƒ P1 A ƒ + ƒ AB ƒ .
= ƒ x2  x1 ƒ 2 + ƒ y2  y1 ƒ 2 + ƒ z2  z1 ƒ 2 = sx2  x1 d2 + sy2  y1 d2 + sz2  z1 d2
Therefore ƒ P1 P2 ƒ = 2sx2  x1 d2 + sy2  y1 d2 + sz2  z1 d2
EXAMPLE 3
Finding the Distance Between Two Points
The distance between P1s2, 1, 5d and P2s 2, 3, 0d is
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ƒ P1 P2 ƒ = 2s 2  2d2 + s3  1d2 + s0  5d2 = 216 + 4 + 25 = 245 L 6.708.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.1 z
851
i
We can use the distance formula to write equations for spheres in space (Figure 12.6). A point P(x, y, z) lies on the sphere of radius a centered at P0sx0 , y0 , z0 d precisely when ƒ P0 P ƒ = a or
P(x, y, z)
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P0(x 0 , y0 , z 0 )
ThreeDimensional Coordinate Systems
sx  x0 d2 + sy  y0 d2 + sz  z0 d2 = a 2 .
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a
The Standard Equation for the Sphere of Radius a and Center sx0 , y0 , z0 d
0
sx  x0 d2 + sy  y0 d2 + sz  z0 d2 = a 2
y x
EXAMPLE 4
Finding the Center and Radius of a Sphere
iaz
FIGURE 12.6 The standard equation of the sphere of radius a centered at the point sx0 , y0 , z0 d is
Find the center and radius of the sphere
x 2 + y 2 + z 2 + 3x  4z + 1 = 0.
nR
sx  x0 d2 + s y  y0 d2 + sz  z0 d2 = a 2 .
ass a
Solution We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the x, y, and zterms as necessary and write each quadratic as a squared linear expression. Then, from the equation in standard form, read off the center and radius. For the sphere here, we have
x 2 + y 2 + z 2 + 3x  4z + 1 = 0 sx 2 + 3xd + y 2 + sz 2  4zd = 1
2
2
2
dH
3 3 4 4 ax 2 + 3x + a b b + y 2 + az 2  4z + a b b = 1 + a b + a b 2 2 2 2 ax +
2
2
3 9 21 b + y 2 + sz  2d2 = 1 + + 4 = . 2 4 4
ma
From this standard form, we read that x0 = 3>2, y0 = 0, z0 = 2, and a = 221>2. The center is s 3>2, 0, 2d. The radius is 221>2.
Mu
ham
EXAMPLE 5
Interpreting Equations and Inequalities
(a) x 2 + y 2 + z 2 6 4 (b) x 2 + y 2 + z 2 … 4
(c) x 2 + y 2 + z 2 7 4 (d) x 2 + y 2 + z 2 = 4, z … 0
The interior of the sphere x 2 + y 2 + z 2 = 4. The solid ball bounded by the sphere x 2 + y 2 + z 2 = 4. Alternatively, the sphere x 2 + y 2 + z 2 = 4 together with its interior. The exterior of the sphere x 2 + y 2 + z 2 = 4. The lower hemisphere cut from the sphere x 2 + y 2 + z 2 = 4 by the xyplane (the plane z = 0).
Just as polar coordinates give another way to locate points in the xyplane (Section 10.5), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for threedimensional space. We examine two of these coordinate systems in Section 15.6.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 852
Chapter 12: Vectors and the Geometry of Space
Sets, Equations, and Inequalities
25. The line through the point s1, 3, 1d parallel to the
2
2
2
z = 0
4. x = 1,
5. x + y = 4, 7. x + z = 4,
y = 0
2
2
z = 2
2
2
x = 0
6. x + y = 4,
z = 0
8. y + z = 1,
y = 0
9. x 2 + y 2 + z 2 = 1,
z = 0
x = 0
10. x 2 + y 2 + z 2 = 25,
12. x 2 + s y  1d2 + z 2 = 4,
29. The slab bounded by the planes z = 0 and z = 1 (planes included)
y = 0
z = 0
b. x Ú 0,
b. 0 … x … 1,
14. a. 0 … x … 1 c. 0 … x … 1,
0 … y … 1,
15. a. x 2 + y 2 + z 2 … 1 z = 0
2
2
no restriction on z
2
2
b. x + y + z … 1, 18. a. x = y,
2
2
b. x + y … 1,
z = 3
z Ú 0
dH
c. x + y … 1,
17. a. x 2 + y 2 + z 2 = 1,
0 … y … 1
b. x 2 + y 2 + z 2 7 1
2
16. a. x + y … 1,
z = 0
0 … z … 1
2
2
y … 0,
z Ú 0
z = 0
b. x = y,
30. The solid cube in the first octant bounded by the coordinate planes and the planes x = 2, y = 2 , and z = 2 31. The halfspace consisting of the points on and below the xyplane 32. The upper hemisphere of the sphere of radius 1 centered at the origin 33. The (a) interior and (b) exterior of the sphere of radius 1 centered at the point (1, 1, 1)
ass a
y Ú 0,
28. The set of points in space that lie 2 units from the point (0, 0, 1) and, at the same time, 2 units from the point s0, 0, 1d Write inequalities to describe the sets in Exercises 29–34.
z = 0
In Exercises 13–18, describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. 13. a. x Ú 0,
27. The circle in which the plane through the point (1, 1, 3) perpendicular to the zaxis meets the sphere of radius 5 centered at the origin
y = 4
11. x 2 + y 2 + sz + 3d2 = 25,
c. zaxis
iaz
2
2. x = 1,
b. yaxis
nR
3. y = 0,
y = 3
a. xaxis
26. The set of points in space equidistant from the origin and the point (0, 2, 0)
You
In Exercises 1–12, give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. 1. x = 2,
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EXERCISES 12.1
no restriction on z
34. The closed region bounded by the spheres of radius 1 and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open region bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out.)
Distance
19. The plane perpendicular to the
35. P1s1, 1, 1d,
P2s3, 3, 0d
36. P1s 1, 1, 5d,
P2s2, 5, 0d
37. P1s1, 4, 5d,
P2s4, 2, 7d
38. P1s3, 4, 5d,
P2s2, 3, 4d
39. P1s0, 0, 0d,
P2s2, 2, 2d
40. P1s5, 3, 2d,
P2s0, 0, 0d
a. xaxis at (3, 0, 0) c. zaxis at s0, 0, 2d
ma
In Exercises 19–28, describe the given set with a single equation or with a pair of equations. b. yaxis at s0, 1, 0d
20. The plane through the point s3, 1, 2d perpendicular to the b. yaxis
c. zaxis
ham
a. xaxis
21. The plane through the point s3, 1, 1d parallel to the a. xyplane
b. yzplane
c. xzplane
22. The circle of radius 2 centered at (0, 0, 0) and lying in the a. xyplane
b. yzplane
c. xzplane
Mu
23. The circle of radius 2 centered at (0, 2, 0) and lying in the a. xyplane
b. yzplane
c. plane y = 2
24. The circle of radius 1 centered at s 3, 4, 1d and lying in a plane parallel to the a. xyplane
b. yzplane
In Exercises 35–40, find the distance between points P1 and P2.
c. xzplane
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Spheres Find the centers and radii of the spheres in Exercises 41–44. 41. sx + 2d2 + y 2 + sz  2d2 = 8 42. ax +
2
2
2
1 1 1 21 b + ay + b + az + b = 2 2 2 4
43. A x  22 B 2 + A y  22 B 2 + A z + 22 B 2 = 2 44. x 2 + ay +
2
2
29 1 1 b + az  b = 3 3 9
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.1 ThreeDimensional Coordinate Systems
Radius
45. (1, 2, 3)
214
46. s0, 1, 5d
2
47. s 2, 0, 0d
23
48. s0, 7, 0d
7
Theory and Examples
2
i
53. Find a formula for the distance from the point P(x, y, z) to the a. xaxis
b. yaxis
c. zaxis
54. Find a formula for the distance from the point P(x, y, z) to the a. xyplane
Find the centers and radii of the spheres in Exercises 49â€“52. 2
52. 3x 2 + 3y 2 + 3z 2 + 2y  2z = 9
You
Center
51. 2x 2 + 2y 2 + 2z 2 + x + y + z = 9
suf
Find equations for the spheres whose centers and radii are given in Exercises 45â€“48.
853
2
49. x + y + z + 4x  4z = 0
b. yzplane
c. xzplane
55. Find the perimeter of the triangle with vertices As 1, 2, 1d, Bs1, 1, 3d , and C(3, 4, 5). 56. Show that the point P(3, 1, 2) is equidistant from the points As2, 1, 3d and B(4, 3, 1).
Mu
ham
ma
dH
ass a
nR
iaz
50. x 2 + y 2 + z 2  6y + 8z = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o 12.2 Vectors
12.2
Vectors
853
Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit of measure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body’s displacement, we have to say in what direction it moved as well as how far. To describe a body’s velocity, we have to know where the body is headed as well as how fast it is going.
Y z
Component Form Terminal point B
AB
Initial point A
FIGURE 12.7 1 AB .
The directed line segment
A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Figure 12.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen unit. For example, a force vector points in the direction in which the force acts; its length is a measure of the force’s strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 12.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 13.)
a m m
a h u M
n a sa s H d
iR a
y
z v
v
0
x
0
(a) two dimensions
x
y
(b) three dimensions
FIGURE 12.8 The velocity vector of a particle moving along a path (a) in the plane (b) in space. The arrowhead on the path indicates the direction of motion of the particle.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 854
Chapter 12: Vectors and the Geometry of Space y B
i
DEFINITIONS Vector, Initial and Terminal Point, Length 1 A vector in the plane is a directed line segment. The directed line segment AB 1 has initial point A and terminal point B; its length is denoted by ƒ AB ƒ . Two vectors are equal if they have the same length and direction.
D C P
0
Q(x 2 , y 2 , z 2 )
Position vector of PQ v 具v1, v 2, v3 典 v2
x
(v1, v 2 , v3)
v3 v1
y
ass a
z
nR
FIGURE 12.9 The four arrows in the plane (directed line segments) shown here have the same length and direction. They therefore represent the same vector, and we 1 1 1 1 write AB = CD = OP = EF .
The arrows we use when we draw vectors are understood to represent the same vector if they have the same length, are parallel, and point in the same direction (Figure 12.9) regardless of the initial point. In textbooks, vectors are usually written in lowercase, boldface letters, for example u, v, and w. Sometimes we use uppercase boldface letters, such as F, to denote a force vector. u, In handwritten form, it is customary to draw small arrows above the letters, for example s s. y, w s s , and F We need a way to represent vectors algebraically so that we can be more precise about the direction of a vector. 1 1 Let v = PQ . There is one directed line segment equal to PQ whose initial point is the origin (Figure 12.10). It is the representative of v in standard position and is the vector we normally use to represent v. We can specify v by writing the coordinates of its terminal point sv1, v2 , v3 d when v is in standard position. If v is a vector in the plane its terminal point sv1, v2 d has two coordinates.
iaz
E
P(x 1, y1, z 1)
You
x F
DEFINITION Component Form If v is a twodimensional vector in the plane equal to the vector with initial point at the origin and terminal point sv1, v2 d, then the component form of v is
dH
O
suf
A
v = 8v1, v29.
If v is a threedimensional vector equal to the vector with initial point at the origin and terminal point sv1, v2, v3 d, then the component form of v is v = 8v1, v2 , v39.
Mu
ham
ma
1 FIGURE 12.10 A vector PQ in standard position has its initial point at the origin. 1 The directed line segments PQ and v are parallel and have the same length.
So a twodimensional vector is an ordered pair v = 8v1, v29 of real numbers, and a threedimensional vector is an ordered triple v = 8v1, v2 , v39 of real numbers. The numbers v1, v2 , and v3 are called the components of v. 1 Observe that if v = 8v1, v2 , v39 is represented by the directed line segment PQ , where the initial point is Psx1, y1, z1 d and the terminal point is Qsx2 , y2 , z2 d, then x1 + v1 = x2 , y1 + v2 = y2 , and z1 + v3 = z2 (see Figure 12.10). Thus, v1 = x2  x1, v2 = y2  y1 , and 1 v3 = z2  z1 are the components of PQ . In summary, given the points Psx1, y1, z1 d and Qsx2 , y2 , z2 d, the standard position 1 vector v = 8v1, v2 , v39 equal to PQ is v = 8x2  x1, y2  y1, z2 , z19.
If v is twodimensional with Psx1, y1 d and Qsx2 , y2 d as points in the plane, then v = 8x2  x1, y2  y19. There is no third component for planar vectors. With this understanding, we will develop the algebra of threedimensional vectors and simply drop the third component when the vector is twodimensional (a planar vector).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2
Vectors
855
You
suf
i
Two vectors are equal if and only if their standard position vectors are identical. Thus 8u1, u2 , u39 and 8v1, v2 , v39 are equal if and only if u1 = v1, u2 = v2 , and u3 = v3 . 1 The magnitude or length of the vector PQ is the length of any of its equivalent directed line segment representations. In particular, if v = 8x2  x1, y2  y1, z2  z19 is the 1 standard position vector for PQ , then the distance formula gives the magnitude or length of v, denoted by the symbol ƒ v ƒ or ƒ ƒ v ƒ ƒ .
1 The magnitude or length of the vector v = PQ is the nonnegative number
(See Figure 12.10.)
iaz
ƒ v ƒ = 2v12 + v22 + v32 = 2sx2  x1 d2 + s y2  y1 d2 + sz2  z1 d2
EXAMPLE 1
nR
The only vector with length 0 is the zero vector 0 = 80, 09 or 0 = 80, 0, 09. This vector is also the only vector with no specific direction.
Component Form and Length of a Vector
ass a
Find the (a) component form and (b) length of the vector with initial point Ps 3, 4, 1d and terminal point Qs 5, 2, 2d. Solution
1 (a) The standard position vector v representing PQ has components
dH
v1 = x2  x1 = 5  s 3d = 2,
v2 = y2  y1 = 2  4 = 2,
and
v3 = z2  z1 = 2  1 = 1.
ma
1 The component form of PQ is v = 82, 2, 19.
1 (b) The length or magnitude of v = PQ is
ham
y
F = 具a, b典
Mu
45°
x
FIGURE 12.11 The force pulling the cart forward is represented by the vector F of magnitude 20 (pounds) making an angle of 45° with the horizontal ground (positive xaxis) (Example 2).
ƒ v ƒ = 2s 2d2 + s 2d2 + s1d2 = 29 = 3.
EXAMPLE 2
Force Moving a Cart
A small cart is being pulled along a smooth horizontal floor with a 20lb force F making a 45° angle to the floor (Figure 12.11). What is the effective force moving the cart forward? Solution
The effective force is the horizontal component of F = 8a, b9, given by 22 a = ƒ F ƒ cos 45° = s20d a b L 14.14 lb. 2
Notice that F is a twodimensional vector.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 856
Chapter 12: Vectors and the Geometry of Space
i
Vector Algebra Operations
You
suf
Two principal operations involving vectors are vector addition and scalar multiplication. A scalar is simply a real number, and is called such when we want to draw attention to its differences from vectors. Scalars can be positive, negative, or zero.
DEFINITIONS Vector Addition and Multiplication of a Vector by a Scalar Let u = 8u1, u2 , u39 and v = 8v1, v2 , v39 be vectors with k a scalar.
iaz
Addition: u + v = 8u1 + v1, u2 + v2 , u3 + v39 Scalar multiplication: ku = 8ku1, ku2 , ku39
y
ass a
nR
We add vectors by adding the corresponding components of the vectors. We multiply a vector by a scalar by multiplying each component by the scalar. The definitions apply to planar vectors except there are only two components, 8u1, u29 and 8v1, v29. The definition of vector addition is illustrated geometrically for planar vectors in Figure 12.12a, where the initial point of one vector is placed at the terminal point of the other. Another interpretation is shown in Figure 12.12b (called the parallelogram law of addition), where the sum, called the resultant vector, is the diagonal of the parallelogram. In physics, forces add vectorially as do velocities, accelerations, and so on. So the force acting on a particle subject to electric and gravitational forces is obtained by adding the two force vectors. y
dH
具u1 v1, u 2 v 2 典
u+v
v2
v
u
u+v v
v1
u
ma
u2
Mu
1.5u
ham
0
u
2u
(a)
x
0 (b)
FIGURE 12.12 (a) Geometric interpretation of the vector sum. (b) The parallelogram law of vector addition.
Figure 12.13 displays a geometric interpretation of the product ku of the scalar k and vector u. If k 7 0, then ku has the same direction as u; if k 6 0, then the direction of ku is opposite to that of u. Comparing the lengths of u and ku, we see that 2 2 2 ƒ ku ƒ = 2sku1 d2 + sku2 d2 + sku3 d2 = 2k 2su 1 + u 2 + u 3 d
= 2k 2 2u 12 + u 22 + u 32 = ƒ k ƒ ƒ u ƒ .
–2u
FIGURE 12.13 Scalar multiples of u.
x
u1
The length of ku is the absolute value of the scalar k times the length of u. The vector s 1du = u has the same length as u but points in the opposite direction.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2
Vectors
857
If u = 8u1, u2 , u39 and v = 8v1, v2 , v39, then
suf
u  v = u + s vd.
v
i
By the difference u  v of two vectors, we mean
uv
(a)
You
u  v = 8u1  v1, u2  v2, u3  v39.
u
Note that su  vd + v = u, so adding the vector su  vd to v gives u (Figure 12.14a). Figure 12.14b shows the difference u  v as the sum u + s vd.
EXAMPLE 3
Performing Operations on Vectors
Let u = 81, 3, 19 and v = 84, 7, 09. Find u
(a) 2u + 3v
–v u (–v)
(b) u  v
Solution
(c) `
1 u` . 2
iaz
v
(b) u  v = 81, 3, 19  84, 7, 09 = 81  4, 3  7, 1  09 = 85, 4, 19 (c)
`
2
2
2
3 1 1 3 1 1 1 1 u ` = ` h , , i ` = a b + a b + a b = 211 . 2 2 2 2 2 2 2 2 C
ass a
FIGURE 12.14 (a) The vector u  v, when added to v, gives u. (b) u  v = u + s vd .
nR
(a) 2u + 3v = 281, 3, 19 + 384, 7, 09 = 82, 6, 29 + 812, 21, 09 = 810, 27, 29
(b)
dH
Vector operations have many of the properties of ordinary arithmetic. These properties are readily verified using the definitions of vector addition and multiplication by a scalar.
Properties of Vector Operations Let u, v, w be vectors and a, b be scalars.
ma
1.
Mu
ham
3. 5. 7. 9.
u + v = v + u u + 0 = u 0u = 0 asbud = sabdu sa + bdu = au + bu
2. su + vd + w = u + sv + wd 4. u + s ud = 0 6. 1u = u 8. asu + vd = au + av
An important application of vectors occurs in navigation.
EXAMPLE 4
Finding Ground Speed and Direction
A Boeing® 767® airplane, flying due east at 500 mph in still air, encounters a 70mph tailwind blowing in the direction 60° north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 858
Chapter 12: Vectors and the Geometry of Space
If u = the velocity of the airplane alone and v = the velocity of the tailwind, then ƒ u ƒ = 500 and ƒ v ƒ = 70 (Figure 12.15). The velocity of the airplane with respect to the ground is given by the magnitude and direction of the resultant vector u + v. If we let the positive xaxis represent east and the positive yaxis represent north, then the component forms of u and v are
Solution
500
u = 8500, 09
E
u
v = 870 cos 60°, 70 sin 60°9 = 835, 35239.
and
Therefore,
NOT TO SCALE
u + v = 8535, 35239
FIGURE 12.15 Vectors representing the velocities of the airplane u and tailwind v in Example 4.
You
uv
ƒ u + v ƒ = 25352 + s3513d2 L 538.4 and
3523 L 6.5°. 535
Figure 12.15
nR
u = tan1
iaz
v 30˚ 70
suf
i
N
The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5° north of east.
Unit Vectors
ass a
A vector v of length 1 is called a unit vector. The standard unit vectors are i = 81, 0, 09,
j = 80, 1, 09,
and
k = 80, 0, 19.
Any vector v = 8v1, v2 , v39 can be written as a linear combination of the standard unit vectors as follows:
z
= v181, 0, 09 + v280, 1, 09 + v380, 0, 19
OP2 x 2 i y2 j z 2 k
k
We call the scalar (or number) v1 the icomponent of the vector v, v2 the jcomponent, and v3 the kcomponent. In component form, the vector from P1sx1, y1, z1 d to P2sx2 , y2 , z2 d is 1 P1 P2 = sx2  x1 di + s y2  y1 dj + sz2  z1 dk
P1P2
j
ham
i
y
x
= v1 i + v2 j + v3 k.
ma
P2(x2, y 2 , z 2 )
O
dH
v = 8v1, v2 , v39 = 8v1, 0, 09 + 80, v2 , 09 + 80, 0, v39
P1(x 1, y 1 , z 1 )
(Figure 12.16). Whenever v Z 0, its length ƒ v ƒ is not zero and
`
OP1 x 1i y 1 j z 1k
Mu
FIGURE 12.16 The vector from P1 to P2 1 is P1 P2 = sx2  x1 di + s y2  y1 dj + sz2  z1 dk.
1 1 v` = ƒ v ƒ = 1. v v ƒ ƒ ƒ ƒ
That is, v> ƒ v ƒ is a unit vector in the direction of v, called the direction of the nonzero vector v.
EXAMPLE 5
Finding a Vector’s Direction
Find a unit vector u in the direction of the vector from P1s1, 0, 1d to P2s3, 2, 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2
859
1 We divide P1 P2 by its length: 1 P1 P2 = s3  1di + s2  0dj + s0  1dk = 2i + 2j  k
suf
i
Solution
Vectors
EXAMPLE 6
You
1 ƒ P1 P2 ƒ = 2s2d2 + s2d2 + s 1d2 = 24 + 4 + 1 = 29 = 3 1 2i + 2j  k P1 P2 2 2 1 = i + j  k. u = 1 = 3 3 3 3 ƒ P1 P2 ƒ 1 The unit vector u is the direction of P1 P2 .
Expressing Velocity as Speed Times Direction
Solution
Speed is the magnitude (length) of v:
ƒ v ƒ = 2s3d2 + s 4d2 = 29 + 16 = 5.
nR
HISTORICAL BIOGRAPHY
iaz
If v = 3i  4j is a velocity vector, express v as a product of its speed times a unit vector in the direction of motion.
The unit vector v> ƒ v ƒ has the same direction as v:
Hermann Grassmann (1809–1877)
3i  4j v 3 4 = = i  j. 5 5 5 ƒvƒ
ass a
So
3 4 v = 3i  4j = 5 a i  jb . 5 5 (')'* Length (speed)
Direction of motion
dH
In summary, we can express any nonzero vector v in terms of its two important features, v . length and direction, by writing v = ƒ v ƒ ƒvƒ
Mu
ham
ma
If v Z 0, then v 1. is a unit vector in the direction of v; ƒvƒ v 2. the equation v = ƒ v ƒ expresses v in terms of its length and direction. ƒvƒ
EXAMPLE 7
A Force Vector
A force of 6 newtons is applied in the direction of the vector v = 2i + 2j  k. Express the force F as a product of its magnitude and direction. v Solution The force vector has magnitude 6 and direction , so ƒvƒ 2i + 2j  k 2i + 2j  k v F = 6 = 6 = 6 2 2 2 3 ƒvƒ 22 + 2 + s 1d
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2 2 1 = 6 a i + j  kb . 3 3 3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 860
Chapter 12: Vectors and the Geometry of Space
i
Midpoint of a Line Segment
suf
Vectors are often useful in geometry. For example, the coordinates of the midpoint of a line segment are found by averaging.
P1(x 1, y1, z 1)
x1 + x2 y1 + y2 z1 + z2 , , b. 2 2 2
iaz
a
You
The midpoint M of the line segment joining points P1sx1, y1, z1 d and P2sx2 , y2 , z2 d is the point
To see why, observe (Figure 12.17) that
1 1 1 1 1 1 1 1 OM = OP1 + sP1 P2 d = OP1 + sOP2  OP1 d 2 2
x x 2 y1 y 2 z1 z 2 , , M 1 2 2 2
ass a
P2(x 2, y 2 , z 2 )
EXAMPLE 8
O
Finding Midpoints
The midpoint of the segment joining P1s3, 2, 0d and P2s7, 4, 4d is a
3 + 7 2 + 4 0 + 4 , , b = s5, 1, 2d. 2 2 2
Mu
ham
ma
dH
FIGURE 12.17 The coordinates of the midpoint are the averages of the coordinates of P1 and P2 .
nR
1 1 1 sOP1 + OP2 d 2 y1 + y2 x1 + x2 z1 + z2 = i + j + k. 2 2 2 =
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 860
i f su
Chapter 12: Vectors and the Geometry of Space
u o zY
EXERCISES 12.2 Vectors in the Plane
In Exercises 1–8, let u = 83, 29 and v = 82, 59 . Find the (a) component form and (b) magnitude (length) of the vector. 1. 3u
2. 2v
3. u + v
4. u  v
5. 2u  3v
6. 2u + 5v
7.
8. 
5 12 u + v 13 13
In Exercises 9–16, find the component form of the vector. 1 9. The vector PQ , where P = s1, 3d and Q = s2, 1d 1 10. The vector OP where O is the origin and P is the midpoint of segment RS, where R = s2, 1d and S = s 4, 3d
h u M
11. The vector from the point A = s2, 3d to the origin 1 1 12. The sum of AB and CD , where A = s1, 1d, B = s2, 0d, C = s 1, 3d , and D = s 2, 2d
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14. The unit vector that makes an angle u = 3p>4 with the positive xaxis
n a s s a
H d a m am
3 4 u + v 5 5
a i R
13. The unit vector that makes an angle u = 2p>3 with the positive xaxis
15. The unit vector obtained by rotating the vector 80, 19 120° counterclockwise about the origin 16. The unit vector obtained by rotating the vector 81, 09 135° counterclockwise about the origin
Vectors in Space In Exercises 17–22, express each vector in the form v = v1 i + v2 j + v3 k . 1 17. P1 P2 if P1 is the point s5, 7, 1d and P2 is the point s2, 9, 2d 1 18. P1 P2 if P1 is the point (1, 2, 0) and P2 is the point s 3, 0, 5d 1 19. AB if A is the point s 7, 8, 1d and B is the point s 10, 8, 1d 1 20. AB if A is the point (1, 0, 3) and B is the point s 1, 4, 5d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2 Vectors 21. 5u  v if u = 81, 1, 19 and v = 82, 0, 39
861
Length
Geometry and Calculation
Direction
a. 7
In Exercises 23 and 24, copy vectors u, v, and w head to tail as needed to sketch the indicated vector.
j 3 4  i  k b. 22 5 5 13 3 4 12 c. i j k 12 13 13 13 1 1 1 d. a 7 0 i + j k 22 23 26 33. Find a vector of magnitude 7 in the direction of v = 12i  5k .
You
23.
suf
22. 2u + 3v if u = 81, 0, 29 and v = 81, 1, 19
i
32. Find the vectors whose lengths and directions are given. Try to do the calculations without writing.
b. u + v + w
c. u  v
d. u  w
34. Find a vector of magnitude 3 in the direction opposite to the direction of v = s1>2di  s1>2dj  s1>2dk .
iaz
a. u + v
Vectors Determined by Points; Midpoints
24.
In Exercises 35–38, find
1
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a. the direction of P1 P2 and
b. the midpoint of line segment P1 P2 . P2s2, 5, 0d
35. P1s 1, 1, 5d 36. P1s1, 4, 5d
P2s4, 2, 7d
b. u  v + w
c. 2u  v
d. u + v + w
Length and Direction
dH
a. u  v
ass a
P2s2, 3, 4d 37. P1s3, 4, 5d P2s2, 2, 2d 38. P1s0, 0, 0d 1 39. If AB = i + 4j  2k and B is the point (5, 1, 3), find A. 1 40. If AB = 7i + 3j + 8k and A is the point s 2, 3, 6d , find B.
In Exercises 25–30, express each vector as a product of its length and direction. 26. 9i  2j + 6k 3 4 27. 5k 28. i + k 5 5 j 1 i 1 1 k i j k + + 29. 30. 26 26 26 23 23 23 31. Find the vectors whose lengths and directions are given. Try to do the calculations without writing.
ham
ma
25. 2i + j  2k
Length
Direction
a. 2
i
k 3 j + 5 6 i 7
Mu
b. 23 1 c. 2 d. 7
4 k 5 3 2 j + k 7 7
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Theory and Applications 41. Linear combination Let u = 2i + j, v = i + j , and w = i  j . Find scalars a and b such that u = av + bw .
42. Linear combination Let u = i  2j , v = 2i + 3j , and w = i + j . Write u = u1 + u2 , where u1 is parallel to v and u2 is parallel to w. (See Exercise 41.) 43. Force vector You are pulling on a suitcase with a force F (pictured here) whose magnitude is ƒ F ƒ = 10 lb . Find the i and jcomponents of F. F 30°
44. Force vector A kite string exerts a 12lb pull s ƒ F ƒ = 12d on a kite and makes a 45° angle with the horizontal. Find the horizontal and vertical components of F.
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4100 AWL/Thomas_ch12p848905 8/25/04 2:43 PM Page 862
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space c. Find the coordinates of the point in which the medians of ¢ABC intersect. According to Exercise 29, Section 6.4, this point is the plate’s center of mass.
i
45° F
z
You
C(1, 1, 3)
45. Velocity An airplane is flying in the direction 25° west of north at 800 km> h. Find the component form of the velocity of the airplane, assuming that the positive xaxis represents due east and the positive yaxis represents due north.
a. At what point is the tree located? b. At what point is the telephone pole?
y
B(1, 3, 0)
M
iaz
x
A(4, 2, 0)
50. Find the vector from the origin to the point of intersection of the medians of the triangle whose vertices are As1, 1, 2d,
Bs2, 1, 3d,
and
Cs 1, 2, 1d .
51. Let ABCD be a general, not necessarily planar, quadrilateral in space. Show that the two segments joining the midpoints of opposite sides of ABCD bisect each other. (Hint: Show that the segments have the same midpoint.)
ass a
47. Location A bird flies from its nest 5 km in the direction 60° north of east, where it stops to rest on a tree. It then flies 10 km in the direction due southeast and lands atop a telephone pole. Place an xycoordinate system so that the origin is the bird’s nest, the xaxis points east, and the yaxis points north.
c.m.
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46. Velocity An airplane is flying in the direction 10° east of south at 600 km> h. Find the component form of the velocity of the airplane, assuming that the positive xaxis represents due east and the positive yaxis represents due north.
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862
48. Use similar triangles to find the coordinates of the point Q that divides the segment from P1sx1, y1, z1 d to P2sx2 , y2 , z2 d into two lengths whose ratio is p>q = r .
dH
49. Medians of a triangle Suppose that A, B, and C are the corner points of the thin triangular plate of constant density shown here. a. Find the vector from C to the midpoint M of side AB.
53. Suppose that A, B, and C are vertices of a triangle and that a, b, and c are, respectively, the midpoints of the opposite sides. Show 1 1 1 that Aa + Bb + Cc = 0 .
54. Unit vectors in the plane Show that a unit vector in the plane can be expressed as u = scos udi + ssin udj , obtained by rotating i through an angle u in the counterclockwise direction. Explain why this form gives every unit vector in the plane.
Mu
ham
ma
b. Find the vector from C to the point that lies twothirds of the way from C to M on the median CM.
52. Vectors are drawn from the center of a regular nsided polygon in the plane to the vertices of the polygon. Show that the sum of the vectors is zero. (Hint: What happens to the sum if you rotate the polygon about its center?)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 862
12.3
The Dot Product
F
v
m a h u M
FIGURE 12.18 The magnitude of the force F in the direction of vector v is the length ƒ F ƒ cos u of the projection of F onto v.
a i R an
If a force F is applied to a particle moving along a path, we often need to know the magnitude of the force in the direction of motion. If v is parallel to the tangent line to the path at the point where F is applied, then we want the magnitude of F in the direction of v. Figure 12.18 shows that the scalar quantity we seek is the length ƒ F ƒ cos u, where u is the angle between the two vectors F and v. In this section, we show how to calculate easily the angle between two vectors directly from their components. A key part of the calculation is an expression called the dot product. Dot products are also called inner or scalar products because the product results in a scalar, not a vector. After investigating the dot product, we apply it to finding the projection of one vector onto another (as displayed in Figure 12.18) and to finding the work done by a constant force acting through a displacement.
d a m
Length F cos
i f su
u o zY
Chapter 12: Vectors and the Geometry of Space
s s Ha
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3
863
The Dot Product
i
Angle Between Vectors
v
u
You
suf
When two nonzero vectors u and v are placed so their initial points coincide, they form an angle u of measure 0 … u … p (Figure 12.19). If the vectors do not lie along the same line, the angle u is measured in the plane containing both of them. If they do lie along the same line, the angle between them is 0 if they point in the same direction, and p if they point in opposite directions. The angle u is the angle between u and v. Theorem 1 gives a formula to determine this angle.
FIGURE 12.19 The angle between u and v.
iaz
THEOREM 1 Angle Between Two Vectors The angle u between two nonzero vectors u = 8u1, u2 , u39 8v1, v2 , v39 is given by
v =
u1 v1 + u2 v2 + u3 v3 b. ƒuƒ ƒvƒ
nR
u = cos1 a
and
ass a
Before proving Theorem 1 (which is a consequence of the law of cosines), let’s focus attention on the expression u1 v1 + u2 v2 + u3 v3 in the calculation for u.
dH
DEFINITION Dot Product The dot product u # v s“u dot v”d of vectors u = 8u1, u2 , u39 and v = 8v1, v2 , v39 is
ma
EXAMPLE 1
u # v = u1 v1 + u2 v2 + u3 v3 .
Finding Dot Products
ham
(a) 81, 2, 19 # 86, 2, 39 = s1ds 6d + s 2ds2d + s 1ds 3d = 6  4 + 3 = 7
w
Mu
u
v
1 1 (b) a i + 3j + kb # s4i  j + 2kd = a bs4d + s3ds 1d + s1ds2d = 1 2 2 The dot product of a pair of twodimensional vectors is defined in a similar fashion: 8u1, u29 # 8v1, v29 = u1 v1 + u2 v2 . Proof of Theorem 1 Applying the law of cosines (Equation (6), Section 1.6) to the triangle in Figure 12.20, we find that
FIGURE 12.20 The parallelogram law of addition of vectors gives w = u  v.
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ƒ w ƒ 2 = ƒ u ƒ 2 + ƒ v ƒ 2  2 ƒ u ƒ ƒ v ƒ cos u 2 ƒ u ƒ ƒ v ƒ cos u = ƒ u ƒ 2 + ƒ v ƒ 2  ƒ w ƒ 2 .
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Law of cosines
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 864
Chapter 12: Vectors and the Geometry of Space
ƒvƒ2 =
A 2v12 + v22 + v32 B 2 = v12 + v22 + v32
A 2su1  v1 d2 + su2  v2 d2 + su3  v3 d2 B 2
You
ƒwƒ2 =
A 2u 12 + u 22 + u 32 B 2 = u 12 + u 22 + u 32
suf
ƒuƒ2 =
i
Because w = u  v, the component form of w is 8u1  v1, u2  v2 , u3  v39. So
= su1  v1 d2 + su2  v2 d2 + su3  v3 d2
= u 12  2u1v1 + v12 + u 22  2u2v2 + v22 + u 32  2u3v3 + v32 and
iaz
ƒ u ƒ 2 + ƒ v ƒ 2  ƒ w ƒ 2 = 2su1 v1 + u2v2 + u3 v3). Therefore,
2 ƒ u ƒ ƒ v ƒ cos u = ƒ u ƒ 2 + ƒ v ƒ 2  ƒ w ƒ 2 = 2su1 v1 + u2 v2 + u3 v3 d
nR
ƒ u ƒ ƒ v ƒ cos u = u1 v1 + u2 v2 + u3 v3 cos u =
ass a
So
u1 v1 + u2 v2 + u3 v3 ƒuƒ ƒvƒ
u = cos1 a
u1 v1 + u2 v2 + u3 v3 b ƒuƒ ƒvƒ
With the notation of the dot product, the angle between two vectors u and v can be written as
dH
u = cos1 a
EXAMPLE 2
u#v b. ƒuƒ ƒvƒ
Finding the Angle Between Two Vectors in Space
ma
Find the angle between u = i  2j  2k and v = 6i + 3j + 2k.
Mu
ham
Solution
We use the formula above: u # v = s1ds6d + s 2ds3d + s 2ds2d = 6  6  4 = 4 ƒ u ƒ = 2s1d2 + s 2d2 + s 2d2 = 29 = 3 ƒ v ƒ = 2s6d2 + s3d2 + s2d2 = 249 = 7 u = cos1 a = cos1 a
u#v b ƒuƒ ƒvƒ 4 b L 1.76 radians. s3ds7d
The angle formula applies to twodimensional vectors as well.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3 y
EXAMPLE 3
865
Finding an Angle of a Triangle
i
B(3, 5)
The Dot Product
C(5, 2)
A
x
1
FIGURE 12.21 Example 3.
1 1 The angle u is the angle between the vectors CA and CB . The component forms of these two vectors are 1 1 CA = 85, 29 and CB = 82, 39.
Solution
1
The triangle in
You
suf
Find the angle u in the triangle ABC determined by the vertices A = s0, 0d, B = s3, 5d, and C = s5, 2d (Figure 12.21).
First we calculate the dot product and magnitudes of these two vectors. 1 1 CA # CB = s 5ds 2d + s 2ds3d = 4
iaz
1 ƒ CA ƒ = 2s 5d2 + s 2d2 = 229 1 ƒ CB ƒ = 2s 2d2 + s3d2 = 213 Then applying the angle formula, we have
nR
1 1 CA # CB u = cos1 £ 1 1 ≥ ƒ CA ƒ ƒ CB ƒ
ass a
= cos1 £
4
A 229 B A 213 B
L 78.1°
or
≥
1.36 radians.
dH
Perpendicular (Orthogonal) Vectors
ma
Two nonzero vectors u and v are perpendicular or orthogonal if the angle between them is p>2. For such vectors, we have u # v = 0 because cos sp>2d = 0. The converse is also true. If u and v are nonzero vectors with u # v = ƒ u ƒ ƒ v ƒ cos u = 0, then cos u = 0 and u = cos1 0 = p>2.
Mu
ham
DEFINITION Orthogonal Vectors Vectors u and v are orthogonal (or perpendicular) if and only if u # v = 0.
EXAMPLE 4
Applying the Definition of Orthogonality
(a) u = 83, 29 and v = 84, 69 are orthogonal because u # v = s3ds4d + s 2ds6d = 0. (b) u = 3i  2j + k and v = 2j + 4k are orthogonal because u # v = s3ds0d + s 2ds2d + s1ds4d = 0. (c) 0 is orthogonal to every vector u since
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0 # u = 80, 0, 09 # 8u1, u2, u39 = s0dsu1 d + s0dsu2 d + s0dsu3 d = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 866
Chapter 12: Vectors and the Geometry of Space
i
Dot Product Properties and Vector Projections
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The dot product obeys many of the laws that hold for ordinary products of real numbers (scalars).
HISTORICAL BIOGRAPHY
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You
Properties of the Dot Product If u, v, and w are any vectors and c is a scalar, then 1. u # v = v # u 2. scud # v = u # scvd = csu # vd 3. u # sv + wd = u # v + u # w 4. u # u = ƒ u ƒ 2 5. 0 # u = 0.
Proofs of Properties 1 and 3 The properties are easy to prove using the definition. For instance, here are the proofs of Properties 1 and 3.
Carl Friedrich Gauss (1777–1855) Q
u # v = u1 v1 + u2 v2 + u3 v3 = v1 u1 + v2 u2 + v3 u3 = v # u u # sv + wd = 8u1, u2 , u39 # 8v1 + w1, v2 + w2 , v3 + w39
ass a
1. 3.
= u1sv1 + w1 d + u2sv2 + w2 d + u3sv3 + w3 d
u
= u1 v1 + u1 w1 + u2 v2 + u2 w2 + u3 v3 + u3 w3
v P
R
S
= su1 v1 + u2 v2 + u3 v3 d + su1 w1 + u2 w2 + u3 w3 d
u v P
S
We now return to the problem of projecting one vector onto another, posed in the 1 1 opening to this section. The vector projection of u = PQ onto a nonzero vector v = PS 1 (Figure 12.22) is the vector PR determined by dropping a perpendicular from Q to the line PS. The notation for this vector is
ma
R
dH
= u#v + u#w
Q
ham
FIGURE 12.22 The vector projection of u onto v.
Force u
projv u
s“the vector projection of u onto v”d.
If u represents a force, then projv u represents the effective force in the direction of v (Figure 12.23). If the angle u between u and v is acute, projv u has length ƒ u ƒ cos u and direction v> ƒ v ƒ (Figure 12.24). If u is obtuse, cos u 6 0 and projv u has length  ƒ u ƒ cos u and direction v> ƒ v ƒ . In both cases, projv u = s ƒ u ƒ cos ud
Mu
v
FIGURE 12.23 If we pull on the box with force u, the effective force moving the box forward in the direction v is the projection of u onto v.
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= a
u#v v b ƒvƒ ƒvƒ
= a
u#v bv. ƒvƒ2
v ƒvƒ ƒ u ƒ cos u =
ƒ u ƒ ƒ v ƒ cos u u#v = ƒvƒ ƒvƒ
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH The Dot Product
867
u
u
proj v u
proj v u
v
v
You
suf
i
12.3
Length u cos (a)
Length –u cos
(b)
iaz
FIGURE 12.24 The length of projv u is (a) ƒ u ƒ cos u if cos u Ú 0 and (b)  ƒ u ƒ cos u if cos u 6 0 .
nR
The number ƒ u ƒ cos u is called the scalar component of u in the direction of v. To summarize,
Vector projection of u onto v:
projv u = a
u#v bv ƒvƒ2
(1)
ass a
Scalar component of u in the direction of v: ƒ u ƒ cos u =
v u#v = u# v v ƒ ƒ ƒ ƒ
(2)
dH
Note that both the vector projection of u onto v and the scalar component of u onto v depend only on the direction of the vector v and not its length (because we dot u with v> ƒ v ƒ , which is the direction of v).
ma
EXAMPLE 5
Finding the Vector Projection
Find the vector projection of u = 6i + 3j + 2k onto v = i  2j  2k and the scalar component of u in the direction of v.
Mu
ham
Solution
We find projv u from Equation (1): u#v 6  6  4 projv u = v # v v = si  2j  2kd 1 + 4 + 4 = 
8 8 4 4 si  2j  2kd =  i + j + k. 9 9 9 9
We find the scalar component of u in the direction of v from Equation (2): ƒ u ƒ cos u = u #
v 1 2 2 = s6i + 3j + 2kd # a i  j  kb 3 3 3 ƒvƒ
= 2  2 
4 4 =  . 3 3
Equations (1) and (2) also apply to twodimensional vectors.
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Chapter 12: Vectors and the Geometry of Space
Finding Vector Projections and Scalar Components
i
EXAMPLE 6
projv F = ¢ =
F#v ≤v ƒvƒ2
You
The vector projection is
5  6 1 si  3jd = si  3jd 1 + 9 10
= 
3 1 i + j. 10 10
iaz
Solution
suf
Find the vector projection of a force F = 5i + 2j onto v = i  3j and the scalar component of F in the direction of v.
The scalar component of F in the direction of v is
Work
D
Q
F cos
FIGURE 12.25 The work done by a constant force F during a displacement D is s ƒ F ƒ cos ud ƒ D ƒ .
ass a
P
In Chapter 6, we calculated the work done by a constant force of magnitude F in moving an object through a distance d as W = Fd. That formula holds only if the force is directed 1 along the line of motion. If a force F moving an object through a displacement D = PQ has some other direction, the work is performed by the component of F in the direction of D. If u is the angle between F and D (Figure 12.25), then scalar component of F Work = ain the direction of D bslength of Dd
dH
F
5  6 F#v 1 = = . ƒvƒ 21 + 9 210
nR
ƒ F ƒ cos u =
= s ƒ F ƒ cos ud ƒ D ƒ = F # D.
Mu
ham
ma
DEFINITION Work by Constant Force 1 The work done by a constant force F acting through a displacement D = PQ is W = F # D = ƒ F ƒ ƒ D ƒ cos u,
where u is the angle between F and D.
EXAMPLE 7
Applying the Definition of Work
If ƒ F ƒ = 40 N (newtons), ƒ D ƒ = 3 m, and u = 60°, the work done by F in acting from P to Q is
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Work = = = =
ƒ F ƒ ƒ D ƒ cos u s40ds3d cos 60° s120ds1>2d 60 J s joulesd.
Definition Given values
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3
The Dot Product
869
suf
i
We encounter more challenging work problems in Chapter 16 when we learn to find the work done by a variable force along a path in space.
Writing a Vector as a Sum of Orthogonal Vectors
u = u1 i + u2 j
u = u1 i + su2 j + u3 kd
u proj v u
v proj v u
FIGURE 12.26 Writing u as the sum of vectors parallel and orthogonal to v.
nR
iaz
(since i # j = i # k = j # k = 0). Sometimes, however, it is more informative to express u as a different sum. In mechanics, for instance, we often need to write a vector u as a sum of a vector parallel to a given vector v and a vector orthogonal to v. As an example, in studying the motion of a particle moving along a path in the plane (or space), it is desirable to know the components of the acceleration vector in the direction of the tangent to the path (at a point) and of the normal to the path. (These tangential and normal components of acceleration are investigated in Section 13.4.) The acceleration vector can then be expressed as the sum of its (vector) tangential and normal components (which reflect important geometric properties about the nature of the path itself, such as curvature). Velocity and acceleration vectors are studied in the next chapter. Generally, for vectors u and v, it is easy to see from Figure 12.26 that the vector
ass a
u
or
You
We know one way to write a vector u = 8u1, u29 or u = 8u1, u2, u39 as a sum of two orthogonal vectors:
u  projv u
is orthogonal to the projection vector projv u (which has the same direction as v). The following calculation verifies this observation:
ma
dH
su  projv ud # projv u = ¢ u  ¢ = ¢ =
u#v u#v v # ≤v 2≤ ≤ ¢ ƒvƒ ƒvƒ2 2
u#v u#v su # vd  ¢ ≤ sv # vd 2≤ ƒvƒ ƒvƒ2
su # vd2 su # vd2 ƒvƒ2 ƒvƒ2
Equation (1)
Dot product properties 2 and 3 v # v = ƒ v ƒ 2 cancels
= 0.
u = projv u + su  projv ud
expresses u as a sum of orthogonal vectors.
How to Write u as a Vector Parallel to v Plus a Vector Orthogonal to v
Mu
ham
So the equation
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u = projv u + su  projv ud u#v u#v v + ¢u  ¢ ≤v≤ 2≤ ƒvƒ ƒvƒ2 (')'* ('')''*
= ¢
Parallel to v
Orthogonal to v
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 870
Chapter 12: Vectors and the Geometry of Space
Force on a Spacecraft
i
EXAMPLE 8
suf
A force F = 2i + j  3k is applied to a spacecraft with velocity vector v = 3i  j. Express F as a sum of a vector parallel to v and a vector orthogonal to v.
F = projv F + sF  projv Fd F#v F#v = v # v v + aF  v # v vb
=
6  1 6  1 bv + aF  a bvb 9 + 1 9 + 1
iaz
= a
You
Solution
5 5 s3i  jd + a2i + j  3k s3i  jdb 10 10
nR
3 3 1 1 = a i  jb + a i + j  3kb . 2 2 2 2
ass a
The force s3>2di  s1/2dj is the effective force parallel to the velocity v. The force s1>2di + s3/2dj  3k is orthogonal to v. To check that this vector is orthogonal to v, we find the dot product:
Mu
ham
ma
dH
3 3 3 1 a i + j  3kb # s3i  jd =  = 0. 2 2 2 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 870
i f su
Chapter 12: Vectors and the Geometry of Space
EXERCISES 12.3 Dot Product and Projections
T
In Exercises 1–8, find a. ƒuƒ b. the cosine of the angle between v and u
u = 2i + 4j  25k
2. v = s3>5di + s4>5dk,
u = 5i + 12j
3. v = 10i + 11j  2k,
u = 3j + 4k
4. v = 2i + 10j  11k,
u = 2i + 2j + k
5. v = 5j  3k,
d a m m a h
u = 2i + 217j
8. v = h
1
i,
u M 1
,
a H
u = 22i + 23j + 2k
7. v = 5i + j,
22 23
u = h
1
22
,
1
23
v = i + 2j  k
10. u = 2i  2j + k,
u = i + j + k
6. v = i + j,
n a ss
9. u = 2i + j,
c. the scalar component of u in the direction of v 1. v = 2i  4j + 25k,
Angles Between Vectors
Find the angles between the vectors in Exercises 9–12 to the nearest hundredth of a radian.
v # u, ƒ v ƒ ,
d. the vector projv u .
u o Y z a i R
i
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11. u = 23i  7j,
v = 3i + 4k
v = 23i + j  2k
12. u = i + 22j  22k,
v = i + j + k
13. Triangle Find the measures of the angles of the triangle whose vertices are A = s 1, 0d, B = s2, 1d , and C = s1, 2d . 14. Rectangle Find the measures of the angles between the diagonals of the rectangle whose vertices are A = s1, 0d, B = s0, 3d, C = s3, 4d , and D = s4, 1d .
15. Direction angles and direction cosines The direction angles a, b , and g of a vector v = ai + bj + ck are defined as follows: a is the angle between v and the positive xaxis s0 … a … pd b is the angle between v and the positive yaxis s0 … b … pd g is the angle between v and the positive zaxis s0 … g … pd .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3 The Dot Product z
871
v2
␥
v
v1 v 2
You
0
suf
i
two vectors to be orthogonal to their difference? Give reasons for your answer.

␣
v1
y x
–v 2
v1 v 2
b cos b = , ƒvƒ
c cos g = , ƒvƒ
and cos2 a + cos2 b + cos2 g = 1 . These cosines are called the direction cosines of v. b. Unit vectors are built from direction cosines Show that if v = ai + bj + ck is a unit vector, then a, b, and c are the direction cosines of v.
East
rth
ma
No
Decomposing Vectors
In Exercises 17–19, write u as the sum of a vector parallel to v and a vector orthogonal to v. v = i + j
ham
17. u = 3j + 4k, 18. u = j + k,
v = i + j
19. u = 8i + 4j  12k,
A
v
–u
O
B
u
23. Diagonals of a rhombus Show that the diagonals of a rhombus (parallelogram with sides of equal length) are perpendicular. 24. Perpendicular diagonals Show that squares are the only rectangles with perpendicular diagonals.
dH
C
ass a
16. Water main construction A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the east direction. Determine the angle u required in the water main for the turn from north to east.
22. Orthogonality on a circle Suppose that AB is the diameter of a circle with center O and that C is a point on one of the two arcs 1 1 joining A and B. Show that CA and CB are orthogonal.
iaz
a cos a = , ƒvƒ
nR
a. Show that
v = i + 2j  k
Mu
20. Sum of vectors u = i + sj + kd is already the sum of a vector parallel to i and a vector orthogonal to i. If you use v = i , in the decomposition u = projv u + su  projv ud , do you get projv u = i and su  projv ud = j + k ? Try it and find out.
25. When parallelograms are rectangles Prove that a parallelogram is a rectangle if and only if its diagonals are equal in length. (This fact is often exploited by carpenters.) 26. Diagonal of parallelogram Show that the indicated diagonal of the parallelogram determined by vectors u and v bisects the angle between u and v if ƒ u ƒ = ƒ v ƒ . K L
27. Projectile motion A gun with muzzle velocity of 1200 ft> sec is fired at an angle of 8° above the horizontal. Find the horizontal and vertical components of the velocity. 28. Inclined plane Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the component of the force parallel to the inclined plane equal to 2.5 lb.
M
Geometry and Examples 21. Sums and differences In the accompanying figure, it looks as if v1 + v2 and v1  v2 are orthogonal. Is this mere coincidence, or are there circumstances under which we may expect the sum of
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space
Theory and Examples
Work u#v =
i
872
43. Work along a line Find the work done by a force F = 5i (magnitude 5 N) in moving an object along the line from the origin to the point (1, 1) (distance in meters).
b. Under what circumstances, if any, does ƒ u # v ƒ equal ƒ u ƒ ƒ v ƒ ? Give reasons for your answer.
44. Locomotive The union Pacific’s Big Boy locomotive could pull 6000ton trains with a tractive effort (pull) of 602,148 N (135,375 lb). At this level of effort, about how much work did Big Boy do on the (approximately straight) 605km journey from San Francisco to Los Angeles?
y
You
30. Copy the axes and vector shown here. Then shade in the points (x, y) for which sxi + yjd # v … 0 . Justify your answer.
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29. a. CauchySchwartz inequality Use the fact that ƒ u ƒ ƒ v ƒ cos u to show that the inequality ƒ u # v ƒ … ƒ u ƒ ƒ v ƒ holds for any vectors u and v.
45. Inclined plane How much work does it take to slide a crate 20 m along a loading dock by pulling on it with a 200 N force at an angle of 30° from the horizontal?
L
0
31. Orthogonal unit vectors If u1 and u2 are orthogonal unit vectors and v = au1 + bu2 , find v # u1 .
Equations for Lines in the Plane
60° 1000 lb magnitude force
ass a
32. Cancellation in dot products In realnumber multiplication, if uv1 = uv2 and u Z 0 , we can cancel the u and conclude that v1 = v2 . Does the same rule hold for the dot product: If u # v1 = u # v2 and u Z 0 , can you conclude that v1 = v2 ? Give reasons for your answer.
nR
iaz
46. Sailboat The wind passing over a boat’s sail exerted a 1000lb magnitude force F as shown here. How much work did the wind perform in moving the boat forward 1 mi? Answer in footpounds.
x
dH
33. Line perpendicular to a vector Show that the vector v = ai + bj is perpendicular to the line ax + by = c by establishing that the slope of v is the negative reciprocal of the slope of the given line.
Angles Between Lines in the Plane The acute angle between intersecting lines that do not cross at right angles is the same as the angle determined by vectors normal to the lines or by the vectors parallel to the lines. n1
n2
L2 L2
34. Line parallel to a vector Show that the vector v = ai + bj is parallel to the line bx  ay = c by establishing that the slope of the line segment representing v is the same as the slope of the given line.
ma
F
v2
v1 L1
L1
Use this fact and the results of Exercise 33 or 34 to find the acute angles between the lines in Exercises 47–52.
35. Ps2, 1d,
47. 3x + y = 5,
ham
In Exercises 35–38, use the result of Exercise 33 to find an equation for the line through P perpendicular to v. Then sketch the line. Include v in your sketch as a vector starting at the origin. v = i + 2j
36. Ps 1, 2d,
v = 2i  j
37. Ps 2, 7d, 38. Ps11, 10d,
v = 2i + j
Mu 41. Ps1, 2d,
v = i  j
v = i  2j
40. Ps0, 2d, 42. Ps1, 3d,
y =  23x + 2
49. 23x  y = 2,
v = 2i  3j
50. x + 23y = 1,
In Exercises 39–42, use the result of Exercise 34 to find an equation for the line through P parallel to v. Then sketch the line. Include v in your sketch as a vector starting at the origin. 39. Ps 2, 1d,
2x  y = 4
48. y = 23x  1,
v = 2i + 3j v = 3i  2j
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51. 3x  4y = 3, 52. 12x + 5y = 1,
x  23y = 1
A 1  23 B x + A 1 + 23 B y = 8
x  y = 7
2x  2y = 3
Angles Between Differentiable Curves The angles between two differentiable curves at a point of intersection are the angles between the curves’ tangent lines at these points. Find
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3 The Dot Product
y = x2
stwo points of intersectiond
x = y2 2
56. y = x ,
stwo points of intersectiond
i
55. y = x 3,
y = 1x
stwo points of intersectiond
Mu
ham
ma
dH
ass a
nR
iaz
You
53. y = s3>2d  x 2,
54. x = s3>4d  y 2, x = y 2  s3>4d stwo points of intersectiond
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the angles between the curves in Exercises 53â€“56. Note that if v = ai + bj is a vector in the plane, then the vector has slope b> a provided a Z 0 .
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873
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
12.4
873
i
12.4 The Cross Product
ou suf
The Cross Product
Ria
zY
In studying lines in the plane, when we needed to describe how a line was tilting, we used the notions of slope and angle of inclination. In space, we want a way to describe how a plane is tilting. We accomplish this by multiplying two vectors in the plane together to get a third vector perpendicular to the plane. The direction of this third vector tells us the “inclination” of the plane. The product we use to multiply the vectors together is the vector or cross product, the second of the two vector multiplication methods we study in calculus. Cross products are widely used to describe the effects of forces in studies of electricity, magnetism, fluid flows, and orbital mechanics. This section presents the mathematical properties that account for the use of cross products in these fields.
The Cross Product of Two Vectors in Space uv
an
We start with two nonzero vectors u and v in space. If u and v are not parallel, they determine a plane. We select a unit vector n perpendicular to the plane by the righthand rule. This means that we choose n to be the unit (normal) vector that points the way your right thumb points when your fingers curl through the angle u from u to v (Figure 12.27). Then the cross product u * v (“u cross v”) is the vector defined as follows.
v
n
ass
u
DEFINITION
ad H
u * v = s ƒ u ƒ ƒ v ƒ sin ud n
The construction of
Mu ha
mm
FIGURE 12.27 u * v.
Cross Product
Unlike the dot product, the cross product is a vector. For this reason it’s also called the vector product of u and v, and applies only to vectors in space. The vector u * v is orthogonal to both u and v because it is a scalar multiple of n. Since the sines of 0 and p are both zero, it makes sense to define the cross product of two parallel nonzero vectors to be 0. If one or both of u and v are zero, we also define u * v to be zero. This way, the cross product of two vectors u and v is zero if and only if u and v are parallel or one or both of them are zero.
Parallel Vectors Nonzero vectors u and v are parallel if and only if u * v = 0 .
The cross product obeys the following laws.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 874
Chapter 12: Vectors and the Geometry of Space
You
u
vu
To visualize Property 4, for example, notice that when the fingers of a right hand curl through the angle u from v to u, the thumb points the opposite way and the unit vector we choose in forming v * u is the negative of the one we choose in forming u * v (Figure 12.28). Property 1 can be verified by applying the definition of cross product to both sides of the equation and comparing the results. Property 2 is proved in Appendix 6. Property 3 follows by multiplying both sides of the equation in Property 2 by 1 and reversing the order of the products using Property 4. Property 5 is a definition. As a rule, cross product multiplication is not associative so su * vd * w does not generally equal u * sv * wd. (See Additional Exercise 15.) When we apply the definition to calculate the pairwise cross products of i, j, and k, we find (Figure 12.29)
k i j –( j i)
–i –j
ass a
z
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FIGURE 12.28 The construction of v * u.
suf
–n
i
Properties of the Cross Product If u, v, and w are any vectors and r, s are scalars, then 1. srud * ss vd = srsdsu * vd 2. u * sv + wd = u * v + u * w 3. sv + wd * u = v * u + w * u 4. v * u = su * vd 5. 0 * u = 0
v
j k i –(i k)
–k
i
k * i = si * kd = j
Diagram for recalling these products
and
i * i = j * j = k * k = 0.
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FIGURE 12.29 The pairwise cross products of i, j, and k.
dH
j * k = sk * jd = i
x i j k –(k j)
j
i * j = sj * id = k
y
i
k
ƒ u * v ƒ Is the Area of a Parallelogram
ham
Because n is a unit vector, the magnitude of u * v is
Area base ⋅ height u ⋅ vsin u × v
v
Mu
h v sin
u
FIGURE 12.30 The parallelogram determined by u and v.
ƒ u * v ƒ = ƒ u ƒ ƒ v ƒ ƒ sin u ƒ ƒ n ƒ = ƒ u ƒ ƒ v ƒ sin u.
This is the area of the parallelogram determined by u and v (Figure 12.30), ƒ u ƒ being the base of the parallelogram and ƒ v ƒ ƒ sin u ƒ the height.
Determinant Formula for u * v Our next objective is to calculate u * v from the components of u and v relative to a Cartesian coordinate system.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.4
u * v = su1 i + u2 j + u3 kd * sv1 i + v2 j + v3 kd = u1 v1 i * i + u1 v2 i * j + u1 v3 i * k
 a2 `
a3 b2 b3 3 = a1 ` c2 c3 b1 c1
+ u2 v1 j * i + u2 v2 j * j + u2 v3 j * k
+ u3 v1 k * i + u3 v2 k * j + u3 v3 k * k
= su2 v3  u3 v2 di  su1 v3  u3 v1 dj + su1 v2  u2 v1 dk.
b3 ` c3
b3 b1 ` + a3 ` c3 c1
The terms in the last line are the same as the terms in the expansion of the symbolic determinant
iaz
a2 b2 c2
c1
You
1 ` = s2ds3d  s1ds 4d 3 = 6 + 4 = 10
b2 ` c2
i 3 u1
EXAMPLE
v1
3 1 3
k u3 3 . v3
We therefore have the following rule. 1 ` 3
ass a
1 1 1 1 3 = s 5d ` ` 3 1 1 2 1 2  s3d ` ` + s1d ` 4 1 4 = 5s1  3d  3s2 + 4d + 1s6 + 4d = 10  18 + 10 = 2
5 3 2 4
j u2 v2
nR
a1 3 b1
i
Then the distributive laws and the rules for multiplying i, j, and k tell us that
b ` = ad  bc d
EXAMPLE 2 4
v = v1 i + v2 j + v3 k.
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u = u1 i + u2 j + u3 k,
2 * 2 and 3 * 3 determinants are evaluated as follows:
`
875
Suppose that
Determinants
a ` c
The Cross Product
Calculating Cross Products Using Determinants If u = u1 i + u2 j + u3 k and v = v1 i + v2 j + v3 k, then i u * v = 3 u1 v1
dH
(For more information, see the Web site at www.awbc.com/thomas.)
ma
EXAMPLE 1
j u2 v2
k u3 3 . v3
Calculating Cross Products with Determinants
Find u * v and v * u if u = 2i + j + k and v = 4i + 3j + k.
z
Solution
ham
R(–1, 1, 2)
i 3 u * v = 2 4
0
Mu
P(1, –1, 0)
x
j 1 3
k 1 13 = p 3 1
1 2 pi  p 1 4
1 2 pj + p 1 4
1 pk 3
= 2i  6j + 10k y
Q(2, 1, –1)
FIGURE 12.31 The area of triangle PQR 1 1 is half of ƒ PQ * PR ƒ (Example 2).
v * u = su * vd = 2i + 6j  10k
EXAMPLE 2
Finding Vectors Perpendicular to a Plane
Find a vector perpendicular to the plane of Ps1, 1, 0d, Qs2, 1, 1d, and Rs 1, 1, 2d (Figure 12.31).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 876
Chapter 12: Vectors and the Geometry of Space
1 1 The vector PQ * PR is perpendicular to the plane because it is perpendicular to both vectors. In terms of components, 1 PQ = s2  1di + s1 + 1dj + s 1  0dk = i + 2j  k 1 PR = s 1  1di + s1 + 1dj + s2  0dk = 2i + 2j + 2k j 2 2
k 2 1 3 = ` 2 2
1 1 `i  ` 2 2
1 1 `j + ` 2 2
2 `k 2
iaz
= 6i + 6k.
EXAMPLE 3
You
i 1 1 3 PQ * PR = 1 2
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Solution
Finding the Area of a Triangle
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Find the area of the triangle with vertices Ps1, 1, 0d, Qs2, 1, 1d, and Rs 1, 1, 2d (Figure 12.31). The area of the parallelogram determined by P, Q, and R is 1 1 Values from Example 2. ƒ PQ * PR ƒ = ƒ 6i + 6k ƒ
Solution
ass a
= 2s6d2 + s6d2 = 22 # 36 = 622 .
The triangle’s area is half of this, or 322 .
EXAMPLE 4
Finding a Unit Normal to a Plane
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Find a unit vector perpendicular to the plane of Ps1, 1, 0d, Qs2, 1, 1d, and Rs 1, 1, 2d.
1 1 Since PQ * PR is perpendicular to the plane, its direction n is a unit vector perpendicular to the plane. Taking values from Examples 2 and 3, we have 1 1 PQ * PR 6i + 6k 1 1 n = 1 = i + k. 1 = 6 22 22 22 ƒ PQ * PR ƒ
ma
Solution
n
ham
Torque
For ease in calculating the cross product using determinants, we usually write vectors in the form v = v1 i + v2 j + v3 k rather than as ordered triples v = 8v1, v2 , v39.
r
Mu
Component of F perpendicular to r. Its length is F sin .
F
FIGURE 12.32 The torque vector describes the tendency of the force F to drive the bolt forward.
Torque When we turn a bolt by applying a force F to a wrench (Figure 12.32), the torque we produce acts along the axis of the bolt to drive the bolt forward. The magnitude of the torque depends on how far out on the wrench the force is applied and on how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque’s magnitude is the product of the length of the lever arm r and the scalar component of F perpendicular to r. In the notation of Figure 12.32,
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Magnitude of torque vector = ƒ r ƒ ƒ F ƒ sin u,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.4
The Cross Product
877
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i
or ƒ r * F ƒ . If we let n be a unit vector along the axis of the bolt in the direction of the torque, then a complete description of the torque vector is r * F, or Torque vector = s ƒ r ƒ ƒ F ƒ sin ud n.
EXAMPLE 5
70° F
20 lb magnitude force
FIGURE 12.33 The magnitude of the torque exerted by F at P is about 56.4 ftlb (Example 5).
Finding the Magnitude of a Torque
The magnitude of the torque generated by force F at the pivot point P in Figure 12.33 is 1 1 ƒ PQ * F ƒ = ƒ PQ ƒ ƒ F ƒ sin 70° L s3ds20ds0.94d L 56.4 ftlb.
iaz
Q
nR
3 ft bar P
You
Recall that we defined u * v to be 0 when u and v are parallel. This is consistent with the torque interpretation as well. If the force F in Figure 12.32 is parallel to the wrench, meaning that we are trying to turn the bolt by pushing or pulling along the line of the wrench’s handle, the torque produced is zero.
Triple Scalar or Box Product
The product su * vd # w is called the triple scalar product of u, v, and w (in that order). As you can see from the formula
ass a
ƒ su * vd # w ƒ = ƒ u * v ƒ ƒ w ƒ ƒ cos u ƒ ,
dH
the absolute value of the product is the volume of the parallelepiped (parallelogramsided box) determined by u, v, and w (Figure 12.34). The number ƒ u * v ƒ is the area of the base parallelogram. The number ƒ w ƒ ƒ cos u ƒ is the parallelepiped’s height. Because of this geometry, su * vd # w is also called the box product of u, v, and w.
w
Height w cos
ma ham
Mu
The dot and cross may be interchanged in a triple scalar product without altering its value.
uv
v
Area of base u v
u Volume area of base · height u v w cos (u v) · w
FIGURE 12.34 The number ƒ su * vd # w ƒ is the volume of a parallelepiped.
By treating the planes of v and w and of w and u as the base planes of the parallelepiped determined by u, v, and w, we see that su * vd # w = sv * wd # u = sw * ud # v. Since the dot product is commutative, we also have
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su * vd # w = u # sv * wd.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 878
Chapter 12: Vectors and the Geometry of Space
= w1 `
u2 v2
u3 u1 `j + ` v3 v1
u3 u1 `  w2 ` v3 v1 u2 v2 w2
u2 ` kd # w v2
u3 u1 ` + w3 ` v3 v1
u2 ` v2
u3 v3 3 . w3
iaz
u1 = 3 v1 w1
u3 u1 `i  ` v3 v1
suf
u2 v2
You
su * vd # w = c `
i
The triple scalar product can be evaluated as a determinant:
Calculating the Triple Scalar Product
EXAMPLE 6
u2 v2 w2
nR
u1 su * vd # w = 3 v1 w1
u3 v3 3 w3
Finding the Volume of a Parallelepiped
Solution
ass a
Find the volume of the box (parallelepiped) determined by u = i + 2j  k, v = 2i + 3k, and w = 7j  4k. Using the rule for calculating determinants, we find
dH
su * vd # w =
1 3 2 0
2 0 7
1 3 3 = 23. 4
Mu
ham
ma
The volume is ƒ su * vd # w ƒ = 23 units cubed.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
878
EXERCISES 12.4 Cross Product Calculations In Exercises 1–8, find the length and direction (when defined) of u * v and v * u. 1. u = 2i  2j  k, 2. u = 2i + 3j,
3. u = 2i  2j + 4k, 4. u = i + j  k,
v = i + j  2k
v = 0
m m a h u M
5. u = 2i,
v = 3j
6. u = i * j,
v = j * k
7. u = 8i  2j  4k, 3 1 8. u = i  j + k, 2 2
v = 2i + 2j + k
H ad
v = i + j + 2k
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a i R an
In Exercises 9–14, sketch the coordinate axes and then include the vectors u, v and u * v as vectors starting at the origin.
s s a
v = i  k
v = i + j
i f su
u o zY
Chapter 12: Vectors and the Geometry of Space
9. u = i,
v = j
10. u = i  k,
v = j
11. u = i  k,
v = j + k
12. u = 2i  j, 13. u = i + j,
v = i + 2j v = i  j
14. u = j + 2k,
v = i
Triangles in Space In Exercises 15–18, a. Find the area of the triangle determined by the points P, Q, and R. b. Find a unit vector perpendicular to plane PQR.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.4 The Cross Product sany number cd
e. csu * vd = scud * v = u * scvd
Rs3, 1, 1d
17. Ps2, 2, 1d,
Qs3, 1, 2d,
Rs3, 1, 1d
18. Ps 2, 2, 0d,
Qs0, 1, 1d,
Rs 1, 2, 2d
2
u#u =
g. su * ud # u = 0
ƒuƒ h. su * vd # u = v # su * vd f.
sany number cd
i
Qs2, 1, 3d,
d. scud # v = u # scvd = csu # vd
Rs0, 2, 1d
suf
16. Ps1, 1, 1d,
Qs2, 0, 1d,
29. Given nonzero vectors u, v, and w, use dot product and cross product notation, as appropriate, to describe the following.
Triple Scalar Products In Exercises 19–22, verify that su * vd # w = sv * wd # u = sw * ud # v and find the volume of the parallelepiped (box) determined by u, v, and w.
You
15. Ps1, 1, 2d,
a. The vector projection of u onto v b. A vector orthogonal to u and v
c. A vector orthogonal to u * v and w
d. The volume of the parallelepiped determined by u, v, and w
u
v
w
19. 2i
2j
2k
20. i  j + k
2i + j  2k
i + 2j  k
a. A vector orthogonal to u * v and u * w
21. 2i + j
2i  j + k
i + 2k
b. A vector orthogonal to u + v and u  v
22. i + j  2k
i  k
2i + 4j  2k
c. A vector of length ƒ u ƒ in the direction of v d. The area of the parallelogram determined by u and w
23. Parallel and perpendicular vectors Let u = 5i  j + k, v = j  5k, w = 15i + 3j  3k. Which vectors, if any, are (a) perpendicular? (b) Parallel? Give reasons for your answers.
31. Let u, v, and w be vectors. Which of the following make sense, and which do not? Give reasons for your answers. a. su * vd # w
b. u * sv # wd
c. u * sv * wd
d. u # sv # wd
ass a
24. Parallel and perpendicular vectors Let u = i + 2j  k, v = i + j + k, w = i + k, r = sp>2di  pj + sp>2dk. Which vectors, if any, are (a) perpendicular? (b) Parallel? Give reasons for your answers.
nR
iaz
30. Given nonzero vectors u, v, and w, use dot product and cross product notation to describe the following.
Theory and Examples
dH
In Exercises 39 and 40, find the magnitude of the torque exerted by F 1 on the bolt at P if ƒ PQ ƒ = 8 in . and ƒ F ƒ = 30 lb . Answer in footpounds. 25.
26. F
F
135°
Q
ma
60°
Q
P
P
ham
27. Which of the following are always true, and which are not always true? Give reasons for your answers. a. ƒ u ƒ = 2u # u c. u * 0 = 0 * u = 0
b. u # u = ƒ u ƒ d. u * s ud = 0
32. Cross products of three vectors Show that except in degenerate cases, su * vd * w lies in the plane of u and v, whereas u * sv * wd lies in the plane of v and w. What are the degenerate cases? 33. Cancellation in cross products If u * v = u * w and u Z 0, then does v = w? Give reasons for your answer. 34. Double cancellation If u Z 0 and if u * v = u * w and u # v = u # w, then does v = w? Give reasons for your answer.
Area in the Plane Find the areas of the parallelograms whose vertices are given in Exercises 35–38. 35. As1, 0d,
Bs0, 1d,
Cs 1, 0d,
36. As0, 0d,
Bs7, 3d,
Cs9, 8d,
37. As 1, 2d,
Bs2, 0d,
38. As 6, 0d,
Bs1, 4d,
g. su * vd # v = 0
41. As 5, 3d,
Bs1, 2d,
h. su * vd # w = u # sv * wd
42. As 6, 0d,
Bs10, 5d,
Mu
40. As 1, 1d,
c. s ud * v = su * vd
To Read it Online & Download:
Ds 4, 5d
Find the areas of the triangles whose vertices are given in Exercises 39–42. 39. As0, 0d,
b. u * v = sv * ud
Ds4, 3d
Cs3, 1d,
f. u * sv + wd = u * v + u * w
28. Which of the following are always true, and which are not always true? Give reasons for your answers.
Ds0, 1d Ds2, 5d
Cs7, 1d,
e. u * v = v * u
a. u # v = v # u
879
Bs 2, 3d,
Cs3, 1d
Bs3, 3d,
Cs2, 1d Cs6, 2d Cs 2, 4d
43. Triangle area Find a formula for the area of the triangle in the xyplane with vertices at s0, 0d, sa1, a2 d , and sb1, b2 d . Explain your work. 44. Triangle area Find a concise formula for the area of a triangle with vertices sa1, a2 d, sb1, b2 d , and sc1, c2 d .
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Chapter 12: Vectors and the Geometry of Space
12.5
suf
i
Lines and Planes in Space
Lines and Line Segments in Space
P(x, y, z)
L v
nR
P0(x 0 , y0 , z 0 )
In the plane, a line is determined by a point and a number giving the slope of the line. In space a line is determined by a point and a vector giving the direction of the line. Suppose that L is a line in space passing through a point P0sx0 , y0 , z0 d parallel to a 1 vector v = v1 i + v2 j + v3 k. Then L is the set of all points P(x, y, z) for which P0 P is 1 parallel to v (Figure 12.35). Thus, P0 P = tv for some scalar parameter t. The value of t depends on the location of the point P along the line, and the domain of t is s  q , q d. The 1 expanded form of the equation P0 P = tv is
ass a
z
iaz
You
In the calculus of functions of a single variable, we used our knowledge of lines to study curves in the plane. We investigated tangents and found that, when highly magnified, differentiable curves were effectively linear. To study the calculus of functions of more than one variable in the next chapter, we start with planes and use our knowledge of planes to study the surfaces that are the graphs of functions in space. This section shows how to use scalar and vector products to write equations for lines, line segments, and planes in space.
sx  x0 di + s y  y0 dj + sz  z0 dk = tsv1 i + v2 j + v3 kd,
0
which can be rewritten as
x
(1)
If r(t) is the position vector of a point P(x, y, z) on the line and r0 is the position vector of the point P0sx0, y0, z0 d, then Equation (1) gives the following vector form for the equation of a line in space.
ma
FIGURE 12.35 A point P lies on L 1 through P0 parallel to v if and only if P0 P is a scalar multiple of v.
xi + yj + zk = x0 i + y0 j + z0 k + tsv1 i + v2 j + v3 kd.
dH
y
Mu
ham
Vector Equation for a Line A vector equation for the line L through P0sx0 , y0 , z0 d parallel to v is rstd = r0 + tv,
q 6 t 6 q,
(2)
where r is the position vector of a point P(x, y, z) on L and r0 is the position vector of P0sx0 , y0 , z0 d.
Equating the corresponding components of the two sides of Equation (1) gives three scalar equations involving the parameter t: x = x0 + tv1,
y = y0 + tv2,
z = z0 + tv3 .
These equations give us the standard parametrization of the line for the parameter interval q 6 t 6 q.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.5
Lines and Planes in Space
881
z
4
y = y0 + tv2,
4
v 2i 4j 2k
FIGURE 12.36 Selected points and parameter values on the line x = 2 + 2t, y = 4t, z = 4  2t . The arrows show the direction of increasing t (Example 1).
EXAMPLE 1
(3)
Parametrizing a Line Through a Point Parallel to a Vector
Find parametric equations for the line through s 2, 0, 4d parallel to v = 2i + 4j  2k (Figure 12.36). Solution With P0sx0 , y0 , z0 d equal to s 2, 0, 4d and v1 i + v2 j + v3 k equal to 2i + 4j  2k, Equations (3) become
iaz
t2
y P2(2, 8, 0)
q 6 t 6 q
x = 2 + 2t,
EXAMPLE 2
y = 4t,
z = 4  2t.
Parametrizing a Line Through Two Points
nR
8
4
z = z0 + tv3,
You
t1
0 2
x
x = x0 + tv1,
P1(0, 4, 2)
2
suf
i
Parametric Equations for a Line The standard parametrization of the line through P0sx0 , y0 , z0 d parallel to v = v1 i + v2 j + v3 k is
P0(–2, 0, 4) t0
Find parametric equations for the line through Ps 3, 2, 3d and Qs1, 1, 4d. The vector 1 PQ = s1  s 3ddi + s 1  2dj + s4  s 3ddk = 4i  3j + 7k
ass a
Solution
is parallel to the line, and Equations (3) with sx0 , y0 , z0 d = s 3, 2, 3d give x = 3 + 4t,
y = 2  3t,
z = 3 + 7t.
dH
We could have chosen Qs1, 1, 4d as the “base point” and written x = 1 + 4t,
y = 1  3t,
z = 4 + 7t.
Q(1, –1, 4)
z
ham
t1
ma
These equations serve as well as the first; they simply place you at a different point on the line for a given value of t.
Notice that parametrizations are not unique. Not only can the “base point” change, but so can the parameter. The equations x = 3 + 4t 3, y = 2  3t 3 , and z = 3 + 7t 3 also parametrize the line in Example 2. To parametrize a line segment joining two points, we first parametrize the line through the points. We then find the tvalues for the endpoints and restrict t to lie in the closed interval bounded by these values. The line equations together with this added restriction parametrize the segment.
–3
–1
0
Mu x
EXAMPLE 3
2
1
y
t0 P(–3, 2, –3)
FIGURE 12.37 Example 3 derives a parametrization of line segment PQ. The arrow shows the direction of increasing t.
Parametrizing a Line Segment
Parametrize the line segment joining the points Ps 3, 2, 3d and Qs1, 1, 4d (Figure 12.37). Solution We begin with equations for the line through P and Q, taking them, in this case, from Example 2:
To Read it Online & Download:
x = 3 + 4t,
y = 2  3t,
z = 3 + 7t.
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Chapter 12: Vectors and the Geometry of Space
suf
sx, y, zd = s 3 + 4t, 2  3t, 3 + 7td
i
We observe that the point
on the line passes through Ps 3, 2, 3d at t = 0 and Qs1, 1, 4d at t = 1. We add the restriction 0 … t … 1 to parametrize the segment: y = 2  3t,
z = 3 + 7t,
0 … t … 1.
You
x = 3 + 4t,
The vector form (Equation (2)) for a line in space is more revealing if we think of a line as the path of a particle starting at position P0sx0 , y0 , z0 d and moving in the direction of vector v. Rewriting Equation (2), we have rstd = r0 + tv
æ Speed
æ
Time
(4)
Direction
nR
Initial position
iaz
v = r0 + t ƒ v ƒ . ƒvƒ æ æ
In other words, the position of the particle at time t is its initial position plus its distance moved sspeed * timed in the direction v> ƒ v ƒ of its straightline motion.
Flight of a Helicopter
ass a
EXAMPLE 4
A helicopter is to fly directly from a helipad at the origin in the direction of the point (1, 1, 1) at a speed of 60 ft> sec. What is the position of the helicopter after 10 sec? We place the origin at the starting position (helipad) of the helicopter. Then the unit vector
dH
Solution
u =
1 1 1 i + j + k 23 23 23
Mu
ham
ma
gives the flight direction of the helicopter. From Equation (4), the position of the helicopter at any time t is rstd = r0 + tsspeeddu = 0 + ts60d ¢
1 1 1 i + j + k≤ 23 23 23
= 2023tsi + j + kd. When t = 10 sec, rs10d = 20023 si + j + kd = h 200 23, 20023, 20023 i . After 10 sec of flight from the origin toward (1, 1, 1), the helicopter is located at the point s20023, 20023, 20023d in space. It has traveled a distance of s60 ft>secds10 secd = 600 ft, which is the length of the vector r(10).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.5
Lines and Planes in Space
883
The Distance from a Point to a Line in Space
i
S
You
suf
To find the distance from a point S to a line that passes through a point P parallel to a vec1 tor v, we find the absolute value of the scalar component of PS in the direction of a vector normal to the line (Figure 12.38). In the notation of the figure, the absolute value of the 1 ƒ PS * v ƒ 1 scalar component is, ƒ PS ƒ sin u, which is . ƒvƒ
PS sin
v
FIGURE 12.38 The distance from S to the line through P parallel to v is 1 ƒ PS ƒ sin u , where u is the angle between 1 PS and v.
EXAMPLE 5
(5)
iaz
Distance from a Point S to a Line Through P Parallel to v 1 ƒ PS * v ƒ d = ƒvƒ
Finding Distance from a Point to a Line
nR
P
Find the distance from the point S(1, 1, 5) to the line x = 1 + t,
ass a
L:
y = 3  t,
z = 2t.
We see from the equations for L that L passes through P(1, 3, 0) parallel to v = i  j + 2k. With 1 PS = s1  1di + s1  3dj + s5  0dk = 2j + 5k Solution
i 1 3 PS * v = 0 1
dH
and
j 2 1
k 5 3 = i + 5j + 2k, 2
ma
Equation (5) gives d =
1 ƒ PS * v ƒ 21 + 25 + 4 230 = = = 25. ƒvƒ 21 + 1 + 4 26
An Equation for a Plane in Space
Plane M
ham
n
P(x, y, z)
Mu
P0(x 0 , y0 , z 0 )
FIGURE 12.39 The standard equation for a plane in space is defined in terms of a vector normal to the plane: A point P lies in the plane through P0 normal to n if and 1 only if n # P0P = 0 .
A plane in space is determined by knowing a point on the plane and its “tilt” or orientation. This “tilt” is defined by specifying a vector that is perpendicular or normal to the plane. Suppose that plane M passes through a point P0sx0 , y0 , z0 d and is normal to the nonzero vector n = Ai + Bj + Ck. Then M is the set of all points P(x, y, z) for which 1 1 P0 P is orthogonal to n (Figure 12.39). Thus, the dot product n # P0 P = 0. This equation is equivalent to sAi + Bj + Ckd # [sx  x0 di + s y  y0 dj + sz  z0 dk] = 0 or
To Read it Online & Download:
Asx  x0 d + Bs y  y0 d + Csz  z0 d = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space
suf
Equation for a Plane The plane through P0sx0 , y0 , z0 d normal to n = Ai + Bj + Ck has
Finding an Equation for a Plane
You
1 n # P0 P = 0 Asx  x0 d + Bsy  y0 d + Csz  z0 d = 0 Ax + By + Cz = D, where D = Ax0 + By0 + Cz0
Vector equation: Component equation: Component equation simplified:
EXAMPLE 6
i
884
Solution
The component equation is
iaz
Find an equation for the plane through P0s 3, 0, 7d perpendicular to n = 5i + 2j  k.
nR
5sx  s 3dd + 2s y  0d + s 1dsz  7d = 0. Simplifying, we obtain
5x + 15 + 2y  z + 7 = 0
ass a
5x + 2y  z = 22.
Notice in Example 6 how the components of n = 5i + 2j  k became the coefficients of x, y, and z in the equation 5x + 2y  z = 22. The vector n = Ai + Bj + Ck is normal to the plane Ax + By + Cz = D.
Finding an Equation for a Plane Through Three Points
dH
EXAMPLE 7
Find an equation for the plane through A(0, 0, 1), B(2, 0, 0), and C(0, 3, 0). We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equation for the plane. The cross product
Mu
ham
ma
Solution
i 1 1 AB * AC = 3 2 0
j 0 3
k 1 3 = 3i + 2j + 6k 1
is normal to the plane. We substitute the components of this vector and the coordinates of A(0, 0, 1) into the component form of the equation to obtain 3sx  0d + 2s y  0d + 6sz  1d = 0 3x + 2y + 6z = 6.
Lines of Intersection Just as lines are parallel if and only if they have the same direction, two planes are parallel if and only if their normals are parallel, or n1 = kn2 for some scalar k. Two planes that are not parallel intersect in a line.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.5
885
Finding a Vector Parallel to the Line of Intersection of Two Planes
i
EXAMPLE 8
Lines and Planes in Space
suf
Find a vector parallel to the line of intersection of the planes 3x  6y  2z = 15 and 2x + y  2z = 5. The line of intersection of two planes is perpendicular to both planes’ normal vectors n1 and n2 (Figure 12.40) and therefore parallel to n1 * n2 . Turning this around, n1 * n2 is a vector parallel to the planes’ line of intersection. In our case,
n2
AN PL
i n1 * n2 = 3 3 2
E1
FIGURE 12.40 How the line of intersection of two planes is related to the planes’ normal vectors (Example 8).
j 6 1
k 2 3 = 14i + 2j + 15k. 2
Any nonzero scalar multiple of n1 * n2 will do as well.
EXAMPLE 9
iaz
n1 n 2
You
PL
A
N
E
2
Solution n1
Parametrizing the Line of Intersection of Two Planes
nR
Find parametric equations for the line in which the planes 3x  6y  2z = 15 and 2x + y  2z = 5 intersect. We find a vector parallel to the line and a point on the line and use Equations (3). Example 8 identifies v = 14i + 2j + 15k as a vector parallel to the line. To find a point on the line, we can take any point common to the two planes. Substituting z = 0 in the plane equations and solving for x and y simultaneously identifies one of these points as s3, 1, 0d. The line is
ass a
Solution
y = 1 + 2t,
z = 15t.
dH
x = 3 + 14t,
The choice z = 0 is arbitrary and we could have chosen z = 1 or z = 1 just as well. Or we could have let x = 0 and solved for y and z. The different choices would simply give different parametrizations of the same line.
Mu
ham
ma
Sometimes we want to know where a line and a plane intersect. For example, if we are looking at a flat plate and a line segment passes through it, we may be interested in knowing what portion of the line segment is hidden from our view by the plate. This application is used in computer graphics (Exercise 74).
EXAMPLE 10
Finding the Intersection of a Line and a Plane
Find the point where the line x =
8 + 2t, 3
y = 2t,
z = 1 + t
intersects the plane 3x + 2y + 6z = 6. Solution
The point
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a
8 + 2t, 2t, 1 + tb 3
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Chapter 12: Vectors and the Geometry of Space
8 + 2tb + 2s 2td + 6s1 + td = 6 3 8 + 6t  4t + 6 + 6t = 6
You
8t = 8
suf
3a
i
lies in the plane if its coordinates satisfy the equation of the plane, that is, if
t = 1.
The point of intersection is
8 2  2, 2, 1  1b = a , 2, 0b . 3 3
iaz
sx, y, zd ƒ t = 1 = a
The Distance from a Point to a Plane
nR
If P is a point on a plane with normal n, then the distance from any point S to the plane is 1 the length of the vector projection of PS onto n. That is, the distance from S to the plane is 1 n d = ` PS # ` ƒnƒ
(6)
ass a
where n = Ai + Bj + Ck is normal to the plane.
EXAMPLE 11
Finding the Distance from a Point to a Plane
Find the distance from S(1, 1, 3) to the plane 3x + 2y + 6z = 6. We find a point P in the plane and calculate the length of the vector projection 1 of PS onto a vector n normal to the plane (Figure 12.41). The coefficients in the equation 3x + 2y + 6z = 6 give
dH
Solution
Mu
ham
ma
n = 3i + 2j + 6k.
z n 3i 2j 6k S(1, 1, 3)
3x 2y 6z 6
(0, 0, 1) Distance from S to the plane
0
(2, 0, 0)
P(0, 3, 0)
y
x
FIGURE 12.41 The distance from S to the plane is the 1 length of the vector projection of PS onto n (Example 11).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.5 Lines and Planes in Space
887
suf
i
The points on the plane easiest to find from the plane’s equation are the intercepts. If we take P to be the yintercept (0, 3, 0), then 1 PS = s1  0di + s1  3dj + s3  0dk = i  2j + 3k,
You
n2
ƒ n ƒ = 2s3d2 + s2d2 + s6d2 = 249 = 7. The distance from S to the plane is n1
1 n d = ` PS # ` ƒnƒ
1 length of projn PS
3 18 17 4 ` =  + . 7 7 7 7
nR
= `
iaz
3 6 2 = ` si  2j + 3kd # a i + j + kb ` 7 7 7
Angles Between Planes FIGURE 12.42 The angle between two planes is obtained from the angle between their normals.
The angle between two intersecting planes is defined to be the (acute) angle determined by their normal vectors (Figure 12.42). Find the angle between the planes 3x  6y  2z = 15 and
ass a
EXAMPLE 12
2x + y  2z = 5. Solution
The vectors
dH
n1 = 3i  6j  2k,
n2 = 2i + j  2k
u = cos1 a = cos1 a
n1 # n2 b ƒ n1 ƒ ƒ n2 ƒ 4 b 21
L 1.38 radians.
About 79 deg
Mu
ham
ma
are normals to the planes. The angle between them is
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o Y iaz 12.5 Lines and Planes in Space
EXERCISES 12.5
R n ssa
5. The line through the origin parallel to the vector 2j + k
Lines and Line Segments Find parametric equations for the lines in Exercises 1â€“12.
a H d
1. The line through the point Ps3, 4, 1d parallel to the vector i + j + k
a m m a h u M
2. The line through Ps1, 2, 1d and Qs 1, 0, 1d 3. The line through Ps 2, 0, 3d and Qs3, 5, 2d 4. The line through P(1, 2, 0) and Qs1, 1, 1d
887
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6. The line through the point s3, 2, 1d parallel to the line x = 1 + 2t, y = 2  t, z = 3t 7. The line through (1, 1, 1) parallel to the zaxis 8. The line through (2, 4, 5) perpendicular to the plane 3x + 7y  5z = 21 9. The line through s0, 7, 0d perpendicular to the plane x + 2y + 2z = 13
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space 36. s2, 1, 1d;
x = 2t,
37. s3, 1, 4d;
x = 4  t,
11. The xaxis
38. s 1, 4, 3d;
x = 10 + 4t,
12. The zaxis
Find parametrizations for the line segments joining the points in Exercises 13–20. Draw coordinate axes and sketch each segment, indicating the direction of increasing t for your parametrization. 15. (1, 0, 0),
(1, 1, 3> 2)
(1, 1, 0)
z = 2t
y = 3 + 2t,
z = 5 + 3t
y = 3,
z = 4t
In Exercises 39–44, find the distance from the point to the plane. 39. s2, 3, 4d,
x + 2y + 2z = 13
14. (0, 0, 0),
(1, 0, 0)
40. s0, 0, 0d,
16. (1, 1, 0),
(1, 1, 1)
41. s0, 1, 1d,
4y + 3z = 12
2x + y + 2z = 4
3x + 2y + 6z = 6
You
13. (0, 0, 0),
y = 1 + 2t,
i
10. The line through (2, 3, 0) perpendicular to the vectors u = i + 2j + 3k and v = 3i + 4j + 5k
suf
888
17. s0, 1, 1d,
s0, 1, 1d
18. (0, 2, 0),
(3, 0, 0)
42. s2, 2, 3d,
19. (2, 0, 2),
(0, 2, 0)
20. s1, 0, 1d,
s0, 3, 0d
43. s0, 1, 0d,
2x + y + 2z = 4
44. s1, 0, 1d,
4x + y + z = 4
Find equations for the planes in Exercises 21–26. 21. The plane through P0s0, 2, 1d normal to n = 3i  2j  k
iaz
45. Find the distance from the plane x + 2y + 6z = 1 to the plane x + 2y + 6z = 10 .
Planes
46. Find the distance from the line x = 2 + t, y = 1 + t, z = s1>2d  s1>2dt to the plane x + 2y + 6z = 10 .
3x + y + z = 7
Angles
23. The plane through s1, 1, 1d, s2, 0, 2d , and s0, 2, 1d 24. The plane through (2, 4, 5), (1, 5, 7), and s 1, 6, 8d
z = 4t
2x + y  2z = 2
48. 5x + y  z = 10,
ass a
y = 1 + 3t,
Find the angles between the planes in Exercises 47 and 48. 47. x + y = 1,
25. The plane through P0s2, 4, 5d perpendicular to the line x = 5 + t,
nR
22. The plane through s1, 1, 3d parallel to the plane
26. The plane through As1, 2, 1d perpendicular to the vector from the origin to A
dH
27. Find the point of intersection of the lines x = 2t + 1, y = 3t + 2, z = 4t + 3 , and x = s + 2, y = 2s + 4, z = 4s  1 , and then find the plane determined by these lines.
28. Find the point of intersection of the lines x = t, y = t + 2, z = t + 1 , and x = 2s + 2, y = s + 3, z = 5s + 6 , and then find the plane determined by these lines.
ma
In Exercises 29 and 30, find the plane determined by the intersecting lines. 29. L1: x = 1 + t, y = 2 + t, z = 1  t;  q 6 t 6 q
ham
L2: x = 1  4s, y = 1 + 2s, z = 2  2s;  q 6 s 6 q 30. L1: x = t, y = 3  3t, z = 2  t;  q 6 t 6 q L2: x = 1 + s, y = 4 + s, z = 1 + s;  q 6 s 6 q
x  2y + 3z = 1
T Use a calculator to find the acute angles between the planes in Exercises 49–52 to the nearest hundredth of a radian. 2x  2y  z = 5
49. 2x + 2y + 2z = 3, 50. x + y + z = 1,
z = 0
51. 2x + 2y  z = 3,
sthe xyplaned
x + 2y + z = 2
52. 4y + 3z = 12,
3x + 2y + 6z = 6
Intersecting Lines and Planes In Exercises 53–56, find the point in which the line meets the plane. 53. x = 1  t, 54. x = 2,
y = 3t,
y = 3 + 2t,
55. x = 1 + 2t,
z = 1 + t; z = 2  2t;
y = 1 + 5t,
56. x = 1 + 3t,
y = 2,
2x  y + 3z = 6 6x + 3y  4z = 12
z = 3t; z = 5t;
x + y + z = 2 2x  3z = 7
Find parametrizations for the lines in which the planes in Exercises 57–60 intersect.
32. Find a plane through the points P1s1, 2, 3d, P2s3, 2, 1d and perpendicular to the plane 4x  y + 2z = 7 .
57. x + y + z = 1,
Mu
31. Find a plane through P0s2, 1, 1d and perpendicular to the line of intersection of the planes 2x + y  z = 3, x + 2y + z = 2 .
x + y = 2
58. 3x  6y  2z = 3, 59. x  2y + 4z = 2,
2x + y  2z = 2 x + y  2z = 5
4y  5z = 17
Distances
60. 5x  2y = 11,
In Exercises 33–38, find the distance from the point to the line.
Given two lines in space, either they are parallel, or they intersect, or they are skew (imagine, for example, the flight paths of two planes in the sky). Exercises 61 and 62 each give three lines. In each exercise, determine whether the lines, taken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersection.
33. s0, 0, 12d;
x = 4t,
y = 2t,
z = 2t
34. s0, 0, 0d;
x = 5 + 3t,
y = 5 + 4t,
z = 3  5t
35. s2, 1, 3d;
x = 2 + 2t,
y = 1 + 6t,
z = 3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.5 Lines and Planes in Space
889
72. Suppose L1 and L2 are disjoint (nonintersecting) nonparallel lines. Is it possible for a nonzero vector to be perpendicular to both L1 and L2 ? Give reasons for your answer.
62. L1: x = 1 + 2t,
Computer Graphics
Theory and Examples 63. Use Equations (3) to generate a parametrization of the line through Ps2, 4, 7d parallel to v1 = 2i  j + 3k . Then generate another parametrization of the line using the point P2s 2, 2, 1d and the vector v2 = i + s1>2dj  s3>2dk . 64. Use the component form to generate an equation for the plane through P1s4, 1, 5d normal to n1 = i  2j + k . Then generate another equation for the same plane using the point P2s3, 2, 0d and the normal vector n2 =  22i + 2 22j  22k . 65. Find the points in which the line x = 1 + 2t, y = 1  t, z = 3t meets the coordinate planes. Describe the reasoning behind your answer.
73. Perspective in computer graphics In computer graphics and perspective drawing, we need to represent objects seen by the eye in space as images on a twodimensional plane. Suppose that the eye is at Esx0, 0, 0d as shown here and that we want to represent a point P1sx1, y1, z1 d as a point on the yzplane. We do this by projecting P1 onto the plane with a ray from E. The point P1 will be portrayed as the point P(0, y, z). The problem for us as graphics designers is to find y and z given E and P1 . 1 1 a. Write a vector equation that holds between EP and EP1 . Use the equation to express y and z in terms of x0 , x1, y1 , and z1 . b. Test the formulas obtained for y and z in part (a) by investigating their behavior at x1 = 0 and x1 = x0 and by seeing what happens as x0 : q . What do you find?
ass a
66. Find equations for the line in the plane z = 3 that makes an angle of p>6 rad with i and an angle of p>3 rad with j. Describe the reasoning behind your answer.
suf
y = 1  r,
q 6 t 6 q q 6 s 6 q z = 8 + 3r;  q 6 r 6 q
You
L3: x = 5 + 2r,
z = 3t;
z = 1 + s;
iaz
y = 3s,
nR
y = 1  t,
L2: x = 2  s,
i
61. L1: x = 3 + 2t, y = 1 + 4t, z = 2  t;  q 6 t 6 q L2: x = 1 + 4s, y = 1 + 2s, z = 3 + 4s;  q 6 s 6 q L3: x = 3 + 2r, y = 2 + r, z = 2 + 2r;  q 6 r 6 q
67. Is the line x = 1  2t, y = 2 + 5t, z = 3t parallel to the plane 2x + y  z = 8 ? Give reasons for your answer.
z
P(0, y, z) P1(x1, y1, z1) 0
68. How can you tell when two planes A1 x + B1 y + C1 z = D1 and A2 x + B2 y + C2 z = D2 are parallel? Perpendicular? Give reasons for your answer.
dH
69. Find two different planes whose intersection is the line x = 1 + t, y = 2  t, z = 3 + 2t . Write equations for each plane in the form Ax + By + Cz = D .
ma
70. Find a plane through the origin that meets the plane M: 2x + 3y + z = 12 in a right angle. How do you know that your plane is perpendicular to M?
(x1, y1, 0)
E(x 0, 0, 0) x
74. Hidden lines Here is another typical problem in computer graphics. Your eye is at (4, 0, 0). You are looking at a triangular plate whose vertices are at (1, 0, 1), (1, 1, 0), and s 2, 2, 2d . The line segment from (1, 0, 0) to (0, 2, 2) passes through the plate. What portion of the line segment is hidden from your view by the plate? (This is an exercise in finding intersections of lines and planes.)
Mu
ham
71. For any nonzero numbers a, b, and c, the graph of sx>ad + s y>bd + sz>cd = 1 is a plane. Which planes have an equation of this form?
y
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o Y iaz 12.5 Lines and Planes in Space
12.6
Cylinders and Quadric Surfaces
R n ssa
889
Up to now, we have studied two special types of surfaces: spheres and planes. In this section, we extend our inventory to include a variety of cylinders and quadric surfaces. Quadric surfaces are surfaces defined by seconddegree equations in x, y, and z. Spheres are quadric surfaces, but there are others of equal interest.
a H d a m m a h u M Cylinders
A cylinder is a surface that is generated by moving a straight line along a given planar curve while holding the line parallel to a given fixed line. The curve is called a generating curve for the cylinder (Figure 12.43). In solid geometry, where cylinder means circular
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Chapter 12: Vectors and the Geometry of Space
The Parabolic Cylinder y = x2
You
EXAMPLE 1
suf
i
cylinder, the generating curves are circles, but now we allow generating curves of any kind. The cylinder in our first example is generated by a parabola. When graphing a cylinder or other surface by hand or analyzing one generated by a computer, it helps to look at the curves formed by intersecting the surface with planes parallel to the coordinate planes. These curves are called crosssections or traces.
Find an equation for the cylinder made by the lines parallel to the zaxis that pass through the parabola y = x 2, z = 0 (Figure 12.44).
nR
Generating curve (in the yzplane)
iaz
z
z
ass a
x
y
x
dH
Q 0(x 0, x 02, z)
Generating curve y x 2, z 0
Lines through generating curve parallel to xaxis
FIGURE 12.43 A cylinder and generating curve.
z
y
FIGURE 12.44 The cylinder of lines passing through the parabola y = x 2 in the xyplane parallel to the zaxis (Example 1).
Suppose that the point P0sx0 , x 02, 0d lies on the parabola y = x 2 in the xyplane. Then, for any value of z, the point Qsx0 , x 02, zd will lie on the cylinder because it lies on the line x = x0 , y = x 02 through P0 parallel to the zaxis. Conversely, any point Qsx0 , x 02, zd whose ycoordinate is the square of its xcoordinate lies on the cylinder because it lies on the line x = x0 , y = x 02 through P0 parallel to the zaxis (Figure 12.45). Regardless of the value of z, therefore, the points on the surface are the points whose coordinates satisfy the equation y = x 2 . This makes y = x 2 an equation for the cylinder. Because of this, we call the cylinder “the cylinder y = x 2 .”
ma
Solution
P0(x 0, x 02, 0)
x
ham
0
y
PA
BO RA
y x2
Mu
LA
FIGURE 12.45 Every point of the cylinder in Figure 12.44 has coordinates of the form sx0 , x 02, zd . We call it “the cylinder y = x 2 .”
As Example 1 suggests, any curve ƒsx, yd = c in the xyplane defines a cylinder parallel to the zaxis whose equation is also ƒsx, yd = c. The equation x 2 + y 2 = 1 defines the circular cylinder made by the lines parallel to the zaxis that pass through the circle x 2 + y 2 = 1 in the xyplane. The equation x 2 + 4y 2 = 9 defines the elliptical cylinder made by the lines parallel to the zaxis that pass through the ellipse x 2 + 4y 2 = 9 in the xyplane. In a similar way, any curve gsx, zd = c in the xzplane defines a cylinder parallel to the yaxis whose space equation is also gsx, zd = c (Figure 12.46). Any curve hs y, zd = c
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.6
1
defines a cylinder parallel to the xaxis whose space equation is also hs y, zd = c (Figure 12.47). The axis of a cylinder need not be parallel to a coordinate axis, however.
Generating ellipse: x 2 4z 2 4
i
z
LI
The generating hyperbola: y2 z2 1
EL
z
y 2 z2 1
y –1 z
1
–1 A
x 2 4z 2 4
OL
RB
x
PE
BE
1
y
HY
OL
nR
A
y
y
x
RB
HY
–1
iaz
2
z
You
PS
E
–2
x
891
suf
Elliptical trace (crosssection)
Cylinders and Quadric Surfaces
Cross sections perpendicular to xaxis
x
ass a
FIGURE 12.47 The hyperbolic cylinder y 2  z 2 = 1 is made of lines parallel to the xaxis and passing through the hyperbola y 2  z 2 = 1 in the yzplane. The crosssections of the cylinder in planes perpendicular to the xaxis are hyperbolas congruent to the generating hyperbola.
Quadric Surfaces
The next type of surface we examine is a quadric surface. These surfaces are the threedimensional analogues of ellipses, parabolas, and hyperbolas. A quadric surface is the graph in space of a seconddegree equation in x, y, and z. The most general form is
dH
FIGURE 12.46 The elliptical cylinder x 2 + 4z 2 = 4 is made of lines parallel to the yaxis and passing through the ellipse x 2 + 4z 2 = 4 in the xzplane. The crosssections or “traces” of the cylinder in planes perpendicular to the yaxis are ellipses congruent to the generating ellipse. The cylinder extends along the entire yaxis.
Mu
ham
ma
Ax 2 + By 2 + Cz 2 + Dxy + Eyz + Fxz + Gx + Hy + Jz + K = 0,
where A, B, C, and so on are constants. However, this equation can be simplified by translation and rotation, as in the twodimensional case. We will study only the simpler equations. Although defined differently, the cylinders in Figures 12.45 through 12.47 were also examples of quadric surfaces. The basic quadric surfaces are ellipsoids, paraboloids, elliptical cones, and hyperboloids. (We think of spheres as special ellipsoids.) We now present examples of each type.
EXAMPLE 2
Ellipsoids
The ellipsoid y2 x2 z2 + + = 1 a2 b2 c2
(1)
(Figure 12.48) cuts the coordinate axes at s ; a, 0, 0d, s0, ; b, 0d, and s0, 0, ; cd. It lies within the rectangular box defined by the inequalities ƒ x ƒ … a, ƒ y ƒ … b, and ƒ z ƒ … c. The surface is symmetric with respect to each of the coordinate planes because each variable in the defining equation is squared.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 892
Chapter 12: Vectors and the Geometry of Space z
z
Elliptical crosssection in the plane z z 0
suf
i
c y2 x2 2 1 2 a b in the xyplane The ellipse
y
b SE
ELLIPSE
You
a
EL LIPSE
z0
EL
The ellipse x2 z2 2 1 2 a c in the xzplane
y2
z2 1 c2
iaz
The ellipse
b2 in the yzplane
nR
FIGURE 12.48 The ellipsoid
y
x
LIP
x
ass a
y2 x2 z2 + + 2 = 1 2 2 a b c in Example 2 has elliptical crosssections in each of the three coordinate planes.
The curves in which the three coordinate planes cut the surface are ellipses. For example, y2 x2 + = 1 a2 b2
when
z = 0.
The section cut from the surface by the plane z = z0 , ƒ z0 ƒ 6 c, is the ellipse
dH
y2 x2 + 2 = 1. 2 a s1  sz0>cd d b s1  sz0>cd2 d 2
ma
If any two of the semiaxes a, b, and c are equal, the surface is an ellipsoid of revolution. If all three are equal, the surface is a sphere.
EXAMPLE 3
Paraboloids y2 x2 z + = c a2 b2
(2)
is symmetric with respect to the planes x = 0 and y = 0 (Figure 12.49). The only intercept on the axes is the origin. Except for this point, the surface lies above (if c 7 0) or entirely below (if c 6 0) the xyplane, depending on the sign of c. The sections cut by the coordinate planes are
Mu
ham
The elliptical paraboloid
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x = 0:
the parabola z =
c 2 y b2
y = 0:
the parabola z =
c 2 x a2
z = 0:
the point s0, 0, 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 893
12.6 Cylinders and Quadric Surfaces z
z
c 2 x a2
in the xzplane
i
y2 x2 The ellipse 2 2 1 a b in the plane z c
suf
The parabola z
zc
PAR AB
ELLIPSE
OL A
The parabola z in the yzplane
c 2 y b2
y
iaz
y
You
b
a
x
x
nR
FIGURE 12.49 The elliptical paraboloid sx 2>a 2 d + s y 2>b 2 d = z>c in Example 3, shown for c 7 0 . The crosssections perpendicular to the zaxis above the xyplane are ellipses. The crosssections in the planes that contain the zaxis are parabolas.
Each plane z = z0 above the xyplane cuts the surface in the ellipse
ass a
y2 z0 x2 + = c. 2 2 a b
EXAMPLE 4
Cones
dH
The elliptical cone
y2 x2 z2 + 2 = 2 2 a b c
(3)
Mu
ham
ma
is symmetric with respect to the three coordinate planes (Figure 12.50). The sections cut
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The line z â€“ c y b in the yzplane z c a
The line z ac x in the xzplane
2
z
x2 y 1 a2 b2 in the plane z c
The ellipse
z
b
ELLIP
SE
y
y x
x
ELLI
PSE
FIGURE 12.50 The elliptical cone sx 2>a 2 d + sy 2>b 2 d = sz 2>c 2 d in Example 4. Planes perpendicular to the zaxis cut the cone in ellipses above and below the xyplane. Vertical planes that contain the zaxis cut it in pairs of intersecting lines.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 894
Chapter 12: Vectors and the Geometry of Space
z = 0:
the point s0, 0, 0d.
You
the lines z = ;
suf
y = 0:
c y b c the lines z = ; a x
x = 0:
i
by the coordinate planes are
The sections cut by planes z = z0 above and below the xyplane are ellipses whose centers lie on the zaxis and whose vertices lie on the lines given above. If a = b, the cone is a right circular cone.
Hyperboloids
iaz
EXAMPLE 5
The hyperboloid of one sheet
nR
y2 x2 z2 +  2 = 1 2 2 a b c
(4)
is symmetric with respect to each of the three coordinate planes (Figure 12.51).
ass a
2 2 Part of the hyperbola x 2 z 2 1 in the xzplane a c z
zc
z
2
x2 y The ellipse 2 2 2 a b in the plane z c
b兹2
ma
a
x
2 x2 y 2 1 2 a b in the xyplane
The ellipse
b ELLIPSE
y
y
x ELLIPSE
FIGURE 12.51 The hyperboloid sx 2>a 2 d + sy 2>b 2 d  sz 2>c 2 d = 1 in Example 5. Planes perpendicular to the zaxis cut it in ellipses. Vertical planes containing the zaxis cut it in hyperbolas.
The sections cut out by the coordinate planes are
Mu
ham
2 y2 Part of the hyperbola 2 z 2 1 b c in the yzplane
ELLIPSE
HYPERBOLA
dH
HY PERBOLA
a兹2
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x = 0:
the hyperbola
y = 0:
the hyperbola
z = 0:
y2 b2

z2 = 1 c2
x2 z2  2 = 1 2 a c 2 2 y x the ellipse 2 + 2 = 1. a b
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12.6 Cylinders and Quadric Surfaces
EXAMPLE 6
You
suf
i
The plane z = z0 cuts the surface in an ellipse with center on the zaxis and vertices on one of the hyperbolic sections above. The surface is connected, meaning that it is possible to travel from one point on it to any other without leaving the surface. For this reason, it is said to have one sheet, in contrast to the hyperboloid in the next example, which has two sheets. If a = b, the hyperboloid is a surface of revolution.
Hyperboloids
The hyperboloid of two sheets
y2 x2 z2 = 1 c2 a2 b2
(5)
iaz
is symmetric with respect to the three coordinate planes (Figure 12.52). The plane z = 0 does not intersect the surface; in fact, for a horizontal plane to intersect the surface, we must have ƒ z ƒ Ú c. The hyperbolic sections y2 z2 = 1 c2 b2
y = 0:
x2 z2  2 = 1 2 c a
nR
x = 0:
ass a
have their vertices and foci on the zaxis. The surface is separated into two portions, one above the plane z = c and the other below the plane z = c. This accounts for its name.
PE
RB
OL
ELLIPSE
ma ham
z
b
a
Y
The hyperbola z2 x2 1 c2 a2 in the xzplane
H
0
The hyperbola y2 z2 2 1 2 c b in the yzplane
(0, 0, c) Vertex
y
y
O
LA
(0, 0, –c) Vertex
x
OLA
HY
PE
RB
x
HYPERB
Mu
2
x2 y 1 a2 b2 in the plane z c兹2 The ellipse
A
dH
z
ELLIPSE
FIGURE 12.52 The hyperboloid sz 2>c 2 d  sx 2>a 2 d  sy 2>b 2 d = 1 in Example 6. Planes perpendicular to the zaxis above and below the vertices cut it in ellipses. Vertical planes containing the zaxis cut it in hyperbolas.
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Chapter 12: Vectors and the Geometry of Space
Equations (4) and (5) have different numbers of negative terms. The number in each case is the same as the number of sheets of the hyperboloid. If we replace the 1 on the right side of either Equation (4) or Equation (5) by 0, we obtain the equation
suf
i
z
y2 x2 z2 + = a2 b2 c2
y2 x2 = ;1 a2 b2
0
are asymptotic to the lines
y
You
for an elliptical cone (Equation 3). The hyperboloids are asymptotic to this cone (Figure 12.53) in the same way that the hyperbolas
iaz
y2 x2 = 0 a2 b2
nR
in the xyplane.
x
EXAMPLE 7
A Saddle Point
The hyperbolic paraboloid
FIGURE 12.53 Both hyperboloids are asymptotic to the cone (Example 6).
y2

x2 z = c, a2
ass a b
2
c 7 0
(6)
ma
dH
has symmetry with respect to the planes x = 0 and y = 0 (Figure 12.54). The sections in these planes are the parabola z =
y = 0:
the parabola z = 
The parabola z c2 y 2 in the yzplane b z
PER
PA
BO
RA
PA
x
RA
LA
Saddle point
The parabola z â€“ c2 x2 a in the xzplane
Mu
Part of the hyperbola in the plane z c
y2 b2
(7)
c 2 x . a2
(8)
z
x2 1 a2
BOLA
ham
HY
c 2 y . b2
x = 0:
B OLA
y 2 y2 Part of the hyperbola x 2 2 1 a b in the plane z â€“c
y
x
FIGURE 12.54 The hyperbolic paraboloid sy 2>b 2 d  sx 2>a 2 d = z>c, c 7 0 . The crosssections in planes perpendicular to the zaxis above and below the xyplane are hyperbolas. The crosssections in planes perpendicular to the other axes are parabolas.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.6 Cylinders and Quadric Surfaces
897
y2 b2

z0 x2 = c, 2 a
suf
i
In the plane x = 0, the parabola opens upward from the origin. The parabola in the plane y = 0 opens downward. If we cut the surface by a plane z = z0 7 0, the section is a hyperbola,
USING TECHNOLOGY
iaz
You
with its focal axis parallel to the yaxis and its vertices on the parabola in Equation (7). If z0 is negative, the focal axis is parallel to the xaxis and the vertices lie on the parabola in Equation (8). Near the origin, the surface is shaped like a saddle or mountain pass. To a person traveling along the surface in the yzplane the origin looks like a minimum. To a person traveling in the xzplane the origin looks like a maximum. Such a point is called a saddle point of a surface. Visualizing in Space
Mu
ham
ma
dH
ass a
nR
A CAS or other graphing utility can help in visualizing surfaces in space. It can draw traces in different planes, and many computer graphing systems can rotate a figure so you can see it as if it were a physical model you could turn in your hand. Hiddenline algorithms (see Exercise 74, Section 12.5) are used to block out portions of the surface that you would not see from your current viewing angle. A system may require surfaces to be entered in parametric form, as discussed in Section 16.6 (see also CAS Exercises 57 through 60 in Section 14.1). Sometimes you may have to manipulate the grid mesh to see all portions of a surface.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
897
12.6 Cylinders and Quadric Surfaces
u o Y z
EXERCISES 12.6 Matching Equations with Surfaces
c.
In Exercises 1–12, match the equation with the surface it defines. Also, identify each surface by type (paraboloid, ellipsoid, etc.) The surfaces are labeled (a)–(1). 2
2
2
2
2
2. z + 4y  4x = 4
3. 9y 2 + z 2 = 16
4. y 2 + z 2 = x 2
2
5. x = y  z
2
2
6. x = y  z
7. x 2 + 2z 2 = 8 2
9. x = z  y
10. z = 4x 2  y 2
11. x 2 + 4z 2 = y 2
m a h u M x
d a m z
b.
y
a i R
x
a H
12. 9x 2 + 4y 2 + 2z 2 = 36 z
a.
2
8. z 2 + x 2  y 2 = 1
2
n a ss
2
1. x + y + 4z = 10
z
e.
d.
y
z
z
y
x
z
f.
x x
y
x
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y
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y
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 898
Chapter 12: Vectors and the Geometry of Space
38. sx 2>4d + sy 2>4d  sz 2>9d = 1 39. z 2  x 2  y 2 = 1
41. x  y  sz >4d = 1 2
x
y
2
z
z
j.
44. x 2  y 2 = z
ASSORTED 45. x 2 + y 2 + z 2 = 4
46. 4x 2 + 4y 2 = z 2
47. z = 1 + y 2  x 2 x
y
48. y 2  z 2 = 4
2
49. y = sx + z d
50. z 2  4x 2  4y 2 = 4
51. 16x 2 + 4y 2 = 1
52. z = x 2 + y 2 + 1
iaz
2
y
x
2
2
2
54. x = 4  y 2
53. x + y  z = 4 z
k.
55. x 2 + z 2 = y
z
l.
2
nR
2
57. x + z = 1
59. 16y 2 + 9z 2 = 4x 2 2
x
y
x
2
64. z 2 + 4y 2 = 9 66. y 2  x 2  z 2 = 1
67. x 2  4y 2 = 1
68. z = 4x 2 + y 2  4
ass a
65. z = sx + y d 2
2
2
2
dH
2
70. z = 1  x 2
72. sx 2>4d + y 2  z 2 = 1
74. 36x 2 + 9y 2 + 4z 2 = 36
73. yz = 1
CYLINDERS
60. z = x 2  y 2  1
63. x 2 + y 2  16z 2 = 16
71. x 2 + y 2 = z
Sketch the surfaces in Exercises 13â€“76.
58. 4x 2 + 4y 2 + z 2 = 4 62. 4x 2 + 9z 2 = y 2
69. 4y + z  4x = 4
Drawing
56. z 2  sx 2>4d  y 2 = 1
61. 9x + 4y + z = 36
2
2
2
2
y
42. sx 2>4d  y 2  sz 2>4d = 1
HYPERBOLIC PARABOLOIDS 43. y 2  x 2 = z
i.
40. sy 2>4d  sx 2>4d  z 2 = 1
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y x
2
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37. sy 2>4d + sz 2>9d  sx 2>4d = 1
z
h.
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z
g.
75. 9x 2 + 16y 2 = 4z 2
76. 4z 2  x 2  y 2 = 4
2
14. x + z = 4
13. x + y = 4
Theory and Examples
17. x + 4z = 16
18. 4x 2 + y 2 = 36
77. a. Express the area A of the crosssection cut from the ellipsoid
19. z 2  y 2 = 1
20. yz = 1
15. z = y  1 2
2
ELLIPSOIDS 21. 9x 2 + y 2 + z 2 = 9 23. 4x 2 + 9y 2 + 4z 2 = 36
25. z = x 2 + 4y 2 2
27. z = 8  x  y
24. 9x 2 + 4y 2 + 36z 2 = 36
26. z = x 2 + 9y 2
2
29. x = 4  4y 2  z 2 CONES
22. 4x 2 + 4y 2 + z 2 = 16
ham
PARABOLOIDS
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16. x = y 2
2
28. z = 18  x 2  9y 2 30. y = 1  x 2  z 2
32. y 2 + z 2 = x 2
33. 4x 2 + 9z 2 = 9y 2
34. 9x 2 + 4y 2 = 36z 2
Mu
31. x 2 + y 2 = z 2
HYPERBOLOIDS
35. x 2 + y 2  z 2 = 1
36. y 2 + z 2  x 2 = 1
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x2 +
y2 z2 + = 1 4 9
by the plane z = c as a function of c. (The area of an ellipse with semiaxes a and b is pab .) b. Use slices perpendicular to the zaxis to find the volume of the ellipsoid in part (a). c. Now find the volume of the ellipsoid y2 x2 z2 + + 2 = 1. 2 2 a b c Does your formula give the volume of a sphere of radius a if a = b = c? 78. The barrel shown here is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the zaxis. The crosssections perpendicular to the zaxis are circular. The
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.6 Cylinders and Quadric Surfaces barrel is 2h units high, its midsection radius is R, and its end radii are both r. Find a formula for the barrel’s volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder of radius R and height 2h. Does your formula give the cylinder’s volume? Second, suppose r = 0 and h = R so the barrel is a sphere. Does your formula give the sphere’s volume?
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82. Suppose you set z = 0 in the equation
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Ax 2 + By 2 + Cz 2 + Dxy + Eyz + Fxz + Gx + Hy + Jz + K = 0
to obtain a curve in the xyplane. What will the curve be like? Give reasons for your answer.
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83. Every time we found the trace of a quadric surface in a plane parallel to one of the coordinate planes, it turned out to be a conic section. Was this mere coincidence? Did it have to happen? Give reasons for your answer.
z
r
84. Suppose you intersect a quadric surface with a plane that is not parallel to one of the coordinate planes. What will the trace in the plane be like? Give reasons for your answer.
R y x
T
Computer Grapher Explorations Plot the surfaces in Exercises 85–88 over the indicated domains. If you can, rotate the surface into different viewing positions.
r
–h
iaz
h
79. Show that the volume of the segment cut from the paraboloid
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85. z = y 2,
2 … x … 2, 2
86. z = 1  y , 2
88. z = x + 2y
2
dH
80. a. Find the volume of the solid bounded by the hyperboloid y2 x2 z2 + = 1 a2 b2 c2 and the planes z = 0 and z = h, h 7 0 .
b. Express your answer in part (a) in terms of h and the areas A0 and Ah of the regions cut by the hyperboloid from the planes z = 0 and z = h . c. Show that the volume in part (a) is also given by the formula h sA + 4Am + Ah d , 6 0
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V =
where Am is the area of the region cut by the hyperboloid from the plane z = h>2 .
2 … y … 2 3 … y … 3
over 3 … y … 3
b. 1 … x … 1,
2 … y … 3
c. 2 … x … 2,
2 … y … 2
d. 2 … x … 2,
1 … y … 1
COMPUTER EXPLORATIONS
Surface Plots Use a CAS to plot the surfaces in Exercises 89–94. Identify the type of quadric surface from your graph. y2 y2 x2 x2 z2 z2 89. 90. + = 1 = 1 9 36 25 9 9 16 y2 x2 2 2 2 91. 5x = z  3y 92. = 1 + z 16 9 2 y x2 z2 93. 94. y  24  z 2 = 0  1 = + 9 16 2
Mu
ham
81. If the hyperbolic paraboloid sy 2>b 2 d  sx 2>a 2 d = z>c is cut by the plane y = y1 , the resulting curve is a parabola. Find its vertex and focus.
3 … x … 3,
a. 3 … x … 3,
ass a
by the plane z = h equals half the segment’s base times its altitude. (Figure 12.49 shows the segment for the special case h = c .)
0.5 … y … 2
2 … x … 2,
87. z = x 2 + y 2,
2
y x z + 2 = c a2 b 2
899
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12
Chapter 12
s s a dH
a m m a h
2. How are vectors added and subtracted geometrically? Algebraically?
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z a i R n a
Questions to Guide Your Review
1. When do directed line segments in the plane represent the same vector?
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Questions to Guide Your Review
899
3. How do you find a vectorâ€™s magnitude and direction? 4. If a vector is multiplied by a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 900
Chapter 12: Vectors and the Geometry of Space 11. How do you find equations for lines, line segments, and planes in space? Give examples. Can you express a line in space by a single equation? A plane?
6. What geometric interpretation does the dot product have? Give examples.
12. How do you find the distance from a point to a line in space? From a point to a plane? Give examples.
7. What is the vector projection of a vector u onto a vector v? How do you write u as the sum of a vector parallel to v and a vector orthogonal to v?
13. What are box products? What significance do they have? How are they evaluated? Give an example.
8. Define the cross product (vector product) of two vectors. Which algebraic laws are satisfied by cross products, and which are not? Give examples. When is the cross product of two vectors equal to zero?
15. How do you find the intersection of two lines in space? A line and a plane? Two planes? Give examples.
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16. What is a cylinder? Give examples of equations that define cylinders in Cartesian coordinates. 17. What are quadric surfaces? Give examples of different kinds of ellipsoids, paraboloids, cones, and hyperboloids (equations and sketches).
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ma
dH
ass a
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10. What is the determinant formula for calculating the cross product of two vectors relative to the Cartesian i, j, kcoordinate system? Use it in an example.
14. How do you find equations for spheres in space? Give examples.
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9. What geometric or physical interpretations do cross products have? Give examples.
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5. Define the dot product (scalar product) of two vectors. Which algebraic laws are satisfied by dot products? Give examples. When is the dot product of two vectors equal to zero?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
900
i f u s u
Chapter 12: Vectors and the Geometry of Space
Chapter 12
Practice Exercises
Vector Calculations in Two Dimensions
In Exercises 1–4, let u = 83, 49 and v = 82, 59 . Find (a) the component form of the vector and (b) its magnitude. 1. 3u  4v
2. u + v
3. 2u
4. 5v
5. The vector obtained by rotating 80, 19 through an angle of 2p>3 radians 6. The unit vector that makes an angle of p>6 radian with the positive xaxis 7. The vector 2 units long in the direction 4i  j
d a
Express the vectors in Exercises 9–12 in terms of their lengths and directions. 9. 22i + 22j
m m
10. i  j
11. Velocity vector v = s 2 sin tdi + s2 cos tdj when t = p>2 . 12. Velocity vector v = se t cos t  e t sin tdi + se t sin t + e t cos tdj when t = ln 2.
a h
Vector Calculations in Three Dimensions
u M
Express the vectors in Exercises 13 and 14 in terms of their lengths and directions. 13. 2i  3j + 6k
n a ss
14. i + 2j  k
18. v = i + j + 2k
u = 2i + j  2k
u = i  k
In Exercises 19 and 20, write u as the sum of a vector parallel to v and a vector orthogonal to v.
a H
8. The vector 5 units long in the direction opposite to the direction of s3>5di + s4>5dj
z a i R
In Exercises 17 and 18, find ƒ v ƒ , ƒ u ƒ , v # u, u # v, v * u, u * v, ƒ v * u ƒ , the angle between v and u, the scalar component of u in the direction of v, and the vector projection of u onto v. 17. v = i + j
In Exercises 5–8, find the component form of the vector.
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16. Find a vector 5 units long in the direction opposite to the direction of v = s3>5di + s4>5dk.
19. v = 2i + j  k
20. u = i  2j
u = i + j  5k
v = i + j + k
In Exercises 21 and 22, draw coordinate axes and then sketch u, v, and u * v as vectors at the origin. 21. u = i,
v = i + j
22. u = i  j,
v = i + j
23. If ƒ v ƒ = 2, ƒ w ƒ = 3 , and the angle between v and w is p>3 , find ƒ v  2w ƒ . 24. For what value or values of a will the vectors u = 2i + 4j  5k and v = 4i  8j + ak be parallel?
In Exercises 25 and 26, find (a) the area of the parallelogram determined by vectors u and v and (b) the volume of the parallelepiped determined by the vectors u, v, and w. 25. u = i + j  k, 26. u = i + j,
v = 2i + j + k,
v = j,
w = i  2j + 3k
w = i + j + k
15. Find a vector 2 units long in the direction of v = 4i  j + 4k.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12
49. Find the distance from the point P(1, 4, 0) to the plane through A(0, 0, 0), Bs2, 0, 1d and Cs2, 1, 0d .
28. Find a vector in the plane parallel to the line ax + by = c . In Exercises 29 and 30, find the distance from the point to the line. x = 2 + t,
50. Find the distance from the point (2, 2, 3) to the plane 2x + 3y + 5z = 0 .
z = 1 + t
y = 2 + t,
z = t
31. Parametrize the line that passes through the point (1, 2, 3) parallel to the vector v = 3i + 7k. 32. Parametrize the line segment joining the points P(1, 2, 0) and Qs1, 3, 1d . In Exercises 33 and 34, find the distance from the point to the plane. 33. s6, 0, 6d,
x  y = 4
34. s3, 0, 10d,
2x + 3y + z = 2
In Exercises 37 and 38, find an equation for the plane through points P, Q, and R. 38. Ps1, 0, 0d,
Qs0, 1, 0d,
Rs 1, 2, 1d Rs0, 0, 1d
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39. Find the points in which the line x = 1 + 2t, y = 1  t, z = 3t meets the three coordinate planes. 40. Find the point in which the line through the origin perpendicular to the plane 2x  y  z = 4 meets the plane 3x  5y + 2z = 6 . 41. Find the acute angle between the planes x = 7 and x + y + 22z = 3 .
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42. Find the acute angle between the planes x + y = 1 and y + z = 1. 43. Find parametric equations for the line in which the planes x + 2y + z = 1 and x  y + 2z = 8 intersect.
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44. Show that the line in which the planes x + 2y  2z = 5
and
5x  2y  z = 0
53. Find a vector of magnitude 2 parallel to the line of intersection of the planes x + 2y + z  1 = 0 and x  y + 2z + 7 = 0 .
y = 3t,
55. Find the point in which the line through P(3, 2, 1) normal to the plane 2x  y + 2z = 2 meets the plane.
56. What angle does the line of intersection of the planes 2x + y  z = 0 and x + y + 2z = 0 make with the positive xaxis? 57. The line
x = 3 + 2t,
y = 2t,
z = t
intersects the plane x + 3y  z = 4 in a point P. Find the coordinates of P and find equations for the line in the plane through P perpendicular to L.
58. Show that for every real number k the plane x  2y + z + 3 + k s2x  y  z + 1d = 0 contains the line of intersection of the planes x  2y + z + 3 = 0
and
2x  y  z + 1 = 0 .
59. Find an equation for the plane through As 2, 0, 3d and Bs1, 2, 1d that lies parallel to the line through Cs 2, 13>5, 26>5d and Ds16>5, 13>5, 0d . 60. Is the line x = 1 + 2t, y = 2 + 3t, z = 5t related in any way to the plane 4x  6y + 10z = 9 ? Give reasons for your answer.
a. s2i  3j + 3kd # ssx + 2di + s y  1dj + zkd = 0
z = 1 + 4t .
45. The planes 3x + 6z = 1 and 2x + 2y  z = 3 intersect in a line.
Mu
L:
61. Which of the following are equations for the plane through the points Ps1, 1, 1d , Q(3, 0, 2), and Rs 2, 1, 0d ?
intersect is parallel to the line x = 3 + 2t,
52. Find a unit vector orthogonal to A in the plane of B and C if A = 2i  j + k, B = i + 2j + k, and C = i + j  2k.
ass a
36. Find an equation for the plane that passes through the point s 1, 6, 0d perpendicular to the line x = 1 + t, y = 6  2t, z = 3t .
Qs2, 1, 3d,
51. Find a vector parallel to the plane 2x  y  z = 4 and orthogonal to i + j + k.
54. Find the point in which the line through the origin perpendicular to the plane 2x  y  z = 4 meets the plane 3x  5y + 2z = 6 .
35. Find an equation for the plane that passes through the point s3, 2, 1d normal to the vector n = 2i + j + k.
37. Ps1, 1, 2d,
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30. (0, 4, 1);
y = t,
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x = t,
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29. (2, 2, 0);
47. Is v = 2i  4j + k related in any special way to the plane 2x + y = 5 ? Give reasons for your answer. 1 48. The equation n # P0 P = 0 represents the plane through P0 normal 1 to n. What set does the inequality n # P0 P 7 0 represent?
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27. Suppose that n is normal to a plane and that v is parallel to the plane. Describe how you would find a vector n that is both perpendicular to v and parallel to the plane.
901
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Lines, Planes, and Distances
Practice Exercises
b. x = 3  t,
y = 11t,
z = 2  3t
c. sx + 2d + 11s y  1d = 3z
a. Show that the planes are orthogonal.
d. s2i  3j + 3kd * ssx + 2di + s y  1dj + zkd = 0
b. Find equations for the line of intersection.
e. s2i  j + 3kd * s 3i + kd # ssx + 2di + s y  1dj + zkd =0
46. Find an equation for the plane that passes through the point (1, 2, 3) parallel to u = 2i + 3j + k and v = i  j + 2k.
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62. The parallelogram shown on page 902 has vertices at As2, 1, 4d, Bs1, 0, 1d, Cs1, 2, 3d , and D. Find
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space z
f. the areas of the orthogonal projections of the parallelogram on the three coordinate planes.
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D
A(2, –1, 4)
64. (Continuation of Exercise 63.) Find the distance between the line through A(4, 0, 2) and B(2, 4, 1) and the line through C(1, 3, 2) and D(2, 2, 4).
C(1, 2, 3)
Quadric Surfaces
y x
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63. Distance between lines Find the distance between the line L1 through the points As1, 0, 1d and Bs 1, 1, 0d and the line L2 through the points Cs3, 1, 1d and Ds4, 5, 2d . The distance is to be measured along the line perpendicular to the two lines. First find 1 a vector n perpendicular to both lines. Then project AC onto n.
Identify and sketch the surfaces in Exercises 65–76.
B(1, 0, –1)
a. the coordinates of D,
65. x 2 + y 2 + z 2 = 4
66. x 2 + s y  1d2 + z 2 = 1
67. 4x 2 + 4y 2 + z 2 = 4
68. 36x 2 + 9y 2 + 4z 2 = 36
2
2
69. z = sx + y d
70. y = sx 2 + z 2 d
71. x 2 + y 2 = z 2
72. x 2 + z 2 = y 2
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b. the cosine of the interior angle at B, 1 1 c. the vector projection of BA onto BC ,
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902
2
2
d. the area of the parallelogram,
73. x + y  z = 4
74. 4y 2 + z 2  4x 2 = 4
e. an equation for the plane of the parallelogram,
75. y 2  x 2  z 2 = 1
76. z 2  x 2  y 2 = 1
Mu
ham
ma
dH
ass a
2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 902
i f su
Chapter 12: Vectors and the Geometry of Space
Chapter 12
Additional and Advanced Exercises
H1:
(0, 5, 0)
x
m a h u M NOT TO SCALE
y
x = 6 + 110t,
y = 3 + 10t,
z = 3 + 2t
y = 3 + 4t,
z = 3 + t .
Time t is measured in hours and all coordinates are measured in miles. Due to system malfunctions, H2 stops its flight at (446, 13, 1) and, in a negligible amount of time, lands at (446, 13, 0). Two hours later, H1 is advised of this fact and heads toward H2 at 150 mph. How long will it take H1 to reach H2 ?
a s as
H d
a m
Ship B (4, 0, 0)
ai z
x = 6 + 40t,
R n
H2:
z
Ship A
u o Y
2. A helicopter rescue Two helicopters, H1 and H2 , are traveling together. At time t = 0 , they separate and follow different straightline paths given by
1. Submarine hunting Two surface ships on maneuvers are trying to determine a submarine’s course and speed to prepare for an aircraft intercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is located at (0, 5, 0). All coordinates are given in thousands of feet. Ship A locates the submarine in the direction of the vector 2i + 3j  s1>3dk, and ship B locates it in the direction of the vector 18i  6j  k. Four minutes ago, the submarine was located at s2, 1, 1>3d . The aircraft is due in 20 min. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?
3. Torque The operator’s manual for the Toro® 21 in. lawnmower says “tighten the spark plug to 15 ftlb s20.4 N # md .” If you are installing the plug with a 10.5in. socket wrench that places the center of your hand 9 in. from the axis of the spark plug, about how hard should you pull? Answer in pounds.
9 in.
Submarine
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12
i
8. In the figure here, D is the midpoint of side AB of triangle ABC, and E is onethird of the way between C and B. Use vectors to prove that F is the midpoint of line segment CD. C E
z
O
A
B(1, 3, 2)
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F
A(1, 1, 1)
903
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4. Rotating body The line through the origin and the point A(1, 1, 1) is the axis of rotation of a right body rotating with a constant angular speed of 3> 2 rad> sec. The rotation appears to be clockwise when we look toward the origin from A. Find the velocity v of the point of the body that is at the position B(1, 3, 2).
Additional and Advanced Exercises
D
B
9. Use vectors to show that the distance from P1sx1, y1 d to the line ax + by = c is
1
1 3
x
d =
ƒ ax1 + by1  c ƒ 2a 2 + b 2
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v
y
.
10. a. Use vectors to show that the distance from P1sx1, y1, z1 d to the plane Ax + By + Cz = D is d =
x1  x y1  y z1  z 3 x2  x y2  y z2  z 3 = 0 x3  x y3  y z3  z is an equation for the plane through the three noncollinear points P1sx1, y1, z1 d, P2sx2 , y2 , z2 d , and P3sx3 , y3 , z3 d .
ass a
z z1 z2 z3
1 14 = 0? 1 1
x = a1 s + b1, y = a2 s + b2 , z = a3 s + b3 ,  q 6 s 6 q ,
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intersect or are parallel if and only if
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c1 b1  d1 c2 b2  d2 3 = 0 . a3 c3 b3  d3 7. Parallelogram The accompanying figure shows parallelogram ABCD and the midpoint P of diagonal BD. 1 1 1 a. Express BD in terms of AB and AD . 1 1 1 b. Express AP in terms of AB and AD .
Mu
c. Prove that P is also the midpoint of diagonal AC.
A
B
ƒ D1  D2 ƒ . ƒ Ai + Bj + C k ƒ
c. Find an equation for the plane parallel to the plane 2x  y + 2z = 4 if the point s3, 2, 1d is equidistant from the two planes. d. Write equations for the planes that lie parallel to and 5 units away from the plane x  2y + z = 3 .
x = c1 t + d1, y = c2 t + d2 , z = c3 t + d3 ,  q 6 t 6 q , a1 3 a2
d =
b. Find the distance between the planes 2x + 3y  z = 6 and 2x + 3y  z = 12 .
6. Determinants and lines Show that the lines
and
.
11. a. Show that the distance between the parallel planes Ax + By + Cz = D1 and Ax + By + Cz = D2 is
dH
y y1 y2 y3
2A2 + B 2 + C 2
b. Find an equation for the sphere that is tangent to the planes x + y + z = 3 and x + y + z = 9 if the planes 2x  y = 0 and 3x  z = 0 pass through the center of the sphere.
b. What set of points in space is described by the equation x x 4 1 x2 x3
ƒ Ax1 + By1 + Cz1  D ƒ
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5. Determinants and planes a. Show that
12. Prove that four points A, B, C, and D are coplanar (lie in a com1 1 1 mon plane) if and only if AD # sAB * BC d = 0 . 13. The projection of a vector on a plane Let P be a plane in space and let v be a vector. The vector projection of v onto the plane P, projP v, can be defined informally as follows. Suppose the sun is shining so that its rays are normal to the plane P. Then projP v is the “shadow” of v onto P. If P is the plane x + 2y + 6z = 6 and v = i + j + k, find projP v. 14. The accompanying figure shows nonzero vectors v, w, and z, with z orthogonal to the line L, and v and w making equal angles b with L. Assuming ƒ v ƒ = ƒ w ƒ , find w in terms of v and z. z
C
P v D
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 904
Chapter 12: Vectors and the Geometry of Space
su * vd * w = su # wdv  sv # wdu.
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where r is the vector from P to Q and G is the universal gravitational constant. Moreover, if Q1, Á , Qk are (point) masses with mass m1, Á , mk , respectively, then the force on P due to all the Qi’s is k
u * sv * wd = su # wdv  su # vdw.
F = a
w
a. 2i
2j
2k
b. i  j + k
2i + j  2k
i + 2j  k
c. 2i + j
2i  j + k
i + 2k
d. i + j  2k
i  k
2i + 4j  2k
16. Cross and dot products tors, then
y
P(0, d )
Show that if u, v, w, and r are any vec
a. u * sv * wd + v * sw * ud + w * su * vd = 0 v#w `. v#r
Prove or disprove the formula
u * su * su * vdd # w =  ƒ u ƒ 2u # v * w.
18. By forming the cross product of two appropriate vectors, derive the trigonometric identity
dH
sin sA  Bd = sin A cos B  cos A sin B . 19. Use vectors to prove that
sa 2 + b 2 dsc 2 + d 2 d Ú sac + bdd2
ma
for any four numbers a, b, c, and d. (Hint: Let u = ai + bj and v = ci + dj.)
20. Suppose that vectors u and v are not parallel and that u = w + r, where w is parallel to v and r is orthogonal to v. Express w and r in terms of u and v.
ham
21. Show that ƒ u + v ƒ … ƒ u ƒ + ƒ v ƒ for any vectors u and v. 22. Show that w = ƒ v ƒ u + ƒ u ƒ v bisects the angle between u and v. 23. Show that ƒ v ƒ u + ƒ u ƒ v and ƒ v ƒ u  ƒ u ƒ v are orthogonal. 24. Dot multiplication is positive definite Show that dot multiplication of vectors is positive definite; that is, show that u # u Ú 0 for every vector u and that u # u = 0 if and only if u = 0.
Mu
25. Point masses and gravitation In physics, the law of gravitation says that if P and Q are (point) masses with mass M and m, respectively, then P is attracted to Q by the force F =
…
–2d
Q–1
–d
Q0
0
Q1
Q2
d
2d
Qn …
nd
x
a. Let point P with mass M be located at the point (0, d), d 7 0 , in the coordinate plane. For i = n, n + 1, Á , 1, 0, 1, Á , n , let Qi be located at the point (id, 0) and have mass mi. Find the magnitude of the gravitational force on P due to all the Qi’s. b. Is the limit as n : q of the magnitude of the force on P finite? Why, or why not?
ass a
17. Cross and dot products
u#w u#r
Q–2
Q–n – nd
b. u * v = su # v * idi + su # v * jdj + su # v * kdk c. su * vd # sw * rd = `
where ri is the vector from P to Qi .
iaz
v
ri ,
nR
u
ƒ ri ƒ 3
You
Verify each formula for the following vectors by evaluating its two sides and comparing the results.
GMmi
i=1
suf
15. Triple vector products The triple vector products su * vd * w and u * sv * wd are usually not equal, although the formulas for evaluating them from components are similar:
26. Relativistic sums Einstein’s special theory of relativity roughly says that with respect to a reference frame (coordinate system) no material object can travel as fast as c, the speed of light. So, if sx and sy are two velocities such that ƒ sx ƒ 6 c and ƒ sy ƒ 6 c , then the relativistic sum sx { sy of sx and sy must have length less than c. Einstein’s special theory of relativity says that sx { sy =
gx sx + sy 1 # sx * sxs *# sy d , + 2# g + 1 sx # sy sx sy x c 1 + 2 1 + 2 c c
where gx =
1 1 
sx # sx c2
.
B It can be shown that if ƒ sx ƒ 6 c and ƒ sy ƒ 6 c , then ƒ sx { sy ƒ 6 c . This exercise deals with two special cases. a. Prove that if sx and sy are orthogonal, ƒ sx ƒ 6 c, ƒ sy ƒ 6 c , then ƒ sx { sy ƒ 6 c . b. Prove that if sx and sy are parallel, ƒ sx ƒ 6 c, ƒ sy ƒ 6 c , then ƒ sx { sy ƒ 6 c . ! ! c. Compute limc: q x { y .
GMmr , ƒrƒ3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12
i
Technology Application Projects
Mathematica / Maple Module Using Vectors to Represent Lines and Find Distances
You
Parts I and II: Learn the advantages of interpreting lines as vectors. Part III: Use vectors to find the distance from a point to a line.
Mathematica / Maple Module Putting a Scene in Three Dimensions onto a TwoDimensional Canvas Use the concept of planes in space to obtain a twodimensional image.
Mathematica / Maple Module
iaz
Getting Started in Plotting in 3D
Part I: Use the vector definition of lines and planes to generate graphs and equations, and to compare different forms for the equations of a single line.
Mu
ham
ma
dH
ass a
nR
Part II: Plot functions that are defined implicitly.
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Chapter 12
Technology Application Projects
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13
VECTORVALUED FUNCTIONS AND MOTION IN SPACE
You
Chapter
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
dH
ass a
nR
iaz
OVERVIEW When a body (or object) travels through space, the equations x = ƒstd, y = gstd, and z = hstd that give the body’s coordinates as functions of time serve as parametric equations for the body’s motion and path. With vector notation, we can condense these into a single equation rstd = ƒstdi + gstdj + hstdk that gives the body’s position as a vector function of time. For an object moving in the xyplane, the component function h(t) is zero for all time (that is, identically zero). In this chapter, we use calculus to study the paths, velocities, and accelerations of moving bodies. As we go along, we will see how our work answers the standard questions about the paths and motions of projectiles, planets, and satellites. In the final section, we use our new vector calculus to derive Kepler’s laws of planetary motion from Newton’s laws of motion and gravitation.
Vector Functions
13.1
ma
When a particle moves through space during a time interval I, we think of the particle’s coordinates as functions defined on I:
ham
z
r
O
x = ƒstd,
y = gstd,
z = hstd,
t H I.
(1)
The points sx, y, zd = sƒstd, gstd, hstdd, t H I, make up the curve in space that we call the particle’s path. The equations and interval in Equation (1) parametrize the curve. A curve in space can also be represented in vector form. The vector 1 rstd = OP = ƒstdi + gstdj + hstdk
(2)
P( f (t), g(t), h(t))
Mu
y
x
FIGURE 13.1 The position vector 1 r = OP of a particle moving through space is a function of time.
from the origin to the particle’s position P(ƒ(t), g(t), h(t)) at time t is the particle’s position vector (Figure 13.1). The functions ƒ, g, and h are the component functions (components) of the position vector. We think of the particle’s path as the curve traced by r during the time interval I. Figure 13.2 displays several space curves generated by a computer graphing program. It would not be easy to plot these curves by hand. Equation (2) defines r as a vector function of the real variable t on the interval I. More generally, a vector function or vectorvalued function on a domain set D is a rule that assigns a vector in space to each element in D. For now, the domains will be intervals of real numbers resulting in a space curve. Later, in Chapter 16, the domains will be regions
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1
Vector Functions
907
y
x
(a)
y r(t) (cos t)i (sin t)j (sin 2t)k
(b)
r(t) (4 sin20t)(cos t)i (4 sin20t)(sint)j (cos20t)k (c)
Computergenerated space curves are defined by the position vectors r(t).
ass a
FIGURE 13.2
y
x
nR
r(t) (sin 3t)(cos t)i (sin 3t)(sin t)j tk
iaz
x
z
You
z
z
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i
in the plane. Vector functions will then represent surfaces in space. Vector functions on a domain in the plane or space also give rise to “vector fields,” which are important to the study of the flow of a fluid, gravitational fields, and electromagnetic phenomena. We investigate vector fields and their applications in Chapter 16.
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We refer to realvalued functions as scalar functions to distinguish them from vector functions. The components of r are scalar functions of t. When we define a vectorvalued function by giving its component functions, we assume the vector function’s domain to be the common domain of the components. z
EXAMPLE 1
Graphing a Helix
2 t
P
t
x2 y2 1
Mu
t0
rstd = scos tdi + ssin tdj + tk.
The vector function rstd = scos tdi + ssin tdj + tk
is defined for all real values of t. The curve traced by r is a helix (from an old Greek word for “spiral”) that winds around the circular cylinder x 2 + y 2 = 1 (Figure 13.3). The curve lies on the cylinder because the i and jcomponents of r, being the x and ycoordinates of the tip of r, satisfy the cylinder’s equation:
t 2
r
0
(1, 0, 0)
Solution
ham
t 2
ma
Graph the vector function
x 2 + y 2 = scos td2 + ssin td2 = 1.
y
x
FIGURE 13.3 The upper half of the helix rstd = scos tdi + ssin tdj + t k (Example 1).
The curve rises as the kcomponent z = t increases. Each time t increases by 2p, the curve completes one turn around the cylinder. The equations x = cos t,
y = sin t,
z = t
parametrize the helix, the interval  q 6 t 6 q being understood. You will find more helices in Figure 13.4.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 908
Chapter 13: VectorValued Functions and Motion in Space z
z
x
x
x
y
y
r(t) (cos t)i (sin t)j tk
y
r(t) (cos t)i (sin t)j 0.3tk
iaz
r(t) (cos 5t)i (sin 5t)j tk
Helices drawn by computer.
Limits and Continuity
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FIGURE 13.4
You
suf
i
z
The way we define limits of vectorvalued functions is similar to the way we define limits of realvalued functions.
ass a
DEFINITION Limit of Vector Functions Let rstd = ƒstdi + gstdj + hstdk be a vector function and L a vector. We say that r has limit L as t approaches t0 and write lim rstd = L t:t0
dH
if, for every number P 7 0, there exists a corresponding number d 7 0 such that for all t 0 6 ƒ t  t0 ƒ 6 d
Q
ƒ rstd  L ƒ 6 P.
ma
If L = L1i + L2 j + L3k, then limt:t0 rstd = L precisely when lim ƒstd = L1,
t:t0
lim gstd = L2,
t:t0
and
lim hstd = L3 .
t:t0
Mu
ham
The equation lim rstd = a lim ƒstdbi + a lim gstdbj + a lim hstdbk
t:t0
t:t0
t:t0
t:t0
(3)
provides a practical way to calculate limits of vector functions.
EXAMPLE 2
Finding Limits of Vector Functions
If rstd = scos tdi + ssin tdj + tk, then lim rstd = a lim cos tbi + a lim sin tbj + a lim tbk
t:p>4
t:p>4
t:p>4
t:p>4
22 22 p i + j + k. 2 2 4 We define continuity for vector functions the same way we define continuity for scalar functions.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1
Vector Functions
909
You
suf
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DEFINITION Continuous at a Point A vector function r(t) is continuous at a point t = t0 in its domain if limt:t0 rstd = rst0 d. The function is continuous if it is continuous at every point in its domain.
From Equation (3), we see that r(t) is continuous at t = t0 if and only if each component function is continuous there.
EXAMPLE 3
Continuity of Space Curves
iaz
(a) All the space curves shown in Figures 13.2 and 13.4 are continuous because their component functions are continuous at every value of t in s  q , q d. (b) The function
nR
gstd = scos tdi + ssin tdj + :t;k
is discontinuous at every integer, where the greatest integer function :t; is discontinuous. z
P
ass a
Derivatives and Motion r(t , t) r(t) Q
Suppose that rstd = ƒstdi + gstdj + hstdk is the position vector of a particle moving along a curve in space and that ƒ, g, and h are differentiable functions of t. Then the difference between the particle’s positions at time t and time t + ¢t is
r(t) r(t , t)
¢r = rst + ¢td  rstd
dH
C
(Figure 13.5a). In terms of components,
O
y
(a)
= [ƒst + ¢tdi + gst + ¢tdj + hst + ¢tdk]
r'(t)
r(t , t) r(t) ,t Q
ham
P
ma
x z
r(t , t)
O
 [ƒstdi + gstdj + hstdk]
= [ƒst + ¢td  ƒstd]i + [gst + ¢td  gstd]j + [hst + ¢td  hstd]k.
As ¢t approaches zero, three things seem to happen simultaneously. First, Q approaches P along the curve. Second, the secant line PQ seems to approach a limiting position tangent to the curve at P. Third, the quotient ¢r>¢t (Figure 13.5b) approaches the limit
r(t)
C
¢r = rst + ¢td  rstd
y
lim ¢t:0
+ c lim
(b)
Mu
x
FIGURE 13.5 As ¢t : 0 , the point Q approaches the point P along the curve C. 1 In the limit, the vector PQ>¢t becomes the tangent vector r¿std .
ƒst + ¢td  ƒstd gst + ¢td  gstd ¢r = c lim di + c lim dj ¢t ¢t ¢t ¢t:0 ¢t:0
¢t:0
= c
hst + ¢td  hstd dk ¢t
dg dƒ dh di + c dj + c dk. dt dt dt
We are therefore led by past experience to the following definition.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 910
Chapter 13: VectorValued Functions and Motion in Space
dƒ dg rst + ¢td  rstd dr dh = lim = i + j + k. dt dt dt dt ¢t ¢t:0
You
r¿std =
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i
DEFINITION Derivative The vector function rstd = ƒstdi + gstdj + hstdk has a derivative (is differentiable) at t if ƒ, g, and h have derivatives at t. The derivative is the vector function
C2
C3
C4 C5
ma
FIGURE 13.6 A piecewise smooth curve made up of five smooth curves connected end to end in continuous fashion.
dH
C1
ass a
nR
iaz
A vector function r is differentiable if it is differentiable at every point of its domain. The curve traced by r is smooth if dr> dt is continuous and never 0, that is, if ƒ, g, and h have continuous first derivatives that are not simultaneously 0. The geometric significance of the definition of derivative is shown in Figure 13.5. 1 The points P and Q have position vectors r(t) and rst + ¢td, and the vector PQ is represented by rst + ¢td  rstd. For ¢t 7 0, the scalar multiple s1>¢tdsrst + ¢td  rstdd 1 points in the same direction as the vector PQ . As ¢t : 0, this vector approaches a vector that is tangent to the curve at P (Figure 13.5b). The vector r¿std, when different from 0, is defined to be the vector tangent to the curve at P. The tangent line to the curve at a point sƒst0 d, gst0 d, hst0 dd is defined to be the line through the point parallel to r¿st0 d. We require d r>dt Z 0 for a smooth curve to make sure the curve has a continuously turning tangent at each point. On a smooth curve, there are no sharp corners or cusps. A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth (Figure 13.6). Look once again at Figure 13.5. We drew the figure for ¢t positive, so ¢r points forward, in the direction of the motion. The vector ¢r>¢t, having the same direction as ¢r, points forward too. Had ¢t been negative, ¢r would have pointed backward, against the direction of motion. The quotient ¢r>¢t, however, being a negative scalar multiple of ¢r, would once again have pointed forward. No matter how ¢r points, ¢r>¢t points forward and we expect the vector dr>dt = lim¢t:0 ¢r>¢t, when different from 0, to do the same. This means that the derivative dr> dt is just what we want for modeling a particle’s velocity. It points in the direction of motion and gives the rate of change of position with respect to time. For a smooth curve, the velocity is never zero; the particle does not stop or reverse direction.
Mu
ham
DEFINITIONS Velocity, Direction, Speed, Acceleration If r is the position vector of a particle moving along a smooth curve in space, then vstd =
dr dt
is the particle’s velocity vector, tangent to the curve. At any time t, the direction of v is the direction of motion, the magnitude of v is the particle’s speed, and the derivative a = dv>dt, when it exists, is the particle’s acceleration vector. In summary, 1. 2. 3. 4.
dr . dt Speed = ƒ v ƒ . Speed is the magnitude of velocity: dv d 2r Acceleration is the derivative of velocity: a = = 2. dt dt The unit vector v> ƒ v ƒ is the direction of motion at time t. Velocity is the derivative of position:
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1
Vector Functions
911
Velocity = ƒ v ƒ a
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i
We can express the velocity of a moving particle as the product of its speed and direction: v b = sspeeddsdirectiond. ƒvƒ
z
EXAMPLE 4
Flight of a Hang Glider
You
In Section 12.5, Example 4 we found this expression for velocity useful in locating, for example, the position of a helicopter moving along a straight line in space. Now let’s look at an example of an object moving along a (nonlinear) space curve.
iaz
A person on a hang glider is spiraling upward due to rapidly rising air on a path having position vector rstd = s3 cos tdi + s3 sin tdj + t 2k. The path is similar to that of a helix (although it’s not a helix, as you will see in Section 13.4) and is shown in Figure 13.7 for 0 … t … 4p. Find
FIGURE 13.7 The path of a hang glider with position vector rstd = s3 cos tdi + s3 sin tdj + t 2k (Example 4).
nR
y
Solution
(a) r = s3 cos tdi + s3 sin tdj + t 2k
ass a
(3, 0, 0) x
(a) the velocity and acceleration vectors, (b) the glider’s speed at any time t, (c) the times, if any, when the glider’s acceleration is orthogonal to its velocity.
v =
dr = s3 sin tdi + s3 cos tdj + 2tk dt
d 2r = s3 cos tdi  s3 sin tdj + 2k dt 2 (b) Speed is the magnitude of v:
ma
dH
a =
ƒ vstd ƒ = 2s 3 sin td2 + s3 cos td2 + s2td2 = 29 sin2 t + 9 cos 2 t + 4t 2 = 29 + 4t 2 .
Mu
ham
The glider is moving faster and faster as it rises along its path. (c) To find the times when v and a are orthogonal, we look for values of t for which v # a = 9 sin t cos t  9 cos t sin t + 4t = 4t = 0.
Thus, the only time the acceleration vector is orthogonal to v is when t = 0. We study acceleration for motions along paths in more detail in Section 13.5. There we discover how the acceleration vector reveals the curving nature and tendency of the path to “twist” out of a certain plane containing the velocity vector.
Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 912
Chapter 13: VectorValued Functions and Motion in Space
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Differentiation Rules for Vector Functions Let u and v be differentiable vector functions of t, C a constant vector, c any scalar, and ƒ any differentiable scalar function. Constant Function Rule:
d C = 0 dt
2.
Scalar Multiple Rules:
d [custd] = cu¿std dt
You
1.
d [ƒstdustd] = ƒ¿stdustd + ƒstdu¿std dt d [ustd + vstd] = u¿std + v¿std dt
4.
Difference Rule:
d [ustd  vstd] = u¿std  v¿std dt
5.
Dot Product Rule:
d [ustd # vstd] = u¿std # vstd + ustd # v¿std dt
6.
Cross Product Rule:
7.
Chain Rule:
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Sum Rule:
d [ustd * vstd] = u¿std * vstd + ustd * v¿std dt d [usƒstdd] = ƒ¿stdu¿sƒstdd dt
ass a
dH
When you use the Cross Product Rule, remember to preserve the order of the factors. If u comes first on the left side of the equation, it must also come first on the right or the signs will be wrong.
3.
We will prove the product rules and Chain Rule but leave the rules for constants, scalar multiples, sums, and differences as exercises.
ma
Proof of the Dot Product Rule Suppose that u = u1stdi + u2stdj + u3stdk
and
v = y1stdi + y2stdj + y3stdk.
Mu
ham
Then
d d su # vd = su y + u2 y2 + u3 y3 d dt dt 1 1 = u 1œ y1 + u 2œ y2 + u 3œ y3 + u1y 1œ + u2y 2œ + u3y 3œ . (''')''''* ('''')'''* u¿ # v u # v¿
Proof of the Cross Product Rule We model the proof after the proof of the Product Rule for scalar functions. According to the definition of derivative, ust + hd * vst + hd  ustd * vstd d su * vd = lim . dt h h:0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1
Vector Functions
913
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To change this fraction into an equivalent one that contains the difference quotients for the derivatives of u and v, we subtract and add ustd * vst + hd in the numerator. Then d su * vd dt h:0
ust + hd * vst + hd  ustd * vst + hd + ustd * vst + hd  ustd * vstd h
= lim c h:0
You
= lim
vst + hd  vstd ust + hd  ustd * vst + hd + ustd * d h h
ust + hd  ustd vst + hd  vstd * lim vst + hd + lim ustd * lim . h h h:0 h:0 h:0 h:0
= lim
iaz
The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 52). As h approaches zero, vst + hd approaches v(t) because v, being differentiable at t, is continuous at t (Exercise 53). The two fractions approach the values of du> dt and dv> dt at t. In short,
d da db dc [ussd] = i + j + k dt dt dt dt da ds db ds dc ds = i + j + k ds dt ds dt ds dt
du d u ds = , dt ds dt
dH
where s = ƒstd .
P
=
ds da db dc a i + j + kb dt ds ds ds
=
ds du dt ds
= ƒ¿stdu¿sƒstdd.
s = ƒstd
Vector Functions of Constant Length
y
When we track a particle moving on a sphere centered at the origin (Figure 13.8), the position vector has a constant length equal to the radius of the sphere. The velocity vector dr> dt, tangent to the path of motion, is tangent to the sphere and hence perpendicular to r. This is always the case for a differentiable vector function of constant length: The vector and its first derivative are orthogonal. With the length constant, the change in the function is a change in direction only, and direction changes take place at right angles. We can also obtain this result by direct calculation:
Mu
x
ham
r(t)
ma
z
dr dt
Proof of the Chain Rule Suppose that ussd = assdi + bssdj + cssdk is a differentiable vector function of s and that s = ƒstd is a differentiable scalar function of t. Then a, b, and c are differentiable functions of t, and the Chain Rule for differentiable realvalued functions gives
ass a
As an algebraic convenience, we sometimes write the product of a scalar c and a vector v as vc instead of cv. This permits us, for instance, to write the Chain Rule in a familiar form:
nR
dv du d * v + u * . su * vd = dt dt dt
FIGURE 13.8 If a particle moves on a sphere in such a way that its position r is a differentiable function of time, then r # sd r>dtd = 0 .
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rstd # rstd = c2 d [rstd # rstd] = 0 dt r¿std # rstd + rstd # r¿std = 0
ƒ rstd ƒ = c is constant. Differentiate both sides. Rule 5 with rstd = ustd = vstd
2r¿std # rstd = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 914
Chapter 13: VectorValued Functions and Motion in Space
suf
i
The vectors r œ std and r(t) are orthogonal because their dot product is 0. In summary,
If r is a differentiable vector function of t of constant length, then dr = 0. dt
(4)
You
r#
We will use this observation repeatedly in Section 13.4.
EXAMPLE 5
Supporting Equation (4)
iaz
Show that rstd = ssin tdi + scos tdj + 23k has constant length and is orthogonal to its derivative. Solution
rstd = ssin tdi + scos tdj + 23k
dr = scos tdi  ssin tdj dt
dr = sin t cos t  sin t cos t = 0 dt
ass a
r#
A 13 B 2 = 21 + 3 = 2
nR
ƒ rstd ƒ = 2ssin td2 + scos td2 +
Integrals of Vector Functions
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A differentiable vector function R(t) is an antiderivative of a vector function r(t) on an interval I if dR>dt = r at each point of I. If R is an antiderivative of r on I, it can be shown, working one component at a time, that every antiderivative of r on I has the form R + C for some constant vector C (Exercise 56). The set of all antiderivatives of r on I is the indefinite integral of r on I.
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DEFINITION Indefinite Integral The indefinite integral of r with respect to t is the set of all antiderivatives of r, denoted by 1 rstd dt. If R is any antiderivative of r, then L
rstd dt = Rstd + C.
The usual arithmetic rules for indefinite integrals apply.
EXAMPLE 6 L
Finding Indefinite Integrals
sscos tdi + j  2tkd dt = a
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L
cos t dtbi + a
L
dtbj  a
L
2t dtb k
= ssin t + C1 di + st + C2 dj  st 2 + C3 dk = ssin tdi + tj  t 2k + C
C = C1i + C2 j  C3k
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(5) (6)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1
Vector Functions
915
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As in the integration of scalar functions, we recommend that you skip the steps in Equations (5) and (6) and go directly to the final form. Find an antiderivative for each component and add a constant vector at the end.
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Definite integrals of vector functions are best defined in terms of components.
DEFINITION Definite Integral If the components of rstd = ƒstdi + gstdj + hstdk are integrable over [a, b], then so is r, and the definite integral of r from a to b is La
EXAMPLE 7 L0
b
La
ƒstd dtbi + a
b
La
gstd dtbj + a
b
La
hstd dtbk.
Evaluating Definite Integrals
sscos tdi + j  2tkd dt = a
p
cos t dtbi + a
nR
p
rstd dt = a
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b
L0
L0
p
dtbj  a
= C sin t D 0 i + C t D 0 j  C t 2 D 0 k p
p
p
L0
2t dtbk
p
ass a
= [0  0]i + [p  0]j  [p 2  0 2]k = pj  p 2k
The Fundamental Theorem of Calculus for continuous vector functions says that rstd dt = Rstd D a = Rsbd  Rsad
b
b
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where R is any antiderivative of r, so that R¿std = rstd (Exercise 57).
EXAMPLE 8
Revisiting the Flight of a Glider
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Suppose that we did not know the path of the glider in Example 4, but only its acceleration vector astd = s3 cos tdi  s3 sin tdj + 2k. We also know that initially (at time t = 0), the glider departed from the point (3, 0, 0) with velocity vs0d = 3j. Find the glider’s position as a function of t.
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Solution
Our goal is to find r(t) knowing The differential equation: The initial conditions:
d 2r = s3 cos tdi  s3 sin tdj + 2k dt 2 vs0d = 3j and rs0d = 3i + 0j + 0k. a =
Integrating both sides of the differential equation with respect to t gives vstd = s3 sin tdi + s3 cos tdj + 2tk + C1. We use vs0d = 3j to find C1 :
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3j = s3 sin 0di + s3 cos 0dj + s0dk + C1 3j = 3j + C1 C1 = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 916
Chapter 13: VectorValued Functions and Motion in Space
Integrating both sides of this last differential equation gives
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rstd = s3 cos tdi + s3 sin tdj + t 2k + C2.
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dr = vstd = s3 sin tdi + s3 cos tdj + 2tk. dt
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The gliderâ€™s velocity as a function of time is
We then use the initial condition rs0d = 3i to find C2 :
The gliderâ€™s position as a function of t is
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3i = s3 cos 0di + s3 sin 0dj + s0 2 dk + C2 3i = 3i + s0dj + s0dk + C2 C2 = 0.
rstd = s3 cos tdi + s3 sin tdj + t 2k.
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This is the path of the glider we know from Example 4 and is shown in Figure 13.7.
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Note: It was peculiar to this example that both of the constant vectors of integration, C1 and C2, turned out to be 0. Exercises 31 and 32 give different results for these constants.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
916
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Chapter 13: VectorValued Functions and Motion in Space
EXERCISES 13.1 Motion in the xyplane
7. Motion on the cycloid x = t  sin t, y = 1  cos t
In Exercises 1–4, r(t) is the position of a particle in the xyplane at time t. Find an equation in x and y whose graph is the path of the particle. Then find the particle’s velocity and acceleration vectors at the given value of t. 1. rstd = st + 1di + st 2  1dj, 2. rstd = st 2 + 1di + s2t  1dj, 2 3. rstd = e i + e 2t j, 9 t
t = 1>2
H d
Exercises 5–8 give the position vectors of particles moving along various curves in the xyplane. In each case, find the particle’s velocity and acceleration vectors at the stated times and sketch them as vectors on the curve.
a m
m a h
rstd = ssin tdi + scos tdj;
t = p>4 and p>2
6. Motion on the circle x 2 + y 2 = 16
u M
t t rstd = a4 cos bi + a4 sin bj; 2 2
R n
t = p and 3p>2
8. Motion on the parabola y = x 2 + 1 rstd = t i + st 2 + 1dj;
t = 1, 0, and 1
Velocity and Acceleration in Space
t = 0
5. Motion on the circle x 2 + y 2 = 1
rstd = st  sin tdi + s1  cos tdj;
a s as
t = 1
t = ln 3
4. rstd = scos 2tdi + s3 sin 2tdj,
ai z
u o Y
t = p and 3p>2
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In Exercises 9–14, r(t) is the position of a particle in space at time t. Find the particle’s velocity and acceleration vectors. Then find the particle’s speed and direction of motion at the given value of t. Write the particle’s velocity at that time as the product of its speed and direction. 9. rstd = st + 1di + st 2  1dj + 2t k, 2
t = 1
3
t t j + k, t = 1 3 22 11. rstd = s2 cos tdi + s3 sin tdj + 4t k, t = p>2 10. rstd = s1 + tdi +
4 t k, t = p>6 3 t2 13. rstd = s2 ln st + 1ddi + t 2 j + k, t = 1 2 12. rstd = ssec tdi + stan tdj +
14. rstd = se t di + s2 cos 3tdj + s2 sin 3tdk,
t = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1 Vector Functions
Initial conditions:
15. rstd = s3t + 1di + 23t j + t 2k
dr ` = 8i + 8j dt t = 0
22 22 tbi + a t  16t 2 b j 2 2
32. Differential equation:
17. rstd = sln st 2 + 1ddi + stan1 tdj + 2t 2 + 1 k
Initial conditions:
4 4 1 s1 + td3>2 i + s1  td3>2 j + t k 9 9 3
dr ` = 0 dt t = 0
In Exercises 19 and 20, r(t) is the position vector of a particle in space at time t. Find the time or times in the given time interval when the velocity and acceleration vectors are orthogonal. 19. rstd = st  sin tdi + s1  cos tdj, 20. rstd = ssin tdi + tj + scos tdk,
0 … t … 2p
t Ú 0
33. rstd = ssin tdi + st 2  cos tdj + e t k,
Evaluate the integrals in Exercises 21–26. [t 3i + 7j + st + 1dk] dt
L0
4 cs6  6tdi + 3 2t j + a 2 b k d dt t L1 p>4
23.
Lp>4
[ssin tdi + s1 + cos tdj + ssec2 tdk] dt
p>3
[ssec t tan tdi + stan tdj + s2 sin t cos tdk] dt
L0 4
25.
L1 1
L0
c
1 1 j + k d dt 5  t 2t
2 21  t 2
i +
23 k d dt 1 + t2
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26.
1 ct i +
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24.
t0 = 4p t0 = 2p p 36. rstd = scos tdi + ssin tdj + ssin 2tdk, t0 = 2
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2
22.
t0 = 0
34. rstd = s2 sin tdi + s2 cos tdj + 5t k, 35. rstd = sa sin tdi + sa cos tdj + bt k,
1
21.
As mentioned in the text, the tangent line to a smooth curve rstd = ƒstdi + gstdj + hstdk at t = t0 is the line that passes through the point sƒst0 d, gst0 d, hst0 dd parallel to vst0 d , the curve’s velocity vector at t0 . In Exercises 33–36, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0 .
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Integrating VectorValued Functions
Tangent Lines to Smooth Curves
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18. rstd =
d 2r = si + j + kd dt 2 rs0d = 10i + 10j + 10k and
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16. rstd = a
i
d 2r = 32k dt 2 rs0d = 100k and
31. Differential equation:
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In Exercises 15–18, r(t) is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t = 0.
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Motion on Circular Paths 37. Each of the following equations in parts (a)–(e) describes the motion of a particle having the same path, namely the unit circle x 2 + y 2 = 1 . Although the path of each particle in parts (a)–(e) is the same, the behavior, or “dynamics,” of each particle is different. For each particle, answer the following questions. i.
Does the particle have constant speed? If so, what is its constant speed?
ii.
Is the particle’s acceleration vector always orthogonal to its velocity vector?
iii. Does the particle move clockwise or counterclockwise around the circle?
Solve the initial value problems in Exercises 27–32 for r as a vector function of t.
iv.
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Initial Value Problems for VectorValued Functions
27. Differential equation: Initial condition:
28. Differential equation:
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Initial condition:
29. Differential equation: Initial condition:
30. Differential equation: Initial condition:
Does the particle begin at the point (1, 0)?
a. rstd = scos tdi + ssin tdj,
dr = t i  t j  t k dt rs0d = i + 2j + 3k
b. rstd = cos s2tdi + sin s2tdj,
dr = s180tdi + s180t  16t 2 dj dt rs0d = 100j
e. rstd = cos st 2 di + sin st 2 dj,
3 dr 1 k = st + 1d1>2 i + e t j + 2 t + 1 dt rs0d = k dr = st 3 + 4tdi + t j + 2t 2 k dt rs0d = i + j
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t Ú 0 t Ú 0
c. rstd = cos st  p>2di + sin st  p>2dj, d. rstd = scos tdi  ssin tdj,
t Ú 0
t Ú 0 t Ú 0
38. Show that the vectorvalued function rstd = s2i + 2j + kd 1 1 1 1 1 i j ≤ + sin t ¢ i + j + k≤ 22 22 23 23 23 describes the motion of a particle moving in the circle of radius 1 centered at the point (2, 2, 1) and lying in the plane x + y  2z = 2 . + cos t ¢
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
40. A particle traveling in a straight line is located at the point s1, 1, 2d and has speed 2 at time t = 0 . The particle moves toward the point (3, 0, 3) with constant acceleration 2i + j + k. Find its position vector r(t) at time t.
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39. At time t = 0 , a particle is located at the point (1, 2, 3). It travels in a straight line to the point (4, 1, 4), has speed 2 at (1, 2, 3) and constant acceleration 3i  j + k. Find an equation for the position vector r(t) of the particle at time t.
Let r(t) be the satellite’s position vector at time t. Show that u = yt>r0 and hence that
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Motion Along a Straight Line
yt yt rstd = ar0 cos r0 bi + ar0 sin r0 bj. b. Find the acceleration of the satellite.
c. According to Newton’s law of gravitation, the gravitational force exerted on the satellite is directed toward M and is given by
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F = a
Theory and Examples
rstd = st  sin tdi + s1  cos tdj. T
43. Motion along an ellipse A particle moves around the ellipse sy>3d2 + sz>2d2 = 1 in the yzplane in such a way that its position at time t is
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rstd = s3 cos tdj + s2 sin tdk.
Find the maximum and minimum values of ƒ v ƒ and ƒ a ƒ . (Hint: Find the extreme values of ƒ v ƒ 2 and ƒ a ƒ 2 first and take square roots later.)
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44. A satellite in circular orbit A satellite of mass m is revolving at a constant speed y around a body of mass M (Earth, for example) in a circular orbit of radius r0 (measured from the body’s center of mass). Determine the satellite’s orbital period T (the time to complete one full orbit), as follows:
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a. Coordinatize the orbital plane by placing the origin at the body’s center of mass, with the satellite on the xaxis at t = 0 and moving counterclockwise, as in the accompanying figure. y
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r0
e. From parts (c) and (d), deduce that T2 =
4p2 3 r . GM 0
45. Let v be a differentiable vector function of t. Show that if v # sdv>dtd = 0 for all t, then ƒ v ƒ is constant.
ass a
b. Find the maximum and minimum values of ƒ v ƒ and ƒ a ƒ . (Hint: Find the extreme values of ƒ v ƒ 2 and ƒ a ƒ 2 first and take square roots later.)
M
d. Show that the orbital period T satisfies yT = 2pr0 .
That is, the square of the period of a satellite in circular orbit is proportional to the cube of the radius from the orbital center.
a. Graph r(t). The resulting curve is a cycloid.
r(t)
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42. Motion along a cycloid A particle moves in the xyplane in such a way that its position at time t is
where G is the universal constant of gravitation. Using Newton’s second law, F = ma, show that y 2 = GM>r0 .
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41. Motion along a parabola A particle moves along the top of the parabola y 2 = 2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point (2, 2).
GmM r b , r 02 r0
m t0 x
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46. Derivatives of triple scalar products a. Show that if u, v, and w are differentiable vector functions of t, then dv du # d su # v * wd = v * w + u# * w + dt dt dt (7) dw u#v * . dt b. Show that Equation (7) is equivalent to
u1 d 3 y1 dt w1
u2 y2 w2
du1 u3 dt 4 y3 3 = y1 w3 w1
du2 dt y2 w2
du3 dt 4 y3 w3
u1 dy1 + 4 dt w1
u2 dy2 dt w2
u3 dy3 4 dt w3
u1 y 1 + 4 dw1 dt
u2 y2 dw2 dt
u3 y3 4 . dw3 dt
(8)
Equation (8) says that the derivative of a 3 by 3 determinant of differentiable functions is the sum of the three determinants obtained from the original by differentiating one row at a time. The result extends to determinants of any order.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.1 Vector Functions
dr dr d d r d r ar # * 2b = r# a * 3b. dt dt dt dt dt 2
b
b
s rstdd dt = 
La
3
(9)
is obtained by taking k = 1 .
48. Constant Function Rule Prove that if u is the vector function with the constant value C, then du>dt = 0.
b
La
rstd dt ,
b
sr1std ; r2stdd dt =
La
b
r1std dt ;
La
r2std dt
c. The Constant Vector Multiple Rules:
49. Scalar Multiple Rules
b
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b
C # rstd dt = C #
La
rstd dt
sany constant vector Cd
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a. Prove that if u is a differentiable function of t and c is any real number, then
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b. The Sum and Difference Rules: (Hint: Differentiate on the left and look for vectors whose products are zero.)
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The Rule for Negatives,
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47. (Continuation of Exercise 46.) Suppose that rstd = ƒstdi + gstdj + hstdk and that ƒ, g, and h have derivatives through order three. Use Equation (7) or (8) to show that
919
and
dsc ud du = c . dt dt
b
b. Prove that if u is a differentiable function of t and ƒ is a differentiable scalar function of t, then
rstd dt
sany constant vector Cd
55. Products of scalar and vector functions Suppose that the scalar function u(t) and the vector function r(t) are both defined for a … t … b . a. Show that ur is continuous on [a, b] if u and r are continuous on [a, b].
ass a
dƒ du d sƒud = u + ƒ . dt dt dt
La
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La
b
C * rstd dt = C *
50. Sum and Difference Rules Prove that if u and v are differentiable functions of t, then
b. If u and r are both differentiable on [a, b], show that ur is differentiable on [a, b] and that
du dv d su + vd = + dt dt dt
dr du d surd = u + r . dt dt dt
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and
du dv d su  vd = . dt dt dt
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51. Component Test for Continuity at a Point Show that the vector function r defined by rstd = ƒstdi + g stdj + hstdk is continuous at t = t0 if and only if ƒ, g, and h are continuous at t0 .
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52. Limits of cross products of vector functions Suppose that r1std = ƒ1stdi + ƒ2stdj + ƒ3stdk, r2std = g1stdi + g2stdj + g3stdk, lim t:t0 r1std = A, and lim t:t0 r2std = B. Use the determinant formula for cross products and the Limit Product Rule for scalar functions to show that
56. Antiderivatives of vector functions a. Use Corollary 2 of the Mean Value Theorem for scalar functions to show that if two vector functions R1std and R2std have identical derivatives on an interval I, then the functions differ by a constant vector value throughout I. b. Use the result in part (a) to show that if R(t) is any antiderivative of r(t) on I, then any other antiderivative of r on I equals Rstd + C for some constant vector C. 57. The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus for scalar functions of a real variable holds for vector functions of a real variable as well. Prove this by using the theorem for scalar functions to show first that if a vector function r(t) is continuous for a … t … b , then
lim sr1std * r2stdd = A * B
t: t0
t
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53. Differentiable vector functions are continuous Show that if rstd = ƒstdi + gstdj + hstdk is differentiable at t = t0 , then it is continuous at t0 as well. 54. Establish the following properties of integrable vector functions. a. The Constant Scalar Multiple Rule: b
d rstd dt = rstd dt La at every point t of (a, b). Then use the conclusion in part (b) of Exercise 56 to show that if R is any antiderivative of r on [a, b] then
b
k rstd dt = k r std dt La La
b
sany scalar kd
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La
rstd dt = Rsbd  Rsad .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space In Exercises 62 and 63, you will explore graphically the behavior of the helix
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COMPUTER EXPLORATIONS
Drawing Tangents to Space Curves
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920
Use a CAS to perform the following steps in Exercises 58–61.
rstd = scos atdi + ssin atdj + bt k.
a. Plot the space curve traced out by the position vector r.
as you change the values of the constants a and b. Use a CAS to perform the steps in each exercise.
d. Plot the tangent line together with the curve over the given interval. 58. rstd = ssin t  t cos tdi + scos t + t sin tdj + t 2k, 0 … t … 6p, t0 = 3p>2 59. rstd = 22t i + e t j + e t k,
2 … t … 3,
60. rstd = ssin 2tdi + sln s1 + tddj + t k, t0 = p>4
t0 = 1
0 … t … 4p,
63. Set a = 1 . Plot the helix r(t) together with the tangent line to the curve at t = 3p>2 for b = 1>4, 1>2, 2 , and 4 over the interval 0 … t … 4p . Describe in your own words what happens to the graph of the helix and the position of the tangent line as b increases through these positive values.
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ass a
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61. rstd = sln st 2 + 2ddi + stan1 3tdj + 2t 2 + 1 k, 3 … t … 5, t0 = 3
62. Set b = 1 . Plot the helix r(t) together with the tangent line to the curve at t = 3p>2 for a = 1, 2, 4 , and 6 over the interval 0 … t … 4p . Describe in your own words what happens to the graph of the helix and the position of the tangent line as a increases through these positive values.
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c. Evaluate dr> dt at the given point t0 and determine the equation of the tangent line to the curve at rst0 d .
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b. Find the components of the velocity vector dr> dt.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 920
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Chapter 13: VectorValued Functions and Motion in Space
13.2
Modeling Projectile Motion
u o Y z a i R
When we shoot a projectile into the air we usually want to know beforehand how far it will go (will it reach the target?), how high it will rise (will it clear the hill?), and when it will land (when do we get results?). We get this information from the direction and magnitude of the projectile’s initial velocity vector, using Newton’s second law of motion.
The Vector and Parametric Equations for Ideal Projectile Motion
n a s s a H
To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. In practice, none of these assumptions really holds. The ground moves beneath the projectile as the earth turns, the air creates a frictional force that varies with the projectile’s speed and altitude, and the force of gravity changes as the projectile moves along. All this must be taken into account by applying corrections to the predictions of the ideal equations we are about to derive. The corrections, however, are not the subject of this section. We assume that the projectile is launched from the origin at time t = 0 into the first quadrant with an initial velocity v0 (Figure 13.9). If v0 makes an angle a with the horizontal, then
d a m
m a h
u M
v0 = s ƒ v0 ƒ cos adi + s ƒ v0 ƒ sin adj.
If we use the simpler notation y0 for the initial speed ƒ v0 ƒ , then v0 = sy0 cos adi + sy0 sin adj.
(1)
The projectile’s initial position is
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r0 = 0i + 0j = 0.
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(2)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.2 y
921
Modeling Projectile Motion
m v0 sin j x
d 2r = g j dt 2
Differential equation: Initial conditions:
r = r0
and
dr = v0 dt
when t = 0
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(a)
The first integration gives
dr = sgtdj + v0 . dt
y v
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(x, y)
A second integration gives
a –gj
r = 
r x i yj
1 2 gt j + v0 t + r0 . 2
Substituting the values of v0 and r0 from Equations (1) and (2) gives x
R
ass a
0
d 2r = gj. dt 2
We find r as a function of t by solving the following initial value problem.
v0 cos i r 0 at time t 0 a –gj
and
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v0
d 2r = mgj dt 2
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Newton’s second law of motion says that the force acting on the projectile is equal to the projectile’s mass m times its acceleration, or msd 2r>dt 2 d if r is the projectile’s position vector and t is time. If the force is solely the gravitational force mgj, then
r = 
Horizontal range (b)
v0 t
Collecting terms, we have
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FIGURE 13.9 (a) Position, velocity, acceleration, and launch angle at t = 0 . (b) Position, velocity, and acceleration at a later time t.
1 2 gt j + sy0 cos adti + sy0 sin adtj + 0 2 ('''''')''''''*
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Ideal Projectile Motion Equation r = sy0 cos adti + asy0 sin adt 
1 2 gt bj. 2
(3)
Equation (3) is the vector equation for ideal projectile motion. The angle a is the projectile’s launch angle (firing angle, angle of elevation), and y0 , as we said before, is the projectile’s initial speed. The components of r give the parametric equations x = sy0 cos adt
and
y = sy0 sin adt 
1 2 gt , 2
(4)
where x is the distance downrange and y is the height of the projectile at time t Ú 0.
EXAMPLE 1
Firing an Ideal Projectile
A projectile is fired from the origin over horizontal ground at an initial speed of 500 m> sec and a launch angle of 60°. Where will the projectile be 10 sec later?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 922
Chapter 13: VectorValued Functions and Motion in Space
We use Equation (3) with y0 = 500, a = 60°, g = 9.8, and t = 10 to find the projectile’s components 10 sec after firing. r = sy0 cos adti + asy0 sin adt 
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Solution
1 2 gt bj 2
You
23 1 1 = s500d a bs10di + as500d a b10  a bs9.8ds100dbj 2 2 2 L 2500i + 3840j.
Height, Flight Time, and Range
iaz
Ten seconds after firing, the projectile is about 3840 m in the air and 2500 m downrange.
nR
Equation (3) enables us to answer most questions about the ideal motion for a projectile fired from the origin. The projectile reaches its highest point when its vertical velocity component is zero, that is, when or
t =
y0 sin a g .
ass a
dy = y0 sin a  gt = 0, dt For this value of t, the value of y is ymax = sy0 sin ad a
sy0 sin ad2 y0 sin a y0 sin a 2 1 b g a b = . g g 2 2g
Mu
ham
ma
dH
To find when the projectile lands when fired over horizontal ground, we set the vertical component equal to zero in Equation (3) and solve for t. sy0 sin adt 
1 2 gt = 0 2
t ay0 sin a 
1 gtb = 0 2 t = 0,
t =
2y0 sin a g
Since 0 is the time the projectile is fired, s2y0 sin ad>g must be the time when the projectile strikes the ground. To find the projectile’s range R, the distance from the origin to the point of impact on horizontal ground, we find the value of the horizontal component when t = s2y0 sin ad>g. x = sy0 cos adt R = sy0 cos ad a
2y0 sin a y02 y02 b = s2 sin a cos ad = g g g sin 2a
The range is largest when sin 2a = 1 or a = 45°.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.2
Modeling Projectile Motion
923
ymax =
Flight time:
t =
2y0 sin a g
y02 R = g sin 2a.
iaz
Range:
EXAMPLE 2
sy0 sin ad2 2g
You
Maximum height:
suf
i
Height, Flight Time, and Range for Ideal Projectile Motion For ideal projectile motion when an object is launched from the origin over a horizontal surface with initial speed y0 and launch angle a:
Investigating Ideal Projectile Motion
nR
Find the maximum height, flight time, and range of a projectile fired from the origin over horizontal ground at an initial speed of 500 m> sec and a launch angle of 60º (same projectile as Example 1). Solution
ass a
Maximum height: ymax =
dH
Flight time:
ma
y 10,000
0 –2000
ham
6000 2000
t
R
2s500d sin 60° L 88.4 sec 9.8
y02 = g sin 2a =
s500d2 sin 120° L 22,092 m 9.8
From Equation (3), the position vector of the projectile is
8000 4000
s500 sin 60°d2 L 9566 m 2s9.8d 2y0 sin a = g =
=
Range:
sy0 sin ad2 2g
5000 10,000 15,000 20,000
Mu
FIGURE 13.10 The graph of the projectile described in Example 2.
r = sy0 cos adti + asy0 sin adt 
1 2 gt bj 2
= s500 cos 60°dti + as500 sin 60°dt 
x
= 250ti +
A A 25023 B t  4.9t 2 B j.
1 s9.8dt 2 bj 2
A graph of the projectile’s path is shown in Figure 13.10.
Ideal Trajectories Are Parabolic It is often claimed that water from a hose traces a parabola in the air, but anyone who looks closely enough will see this is not so. The air slows the water down, and its forward progress is too slow at the end to keep pace with the rate at which it falls.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 924
Chapter 13: VectorValued Functions and Motion in Space
y
y = a
v0 â?Ł
g 2y02
cos2 a
b x 2 + stan adx.
suf
i
What is really being claimed is that ideal projectiles move along parabolas, and this we can see from Equations (4). If we substitute t = x>sy0 cos ad from the first equation into the second, we obtain the Cartesiancoordinate equation
FIGURE 13.11 The path of a projectile fired from sx0 , y0 d with an initial velocity v0 at an angle of a degrees with the horizontal.
Firing from sx0, y0 d
If we fire our ideal projectile from the point sx0, y0 d instead of the origin (Figure 13.11), the position vector for the path of motion is r = sx0 + sy0 cos adtdi + ay0 + sy0 sin adt 
EXAMPLE 3
1 2 gt bj, 2
(5)
nR
as you are asked to show in Exercise 19.
iaz
x
0
You
This equation has the form y = ax 2 + bx, so its graph is a parabola.
(x 0 , y0 )
Firing a Flaming Arrow
Mu
ham
ma
dH
ass a
To open the 1992 Summer Olympics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olympic torch with a flaming arrow (Figure 13.12). Suppose that Rebollo shot the arrow at a height of 6 ft above ground level 90 ft from the 70fthigh cauldron, and he wanted the arrow to reach maximum height exactly 4 ft above the center of the cauldron (Figure 13.12).
Photograph is not available.
FIGURE 13.12 Spanish archer Antonio Rebollo lights the Olympic torch in Barcelona with a flaming arrow.
(a) Express ymax in terms of the initial speed y0 and firing angle a. (b) Use ymax = 74 ft (Figure 13.13) and the result from part (a) to find the value of y0 sin a. (c) Find the value of y0 cos a. (d) Find the initial firing angle of the arrow.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.2
Modeling Projectile Motion
925
i
Solution
ymax 74'
â?Ł
6'
90'
0
y = y0 + sy0 sin adt 
1 2 gt 2
= 6 + sy0 sin adt 
1 2 gt . 2
NOT TO SCALE
y0 = 6
We find the time when the arrow reaches its highest point by setting dy>dt = 0 and solving for t, obtaining
iaz
FIGURE 13.13 Ideal path of the arrow that lit the Olympic torch (Example 3).
Equation (5), jcomponent
You
x
(a) We use a coordinate system in which the positive xaxis lies along the ground toward the left (to match the second photograph in Figure 13.12) and the coordinates of the flaming arrow at t = 0 are x0 = 0 and y0 = 6 (Figure 13.13). We have
y
suf
v0
y0 sin a g .
t =
For this value of t, the value of y is
y0 sin a y0 sin a 2 1 g b  2 ga g b
nR
ymax = 6 + sy0 sin ad a
sy0 sin ad2 . 2g
ass a
= 6 +
(b) Using ymax = 74 and g = 32, we see from the preceeding equation in part (a) that 74 = 6 +
sy0 sin ad2 2s32d
dH
or
y0 sin a = 2s68ds64d.
Mu
ham
ma
(c) When the arrow reaches ymax , the horizontal distance traveled to the center of the cauldron is x = 90 ft. We substitute the time to reach ymax from part (a) and the horizontal distance x = 90 ft into the icomponent of Equation (5) to obtain x = x0 + sy0 cos adt
Equation (5), icomponent
90 = 0 + sy0 cos adt = sy0 cos ad a
x = 90, x0 = 0
y0 sin a g b.
t = sy0 sin ad>g
Solving this equation for y0 cos a and using g = 32 and the result from part (b), we have y0 cos a =
90g s90ds32d = . y0 sin a 2s68ds64d
(d) Parts (b) and (c) together tell us that
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y0 sin a tan a = y0 cos a =
A 2s68ds64d B 2 s90ds32d
=
68 45
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 926
Chapter 13: VectorValued Functions and Motion in Space
68 b L 56.5°. 45
You
This is Rebollo’s firing angle.
suf
a = tan1 a
i
or
Projectile Motion with Wind Gusts
The next example shows how to account for another force acting on a projectile. We also assume that the path of the baseball in Example 4 lies in a vertical plane.
Hitting a Baseball
iaz
EXAMPLE 4
nR
A baseball is hit when it is 3 ft above the ground. It leaves the bat with initial speed of 152 ft> sec, making an angle of 20° with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of 8.8i sft>secd to the ball’s initial velocity s8.8 ft>sec = 6 mphd.
Solution
ass a
(a) Find a vector equation (position vector) for the path of the baseball. (b) How high does the baseball go, and when does it reach maximum height? (c) Assuming that the ball is not caught, find its range and flight time.
dH
(a) Using Equation (1) and accounting for the gust of wind, the initial velocity of the baseball is v0 = sy0 cos adi + sy0 sin adj  8.8i = s152 cos 20°di + s152 sin 20°dj  s8.8di = s152 cos 20°  8.8di + s152 sin 20°dj.
Mu
ham
ma
The initial position is r0 = 0i + 3j. Integration of d 2r>dt 2 = gj gives dr = sgtdj + v0 . dt
A second integration gives r = 
1 2 gt j + v0 t + r0 . 2
Substituting the values of v0 and r0 into the last equation gives the position vector of the baseball. r = 
1 2 gt j + v0 t + r0 2
= 16t 2j + s152 cos 20°  8.8dti + s152 sin 20°dtj + 3j
= s152 cos 20°  8.8dti + A 3 + (152 sin 20°dt  16t 2 B j.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.2 Modeling Projectile Motion
927
suf
i
(b) The baseball reaches its highest point when the vertical component of velocity is zero, or dy = 152 sin 20°  32t = 0. dt
t =
You
Solving for t we find
152 sin 20° L 1.62 sec. 32
Substituting this time into the vertical component for r gives the maximum height
iaz
ymax = 3 + s152 sin 20°ds1.62d  16s1.62d2 L 45.2 ft.
nR
That is, the maximum height of the baseball is about 45.2 ft, reached about 1.6 sec after leaving the bat. (c) To find when the baseball lands, we set the vertical component for r equal to 0 and solve for t: 3 + s152 sin 20°dt  16t 2 = 0
ass a
3 + s51.99dt  16t 2 = 0.
The solution values are about t = 3.3 sec and t = 0.06 sec. Substituting the positive time into the horizontal component for r, we find the range R = s152 cos 20°  8.8ds3.3d
dH
L 442 ft.
Thus, the horizontal range is about 442 ft, and the flight time is about 3.3 sec.
Mu
ham
ma
In Exercises 29 through 31, we consider projectile motion when there is air resistance slowing down the flight.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
u o zY
13.2 Modeling Projectile Motion
EXERCISES 13.2 Projectile flights in the following exercises are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal surface unless stated otherwise.
s s Ha
1. Travel time A projectile is fired at a speed of 840 m> sec at an angle of 60째. How long will it take to get 21 km downrange?
d a m
2. Finding muzzle speed Find the muzzle speed of a gun whose maximum range is 24.5 km.
m a h u M
a i R an
3. Flight time and height A projectile is fired with an initial speed of 500 m> sec at an angle of elevation of 45째. a. When and how far away will the projectile strike?
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927
b. How high overhead will the projectile be when it is 5 km downrange? c. What is the greatest height reached by the projectile?
4. Throwing a baseball A baseball is thrown from the stands 32 ft above the field at an angle of 30째 up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 32 ft> sec?
5. Shot put An athlete puts a 16lb shot at an angle of 45째 to the horizontal from 6.5 ft above the ground at an initial speed of 44 ft> sec as suggested in the accompanying figure. How long after launch and how far from the inner edge of the stopboard does the shot land?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 928
Chapter 13: VectorValued Functions and Motion in Space y
i
elevated 45 ft above the tee as shown in the diagram. Assuming that the pin, 369 ft downrange, does not get in the way, where will the ball land in relation to the pin?
suf
v0
Pin
45°
116 ft /sec 45° Tee
6.5 ft
You
Green
45 ft
369 ft
iaz
NOT TO SCALE
x Stopboard
nR
13. The Green Monster A baseball hit by a Boston Red Sox player at a 20° angle from 3 ft above the ground just cleared the left end of the “Green Monster,” the leftfield wall in Fenway Park. This wall is 37 ft high and 315 ft from home plate (see the accompanying figure). a. What was the initial speed of the ball?
6. (Continuation of Exercise 5.) Because of its initial elevation, the shot in Exercise 5 would have gone slightly farther if it had been launched at a 40° angle. How much farther? Answer in inches.
a. What was the ball’s initial speed?
ass a
7. Firing golf balls A spring gun at ground level fires a golf ball at an angle of 45°. The ball lands 10 m away.
b. How long did it take the ball to reach the wall? “Gr
een
315'
Mon ster” 37' wall
379'
b. For the same initial speed, find the two firing angles that make the range 6 m.
dH
8. Beaming electrons An electron in a TV tube is beamed horizontally at a speed of 5 * 10 6 m>sec toward the face of the tube 40 cm away. About how far will the electron drop before it hits?
ma
9. Finding golf ball speed Laboratory tests designed to find how far golf balls of different hardness go when hit with a driver showed that a 100compression ball hit with a clubhead speed of 100 mph at a launch angle of 9° carried 248.8 yd. What was the launch speed of the ball? (It was more than 100 mph. At the same time the club head was moving forward, the compressed ball was kicking away from the club face, adding to the ball’s forward speed.)
ham
10. A human cannonball is to be fired with an initial speed of
Mu
y0 = 80 210>3 ft/sec . The circus performer (of the right caliber, naturally) hopes to land on a special cushion located 200 ft downrange at the same height as the muzzle of the cannon. The circus is being held in a large room with a flat ceiling 75 ft higher than the muzzle. Can the performer be fired to the cushion without striking the ceiling? If so, what should the cannon’s angle of elevation be?
11. A golf ball leaves the ground at a 30° angle at a speed of 90 ft> sec. Will it clear the top of a 30ft tree that is in the way, 135 ft down the fairway? Explain. 12. Elevated green A golf ball is hit with an initial speed of 116 ft> sec at an angle of elevation of 45° from the tee to a green that is
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17' wall 420' 5' wall
380' 302'
3' wall
14. Equalrange firing angles Show that a projectile fired at an angle of a degrees, 0 6 a 6 90 , has the same range as a projectile fired at the same speed at an angle of s90  ad degrees. (In models that take air resistance into account, this symmetry is lost.) 15. Equalrange firing angles What two angles of elevation will enable a projectile to reach a target 16 km downrange on the same level as the gun if the projectile’s initial speed is 400 m> sec?
16. Range and height versus speed
a. Show that doubling a projectile’s initial speed at a given launch angle multiplies its range by 4. b. By about what percentage should you increase the initial speed to double the height and range? 17. Shot put In Moscow in 1987, Natalya Lisouskaya set a women’s world record by putting an 8 lb 13 oz shot 73 ft 10 in. Assuming that she launched the shot at a 40° angle to the horizontal from 6.5 ft above the ground, what was the shot’s initial speed?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.2 Modeling Projectile Motion
i
regardless of the value of y0 . Was this mere coincidence, or must this happen? Give reasons for your answer.
suf
18. Height versus time Show that a projectile attains threequarters of its maximum height in half the time it takes to reach the maximum height.
B
19. Firing from sx0 , y0 d Derive the equations x = x0 + sy0 cos adt,
1 2 gt 2
1 y = y0 + sy0 sin adt  gt 2, 2 A
25. Launching downhill An ideal projectile is launched straight down an inclined plane as shown in the accompanying figure.
dr s0d = sy0 cos adi + sy0 sin adj dt 20. Flaming arrow Using the firing angle found in Example 3, find the speed at which the flaming arrow left Rebollo’s bow. See Figure 13.13.
a. Show that the greatest downhill range is achieved when the initial velocity vector bisects angle AOR. b. If the projectile were fired uphill instead of down, what launch angle would maximize its range? Give reasons for your answer. A
ass a
21. Flaming arrow The cauldron in Example 3 is 12 ft in diameter. Using Equation (5) and Example 3c, find how long it takes the flaming arrow to cover the horizontal distance to the rim. How high is the arrow at this time?
iaz
rs0d = x0 i + y0 j
R
22. Describe the path of a projectile given by Equations (4) when a = 90° .
Mu
ham
ma
dH
23. Model train The accompanying multiflash photograph shows a model train engine moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fired into the air by a spring in the engine’s smokestack. The marble, which continued to move with the same forward speed as the engine, rejoined the engine 1 sec after it was fired. Measure the angle the marble’s path made with the horizontal and use the information to find how high the marble went and how fast the engine was moving.
24. Colliding marbles The figure shows an experiment with two marbles. Marble A was launched toward marble B with launch angle a and initial speed y0 . At the same instant, marble B was released to fall from rest at R tan a units directly above a spot R units downrange from A. The marbles were found to collide
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Vertical
Initial conditions:
␣
nR
d 2r = g j dt 2
You
R tan ␣
v0
(see Equation (5) in the text) by solving the following initial value problem for a vector r in the plane. Differential equation:
929
O
v0 ␣ Hi
ll
R
26. Hitting a baseball under a wind gust A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 ft> sec at a launch angle of 23°. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of 14i sft>secd to the ball’s initial velocity. A 15fthigh fence lies 300 ft from home plate in the direction of the flight. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height? c. Find the range and flight time of the baseball, assuming that the ball is not caught. d. When is the baseball 20 ft high? How far (ground distance) is the baseball from home plate at that height? e. Has the batter hit a home run? Explain. 27. Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6fthigh net. It leaves the point of impact with an initial velocity of 35 ft> sec at an angle of 27° and slips by the opposing team untouched.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
b. How high does the volleyball go, and when does it reach maximum height?
x =
y0 s1  e k t d cos a k
y =
g y0 s1  e k t dssin ad + 2 s1  k t  e k t d k k
c. Find its range and flight time.
e. Suppose that the net is raised to 8 ft. Does this change things? Explain. 28. Where trajectories crest For a projectile fired from the ground at launch angle a with initial speed y0 , consider a as a variable and y0 as a fixed constant. For each a, 0 6 a 6 p>2 , we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse y02 2 y04 , b = 4g 4g 2
d 2r dr = gj  kv = gj  k 2 dt dt
Initial conditions:
rs0d = 0
dr = v0 = sy0 cos adi + sy0 sin adj ` dt t = 0
The drag coefficient k is a positive constant representing resistance due to air density, y0 and a are the projectile’s initial speed and launch angle, and g is the acceleration of gravity.
where x Ú 0 .
30. Hitting a baseball with linear drag Consider the baseball problem in Example 4 when there is linear drag (see Exercise 29). Assume a drag coefficient k = 0.12 , but no gust of wind.
y
ass a
a. From Exercise 29, find a vector form for the path of the baseball.
Ellipse 1 R, y max 2
b. How high does the baseball go, and when does it reach maximum height?
Parabolic trajectory
dH
c. Find the range and flight time of the baseball.
x
0
ma
Projectile Motion with Linear Drag
ham
The main force affecting the motion of a projectile, other than gravity, is air resistance. This slowing down force is drag force, and it acts in a direction opposite to the velocity of the projectile (see accompanying figure). For projectiles moving through the air at relatively low speeds, however, the drag force is (very nearly) proportional to the speed (to the first power) and so is called linear. y
d. When is the baseball 30 ft high? How far (ground distance) is the baseball from home plate at that height? e. A 10fthigh outfield fence is 340 ft from home plate in the direction of the flight of the baseball. The outfielder can jump and catch any ball up to 11 ft off the ground to stop it from going over the fence. Has the batter hit a home run?
31. Hitting a baseball with linear drag under a wind gust Consider again the baseball problem in Example 4. This time assume a drag coefficient of 0.08 and an instantaneous gust of wind that adds a component of 17.6i sft>secd to the initial velocity at the instant the baseball is hit. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height? c. Find the range and flight time of the baseball.
Velocity
d. When is the baseball 35 ft high? How far (ground distance) is the baseball from home plate at that height?
Gravity
e. A 20fthigh outfield fence is 380 ft from home plate in the direction of the flight of the baseball. Has the batter hit a home run? If “yes,” what change in the horizontal component of the ball’s initial velocity would have kept the ball in the park? If “no,” what change would have allowed it to be a home run?
Drag force
Mu
Differential equation:
nR
x 2 + 4 ay 
by solving the following initial value problem for a vector r in the plane.
iaz
d. When is the volleyball 7 ft above the ground? How far (ground distance) is the volleyball from where it will land?
Derive the equations
i
29. Linear drag
suf
a. Find a vector equation for the path of the volleyball.
You
930
x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.3
Arc Length and the Unit Tangent Vector T
suf
i
Arc Length and the Unit Tangent Vector T
13.3
931
iaz
You
Imagine the motions you might experience traveling at high speeds along a path through the air or space. Specifically, imagine the motions of turning to your left or right and the upanddown motions tending to lift you from, or pin you down to, your seat. Pilots flying through the atmosphere, turning and twisting in flight acrobatics, certainly experience these motions. Turns that are too tight, descents or climbs that are too steep, or either one coupled with high and increasing speed can cause an aircraft to spin out of control, possibly even to break up in midair, and crash to Earth. In this and the next two sections, we study the features of a curve’s shape that describe mathematically the sharpness of its turning and its twisting perpendicular to the forward motion.
Arc Length Along a Space Curve
–2
0 1
2
s
FIGURE 13.14 Smooth curves can be scaled like number lines, the coordinate of each point being its directed distance along the curve from a preselected base point.
One of the features of smooth space curves is that they have a measurable length. This enables us to locate points along these curves by giving their directed distance s along the curve from some base point, the way we locate points on coordinate axes by giving their directed distance from the origin (Figure 13.14). Time is the natural parameter for describing a moving body’s velocity and acceleration, but s is the natural parameter for studying a curve’s shape. Both parameters appear in analyses of space flight. To measure distance along a smooth curve in space, we add a zterm to the formula we use for curves in the plane.
nR
3
–1
4
ass a
Base point
b
L =
2 2 dy 2 dx dz b + a b + a b dt. dt dt La C dt
a
(1)
Mu
ham
ma
dH
DEFINITION Length of a Smooth Curve The length of a smooth curve rstd = xstdi + ystdj + zstdk, a … t … b, that is traced exactly once as t increases from t = a to t = b, is
Just as for plane curves, we can calculate the length of a curve in space from any convenient parametrization that meets the stated conditions. We omit the proof. The square root in Equation (1) is ƒ v ƒ , the length of a velocity vector dr> dt. This enables us to write the formula for length a shorter way.
Arc Length Formula b
L =
EXAMPLE 1
La
ƒ v ƒ dt
(2)
Distance Traveled by a Glider
A glider is soaring upward along the helix rstd = scos tdi + ssin tdj + tk. How far does the glider travel along its path from t = 0 to t = 2p L 6.28 sec?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 932
Chapter 13: VectorValued Functions and Motion in Space z
The path segment during this time corresponds to one full turn of the helix (Figure 13.15). The length of this portion of the curve is b
L = 2
La
t
2p
ƒ v ƒ dt =
L0
2s sin td2 + scos td2 + s1d2 dt
2p
=
L0
22 dt = 2p22 units of length.
You
t 2
suf
i
Solution
This is 22 times the length of the circle in the xyplane over which the helix stands. r
0 (1, 0, 0)
t 2 P
t0
If we choose a base point Pst0 d on a smooth curve C parametrized by t, each value of t determines a point Pstd = sxstd, ystd, zstdd on C and a “directed distance” y
t
ƒ vstd ƒ dt,
iaz
sstd =
x
Lt0
measured along C from the base point (Figure 13.16). If t 7 t0 , s(t) is the distance from Pst0 d to P(t). If t 6 t0 , s(t) is the negative of the distance. Each value of s determines a point on C and this parametrizes C with respect to s. We call s an arc length parameter for the curve. The parameter’s value increases in the direction of increasing t. The arc length parameter is particularly effective for investigating the turning and twisting nature of a space curve. We use the Greek letter t (“tau”) as the variable of integration because the letter t is already in use as the upper limit.
z
r
P(t)
0
ass a
nR
FIGURE 13.15 The helix rstd = scos tdi + ssin tdj + tk in Example 1.
Arc Length Parameter with Base Point Pst0 d t
Base point y
s(t) P(t0) x
EXAMPLE 2
Lt0
ƒ vstd ƒ dt
Finding an Arc Length Parametrization
If t0 = 0, the arc length parameter along the helix rstd = scos tdi + ssin tdj + tk from t0 to t is
Mu
ham
Lt0
t
2[x¿std]2 + [y¿std]2 + [z¿std]2 dt =
(3)
If a curve r(t) is already given in terms of some parameter t and s(t) is the arc length function given by Equation (3), then we may be able to solve for t as a function of s: t = tssd. Then the curve can be reparametrized in terms of s by substituting for t: r = rstssdd.
ma
FIGURE 13.16 The directed distance along the curve from Pst0 d to any point P(t) is t sstd = ƒ vstd ƒ dt. Lt0
dH
sstd =
To Read it Online & Download:
t
sstd =
Lt0
ƒ vstd ƒ dt
Equation (3)
22 dt
Value from Example 1
t
=
L0
= 22 t.
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4100 AWL/Thomas_ch13p906964 8/25/04 2:48 PM Page 933
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.3
Arc Length and the Unit Tangent Vector T
933
rstssdd = ¢ cos
s 22
≤ i + ¢ sin
s 22
suf
i
Solving this equation for t gives t = s> 22 . Substituting into the position vector r gives the following arc length parametrization for the helix:
≤j +
s
22
k.
EXAMPLE 3
You
Unlike Example 2, the arc length parametrization is generally difficult to find analytically for a curve already given in terms of some other parameter t. Fortunately, however, we rarely need an exact formula for s(t) or its inverse t(s).
Distance Along a Line
iaz
Show that if u = u1i + u2 j + u3k is a unit vector, then the arc length parameter along the line rstd = sx0 + tu1 di + s y0 + tu2 dj + sz0 + tu3 dk
Solution
so
d d d sx + tu1 di + s y + tu2 dj + sz + tu3 dk = u1i + u2 j + u3k = u, dt 0 dt 0 dt 0
ass a
v =
nR
from the point P0sx0, y0, z0 d where t = 0 is t itself.
t
sstd =
Josiah Willard Gibbs (1839–1903)
L0
1 dt = t.
Since the derivatives beneath the radical in Equation (3) are continuous (the curve is smooth), the Fundamental Theorem of Calculus tells us that s is a differentiable function of t with derivative ds = ƒ vstd ƒ . dt
ma ham Mu
L0
t
ƒ u ƒ dt =
Speed on a Smooth Curve
dH
HISTORICAL BIOGRAPHY
L0
t
ƒ v ƒ dt =
(4)
As we already knew, the speed with which a particle moves along its path is the magnitude of v. Notice that although the base point Pst0 d plays a role in defining s in Equation (3), it plays no role in Equation (4). The rate at which a moving particle covers distance along its path is independent of how far away it is from the base point. Notice also that ds>dt 7 0 since, by definition, ƒ v ƒ is never zero for a smooth curve. We see once again that s is an increasing function of t.
Unit Tangent Vector T We already know the velocity vector v = dr>dt is tangent to the curve and that the vector
To Read it Online & Download:
T =
v ƒvƒ
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4100 AWL/Thomas_ch13p906964 8/25/04 2:48 PM Page 934
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space z
i
is therefore a unit vector tangent to the (smooth) curve. Since ds>dt 7 0 for the curves we are considering, s is onetoone and has an inverse that gives t as a differentiable function of s (Section 7.1). The derivative of the inverse is
v
dt 1 1 = = . ds v ds>dt ƒ ƒ 0 s
T v v
This makes r a differentiable function of s whose derivative can be calculated with the Chain Rule to be
You
r
suf
934
v dr dr dt 1 = = v = = T. ds dt ds ƒvƒ ƒvƒ
y
P(t 0 )
iaz
This equation says that dr> ds is the unit tangent vector in the direction of the velocity vector v (Figure 13.17).
x
nR
FIGURE 13.17 We find the unit tangent vector T by dividing v by ƒ v ƒ .
DEFINITION Unit Tangent Vector The unit tangent vector of a smooth curve r(t) is dr>dt v dr = = . ds ds>dt ƒvƒ
ass a
T =
(5)
dH
The unit tangent vector T is a differentiable function of t whenever v is a differentiable function of t. As we see in Section 13.5, T is one of three unit vectors in a traveling reference frame that is used to describe the motion of space vehicles and other bodies traveling in three dimensions.
EXAMPLE 4
Finding the Unit Tangent Vector T
ma
Find the unit tangent vector of the curve rstd = s3 cos tdi + s3 sin tdj + t 2k
Mu
ham
representing the path of the glider in Example 4, Section 13.1. Solution
In that example, we found v =
dr = s3 sin tdi + s3 cos tdj + 2tk dt
and ƒ v ƒ = 29 + 4t 2 . Thus, T =
To Read it Online & Download:
v 3 cos t 3 sin t 2t i + j + k. = 2 2 ƒvƒ 29 + 4t 29 + 4t 29 + 4t 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.3 Arc Length and the Unit Tangent Vector T
x2 y2 1
Motion on the Unit Circle
For the counterclockwise motion rstd = scos tdi + ssin tdj
P(x, y) r t 0
i
EXAMPLE 5 Tv
(1, 0)
x
around the unit circle,
You
v = s sin tdi + scos tdj
suf
y
is already a unit vector, so T = v (Figure 13.18).
Mu
ham
ma
dH
ass a
nR
iaz
FIGURE 13.18 The motion rstd = (cos tdi + ssin tdj (Example 5).
To Read it Online & Download:
935
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
13.3 Arc Length and the Unit Tangent Vector T
EXERCISES 13.3 Finding Unit Tangent Vectors and Lengths of Curves 1. rstd = s2 cos tdi + s2 sin tdj + 25t k,
0 … t … p
2. rstd = s6 sin 2tdi + s6 cos 2tdj + 5t k,
0 … t … p
3. rstd = t i + s2>3dt
4. rstd = s2 + tdi  st + 1dj + t k, 5. rstd = scos3 t dj + ssin3 t dk, 3
3
6. rstd = 6t i  2t j  3t k,
15. Arc length
0 … t … p
H d
at a distance 26p units along the curve from the origin in the direction of increasing arc length.
a m
rstd = s12 sin tdi  s12 cos tdj + 5t k
at a distance 13p units along the curve from the origin in the direction opposite to the direction of increasing arc length.
m a
Arc Length Parameter
In Exercises 11–14, find the arc length parameter along the curve from the point where t = 0 by evaluating the integral s =
M
t
L0
ƒ vstd ƒ dt
from Equation (3). Then find the length of the indicated portion of the curve.
To Read it Online & Download:
16. Length of helix The length 2p 22 of the turn of the helix in Example 1 is also the length of the diagonal of a square 2p units on a side. Show how to obtain this square by cutting away and flattening a portion of the cylinder around which the helix winds.
sa
22 … t … 2
rstd = s5 sin tdi + s5 cos tdj + 12t k
10. Find the point on the curve
A 22t B i + A 22t B j + s1  t 2 dk
from (0, 0, 1) to A 22, 22, 0 B .
9. Find the point on the curve
h u
n sa
1 … t … 0
Find the length of the curve
rstd =
1 … t … 2
8. rstd = st sin t + cos tdi + st cos t  sin tdj,
iR a
p>2 … t … p
ln 4 … t … 0
14. rstd = s1 + 2tdi + s1 + 3tdj + s6  6tdk,
0 … t … 3
7. rstd = st cos tdi + st sin tdj + A 2 22>3 B t 3>2 k,
t
13. rstd = se cos tdi + se sin tdj + e k,
0 … t … p>2
3
t
Theory and Examples
0 … t … 8
k,
0 … t … p>2
12. rstd = scos t + t sin tdi + ssin t  t cos tdj,
In Exercises 1–8, find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve.
3>2
u o Y z
11. rstd = s4 cos tdi + s4 sin tdj + 3t k, t
935
17. Ellipse
a. Show that the curve rstd = scos tdi + ssin tdj + s1  cos tdk, 0 … t … 2p , is an ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find equations for the cylinder and plane.
b. Sketch the ellipse on the cylinder. Add to your sketch the unit tangent vectors at t = 0, p>2, p , and 3p>2 . c. Show that the acceleration vector always lies parallel to the plane (orthogonal to a vector normal to the plane). Thus, if you draw the acceleration as a vector attached to the ellipse, it will lie in the plane of the ellipse. Add the acceleration vectors for t = 0, p>2, p , and 3p>2 to your sketch. d. Write an integral for the length of the ellipse. Do not try to evaluate the integral; it is nonelementary. T e. Numerical integrator Estimate the length of the ellipse to two decimal places. 18. Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 936
Chapter 13: VectorValued Functions and Motion in Space
0 … t … p>2
b. rstd = [cos st>2d]i + [sin st>2d] j + st>2dk,
0 … t … 4p
y
2p … t … 0
19. The involute of a circle If a string wound around a fixed circle is unwound while held taut in the plane of the circle, its end P traces an involute of the circle. In the accompanying figure, the circle in question is the circle x 2 + y 2 = 1 and the tracing point starts at (1, 0). The unwound portion of the string is tangent to the circle at Q, and t is the radian measure of the angle from the positive xaxis to segment OQ. Derive the parametric equations y = sin t  t cos t,
t 7 0
P(x, y)
t
O
1
(1, 0)
x
iaz
x = cos t + t sin t,
String
Q
You
c. rstd = scos tdi  ssin td j  t k,
suf
a. rstd = scos 4tdi + ssin 4tdj + 4t k,
i
the parametrization you use to compute it, calculate the length of one turn of the helix in Example 1 with the following parametrizations.
20. (Continuation of Exercise 19.) Find the unit tangent vector to the involute of the circle at the point P(x, y).
Mu
ham
ma
dH
ass a
nR
of the point P(x, y) for the involute.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 936
i f u s u
Chapter 13: VectorValued Functions and Motion in Space
Curvature and the Unit Normal Vector N
13.4
o Y
In this section we study how a curve turns or bends. We look first at curves in the coordinate plane, and then at curves in space.
y
Curvature of a Plane Curve
T
P s P0
n a ss
T x
0
FIGURE 13.19 As P moves along the curve in the direction of increasing arc length, the unit tangent vector turns. The value of ƒ d T>ds ƒ at P is called the curvature of the curve at P.
DEFINITION Curvature If T is the unit vector of a smooth curve, the curvature function of the curve is
d a
a H
k = `
dT `. ds
If ƒ dT>ds ƒ is large, T turns sharply as the particle passes through P, and the curvature at P is large. If ƒ dT>ds ƒ is close to zero, T turns more slowly and the curvature at P is smaller. If a smooth curve r(t) is already given in terms of some parameter t other than the arc length parameter s, we can calculate the curvature as
m m
a h
u M
z a i R
As a particle moves along a smooth curve in the plane, T = dr>ds turns as the curve bends. Since T is a unit vector, its length remains constant and only its direction changes as the particle moves along the curve. The rate at which T turns per unit of length along the curve is called the curvature (Figure 13.19). The traditional symbol for the curvature function is the Greek letter k (“kappa”).
T
To Read it Online & Download:
k = `
dT dT dt ` = ` ` ds dt ds
=
dT 1 ` ` ƒ ds>dt ƒ dt
=
1 dT ` `. ƒ v ƒ dt
Chain Rule
ds = ƒvƒ dt
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4100 AWL/Thomas_ch13p906964 8/25/04 2:48 PM Page 937
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Formula for Calculating Curvature If r(t) is a smooth curve, then the curvature is k =
1 dT ` `, ƒ v ƒ dt
(1)
You
where T = v> ƒ v ƒ is the unit tangent vector.
937
i
Curvature and the Unit Normal Vector N
suf
13.4
EXAMPLE 1
The Curvature of a Straight Line Is Zero
On a straight line, the unit tangent vector T always points in the same direction, so its components are constants. Therefore, ƒ dT>ds ƒ = ƒ 0 ƒ = 0 (Figure 13.20).
EXAMPLE 2
nR
FIGURE 13.20 Along a straight line, T always points in the same direction. The curvature, ƒ d T>ds ƒ , is zero (Example 1).
iaz
Testing the definition, we see in Examples 1 and 2 below that the curvature is constant for straight lines and circles.
T
The Curvature of a Circle of Radius a is 1> a
To see why, we begin with the parametrization
ass a
rstd = sa cos tdi + sa sin tdj
of a circle of radius a. Then, v =
dr = sa sin tdi + sa cos tdj dt
dH
2 2 2 ƒ v ƒ = 2s a sin td + sa cos td = 2a = ƒ a ƒ = a.
Mu
ham
ma
From this we find
T =
Since a 7 0, ƒ a ƒ = a.
v = ssin tdi + scos tdj ƒvƒ
dT = scos tdi  ssin tdj dt
`
dT ` = 2cos2 t + sin2 t = 1. dt
Hence, for any value of the parameter t, k =
1 dT 1 1 ` ` = a s1d = a . ƒ v ƒ dt
Although the formula for calculating k in Equation (1) is also valid for space curves, in the next section we find a computational formula that is usually more convenient to apply. Among the vectors orthogonal to the unit tangent vector T is one of particular significance because it points in the direction in which the curve is turning. Since T has constant length (namely, 1), the derivative d T> ds is orthogonal to T (Section 13.1). Therefore, if we divide d T> ds by its length k, we obtain a unit vector N orthogonal to T (Figure 13.21).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 4: Applications of Derivatives
T
T P2
P1
DEFINITION Principal Unit Normal At a point where k Z 0, the principal unit normal vector for a smooth curve in the plane is
i
N 1 dT κ ds
1 dT . N = k ds
The vector dT> ds points in the direction in which T turns as the curve bends. Therefore, if we face in the direction of increasing arc length, the vector dT> ds points toward the right if T turns clockwise and toward the left if T turns counterclockwise. In other words, the principal normal vector N will point toward the concave side of the curve (Figure 13.21). If a smooth curve r(t) is already given in terms of some parameter t other than the arc length parameter s, we can use the Chain Rule to calculate N directly: dT>ds ƒ dT>ds ƒ
nR
FIGURE 13.21 The vector dT> ds, normal to the curve, always points in the direction in which T is turning. The unit normal vector N is the direction of dT> ds.
iaz
N 1 dT κ ds
You
s P0
suf
938
N =
sdT>dtdsdt>dsd ƒ dT>dt ƒ ƒ dt>ds ƒ
ass a
=
=
dT>dt . ƒ dT>dt ƒ
dt 1 7 0 cancels = ds ds>dt
dH
This formula enables us to find N without having to find k and s first.
ma
Formula for Calculating N If r(t) is a smooth curve, then the principal unit normal is N =
d T>dt , ƒ dT>dt ƒ
EXAMPLE 3
Finding T and N
Find T and N for the circular motion rstd = scos 2tdi + ssin 2tdj. Solution
We first find T:
Mu
ham
where T = v> ƒ v ƒ is the unit tangent vector.
To Read it Online & Download:
v = s2 sin 2tdi + s2 cos 2tdj ƒ v ƒ = 24 sin2 2t + 4 cos2 2t = 2 T =
v = ssin 2tdi + scos 2tdj. ƒvƒ
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(2)
4100 AWL/Thomas_ch13p906964 8/25/04 2:48 PM Page 939
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.4
Curvature and the Unit Normal Vector N
939
dT ` = 24 cos2 2t + 4 sin2 2t = 2 dt
You
`
suf
dT = s2 cos 2tdi  s2 sin 2tdj dt
and N =
i
From this we find
dT>dt ƒ dT>dt ƒ
= scos 2tdi  ssin 2tdj.
Equation (2)
Curve Radius of curvature N
T
nR
Center of curvature
Circle of Curvature for Plane Curves
The circle of curvature or osculating circle at a point P on a plane curve where k Z 0 is the circle in the plane of the curve that 1. 2. 3.
is tangent to the curve at P (has the same tangent line the curve has) has the same curvature the curve has at P lies toward the concave or inner side of the curve (as in Figure 13.22).
ass a
Circle of curvature
iaz
Notice that T # N = 0, verifying that N is orthogonal to T. Notice too, that for the circular motion here, N points from r(t) towards the circle’s center at the origin.
The radius of curvature of the curve at P is the radius of the circle of curvature, which, according to Example 2, is
P(x, y)
1 Radius of curvature = r = k .
dH
FIGURE 13.22 The osculating circle at P(x, y) lies toward the inner side of the curve.
To find r, we find k and take the reciprocal. The center of curvature of the curve at P is the center of the circle of curvature.
ma
EXAMPLE 4
Finding the Osculating Circle for a Parabola
Find and graph the osculating circle of the parabola y = x 2 at the origin.
Mu
ham
Solution
We parametrize the parabola using the parameter t = x (Section 10.4,
Example 1) rstd = ti + t 2j. First we find the curvature of the parabola at the origin, using Equation (1): v =
dr = i + 2tj dt
ƒ v ƒ = 21 + 4t 2 so that
To Read it Online & Download:
T =
v = s1 + 4t 2 d1>2 i + 2ts1 + 4t 2 d1>2 j. ƒvƒ
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 940
Chapter 13: VectorValued Functions and Motion in Space y
From this we find
i
y x2
suf
dT = 4ts1 + 4t 2 d3>2 i + [2s1 + 4t 2 d1>2  8t 2s1 + 4t 2 d3>2] j. dt
Osculating circle
At the origin, t = 0, so the curvature is
0
ks0d = x
1
= FIGURE 13.23 The osculating circle for the parabola y = x 2 at the origin (Example 4).
dT 1 ` s0d ` vs0d ƒ ƒ dt
You
1 2
Equation (1)
1 ƒ 0i + 2j ƒ 21
iaz
= s1d20 2 + 22 = 2.
Therefore, the radius of curvature is 1>k = 1>2 and the center of the circle is (0, 1>2) (see Figure 13.23). The equation of the osculating circle is 2
1 1 b = a b 2 2
nR
sx  0d2 + ay 
or
ass a
x 2 + ay 
2
2
1 1 b = . 2 4
dH
You can see from Figure 13.23 that the osculating circle is a better approximation to the parabola at the origin than is the tangent line approximation y = 0.
Curvature and Normal Vectors for Space Curves
z
2b t
0 (a, 0, 0)
r
t 2 P
x 2 y 2 a2
Mu
x
t0
ham
t 2
ma
If a smooth curve in space is specified by the position vector r(t) as a function of some parameter t, and if s is the arc length parameter of the curve, then the unit tangent vector T is dr>ds = v> ƒ v ƒ . The curvature in space is then defined to be
FIGURE 13.24
k = `
rstd = sa cos tdi + sa sin tdj + bt k,
drawn with a and b positive and t Ú 0 (Example 5).
(3)
just as for plane curves. The vector dT> ds is orthogonal to T, and we define the principal unit normal to be dT>dt 1 dT N = k = . ds dT>dt ƒ ƒ
y
The helix
dT 1 dT ` = ` ` ds ƒ v ƒ dt
EXAMPLE 5
(4)
Finding Curvature
Find the curvature for the helix (Figure 13.24) rstd = sa cos tdi + sa sin tdj + btk,
To Read it Online & Download:
a, b Ú 0,
a 2 + b 2 Z 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.4
941
We calculate T from the velocity vector v:
i
Solution
Curvature and the Unit Normal Vector N
suf
v = sa sin tdi + sa cos tdj + bk
ƒ v ƒ = 2a 2 sin2 t + a 2 cos2 t + b 2 = 2a 2 + b 2
v 1 [sa sin tdi + sa cos tdj + bk]. = ƒvƒ 2a 2 + b 2
You
T =
Then using Equation (3),
=
1 dT ` ` ƒ v ƒ dt 1 2a + b
=
`
1
2a + b 2 2
[sa cos tdi  sa sin tdj] `
a ƒ scos tdi  ssin tdj ƒ a2 + b2 a a 2scos td2 + ssin td2 = 2 . 2 a + b a + b2 2
ass a
=
2
nR
2
iaz
k =
dH
From this equation, we see that increasing b for a fixed a decreases the curvature. Decreasing a for a fixed b eventually decreases the curvature as well. Stretching a spring tends to straighten it. If b = 0, the helix reduces to a circle of radius a and its curvature reduces to 1> a, as it should. If a = 0, the helix becomes the zaxis, and its curvature reduces to 0, again as it should.
EXAMPLE 6
Finding the Principal Unit Normal Vector N
ma
Find N for the helix in Example 5. We have
Mu
ham
Solution
To Read it Online & Download:
dT 1 = [sa cos tdi + sa sin tdj] dt 2a 2 + b 2
`
Example 5
dT a 1 ` = 2a 2 cos2 t + a 2 sin2 t = 2 2 2 dt 2a + b 2a + b 2 N =
d T>dt ƒ dT>dt ƒ
= 
2a 2 + b 2 a
Equatiion (4)
1
#
2a + b 2 2
[sa cos tdi + sa sin tdj]
= scos tdi  ssin tdj.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 942
Chapter 13: VectorValued Functions and Motion in Space
Plane Curves
b. Calculate
Find T, N, and k for the plane curves in Exercises 1–4. p>2 6 t 6 p>2 t 7 0
5. A formula for the curvature of the graph of a function in the xyplane a. The graph y = ƒsxd in the xyplane automatically has the parametrization x = x, y = ƒsxd , and the vector formula rsxd = x i + ƒsxdj. Use this formula to show that if ƒ is a twicedifferentiable function of x, then ksxd =
ƒ ƒ–sxd ƒ
C 1 + sƒ¿sxdd D
2 3>2
t Z 0,
for the curve in part (a). Does N exist at t = 0 ? Graph the curve and explain what is happening to N as t passes from negative to positive values.
3. rstd = s2t + 3di + s5  t 2 dj 4. rstd = scos t + t sin tdi + ssin t  t cos tdj,
,
You
2. rstd = sln sec tdi + t j,
ƒ d T>dt ƒ
Space Curves
Find T, N, and k for the space curves in Exercises 9–16.
iaz
p>2 6 t 6 p>2
9. rstd = s3 sin tdi + s3 cos tdj + 4t k 10. rstd = scos t + t sin tdi + ssin t  t cos tdj + 3k 11. rstd = se t cos tdi + se t sin tdj + 2k
nR
1. rstd = t i + sln cos tdj,
d T>dt
N =
suf
i
EXERCISES 13.4
12. rstd = s6 sin 2tdi + s6 cos 2tdj + 5t k 13. rstd = st 3>3di + st 2>2dj,
.
3
14. rstd = scos tdi + ssin tdj,
0 6 t 6 p>2
15. rstd = t i + sa cosh st>addj,
a 7 0
16. rstd = scosh tdi  ssinh tdj + t k
ass a
b. Use the formula for k in part (a) to find the curvature of y = ln scos xd, p>2 6 x 6 p>2 . Compare your answer with the answer in Exercise 1.
t 7 0
3
c. Show that the curvature is zero at a point of inflection.
6. A formula for the curvature of a parametrized plane curve
dH
a. Show that the curvature of a smooth curve rstd = ƒstdi + g stdj defined by twicedifferentiable functions x = ƒstd and y = g std is given by the formula #$ #$ ƒx y  y xƒ k = #2 . # sx + y 2 d3>2
Apply the formula to find the curvatures of the following curves. 0 6 t 6 p
ma
b. rstd = t i + sln sin tdj,
c. rstd = [tan1 ssinh td]i + sln cosh tdj. 7. Normals to plane curves
ham
a. Show that nstd = g¿stdi + ƒ¿stdj and nstd = g¿stdi ƒ¿stdj are both normal to the curve rstd = ƒstdi + g stdj at the point (ƒ(t), g(t)). To obtain N for a particular plane curve, we can choose the one of n or n from part (a) that points toward the concave side of the curve, and make it into a unit vector. (See Figure 13.21.) Apply this method to find N for the following curves.
Mu
b. rstd = t i + e 2t j
c. rstd = 24  t 2 i + t j,
2 … t … 2
8. (Continuation of Exercise 7.) a. Use the method of Exercise 7 to find N for the curve rstd = ti + s1>3dt 3 j when t 6 0 ; when t 7 0 .
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More on Curvature 17. Show that the parabola y = ax 2, a Z 0 , has its largest curvature at its vertex and has no minimum curvature. (Note: Since the curvature of a curve remains the same if the curve is translated or rotated, this result is true for any parabola.) 18. Show that the ellipse x = a cos t, y = b sin t, a 7 b 7 0 , has its largest curvature on its major axis and its smallest curvature on its minor axis. (As in Exercise 17, the same is true for any ellipse.) 19. Maximizing the curvature of a helix In Example 5, we found the curvature of the helix rstd = sa cos tdi + sa sin tdj + bt k sa, b Ú 0d to be k = a>sa 2 + b 2 d . What is the largest value k can have for a given value of b? Give reasons for your answer. 20. Total curvature We find the total curvature of the portion of a smooth curve that runs from s = s0 to s = s1 7 s0 by integrating k from s0 to s1 . If the curve has some other parameter, say t, then the total curvature is s1
K =
Ls0
t1
k ds =
Lt0
t
k
1 ds dt = k ƒ v ƒ dt , dt Lt0
where t0 and t1 correspond to s0 and s1 . Find the total curvatures of a. The portion of the helix rstd = s3 cos tdi + s3 sin tdj + t k, 0 … t … 4p . b. The parabola y = x 2,  q 6 x 6 q . 21. Find an equation for the circle of curvature of the curve rstd = t i + ssin tdj at the point sp>2, 1d . (The curve parametrizes the graph of y = sin x in the xyplane.)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.4 Curvature and the Unit Normal Vector N
c. Find the unit normal vector N at t0 . Notice that the signs of the components of N depend on whether the unit tangent vector T is turning clockwise or counterclockwise at t = t0 . (See Exercise 7.)
Grapher Explorations
d. If C = a i + b j is the vector from the origin to the center (a, b) of the osculating circle, find the center C from the vector equation
i
22. Find an equation for the circle of curvature of the curve rstd = s2 ln tdi  [t + s1>td] j, e 2 … t … e 2 , at the point s0, 2d , where t = 1 .
The formula
23. y = x 2, 25. y = sin x,
2 … x … 2 0 … x … 2p
24. y = x 4>4, 26. y = e x,
You
,
derived in Exercise 5, expresses the curvature ksxd of a twicedifferentiable plane curve y = ƒsxd as a function of x. Find the curvature function of each of the curves in Exercises 23–26. Then graph ƒ(x) together with ksxd over the given interval. You will find some surprises. 2 … x … 2 1 … x … 2
The point Psx0 , y0 d on the curve is given by the position vector rst0 d . e. Plot implicitly the equation sx  ad2 + s y  bd2 = 1>k2 of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure it is square. 27. rstd = s3 cos tdi + s5 sin tdj, 28. rstd = scos3 tdi + ssin3 tdj, 29. rstd = t 2i + st 3  3tdj,
COMPUTER EXPLORATIONS
Circles of Curvature
0 … t … 2p,
0 … t … 2p,
4 … t … 4,
30. rstd = st 3  2t 2  tdi +
t0 = p>4 t0 = p>4
t0 = 3>5
3t
j, 2 … t … 5, 21 + t 2 31. rstd = s2t  sin tdi + s2  2 cos tdj, 0 … t … 3p, t0 = 3p>2
ass a
In Exercises 27–34 you will use a CAS to explore the osculating circle at a point P on a plane curve where k Z 0 . Use a CAS to perform the following steps:
1 Nst0 d . kst0 d
iaz
C 1 + s ƒ¿sxdd2 D 3>2
C = rst0 d +
nR
ksxd =
ƒ ƒ–sxd ƒ
suf
T
943
a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like.
32. rstd = se t cos tdi + se t sin tdj,
b. Calculate the curvature k of the curve at the given value t0 using the appropriate formula from Exercise 5 or 6. Use the parametrization x = t and y = ƒstd if the curve is given as a function y = ƒsxd .
34. y = xs1  xd2>5,
2 … x … 5,
0 … t … 6p,
x0 = 1
1 … x … 2,
x0 = 1>2
Mu
ham
ma
dH
33. y = x 2  x,
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t0 = 1
t0 = p>4
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
13.5 Torsion and the Unit Binormal Vector B
z
BT×N N 1 dT κ ds r
P
T dr ds
a i R an
If you are traveling along a space curve, the Cartesian i, j, and k coordinate system for representing the vectors describing your motion are not truly relevant to you. What is meaningful instead are the vectors representative of your forward direction (the unit tangent vector T), the direction in which your path is turning (the unit normal vector N), and the tendency of your motion to “twist” out of the plane created by these vectors in the direction perpendicular to this plane (defined by the unit binormal vector B = T * N). Expressing the acceleration vector along the curve as a linear combination of this TNB frame of mutually orthogonal unit vectors traveling with the motion (Figure 13.25) is particularly revealing of the nature of the path and motion along it.
d a m
s P0 x
u o zY
Torsion and the Unit Binormal Vector B
13.5
m a h u M
y
FIGURE 13.25 The TNB frame of mutually orthogonal unit vectors traveling along a curve in space.
943
s s Ha
Torsion
The binormal vector of a curve in space is B = T * N, a unit vector orthogonal to both T and N (Figure 13.26). Together T, N, and B define a moving righthanded vector frame that plays a significant role in calculating the paths of particles moving through space. It is
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 944
Chapter 13: VectorValued Functions and Motion in Space
B
P
dN dB dT = * N + T * . ds ds ds
N
You
T
suf
i
called the Frenet (“frenay”) frame (after JeanFrédéric Frenet, 1816–1900), or the TNB frame. How does dB> ds behave in relation to T, N, and B? From the rule for differentiating a cross product, we have
Since N is the direction of dT> ds, sdT>dsd * N = 0 and
dN dN dB = 0 + T * = T * . ds ds ds
iaz
From this we see that dB> ds is orthogonal to T since a cross product is orthogonal to its factors. Since dB> ds is also orthogonal to B (the latter has constant length), it follows that dB> ds is orthogonal to the plane of B and T. In other words, dB> ds is parallel to N, so dB> ds is a scalar multiple of N. In symbols,
nR
FIGURE 13.26 The vectors T, N, and B (in that order) make a righthanded frame of mutually orthogonal unit vectors in space.
dB = tN. ds
ass a
The negative sign in this equation is traditional. The scalar t is called the torsion along the curve. Notice that dB # N = tN # N = ts1d = t, ds
dH
so that
t = 
dB # N. ds
ma
DEFINITION Torsion Let B = T * N. The torsion function of a smooth curve is
Binormal Rectifying plane
ham
Normal plane
t = 
dB # N. ds
(1)
B
P
Mu
T
Principal normal
N
Unit tangent
Osculating plane
FIGURE 13.27 The names of the three planes determined by T, N, and B.
Unlike the curvature k, which is never negative, the torsion t may be positive, negative, or zero. The three planes determined by T, N, and B are named and shown in Figure 13.27. The curvature k = ƒ dT>ds ƒ can be thought of as the rate at which the normal plane turns as the point P moves along its path. Similarly, the torsion t = sdB>dsd # N is the rate at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists. If we think of the curve as the path of a moving body, then ƒ dT>ds ƒ tells how much the path turns to the left or right as the object moves along; it is called the curvature of the object’s path. The number sdB>dsd # N tells how much a body’s path rotates or
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.5
945
Torsion and the Unit Binormal Vector B
s increases
iaz
B
You
dB ds
The torsion at P is –(d B/ds)⋅ N.
suf
i
twists out of its plane of motion as the object moves along; it is called the torsion of the body’s path. Look at Figure 13.28. If P is a train climbing up a curved track, the rate at which the headlight turns from side to side per unit distance is the curvature of the track. The rate at which the engine tends to twist out of the plane formed by T and N is the torsion.
The curvature at P is (dT/ds).
N
T
P
nR
s0
FIGURE 13.28 Every moving body travels with a TNB frame that characterizes the geometry of its path of motion.
ass a
Tangential and Normal Components of Acceleration
dH
When a body is accelerated by gravity, brakes, a combination of rocket motors, or whatever, we usually want to know how much of the acceleration acts in the direction of motion, in the tangential direction T. We can calculate this using the Chain Rule to rewrite v as v =
ds dr dr ds = = T dt ds dt dt
Mu
ham
ma
and differentiating both ends of this string of equalities to get a =
dv ds ds dT d d 2s aT b = 2 T + = dt dt dt dt dt dt
=
d 2s ds dT ds d 2s ds ds T + a b = 2T + akN b 2 dt ds dt dt dt dt dt
=
ds d 2s T + k a b N. dt dt 2
dT = kN ds
2
DEFINITION
Tangential and Normal Components of Acceleration a = aTT + aNN,
(2)
where aT =
d 2s d = v dt ƒ ƒ dt 2
and
aN = k a
2
ds b = kƒ v ƒ2 dt
are the tangential and normal scalar components of acceleration.
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(3)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
ds a N κ dt
i
T 2 a T d 2s dt
s P0
nR
FIGURE 13.29 The tangential and normal components of acceleration. The acceleration a always lies in the plane of T and N, orthogonal to B.
You
N 2
Notice that the binormal vector B does not appear in Equation (2). No matter how the path of the moving body we are watching may appear to twist and turn in space, the acceleration a always lies in the plane oƒ T and N orthogonal to B. The equation also tells us exactly how much of the acceleration takes place tangent to the motion sd 2s>dt 2 d and how much takes place normal to the motion [ksds>dtd2] (Figure 13.29). What information can we glean from Equations (3)? By definition, acceleration a is the rate of change of velocity v, and in general, both the length and direction of v change as a body moves along its path. The tangential component of acceleration aT measures the rate of change of the length of v (that is, the change in the speed). The normal component of acceleration a N measures the rate of change of the direction of v. Notice that the normal scalar component of the acceleration is the curvature times the square of the speed. This explains why you have to hold on when your car makes a sharp (large k), highspeed (large ƒ v ƒ ) turn. If you double the speed of your car, you will experience four times the normal component of acceleration for the same curvature. If a body moves in a circle at a constant speed, d 2s>dt 2 is zero and all the acceleration points along N toward the circle’s center. If the body is speeding up or slowing down, a has a nonzero tangential component (Figure 13.30). To calculate a N , we usually use the formula a N = 2ƒ a ƒ 2  a T2 , which comes from solving the equation ƒ a ƒ 2 = a # a = a T2 + a N2 for a N . With this formula, we can find aN without having to calculate k first.
suf
a
iaz
946
ass a
Formula for Calculating the Normal Component of Acceleration a N = 2ƒ a ƒ 2  a T2
d 2s T dt 2
Finding the Acceleration Scalar Components aT, aN
EXAMPLE 1
Without finding T and N, write the acceleration of the motion
P v 2 v2 N N
dH
a
(4)
rstd = scos t + t sin tdi + ssin t  t cos tdj,
t 7 0
in the form a = a TT + aNN. (The path of the motion is the involute of the circle in Figure 13.31.)
C
We use the first of Equations (3) to find a T :
ma
Solution
v =
= st cos tdi + st sin tdj
ƒ v ƒ = 2t 2 cos2 t + t 2 sin2 t = 2t 2 = ƒ t ƒ = t aT =
d d v = std = 1. dt ƒ ƒ dt
t 7 0 Equation (3)
Knowing a T , we use Equation (4) to find a N :
Mu
ham
FIGURE 13.30 The tangential and normal components of the acceleration of a body that is speeding up as it moves counterclockwise around a circle of radius r .
dr = s sin t + sin t + t cos tdi + scos t  cos t + t sin tdj dt
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a = scos t  t sin tdi + ssin t + t cos tdj ƒ a ƒ2 = t2 + 1 a N = 2ƒ a ƒ  a T 2
After some algebra 2
= 2st 2 + 1d  s1d = 2t 2 = t.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.5
Torsion and the Unit Binormal Vector B
947
a P(x, y)
tN
Formulas for Computing Curvature and Torsion
v * a = a
t O
= a
x
(1, 0)
It follows that
v = dr>dt = sds>dtdT
3
ds d 2s ds bsT * Td + k a b sT * Nd dt dt 2 dt
= ka
x2 y2 1
3
ds b B. dt
3
ds = ƒvƒ dt
T * T = 0 and T * N = B
and
ƒBƒ = 1
ass a
ds 3 ƒ v * a ƒ = k ` dt ` ƒ B ƒ = k ƒ v ƒ . Solving for k gives the following formula.
dH
FIGURE 13.31 The tangential and normal components of the acceleration of the motion rstd = scos t + t sin tdi + ssin t  t cos tdj, for t 7 0 . If a string wound around a fixed circle is unwound while held taut in the plane of the circle, its end P traces an involute of the circle (Example 1).
2
ds d 2s ds Tb * c 2 T + k a b N d dt dt dt
iaz
r
Q
You
We now give some easytouse formulas for computing the curvature and torsion of a smooth curve. From Equation (2), we have
ing
Str
nR
y
suf
a = aTT + aNN = s1dT + stdN = T + tN. T
i
We then use Equation (2) to find a:
Vector Formula for Curvature ƒv * aƒ ƒ v ƒ3
(5)
ham
ma
k =
Mu
Newton’s Dot Notation for Derivatives The dots in Equation (6) denote differentiation with respect to t, one # derivative for each dot. Thus, x (“x $ dot”) means dx> dt, x (“x double dot”) % means d 2x>dt 2 , and x (“x triple dot”) # means d 3x>dt 3 . Similarly, y = dy>dt , and so on.
Equation (5) calculates the curvature, a geometric property of the curve, from the velocity and acceleration of any vector representation of the curve in which ƒ v ƒ is different from zero. Take a moment to think about how remarkable this really is: From any formula for motion along a curve, no matter how variable the motion may be (as long as v is never zero), we can calculate a physical property of the curve that seems to have nothing to do with the way the curve is traversed. The most widely used formula for torsion, derived in more advanced texts, is
To Read it Online & Download:
t =
# x $ 3x % x
# y $ y % y
# z $3 z % z
ƒ v * a ƒ2
sif v * a Z 0d.
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(6)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 948
Chapter 13: VectorValued Functions and Motion in Space
EXAMPLE 2
suf
i
This formula calculates the torsion directly from the derivatives of the component functions x = ƒstd, y = gstd, z = hstd that make up r. The determinant’s first row comes from # v, the second row comes from a, and the third row comes from a = da>dt.
Finding Curvature and Torsion
rstd = sa cos tdi + sa sin tdj + btk, Solution
You
Use Equations (5) and (6) to find k and t for the helix a, b Ú 0,
a 2 + b 2 Z 0.
We calculate the curvature with Equation (5):
iaz
v = sa sin tdi + sa cos tdj + bk a = sa cos tdi  sa sin tdj j a cos t a sin t
k b3 0
nR
i 3 v * a = a sin t a cos t
ass a
= sab sin tdi  sab cos tdj + a 2k ƒv * aƒ 2a 2b 2 + a 4 a2a 2 + b 2 a = = 2 = 2 . 3 2 2 3>2 a + b2 sa + b d sa + b 2 d3>2 ƒvƒ
k =
(7)
dH
Notice that Equation (7) agrees with the result in Example 5 in Section 13.4, where we calculated the curvature directly from its definition. To evaluate Equation (6) for the torsion, we find the entries in the determinant by differentiating r with respect to t. We already have v and a, and da # a = = sa sin tdi  sa cos tdj. dt
Mu
ham
ma
Hence,
t =
# x $ 3x % x
# y $ y % y
# z $3 z % z
ƒ v * a ƒ2
a sin t 3 a cos t a sin t =
a cos t a sin t a cos t
A a2a 2 + b 2 B 2
=
bsa 2 cos2 t + a 2 sin2 td a 2sa 2 + b 2 d
=
b . a + b2
b 03 0
Value of ƒ v * a ƒ from Equation (7)
2
From this last equation we see that the torsion of a helix about a circular cylinder is constant. In fact, constant curvature and constant torsion characterize the helix among all curves in space.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.5 Torsion and the Unit Binormal Vector B
T =
v ƒvƒ
Principal unit normal vector:
N =
dT>dt ƒ dT>dt ƒ
Binormal vector:
B = T * N
Curvature:
k = `
You
Unit tangent vector:
suf
i
Formulas for Curves in Space
iaz
ƒv * aƒ dT ` = ds ƒ v ƒ3 # x 3 x$ % x dB # t = N = ds ƒv
# y $ y % y
# z $3 z % z
* a ƒ2
nR
Torsion: Tangential and normal scalar components of acceleration:
a = a TT + a NN
aT =
d v dt ƒ ƒ
Mu
ham
ma
dH
ass a
a N = k ƒ v ƒ 2 = 2ƒ a ƒ 2  a T2
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949
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
13.5 Torsion and the Unit Binormal Vector B
EXERCISES 13.5 Finding Torsion and the Binormal Vector
11. rstd = st + 1di + 2tj + t 2k,
n a ss 2
3. rstd = se cos tdi + se sin tdj + 2k 4. rstd = s6 sin 2tdi + s6 cos 2tdj + 5t k t 7 0
7. rstd = ti + sa cosh st>addj,
a 7 0
ad
8. rstd = scosh tdi  ssinh tdj + t k
m m a uh
a H
Tangential and Normal Components of Acceleration
In Exercises 9 and 10, write a in the form a TT + aNN without finding T and N. 9. rstd = sa cos tdi + sa sin tdj + bt k
M
3
14. rstd = se t cos tdi + se t sin tdj + 22e t k,
t
0 6 t 6 p>2
t = 0
13. rstd = t i + st + s1>3dt dj + st  s1>3dt 3 dk,
2. rstd = scos t + t sin tdi + ssin t  t cos tdj + 3k
6. rstd = scos3 tdi + ssin3 tdj,
t = 1
12. rstd = st cos tdi + st sin tdj + t 2k,
1. rstd = s3 sin tdi + s3 cos tdj + 4t k
5. rstd = st 3>3di + st 2>2dj,
u o Y
In Exercises 11–14, write a in the form a = a TT + aNN at the given value of t without finding T and N.
For Exercises 1–8 you found T, N, and k in Section 13.4 (Exercises 9–16). Find now B and t for these space curves.
t
z a i R
10. rstd = s1 + 3tdi + st  2dj  3t k
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949
t = 0
t = 0
In Exercises 15 and 16, find r, T, N, and B at the given value of t. Then find equations for the osculating, normal, and rectifying planes at that value of t. 15. rstd = scos tdi + ssin tdj  k,
t = p>4
16. rstd = scos tdi + ssin tdj + t k,
t = 0
Physical Applications 17. The speedometer on your car reads a steady 35 mph. Could you be accelerating? Explain. 18. Can anything be said about the acceleration of a particle that is moving at a constant speed? Give reasons for your answer. 19. Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your answer.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
Curvature also plays a key role in physics. The magnitude of a force required to move an object at constant speed along a curved path is, according to Newton’s laws, a constant multiple of the curvature of the trajectories. Explain mathematically why the second sentence of the quotation is true. 22. Show that a moving particle will move in a straight line if the normal component of its acceleration is zero. 23. A sometime shortcut to curvature If you already know ƒ aN ƒ and ƒ v ƒ , then the formula aN = k ƒ v ƒ 2 gives a convenient way to find the curvature. Use it to find the curvature and radius of curvature of the curve rstd = scos t + t sin tdi + ssin t  t cos tdj,
t 7 0.
24. Show that k and t are both zero for the line
i
suf
28. A formula that calculates T from B and v If we start with the definition t = sd B>dsd # N and apply the Chain Rule to rewrite dB> ds as dB d B dt dB 1 = = , ds dt ds dt ƒ v ƒ
we arrive at the formula
t = 
1 dB # a Nb . ƒ v ƒ dt
The advantage of this formula over Equation (6) is that it is easier to derive and state. The disadvantage is that it can take a lot of work to evaluate without a computer. Use the new formula to find the torsion of the helix in Example 2.
ass a
(Take aN and ƒ v ƒ from Example 1.)
You
21. The following is a quotation from an article in The American Mathematical Monthly, titled “Curvature in the Eighties” by Robert Osserman (October 1990, page 731):
27. Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fixed vector C moves in a plane perpendicular to C. This, in turn, can be viewed as the solution of the following problem in calculus. Suppose rstd = ƒstdi + gstdj + hstdk is twice differentiable for all t in an interval [a, b], that r = 0 when t = a , and that v # k = 0 for all t in [a, b]. Then hstd = 0 for all t in [a, b]. Solve this problem. (Hint: Start with a = d 2r>dt 2 and apply the initial conditions in reverse order.)
iaz
20. An object of mass m travels along the parabola y = x 2 with a constant speed of 10 units> sec. What is the force on the object due to its acceleration at (0, 0)? at s21>2, 2d ? Write your answers in terms of i and j. (Remember Newton’s law, F = ma.)
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950
rstd = sx0 + Atdi + s y0 + Btdj + sz0 + Ctdk.
COMPUTER EXPLORATIONS
Curvature, Torsion, and the TNB Frame
dH
Theory and Examples
25. What can be said about the torsion of a smooth plane curve rstd = ƒstdi + gstdj? Give reasons for your answer. 26. The torsion of a helix the helix
In Example 2, we found the torsion of a, b Ú 0
ma
rstd = sa cos tdi + sa sin tdj + bt k,
29. rstd = st cos tdi + st sin tdj + t k,
t = 23
30. rstd = se t cos tdi + se t sin tdj + e t k,
t = ln 2
31. rstd = st  sin tdi + s1  cos tdj + 2t k, t = 3p 32. rstd = s3t  t 2 di + s3t 2 dj + s3t + t 3 dk,
t = 1
Mu
ham
to be t = b>sa 2 + b 2 d . What is the largest value t can have for a given value of a? Give reasons for your answer.
Rounding the answers to four decimal places, use a CAS to find v, a, speed, T, N, B, k, t , and the tangential and normal components of acceleration for the curves in Exercises 29–32 at the given values of t.
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Y z a i R n a s as
Chapter 13: VectorValued Functions and Motion in Space
13.6
Planetary Motion and Satellites
M
i f u s u o
In this section, we derive Kepler’s laws of planetary motion from Newton’s laws of motion and gravitation and discuss the orbits of Earth satellites. The derivation of Kepler’s laws from Newton’s is one of the triumphs of calculus. It draws on almost everything we have studied so far, including the algebra and geometry of vectors in space, the calculus of vector functions, the solutions of differential equations and initial value problems, and the polar coordinate description of conic sections.
m a uh
H d ma
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.6
Planetary Motion and Satellites
951
Motion in Polar and Cylindrical Coordinates
i
y
(1) uu = ssin udi + scos udj, 1 shown in Figure 13.32. The vector ur points along the position vector OP , so r = rur . The vector uu , orthogonal to ur , points in the direction of increasing u. We find from Equations (1) that
FIGURE 13.32 The length of r is the positive polar coordinate r of the point P. Thus, ur , which is r> ƒ r ƒ , is also r> r. Equations (1) express ur and uu in terms of i and j.
y v
x
O
(4)
See Figure 13.33. As in the previous section, we use Newton’s dot# notation for time deriva# tives to keep the formulas as simple as we can: ur means dur>dt, u means du>dt, and so on. The acceleration is $ # # # $ # # ## (5) a = v = sr ur + r ur d + sr u uu + ru uu + ru uu d. # # When Equations (3) are used to evaluate ur and uu and the components are separated, the equation for acceleration becomes # $ $ ## (6) a = sr  ru2 dur + sru + 2r u duu .
dH
FIGURE 13.33 In polar coordinates, the velocity vector is # # v = r ur + ru uu
(2)
When we differentiate ur and uu with respect to t to find how they change with time, the Chain Rule gives # # du u # dur # # # (3) ur = u = u uu, uu = u = u ur . du du # d # # # # v = r = ar ur b = r ur + rur = r ur + ru uu . dt
P(r, )
duu = scos udi  ssin udj = ur . du
Hence,
. ru . r ur
r
dur = ssin udi + scos udj = uu du
iaz
x
nR
O
You
ur = scos udi + ssin udj,
ur
ass a
P(r, )
r
suf
When a particle moves along a curve in the polar coordinate plane, we express its position, velocity, and acceleration in terms of the moving unit vectors
u
ma
To extend these equations of motion to space, we add zk to the righthand side of the equation r = rur . Then, in these cylindrical coordinates,
Notice that ƒ r ƒ Z r if z Z 0 .
ham
z
k
Mu
zk
r ur
ur
y
x
(7)
The vectors ur , uu , and k make a righthanded frame (Figure 13.34) in which ur * uu = k,
u
r r ur z k
r = rur + zk # # # v = r ur + ru uu + z k # $ $ ## $ a = sr  ru2 dur + sru + 2r u duu + z k.
uu * k = ur ,
k * ur = uu .
(8)
Planets Move in Planes Newton’s law of gravitation says that if r is the radius vector from the center of a sun of mass M to the center of a planet of mass m, then the force F of the gravitational attraction between the planet and sun is
FIGURE 13.34 Position vector and basic unit vectors in cylindrical coordinates.
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F = 
GmM r 2 ƒrƒ ƒrƒ
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(9)
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Chapter 13: VectorValued Functions and Motion in Space
M r
m
GmM r $ mr = , ƒ r ƒ2 ƒ r ƒ
r F – GmM r2 r
GM r $ r = . ƒ r ƒ2 ƒ r ƒ
The planet is accelerated toward the sun’s center at all times. $ Equation (10) says that r is a scalar multiple of r, so that $ r * r = 0. $ # A routine calculation shows r * r to be the derivative of r * r:
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FIGURE 13.35 The force of gravity is directed along the line joining the centers of mass.
You
r r
suf
i
(Figure 13.35). The number G is the universal gravitational constant. If we measure mass in kilograms, force in newtons, and distance in meters, G is about 6.6726 * 10 11 Nm2 kg2 . $ Combining Equation (9) with Newton’s second law, F = mr, for the force acting on the planet gives
nR
d # # # $ $ sr * r d = r * r + r * r = r * r. dt (')'*
(10)
(11)
(12)
0
Hence Equation (11) is equivalent to
. Cr×r
. r
(13)
# r * r = C
(14)
which integrates to
FIGURE 13.36 A planet that obeys Newton’s laws of gravitation and motion travels in the plane through the sun’s center # of mass perpendicular to C = r * r.
for some constant vector C. # Equation (14) tells us that r and r always lie in a plane perpendicular to C. Hence, the planet moves in a fixed plane through the center of its sun (Figure 13.36).
dH
Planet
d # sr * r d = 0, dt
ass a
Sun
r
Coordinates and Initial Conditions
Perihelion position (point closest to the sun)
Planet P(r, )
0
ham
r
Sun
FIGURE 13.37 The coordinate system for planetary motion. The motion is counterclockwise when viewed from # above, as it is here, and u 7 0 .
Mu
We now introduce coordinates in a way that places the origin at the sun’s center of mass and makes the plane of the planet’s motion the polar coordinate plane. This makes r the planet’s polar coordinate position vector and makes ƒ r ƒ equal to r and r> ƒ r ƒ equal to ur . We also position the zaxis in a way that makes k the direction of C. Thus, k has the same # righthand relation to r * r that C does, and the planet’s motion is counterclockwise when # viewed from the positive zaxis. This makes u increase with t, so that u 7 0 for all t. Finally, we rotate the polar coordinate plane about the zaxis, if necessary, to make the initial ray coincide with the direction r has when the planet is closest to the sun. This runs the ray through the planet’s perihelion position (Figure 13.37). If we measure time so that t = 0 at perihelion, we have the following initial conditions for the planet’s motion.
ma
z
1. 2. 3. 4.
r = r0 , the minimum radius, when t = 0 # r = 0 when t = 0 (because r has a minimum value then) u = 0 when t = 0 ƒ v ƒ = y0 when t = 0
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Planetary Motion and Satellites
953
suf
# r = 0 when t = 0
ƒ uu ƒ = 1 # r and u both positive
iaz
we also know that # 5. ru = y0 when t = 0.
Equation (4)
You
y0 = ƒ v ƒ t = 0 # # = ƒ r ur + ru u u ƒ t = 0 # = ƒ ru u u ƒ t = 0 # = s ƒ ru ƒ ƒ uu ƒ dt = 0 # = ƒ ru ƒ t = 0 # = sru dt = 0,
i
Since
Kepler’s First Law (The Conic Section Law)
Johannes Kepler (1571–1630)
Kepler’s ƒirst law says that a planet’s path is a conic section with the sun at one focus. The eccentricity of the conic is
nR
HISTORICAL BIOGRAPHY
r0y02  1 GM
(15)
s1 + edr0 . 1 + e cos u
(16)
e =
ass a
and the polar equation is
r =
dH
The derivation uses Kepler’s second law, so we will state and prove the second law before proving the first law.
Planet r
Kepler’s second law says that the radius vector from the sun to a planet (the vector r in our model) sweeps out equal areas in equal times (Figure 13.38). To derive the law, we use # Equation (4) to evaluate the cross product C = r * r from Equation (14): # C = r * r = r * v # # = rur * sr u r + ru u u d Equation (4) # # (17) = rrsur * ur d + rsru dsur * uu d
ma
Sun
Kepler’s Second Law (The Equal Area Law)
Mu
ham
FIGURE 13.38 The line joining a planet to its sun sweeps over equal areas in equal times.
('')''*
('')''*
0
k
# = rsru dk. Setting t equal to zero shows that # C = [rsru d]t = 0 k = r0 y0 k. Substituting this value for C in Equation (17) gives # # r0 y0 k = r 2u k, or r 2u = r0 y0 . This is where the area comes in. The area differential in polar coordinates is
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dA =
1 2 r du 2
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(18)
(19)
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Chapter 13: VectorValued Functions and Motion in Space
i
(Section 10.7). Accordingly, dA> dt has the constant value
suf
dA 1 # 1 = r 2u = r0 y0. 2 2 dt
(20)
You
So dA>dt is constant, giving Kepler’s second law. For Earth, r0 is about 150,000,000 km, y0 is about 30 km> sec, and dA> dt is about 2,250,000,000 km2>sec. Every time your heart beats, Earth advances 30 km along its orbit, and the radius joining Earth to the sun sweeps out 2,250,000,000 km2 of area.
Proof of Kepler’s First Law
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To prove that a planet moves along a conic section with one focus at its sun, we need to express the planet’s radius r as a function of u. This requires a long sequence of calculations and some substitutions that are not altogether obvious. We begin with the equation that comes from equating the coefficients of ur = r> ƒ r ƒ in Equations (6) and (10): (21)
r02y02 GM $  2 . r = r3 r
(22)
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# GM $ r  ru2 =  2 . r
ass a
# We eliminate u temporarily by replacing it with r0 y0>r 2 from Equation (19) and rearrange the resulting equation to get
We change this into a firstorder equation by a change of variable. With dr , dt
dH
p =
dp dp dr dp d 2r = = = p , dt dr dt dr dt 2
Chain Rule
Equation (22) becomes
dp r0 2y02 GM  2 . = 3 dr r r
(23)
ma
p
Mu
ham
Multiplying through by 2 and integrating with respect to r gives r0 2y02 2GM # (24) p 2 = sr d2 = + r + C1 . 2 r # The initial conditions that r = r0 and r = 0 when t = 0 determine the value of C1 to be C1 = y02 
2GM r0 .
Accordingly, Equation (24), after a suitable rearrangement, becomes r0 2 # 1 1 r 2 = y02 a1  2 b + 2GM a r  r0 b . r
(25)
The effect of going from Equation (21) to Equation (25) has been to replace a secondorder differential equation in r by a firstorder differential equation in r. Our goal is still to express r in terms of u, so we now bring u back into the picture. To accomplish this, we
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Planetary Motion and Satellites
955
2
2GM 1 1 dr 1 1 1 a b = 2  2 + 2 2 a r  r0 b r 4 du r0 r r0 y0 1 1 1 1 = 2  2 + 2h a r  r0 b . r0 r
2
2
du 1 dr b = 4 a b , du r du
iaz
2
a
du b = u 02  u 2 + 2hu  2hu0 = su0  hd2  su  hd2 , du
(27)
nR
a
du 1 dr =  2 , du r du
1 u0 = r0 ,
obtaining
(26)
GM h = 2 2 r0 y0
You
To simplify further, we substitute 1 u = r,
suf
i
divide both sides of Equation (25) by the squares of the corresponding sides of the equa# # # tion r 2u = r0 y0 (Equation 19) and use the fact that r>u = sdr>dtd>sdu>dtd = dr>du to get
ass a
du = ; 2su0  hd2  su  hd2 . (28) du # Which sign do we take? We know that u = r0y0>r 2 is positive. Also, r starts from a # minimum value at t = 0, so it cannot immediately decrease, and r Ú 0, at least for early positive values of t. Therefore, # dr r = # Ú 0 du u
and
du 1 dr =  2 … 0. du r du
du 1 = 1 2su0  hd2  su  hd2 du cos
ma
dH
The correct sign for Equation (28) is the negative sign. With this determined, we rearrange Equation (28) and integrate both sides with respect to u:
1
u  h a b = u + C2 . u0  h
(29)
Mu
ham
The constant C2 is zero because u = u0 when u = 0 and cos 1 s1d = 0. Therefore, u  h = cos u u0  h
and 1 r = u = h + su0  hd cos u.
(30)
A few more algebraic maneuvers produce the final equation s1 + edr0 , 1 + e cos u
(31)
r0y02 1  1 =  1. GM r0 h
(32)
r = where
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Chapter 13: VectorValued Functions and Motion in Space
suf
i
Together, Equations (31) and (32) say that the path of the planet is a conic section with one focus at the sun and with eccentricity sr0y02>GMd  1. This is the modern formulation of Kepler’s first law.
Kepler’s Third Law (The Time–Distance Law)
4p2 T2 = . 3 GM a
You
The time T it takes a planet to go around its sun once is the planet’s orbital period. Kepler’s third law says that T and the orbit’s semimajor axis a are related by the equation (33)
ass a
nR
iaz
Since the righthand side of this equation is constant within a given solar system, the ratio of T 2 to a 3 is the same ƒor every planet in the system. Kepler’s third law is the starting point for working out the size of our solar system. It allows the semimajor axis of each planetary orbit to be expressed in astronomical units, Earth’s semimajor axis being one unit. The distance between any two planets at any time can then be predicted in astronomical units and all that remains is to find one of these distances in kilometers. This can be done by bouncing radar waves off Venus, for example. The astronomical unit is now known, after a series of such measurements, to be 149,597,870 km. We derive Kepler’s third law by combining two formulas for the area enclosed by the planet’s elliptical orbit: Area = pab
Formula 2:
Area =
dH
The geometry formula in which a is the semimajor axis and b is the semiminor axis
Formula 1:
=
1 r0 y0 dt L0 2
=
1 Tr y . 2 0 0
T
dA
L0
T
Equation (20)
Mu
ham
ma
Equating these gives
2pab 2pa 2 T = r0 y0 = r0 y0 21  e 2 .
For any ellipse, b = a21  e 2
(34)
It remains only to express a and e in terms of r0, y0, G, and M. Equation (32) does this for e. For a, we observe that setting u equal to p in Equation (31) gives rmax = r0
1 + e . 1  e
Hence, 2a = r0 + rmax =
2r0 2r0GM = . 1  e 2GM  r0y02
(35)
Squaring both sides of Equation (34) and substituting the results of Equations (32) and (35) now produces Kepler’s third law (Exercise 15).
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Planetary Motion and Satellites
957
i
Orbit Data Although Kepler discovered his laws empirically and stated them only for the six planets known at the time, the modern derivations of Kepler’s laws show that they apply to any body driven by a force that obeys an inverse square law like Equation (9). They apply to Halley’s comet and the asteroid Icarus. They apply to the moon’s orbit about Earth, and they applied to the orbit of the spacecraft Apollo 8 about the moon. Tables 13.1 through 13.3 give additional data for planetary orbits and for the orbits of seven of Earth’s artificial satellites (Figure 13.39). Vanguard 1 sent back data that revealed differences between the levels of Earth’s oceans and provided the first determination of the precise locations of some of the more isolated Pacific islands. The data also verified that the gravitation of the sun and moon would affect the orbits of Earth’s satellites and that solar radiation could exert enough pressure to deform an orbit.
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Perigee height Apogee height
You
Earth
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FIGURE 13.39 The orbit of an Earth satellite: 2a = diameter of Earth + perigee height + apogee height .
Planet
Semimajor axis a*
Eccentricity e
Period T
57.95 108.11 149.57 227.84 778.14 1427.0 2870.3 4499.9 5909
0.2056 0.0068 0.0167 0.0934 0.0484 0.0543 0.0460 0.0082 0.2481
87.967 days 224.701 days 365.256 days 1.8808 years 11.8613 years 29.4568 years 84.0081 years 164.784 years 248.35 years
dH
ass a
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto
nR
TABLE 13.1 Values of a, e, and T for the major planets
*
ma
Millions of kilometers.
TABLE 13.2 Data on Earth’s satellites
Launch date
Sputnik 1 Vanguard 1 Syncom 3 Skylab 4 Tiros II GOES 4 Intelsat 5
Oct. 1957 Mar. 1958 Aug. 1964 Nov. 1973 Oct. 1978 Sept. 1980 Dec. 1980
57.6 days 300 years 710 6 years 84.06 days 500 years 710 6 years 710 6 years
Mu
ham
Name
Time or expected time aloft
Mass at launch (kg)
Period (min)
Perigee height (km)
Apogee height (km)
Semimajor axis a (km)
Eccentricity
83.6 1.47 39 13,980 734 627 1928
96.2 138.5 1436.2 93.11 102.12 1436.2 1417.67
215 649 35,718 422 850 35,776 35,143
939 4340 35,903 437 866 35,800 35,707
6955 8872 42,189 6808 7236 42,166 41,803
0.052 0.208 0.002 0.001 0.001 0.0003 0.007
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Chapter 13: VectorValued Functions and Motion in Space
You
suf
G = 6.6726 * 10 11 Nm2 kg2 1.99 * 10 30 kg 5.975 * 10 24 kg 6378.533 km 6356.912 km 1436.1 min 1 year = 365.256 days
iaz
Universal gravitational constant: Sun’s mass: Earth’s mass: Equatorial radius of Earth: Polar radius of Earth: Earth’s rotational period: Earth’s orbital period:
i
TABLE 13.3 Numerical data
Mu
ham
ma
dH
ass a
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Syncom 3 is one of a series of U.S. Department of Defense telecommunications satellites. Tiros II (for “television infrared observation satellite”) is one of a series of weather satellites. GOES 4 (for “geostationary operational environmental satellite”) is one of a series of satellites designed to gather information about Earth’s atmosphere. Its orbital period, 1436.2 min, is nearly the same as Earth’s rotational period of 1436.1 min, and its orbit is nearly circular se = 0.0003d. Intelsat 5 is a heavycapacity commercial telecommunications satellite.
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i f su
Chapter 13: VectorValued Functions and Motion in Space
EXERCISES 13.6
1. Period of Skylab 4 Since the orbit of Skylab 4 had a semimajor axis of a = 6808 km , Kepler’s third law with M equal to Earth’s mass should give the period. Calculate it. Compare your result with the value in Table 13.2. 2. Earth’s velocity at perihelion Earth’s distance from the sun at perihelion is approximately 149,577,000 km, and the eccentricity of Earth’s orbit about the sun is 0.0167. Find the velocity y0 of Earth in its orbit at perihelion. (Use Equation (15).)
a H
3. Semimajor axis of Proton I In July 1965, the USSR launched Proton I, weighing 12,200 kg (at launch), with a perigee height of 183 km, an apogee height of 589 km, and a period of 92.25 min. Using the relevant data for the mass of Earth and the gravitational constant G, find the semimajor axis a of the orbit from Equation (3). Compare your answer with the number you get by adding the perigee and apogee heights to the diameter of the Earth.
a uh
m m
ad
4. Semimajor axis of Viking I The Viking I orbiter, which surveyed Mars from August 1975 to June 1976, had a period of 1639 min. Use this and the mass of Mars, 6.418 * 10 23 kg , to find the semimajor axis of the Viking I orbit.
M
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z a i R
u o Y
5. Average diameter of Mars (Continuation of Exercise 4.) The Viking I orbiter was 1499 km from the surface of Mars at its closest point and 35,800 km from the surface at its farthest point. Use this information together with the value you obtained in Exercise 4 to estimate the average diameter of Mars.
Reminder: When a calculation involves the gravitational constant G, express force in newtons, distance in meters, mass in kilograms, and time in seconds.
n a ss
6. Period of Viking 2 The Viking 2 orbiter, which surveyed Mars from September 1975 to August 1976, moved in an ellipse whose semimajor axis was 22,030 km. What was the orbital period? (Express your answer in minutes.)
7. Geosynchronous orbits Several satellites in Earth’s equatorial plane have nearly circular orbits whose periods are the same as Earth’s rotational period. Such orbits are geosynchronous or geostationary because they hold the satellite over the same spot on the Earth’s surface. a. Approximately what is the semimajor axis of a geosynchronous orbit? Give reasons for your answer. b. About how high is a geosynchronous orbit above Earth’s surface? c. Which of the satellites in Table 13.2 have (nearly) geosynchronous orbits?
8. The mass of Mars is 6.418 * 10 23 kg . If a satellite revolving about Mars is to hold a stationary orbit (have the same period as
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 13.6 Planetary Motion and Satellites
i
9. Distance from Earth to the moon The period of the moon’s rotation about Earth is 2.36055 * 10 6 sec . About how far away is the moon?
In Exercises 16 and 17, two planets, planet A and planet B, are orbiting their sun in circular orbits with A being the inner planet and B being farther away from the sun. Suppose the positions of A and B at time t are
suf
the period of Mars’s rotation, which is 1477.4 min), what must the semimajor axis of its orbit be? Give reasons for your answer.
959
rAstd = 2 cos s2ptdi + 2 sin s2ptdj
10. Finding satellite speed A satellite moves around Earth in a circular orbit. Express the satellite’s speed as a function of the orbit’s radius.
and
11. Orbital period If T is measured in seconds and a in meters, what is the value of T 2>a 3 for planets in our solar system? For satellites orbiting Earth? For satellites orbiting the moon? (The moon’s mass is 7.354 * 10 22 kg .)
respectively, where the sun is assumed to be located at the origin and distance is measured in astronomical units. (Notice that planet A moves faster than planet B.) The people on planet A regard their planet, not the sun, as the center of their planetary system (their solar system).
dA 1 # = ƒr * rƒ . 2 dt
You
T 17. Using planet A as the origin, graph the path of planet B. This exercise illustrates the difficulty that people before Kepler’s time, with an earthcentered (planet A) view of our solar system, had in understanding the motions of the planets (i.e., planet B = Mars). See D. G. Saari’s article in the American Mathematical Monthly, Vol. 97 (Feb. 1990), pp. 105–119.
ass a
14. Suppose that r is the position vector of a particle moving along a plane curve and dA> dt is the rate at which the vector sweeps out area. Without introducing coordinates, and assuming the necessary derivatives exist, give a geometric argument based on increments and limits for the validity of the equation
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13. Circular orbits Show that a planet in a circular orbit moves with a constant speed. (Hint: This is a consequence of one of Kepler’s laws.)
16. Using planet A as the origin of a new coordinate system, give parametric equations for the location of planet B at time t. Write your answer in terms of cos sptd and sin sptd .
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12. Type of orbit For what values of y0 in Equation (15) is the orbit in Equation (16) a circle? An ellipse? A parabola? A hyperbola?
rBstd = 3 cos sptdi + 3 sin sptdj,
Mu
ham
ma
dH
15. Kepler’s third law Complete the derivation of Kepler’s third law (the part following Equation (34)).
18. Kepler discovered that the path of Earth around the sun is an ellipse with the sun at one of the foci. Let r(t) be the position vector from the center of the sun to the center of Earth at time t. Let w be the vector from Earth’s South Pole to North Pole. It is known that w is constant and not orthogonal to the plane of the ellipse (Earth’s axis is tilted). In terms of r(t) and w, give the mathematical meaning of (i) perihelion, (ii) aphelion, (iii) equinox, (iv) summer solstice, (v) winter solstice.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13
Chapter 13
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Questions to Guide Your Review
u o Y z a i R n a s s a
Questions to Guide Your Review
959
1. State the rules for differentiating and integrating vector functions. Give examples.
6. How do you measure distance along a smooth curve in space from a preselected base point? Give an example.
2. How do you define and calculate the velocity, speed, direction of motion, and acceleration of a body moving along a sufficiently differentiable space curve? Give an example.
7. What is a differentiable curve’s unit tangent vector? Give an example.
H ad
3. What is special about the derivatives of vector functions of constant length? Give an example.
4. What are the vector and parametric equations for ideal projectile motion? How do you find a projectile’s maximum height, flight time, and range? Give examples.
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5. How do you define and calculate the length of a segment of a smooth space curve? Give an example. What mathematical assumptions are involved in the definition?
h u M
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8. Define curvature, circle of curvature (osculating circle), center of curvature, and radius of curvature for twicedifferentiable curves in the plane. Give examples. What curves have zero curvature? Constant curvature? 9. What is a plane curve’s principal normal vector? When is it defined? Which way does it point? Give an example.
10. How do you define N and k for curves in space? How are these quantities related? Give examples.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
13. State Kepler’s laws. To what phenomena do they apply?
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12. What formulas are available for writing a moving body’s acceleration as a sum of its tangential and normal components? Give an
example. Why might one want to write the acceleration this way? What if the body moves at a constant speed? At a constant speed around a circle?
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11. What is a curve’s binormal vector? Give an example. How is this vector related to the curve’s torsion? Give an example.
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Chapter 13: VectorValued Functions and Motion in Space
Chapter 13
Practice Exercises
u o
Motion in a Cartesian Plane
a. Sketch the curve traced by P during the interval 0 … t … 3 .
In Exercises 1 and 2, graph the curves and sketch their velocity and acceleration vectors at the given values of t. Then write a in the form a = a TT + aNN without finding T and N, and find the value of k at the given values of t.
b. Find v and a at t = 0, 1, 2 , and 3 and add these vectors to your sketch.
A 22 sin t B j, t = 0 and p>4 2. rstd = A 23 sec t B i + A 23 tan t B j, t = 0
1. rstd = s4 cos tdi +
1 21 + t 2
t
i +
21 + t 2
n a s sa
j.
Find the particle’s highest speed. t
t
4. Suppose rstd = se cos tdi + se sin tdj. Show that the angle between r and a never changes. What is the angle? 5. Finding curvature At point P, the velocity and acceleration of a particle moving in the plane are v = 3i + 4j and a = 5i + 15j . Find the curvature of the particle’s path at P.
H d a m
x
6. Find the point on the curve y = e where the curvature is greatest.
7. A particle moves around the unit circle in the xyplane. Its position at time t is r = xi + yj, where x and y are differentiable functions of t. Find dy> dt if v # i = y . Is the motion clockwise, or counterclockwise? 8. You send a message through a pneumatic tube that follows the curve 9y = x 3 (distance in meters). At the point (3, 3), v # i = 4 and a # i = 2 . Find the values of v # j and a # j at (3, 3).
m a
9. Characterizing circular motion A particle moves in the plane so that its velocity and position vectors are always orthogonal. Show that the particle moves in a circle centered at the origin.
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10. Speed along a cycloid A circular wheel with radius 1 ft and center C rolls to the right along the xaxis at a halfturn per second. (See the accompanying figure.) At time t seconds, the position vector of the point P on the wheel’s circumference is r = spt  sin ptdi + s1  cos ptdj.
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iR a y
3. The position of a particle in the plane at time t is r =
Y z
c. At any given time, what is the forward speed of the topmost point of the wheel? Of C?
t
r
C 1
P
x
0
Projectile Motion and Motion in a Plane 11. Shot put A shot leaves the thrower’s hand 6.5 ft above the ground at a 45° angle at 44 ft> sec. Where is it 3 sec later?
12. Javelin A javelin leaves the thrower’s hand 7 ft above the ground at a 45° angle at 80 ft> sec. How high does it go?
13. A golf ball is hit with an initial speed y0 at an angle a to the horizontal from a point that lies at the foot of a straightsided hill that is inclined at an angle f to the horizontal, where 0 6 f 6 a 6
p . 2
Show that the ball lands at a distance 2y02 cos a g cos2 f
sin sa  fd ,
measured up the face of the hill. Hence, show that the greatest range that can be achieved for a given y0 occurs when a = sf>2d + sp>4d , i.e., when the initial velocity vector bisects the angle between the vertical and the hill.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13
where
Mu L0
i (a)
y 2
x
y2
T
a2
s
O
x s 0 at (a, 0)
(b)
FIGURE 13.40 Figures for Exercise 20.
Motion in Space
d $ # # s = 2x 2 + y 2 . dt
Find the lengths of the curves in Exercises 21 and 22. 21. rstd = s2 cos tdi + s2 sin tdj + t 2k,
19. Curvature Express the curvature of the curve rstd = a
x
0
P
ma
ham # # x2 + y2 $ $ $ , 2x 2 + y 2  s 2
suf i
a
18. Radius of curvature Show that the radius of curvature of a twicedifferentiable plane curve rstd = ƒstdi + g stdj is given by the formula
t
1 y = sy0 sin adt  gt 2 , 2
show that x 2 + s y + gt 2>2d2 = y02 t 2 . This shows that projectiles launched simultaneously from the origin at the same initial speed will, at any given instant, all lie on the circle of radius y0 t centered at s0, gt 2>2d , regardless of their launch angle. These circles are the synchronous curves of the launching.
r =
22. rstd = s3 cos tdi + s3 sin tdj + 2t
t
1 1 cos a pu2 b du b i + a sin a pu2 b du bj 2 2 L0
0 … t … p>4
3>2
k,
0 … t … 3
In Exercises 23–26, find T, N, B, k , and t at the given value of t. 23. rstd =
as a function of the directed distance s measured along the curve from the origin. (See the accompanying figure.)
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x
T
By eliminating a from the ideal projectile
x = sy0 cos adt,
0.75
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17. Synchronous curves equations
0.5
20. An alternative definition of curvature in the plane An alternative definition gives the curvature of a sufficiently differentiable plane curve to be ƒ df>ds ƒ , where f is the angle between T and i (Figure 13.40a). Figure 13.40b shows the distance s measured counterclockwise around the circle x 2 + y 2 = a 2 from the point (a, 0) to a point P, along with the angle f at P. Calculate the circle’s curvature using the alternative definition. (Hint: f = u + p>2 .)
a. Assuming that Felke launched the javelin at a 40° angle to the horizontal 6.5 ft above the ground, what was the javelin’s initial speed? b. How high did the javelin go?
0.25
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–0.75 –0.5 –0.25 –0.2 –0.4 –0.6
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T 16. Javelin In Potsdam in 1988, Petra Felke of (then) East Germany set a women’s world record by throwing a javelin 262 ft 5 in.
0.6 0.4 0.2
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b. A new world record of 177 ft. 9 in. was set on June 5, 1988, by Prof. Emeritus Heinrich of Rensselaer Polytechnic Institute, firing from 4 ft. above ground level at the Woodbury Vineyards Winery, New York. Assuming an ideal trajectory, what was the cork’s initial speed?
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T 14. The Dictator The Civil War mortar Dictator weighed so much (17,120 lb) that it had to be mounted on a railroad car. It had a 13in. bore and used a 20lb powder charge to fire a 200lb shell. The mortar was made by Mr. Charles Knapp in his ironworks in Pittsburgh, Pennsylvania, and was used by the Union army in 1864 in the siege of Petersburg, Virginia. How far did it shoot? Here we have a difference of opinion. The ordnance manual claimed 4325 yd, while field officers claimed 4752 yd. Assuming a 45° firing angle, what muzzle speeds are involved here? T 15. The World’s record for popping a champagne cork a. Until 1988, the world’s record for popping a champagne cork was 109 ft. 6 in., once held by Captain Michael Hill of the British Royal Artillery (of course). Assuming Cpt. Hill held the bottle neck at ground level at a 45° angle, and the cork behaved like an ideal projectile, how fast was the cork going as it left the bottle?
Practice Exercises
4 4 1 s1 + td3>2 i + s1  td3>2j + t k, 9 9 3
24. rstd = se t sin 2tdi + se t cos 2tdj + 2e t k,
t = 0
t = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
t = ln 2
26. rstd = s3 cosh 2tdi + s3 sinh 2tdj + 6t k,
35. The view from Skylab 4 What percentage of Earth’s surface area could the astronauts see when Skylab 4 was at its apogee height, 437 km above the surface? To find out, model the visible surface as the surface generated by revolving the circular arc GT, shown here, about the yaxis. Then carry out these steps:
i
1 2t e j, 2
t = ln 2
In Exercises 27 and 28, write a in the form a = a TT + aNN at t = 0 without finding T and N. 27. rstd = s2 + 3t + 3t 2 di + s4t + 4t 2 dj  s6 cos tdk 28. rstd = s2 + tdi + st + 2t 2 dj + s1 + t 2 dk
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25. rstd = ti +
1.
Use similar triangles in the figure to show that y0>6380 = 6380>s6380 + 437d . Solve for y0 .
2.
To four significant digits, calculate the visible area as
29. Find T, N, B, k , and t as functions of t if rstd = ssin tdi +
A 22 cos t B j + ssin tdk.
6380
VA =
31. The position of a particle moving in space at time t Ú 0 is
3.
2px
Ly0
C
1 + a
2
dx b dy . dy
Express the result as a percentage of Earth’s surface area.
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30. At what times in the interval 0 … t … p are the velocity and acceleration vectors of the motion rstd = i + s5 cos tdj + s3 sin tdk orthogonal?
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Find the first time r is orthogonal to the vector i  j. 32. Find equations for the osculating, normal, and rectifying planes of the curve rstd = t i + t 2j + t 3k at the point (1, 1, 1).
0
6380 x
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33. Find parametric equations for the line that is tangent to the curve rstd = e t i + ssin tdj + ln s1  tdk at t = 0 .
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S (Skylab) 437 G x 兹(6380) 2 y 2 T y0
t t rstd = 2i + a4 sin b j + a3  p bk. 2
34. Find parametric equations for the line tangent to the helix r(t) = 2 cos t B i +
A
2 sin t B j + tk at the point where t = p>4 .
Mu
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A
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 962
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Chapter 13: VectorValued Functions and Motion in Space
Chapter 13 Applications
2. A straight river is 20 m wide. The velocity of the river at (x, y) is
1. A straight river is 100 m wide. A rowboat leaves the far shore at time t = 0 . The person in the boat rows at a rate of 20 m> min, always toward the near shore. The velocity of the river at (x, y) is v = a
1 s y  50d2 + 10 b i m>min, 250
Far shore
m a h u M 0
n a ss
0 6 y 6 100 .
d a m
a i R
3xs20  xd j m>min, 100
0 â€Ś x â€Ś 20 .
y
a H
b. How far downstream will the boat land on the near shore? y
v = 
A boat leaves the shore at (0, 0) and travels through the water with a constant velocity. It arrives at the opposite shore at (20, 0). The speed of the boat is always 220 m>min .
a. Given that rs0d = 0i + 100j, what is the position of the boat at time t?
100
u o Y z
Additional and Advanced Exercises
x
Near shore
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0
20
x
a. Find the velocity of the boat. b. Find the location of the boat at time t. c. Sketch the path of the boat.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13
r =
6. A Kepler equation The problem of locating a planet in its orbit at a given time and date eventually leads to solving “Kepler” equations of the form ƒsxd = x  1 
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8. Express the curvature of a twicedifferentiable curve r = ƒsud in the polar coordinate plane in terms of ƒ and its derivatives.
4. Suppose the curve in Exercise 3 is replaced by the conical helix r = au, z = bu shown in the accompanying figure. a. Express the angular velocity du>dt as a function of u .
ma
b. Express the distance the particle travels along the helix as a function of u .
ham
b. With your computer or calculator in radian mode, use Newton’s method to find the solution to as many places as you can.
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x The helix r a, z b
Positive zaxis points down.
a. Show that this particular equation has a solution between x = 0 and x = 2 .
7. In Section 13.6, we found the velocity of a particle moving in the plane to be # # # # v = x i + y j = r ur + r u uu . # # # # a. Express x and y in terms of r and ru by evaluating the dot products v # i and v # j. # # # # b. Express r and ru in terms of x and y by evaluating the dot products v # ur and v # uu .
a
P
1 sin x = 0 . 2
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c. Express the tangential and normal components of the velocity dr> dt and acceleration d 2r>dt 2 as functions of t. Does the acceleration have any nonzero component in the direction of the binormal vector B?
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You
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b. Express the particle’s u and zcoordinates as functions of t.
y
s1 + edr0 1 + e cos u
that a planet is closest to its sun when u = 0 and show that r = r0 at that time.
a. Find the angular velocity du>dt when u = 2p .
r
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5. Deduce from the orbit equation
sa, b 7 0d
under the influence of gravity, as in the accompanying figure. The u in this equation is the cylindrical coordinate u and the helix is the curve r = a, z = bu, u Ú 0 , in cylindrical coordinates. We assume u to be a differentiable function of t for the motion. The law of conservation of energy tells us that the particle’s speed after it has fallen straight down a distance z is 22gz , where g is the constant acceleration of gravity.
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Polar Coordinate Systems and Motion in Space
3. A frictionless particle P, starting from rest at time t = 0 at the point (a, 0, 0), slides down the helix rsud = sa cos udi + sa sin udj + buk
Additional and Advanced Exercises
x
9. A slender rod through the origin of the polar coordinate plane rotates (in the plane) about the origin at the rate of 3 rad> min. A beetle starting from the point (2, 0) crawls along the rod toward the origin at the rate of 1 in.> min. a. Find the beetle’s acceleration and velocity in polar form when it is halfway to (1 in. from) the origin.
T b. To the nearest tenth of an inch, what will be the length of the path the beetle has traveled by the time it reaches the origin? 10. Conservation of angular momentum Let r(t) denote the position in space of a moving object at time t. Suppose the force acting on the object at time t is
y
Fstd = Conical helix r a, z b
Mu
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Cone z ba r Positive zaxis points down. z
To Read it Online & Download:
c rstd , ƒ rstd ƒ 3
where c is a constant. In physics the angular momentum of an object at time t is defined to be Lstd = rstd * mvstd , where m is the mass of the object and v(t) is the velocity. Prove that angular momentum is a conserved quantity; i.e., prove that L(t) is a constant vector, independent of time. Remember Newton’s law F = ma. (This is a calculus problem, not a physics problem.)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
ur = scos udi + ssin udj,
uu = ssin udi + scos udj,
and k (see accompanying figure). The particle’s position vector is then r = r ur + z k, where r is the positive polar distance coordinate of the particle’s position. z k
i
11. Unit vectors for position and motion in cylindrical coordinates When the position of a particle moving in space is given in cylindrical coordinates, the unit vectors we use to describe its position and motion are
a. Show that ur, uu , and k, in this order, form a righthanded frame of unit vectors. b. Show that d ur = uu du
and
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Cylindrical Coordinate Systems
d uu = ur . du
c. Assuming that the necessary derivatives with respect to t # # $ # exist, express v = r and a = r in terms of ur , uu, k, r , and u . # (The dots indicate derivatives with respect to t: r means $ dr>dt, r means d 2r>dt 2 , and so on.) Section 13.6 derives these formulas and shows how the vectors mentioned here are used in describing planetary motion.
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12. Arc length in cylindrical coordinates a. Show that when you express ds 2 = dx 2 + dy 2 + dz 2 in terms of cylindrical coordinates, you get ds 2 = dr 2 + r 2 du2 + dz 2 .
u ur r
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b. Interpret this result geometrically in terms of the edges and a diagonal of a box. Sketch the box.
z 0
c. Use the result in part (a) to find the length of the curve r = e u, z = e u, 0 … u … u ln 8 .
Mu
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ma
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(r, , 0)
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r x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 13: VectorValued Functions and Motion in Space
Technology Application Projects
i
Chapter 13
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Mathematica / Maple Module
You
Radar Tracking of a Moving Object Visualize position, velocity, and acceleration vectors to analyze motion.
Mathematica / Maple Module Parametric and Polar Equations with a Figure Skater Visualize position, velocity, and acceleration vectors to analyze motion.
Mathematica / Maple Module
Mu
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ma
dH
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Moving in Three Dimensions Compute distance traveled, speed, curvature, and torsion for motion along a space curve. Visualize and compute the tangential, normal, and binormal vectors associated with motion along a space curve.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
14
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Chapter
PARTIAL DERIVATIVES
14.1
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OVERVIEW In studying a realworld phenomenon, a quantity being investigated usually depends on two or more independent variables. So we need to extend the basic ideas of the calculus of functions of a single variable to functions of several variables. Although the calculus rules remain essentially the same, the calculus is even richer. The derivatives of functions of several variables are more varied and more interesting because of the different ways in which the variables can interact. Their integrals lead to a greater variety of applications. The studies of probability, statistics, fluid dynamics, and electricity, to mention only a few, all lead in natural ways to functions of more than one variable.
Functions of Several Variables
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Many functions depend on more than one independent variable. The function V = pr 2h calculates the volume of a right circular cylinder from its radius and height. The function ƒsx, yd = x 2 + y 2 calculates the height of the paraboloid z = x 2 + y 2 above the point P(x, y) from the two coordinates of P. The temperature T of a point on Earth’s surface depends on its latitude x and longitude y, expressed by writing T = ƒsx, yd. In this section, we define functions of more than one independent variable and discuss ways to graph them. Realvalued functions of several independent real variables are defined much the way you would imagine from the singlevariable case. The domains are sets of ordered pairs (triples, quadruples, ntuples) of real numbers, and the ranges are sets of real numbers of the kind we have worked with all along.
DEFINITIONS Function of n Independent Variables Suppose D is a set of ntuples of real numbers sx1, x2 , Á , xn d. A realvalued function ƒ on D is a rule that assigns a unique (single) real number w = ƒsx1, x2 , Á , xn d to each element in D. The set D is the function’s domain. The set of wvalues taken on by ƒ is the function’s range. The symbol w is the dependent variable of ƒ, and ƒ is said to be a function of the n independent variables x1 to xn. We also call the xj’s the function’s input variables and call w the function’s output variable.
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Chapter 14: Partial Derivatives
EXAMPLE 1
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If ƒ is a function of two independent variables, we usually call the independent variables x and y and picture the domain of ƒ as a region in the xyplane. If ƒ is a function of three independent variables, we call the variables x, y, and z and picture the domain as a region in space. In applications, we tend to use letters that remind us of what the variables stand for. To say that the volume of a right circular cylinder is a function of its radius and height, we might write V = ƒsr, hd. To be more specific, we might replace the notation ƒ(r, h) by the formula that calculates the value of V from the values of r and h, and write V = pr 2h. In either case, r and h would be the independent variables and V the dependent variable of the function. As usual, we evaluate functions defined by formulas by substituting the values of the independent variables in the formula and calculating the corresponding value of the dependent variable.
Evaluating a Function
The value of ƒsx, y, zd = 2x 2 + y 2 + z 2 at the point (3, 0, 4) is
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ƒs3, 0, 4d = 2s3d2 + s0d2 + s4d2 = 225 = 5.
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From Section 12.1, we recognize ƒ as the distance function from the origin to the point (x, y, z) in Cartesian space coordinates.
Domains and Ranges
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In defining a function of more than one variable, we follow the usual practice of excluding inputs that lead to complex numbers or division by zero. If ƒsx, yd = 2y  x 2, y cannot be less than x 2. If ƒsx, yd = 1>sxyd, xy cannot be zero. The domain of a function is assumed to be the largest set for which the defining rule generates real numbers, unless the domain is otherwise specified explicitly. The range consists of the set of output values for the dependent variable.
EXAMPLE 2(a) Function
Domain
Range
w = 2y  x 2
y Ú x2
[0, q d
1 w = xy
xy Z 0
s  q , 0d ´ s0, q d
w = sin xy
Entire plane
[1, 1]
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Functions of Two Variables
(b)
Functions of Three Variables
Function
Domain
Range
w = 2x 2 + y 2 + z 2
Entire space
[0, q d
1 x2 + y2 + z2 w = xy ln z
sx, y, zd Z s0, 0, 0d
s0, q d
Halfspace z 7 0
s  q, q d
w =
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.1
967
Functions of Several Variables
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Functions of Two Variables
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Regions in the plane can have interior points and boundary points just like intervals on the real line. Closed intervals [a, b] include their boundary points, open intervals (a, b) don’t include their boundary points, and intervals such as [a, b) are neither open nor closed. (x0 , y0)
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DEFINITIONS Interior and Boundary Points, Open, Closed A point sx0 , y0 d in a region (set) R in the xyplane is an interior point of R if it is the center of a disk of positive radius that lies entirely in R (Figure 14.1). A point sx0 , y0 d is a boundary point of R if every disk centered at sx0 , y0 d contains points that lie outside of R as well as points that lie in R. (The boundary point itself need not belong to R.) The interior points of a region, as a set, make up the interior of the region. The region’s boundary points make up its boundary. A region is open if it consists entirely of interior points. A region is closed if it contains all its boundary points (Figure 14.2).
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(a) Interior point
(x0 , y0)
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FIGURE 14.1 Interior points and boundary points of a plane region R. An interior point is necessarily a point of R. A boundary point of R need not belong to R.
y
ass a
(b) Boundary point
nR
R
x
0
dH
{(x, y) x 2 y 2 1} Open unit disk. Every point an interior point.
0
{(x, y) x 2 y 2 1} Boundary of unit disk. (The unit circle.)
y
x
0
x
{(x, y) x 2 y 2 ⱕ 1} Closed unit disk. Contains all boundary points.
Mu
ham
ma
FIGURE 14.2 Interior points and boundary points of the unit disk in the plane.
As with intervals of real numbers, some regions in the plane are neither open nor closed. If you start with the open disk in Figure 14.2 and add to it some of but not all its boundary points, the resulting set is neither open nor closed. The boundary points that are there keep the set from being open. The absence of the remaining boundary points keeps the set from being closed.
DEFINITIONS Bounded and Unbounded Regions in the Plane A region in the plane is bounded if it lies inside a disk of fixed radius. A region is unbounded if it is not bounded.
Examples of bounded sets in the plane include line segments, triangles, interiors of triangles, rectangles, circles, and disks. Examples of unbounded sets in the plane include lines, coordinate axes, the graphs of functions defined on infinite intervals, quadrants, halfplanes, and the plane itself.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 968
Chapter 14: Partial Derivatives
EXAMPLE 3
Interior points, where y x 2 0
Describing the Domain of a Function of Two Variables
i
y
suf
Describe the domain of the function ƒsx, yd = 2y  x 2.
Since ƒ is defined only where y  x 2 Ú 0, the domain is the closed, unbounded region shown in Figure 14.3. The parabola y = x 2 is the boundary of the domain. The points above the parabola make up the domain’s interior.
FIGURE 14.3 The domain of ƒsx, yd = 2y  x 2 consists of the shaded region and its bounding parabola y = x 2 (Example 3).
z 100 f (x, y) 75
The surface z f (x, y) 100 x 2 y 2 is the graph of f.
f (x, y) 51 (a typical level curve in the function’s domain) 10 10
f (x, y) 0
DEFINITIONS Level Curve, Graph, Surface The set of points in the plane where a function ƒ(x, y) has a constant value ƒsx, yd = c is called a level curve of ƒ. The set of all points (x, y, ƒ(x, y)) in space, for (x, y) in the domain of ƒ, is called the graph of ƒ. The graph of ƒ is also called the surface z f sx, yd.
EXAMPLE 4
Mu
ham
FIGURE 14.4 The graph and selected level curves of the function ƒsx, yd = 100  x 2  y 2 (Example 4).
Graphing a Function of Two Variables
Graph ƒsx, yd = 100  x 2  y 2 and plot the level curves ƒsx, yd = 0, ƒsx, yd = 51, and ƒsx, yd = 75 in the domain of ƒ in the plane. Solution The domain of ƒ is the entire xyplane, and the range of ƒ is the set of real numbers less than or equal to 100. The graph is the paraboloid z = 100  x 2  y 2, a portion of which is shown in Figure 14.4. The level curve ƒsx, yd = 0 is the set of points in the xyplane at which
ƒsx, yd = 100  x 2  y 2 = 0,
or
x 2 + y 2 = 100,
which is the circle of radius 10 centered at the origin. Similarly, the level curves ƒsx, yd = 51 and ƒsx, yd = 75 (Figure 14.4) are the circles
ma
x
y
There are two standard ways to picture the values of a function ƒ(x, y). One is to draw and label curves in the domain on which ƒ has a constant value. The other is to sketch the surface z = ƒsx, yd in space.
iaz
1
Graphs, Level Curves, and Contours of Functions of Two Variables
nR
0
–1
x
ass a
The parabola y x2 0 is the boundary.
1
dH
Outside, y x2 0
You
Solution
ƒsx, yd = 100  x 2  y 2 = 51,
or
x 2 + y 2 = 49
ƒsx, yd = 100  x 2  y 2 = 75,
or
x 2 + y 2 = 25.
The level curve ƒsx, yd = 100 consists of the origin alone. (It is still a level curve.) The curve in space in which the plane z = c cuts a surface z = ƒsx, yd is made up of the points that represent the function value ƒsx, yd = c. It is called the contour curve ƒsx, yd = c to distinguish it from the level curve ƒsx, yd = c in the domain of ƒ. Figure 14.5 shows the contour curve ƒsx, yd = 75 on the surface z = 100  x 2  y 2 defined by the function ƒsx, yd = 100  x 2  y 2. The contour curve lies directly above the circle x 2 + y 2 = 25, which is the level curve ƒsx, yd = 75 in the function’s domain. Not everyone makes this distinction, however, and you may wish to call both kinds of curves by a single name and rely on context to convey which one you have in mind. On most maps, for example, the curves that represent constant elevation (height above sea level) are called contours, not level curves (Figure 14.6).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.1 Functions of Several Variables
z 100 x 2 y 2
100
You
Plane z 75
suf
z
i
The contour curve f (x, y) 100 x 2 y 2 75 is the circle x 2 y 2 25 in the plane z 75.
969
75
iaz
0 y
nR
x The level curve f (x, y) 100 x 2 y 2 75 is the circle x 2 y 2 25 in the xyplane.
ass a
FIGURE 14.5 A plane z = c parallel to the xyplane intersecting a surface z = ƒsx, yd produces a contour curve.
FIGURE 14.6 Contours on Mt. Washington in New Hampshire. (Reproduced by permission from the Appalachian Mountain Club.)
Functions of Three Variables
ma
dH
In the plane, the points where a function of two independent variables has a constant value ƒsx, yd = c make a curve in the function’s domain. In space, the points where a function of three independent variables has a constant value ƒsx, y, zd = c make a surface in the function’s domain.
Mu
ham
DEFINITION Level Surface The set of points (x, y, z) in space where a function of three independent variables has a constant value ƒsx, y, zd = c is called a level surface of ƒ.
Since the graphs of functions of three variables consist of points (x, y, z, ƒ(x, y, z)) lying in a fourdimensional space, we cannot sketch them effectively in our threedimensional frame of reference. We can see how the function behaves, however, by looking at its threedimensional level surfaces.
EXAMPLE 5
Describing Level Surfaces of a Function of Three Variables
Describe the level surfaces of the function
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ƒsx, y, zd = 2x 2 + y 2 + z 2 .
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4100 AWL/Thomas_ch14p9651066 9/2/04 11:09 AM Page 970
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 970
Chapter 14: Partial Derivatives
1 y
2 3 x
FIGURE 14.7 The level surfaces of ƒsx, y, zd = 2x 2 + y 2 + z 2 are concentric spheres (Example 5).
z
y
i
suf
DEFINITIONS Interior and Boundary Points for Space Regions A point sx0 , y0 , z0 d in a region R in space is an interior point of R if it is the center of a solid ball that lies entirely in R (Figure 14.8a). A point sx0 , y0 , z0 d is a boundary point of R if every sphere centered at sx0 , y0 , z0 d encloses points that lie outside of R as well as points that lie inside R (Figure 14.8b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary.
(a) Interior point
(x 0 , y0 , z 0 )
ham
z
y
Mu
Examples of open sets in space include the interior of a sphere, the open halfspace z 7 0, the first octant (where x, y, and z are all positive), and space itself. Examples of closed sets in space include lines, planes, the closed halfspace z Ú 0, the first octant together with its bounding planes, and space itself (since it has no boundary points). A solid sphere with part of its boundary removed or a solid cube with a missing face, edge, or corner point would be neither open nor closed. Functions of more than three independent variables are also important. For example, the temperature on a surface in space may depend not only on the location of the point P(x, y, z) on the surface, but also on time t when it is visited, so we would write T = ƒsx, y, z, td.
ma
x
x
The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra dimension, we use solid balls of positive radius instead of disks.
dH
(x 0 , y0 , z 0 )
You
兹x 2 y 2 z 2 3
iaz
兹x 2 y 2 z 2 2
nR
z
Solution The value of ƒ is the distance from the origin to the point (x, y, z). Each level surface 2x 2 + y 2 + z 2 = c, c 7 0 , is a sphere of radius c centered at the origin. Figure 14.7 shows a cutaway view of three of these spheres. The level surface 2x 2 + y 2 + z 2 = 0 consists of the origin alone. We are not graphing the function here; we are looking at level surfaces in the function’s domain. The level surfaces show how the function’s values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from one sphere to another, the function’s value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. The dependence of change on direction is important. We return to it in Section 14.5.
ass a
兹x 2 y 2 z 2 1
(b) Boundary point
FIGURE 14.8 Interior points and boundary points of a region in space.
Computer Graphing Threedimensional graphing programs for computers and calculators make it possible to graph functions of two variables with only a few keystrokes. We can often get information more quickly from a graph than from a formula.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.1
971
Modeling Temperature Beneath Earthâ€™s Surface
i
EXAMPLE 6
Functions of Several Variables
suf
The temperature beneath the Earthâ€™s surface is a function of the depth x beneath the surface and the time t of the year. If we measure x in feet and t as the number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function
You
w = cos s1.7 * 10 2t  0.2xde 0.2x.
iaz
(The temperature at 0 ft is scaled to vary from +1 to 1 , so that the variation at x feet can be interpreted as a fraction of the variation at the surface.) Figure 14.9 shows a computergenerated graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 30 ft, there is almost no variation during the year.
ass a
Temperature
nR
w
, ft
Mu
ham
ma
dH
x
Depth 30
15
FIGURE 14.9 of
Days t
This computergenerated graph
w = cos s1.7 * 102t  0.2xde0.2x shows the seasonal variation of the temperature belowground as a fraction of surface temperature. At x = 15 ft, the variation is only 5% of the variation at the surface. At x = 30 ft, the variation is less than 0.25% of the surface variation (Example 6). (Adapted from art provided by Norton Starr.)
The graph also shows that the temperature 15 ft below the surface is about half a year out of phase with the surface temperature. When the temperature is lowest on the surface (late January, say), it is at its highest 15 ft below. Fifteen feet below the ground, the seasons are reversed. Figure 14.10 shows computergenerated graphs of a number of functions of two variables together with their level curves.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 972
Chapter 14: Partial Derivatives
y
suf
i
z
x
You
x
y (a) z e
– (x 2
y 2 )/8
(sin x 2 cos y 2 )
y
nR
iaz
z
x
y
ass a
x
(b) z sin x 2 sin y
y
ma
dH
z
x
y
x
Mu
ham
(c) z (4x 2 y 2 )e –x
2y 2
y
z
x y
x (d) z xye –y
2
FIGURE 14.10 Computergenerated graphs and level surfaces of typical functions of two variables.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.1 Functions of Several Variables
Domain, Range, and Level Curves
17.
18. y
x
2. ƒsx, yd = 2y  x
2
3. ƒsx, yd = 4x + 9y
2
2
4. ƒsx, yd = x  y
2
216  x 2  y 2
9. ƒsx, yd = ln sx 2 + y 2 d
8. ƒsx, yd = 29  x 2  y 2 10. ƒsx, yd = e sx
2
+ y 2d
a.
y 12. ƒsx, yd = tan1 a x b
11. ƒsx, yd = sin1 s y  xd
Identifying Surfaces and Level Curves
13.
14. y
y
dH
z–
z
xy 2 x2 y2
x
ma
x
ham
x
y
c.
16.
y
y
z (cos x)(cos y) e –兹x 2 y 2 /4
b.
15.
z
x
ass a
Exercises 13–18 show level curves for the functions graphed in (a)–(f). Match each set of curves with the appropriate function.
nR
1
iaz
6. ƒsx, yd = y>x 2
5. ƒsx, yd = xy 7. ƒsx, yd =
y
You
In Exercises 1–12, (a) find the function’s domain, (b) find the function’s range, (c) describe the function’s level curves, (d) find the boundary of the function’s domain, (e) determine if the domain is an open region, a closed region, or neither, and (f ) decide if the domain is bounded or unbounded. 1. ƒsx, yd = y  x
suf
i
EXERCISES 14.1
973
y
Mu
x
To Read it Online & Download:
x
x
y z
1 4x 2 y 2
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x
4100 AWL/Thomas_ch14p9651066 8/25/04 2:52 PM Page 974
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 974
Chapter 14: Partial Derivatives 26. ƒsx, yd = 4x 2 + y 2 + 1
27. ƒsx, yd = 1  ƒ y ƒ
28. ƒsx, yd = 1  ƒ x ƒ  ƒ y ƒ
i
25. ƒsx, yd = 4x 2 + y 2
suf
z
d.
Finding a Level Curve
In Exercises 29–32, find an equation for the level curve of the function ƒ(x, y) that passes through the given point.
x
30. ƒsx, yd = 2x 2  1,
z e – y cos x
s1, 0d
dt , A  22, 22 B 2 Lx 1 + t q n x 32. ƒsx, yd = a a y b , s1, 2d y
31. ƒsx, yd = e.
A 222, 22 B
You
29. ƒsx, yd = 16  x 2  y 2,
y
z
iaz
n=0
Sketching Level Surfaces
nR
In Exercises 33–40, sketch a typical level surface for the function. y x
33. ƒsx, y, zd = x 2 + y 2 + z 2
34. ƒsx, y, zd = ln sx 2 + y 2 + z 2 d
35. ƒsx, y, zd = x + z
36. ƒsx, y, zd = z
2
37. ƒsx, y, zd = x + y
2
2
39. ƒsx, y, zd = z  x  y
xy(x 2 y 2 )
40. ƒsx, y, zd = sx 2>25d + s y 2>16d + sz 2>9d
ass a
z
x2 y2
f.
Finding a Level Surface
z
dH
In Exercises 41–44, find an equation for the level surface of the function through the given point.
y
ham
ma
x
z y2 y4 x2
Identifying Functions of Two Variables Display the values of the functions in Exercises 19–28 in two ways: (a) by sketching the surface z = ƒsx, yd and (b) by drawing an assortment of level curves in the function’s domain. Label each level curve with its function value.
Mu
38. ƒsx, y, zd = y 2 + z 2 2
19. ƒsx, yd = y 2
20. ƒsx, yd = 4  y 2
21. ƒsx, yd = x 2 + y 2
22. ƒsx, yd = 2x 2 + y 2
23. ƒsx, yd = sx 2 + y 2 d
24. ƒsx, yd = 4  x 2  y 2
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41. ƒsx, y, zd = 2x  y  ln z,
s3, 1, 1d
42. ƒsx, y, zd = ln sx 2 + y + z 2 d, s 1, 2, 1d q sx + ydn , sln 2, ln 4, 3d 43. gsx, y, zd = a n!z n n=0 y
44. gsx, y, zd =
du
Lx 21  u2
z
+
dt , L22 t2t 2  1
s0, 1>2, 2d
Theory and Examples 45. The maximum value of a function on a line in space Does the function ƒsx, y, zd = xyz have a maximum value on the line x = 20  t, y = t, z = 20? If so, what is it? Give reasons for your answer. (Hint: Along the line, w = ƒsx, y, zd is a differentiable function of t.) 46. The minimum value of a function on a line in space Does the function ƒsx, y, zd = xy  z have a minimum value on the line x = t  1, y = t  2, z = t + 7? If so, what is it? Give reasons for your answer. (Hint: Along the line, w = ƒsx, y, zd is a differentiable function of t.) 47. The Concorde’s sonic booms Sound waves from the Concorde bend as the temperature changes above and below the altitude at which the plane flies. The sonic boom carpet is the region on the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.1 Functions of Several Variables COMPUTER EXPLORATIONS
i
Explicit Surfaces
suf
ground that receives shock waves directly from the plane, not reflected from the atmosphere or diffracted along the ground. The carpet is determined by the grazing rays striking the ground from the point directly under the plane. (See accompanying figure.)
975
Use a CAS to perform the following steps for each of the functions in Exercises 49–52. a. Plot the surface over the given rectangle.
You
b. Plot several level curves in the rectangle.
c. Plot the level curve of ƒ through the given point. y 49. ƒsx, yd = x sin + y sin 2x, 0 … x … 5p 0 … y … 5p, 2 Ps3p, 3pd 50. ƒsx, yd = ssin xdscos yde 2x 0 … y … 5p, Ps4p, 4pd
2
B w
0.1
T = air temperature at ground level sin degrees Kelvind d = the vertical temperature gradient stemperature drop in degrees Kelvin per kilometerd. 1>2
.
dH
Th b d
0 … x … 5p,
ma
The Washingtonbound Concorde approached the United States from Europe on a course that took it south of Nantucket Island at an altitude of 16.8 km. If the surface temperature is 290 K and the vertical temperature gradient is 5 K> km, how many kilometers south of Nantucket did the plane have to be flown to keep its sonic boom carpet away from the island? (From “Concorde Sonic Booms as an Atmospheric Probe” by N. K. Balachandra, W. L. Donn, and D. H. Rind, Science, Vol. 197 (July 1, 1977), pp. 47–49.)
Mu
ham
48. As you know, the graph of a realvalued function of a single real variable is a set in a twocoordinate space. The graph of a realvalued function of two independent real variables is a set in a threecoordinate space. The graph of a realvalued function of three independent real variables is a set in a fourcoordinate space. How would you define the graph of a realvalued function ƒsx1, x2 , x3 , x4 d of four independent real variables? How would you define the graph of a realvalued function ƒsx1, x2 , x3 , Á , xn d of n independent real variables?
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0 … x … 2p,
Implicit Surfaces
Use a CAS to plot the level surfaces in Exercises 53–56. 53. 4 ln sx 2 + y 2 + z 2 d = 1
ass a
h = the Concorde’s altitude sin kilometersd
52. ƒsx, yd = e sx  yd sin sx 2 + y 2 d, 2p … y … p, Psp, pd
nR
The width w of the region in which people on the ground hear the Concorde’s sonic boom directly, not reflected from a layer in the atmosphere, is a function of
w = 4a
,
51. ƒsx, yd = sin sx + 2 cos yd, 2p … x … 2p, 2p … y … 2p, Psp, pd
Sonic boom carpet
The formula for w is
+ y2>8
iaz
A
54. x 2 + z 2 = 1
55. x + y 2  3z 2 = 1 x 56. sin a b  scos yd2x 2 + z 2 = 2 2
Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations x = ƒstd, y = gstd defined on some parameter interval I, you can sometimes describe surfaces in space with a triple of equations x = ƒsu, yd, y = gsu, yd, z = hsu, yd defined on some parameter rectangle a … u … b, c … y … d. Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.6.) Use a CAS to plot the surfaces in Exercises 57–60. Also plot several level curves in the xyplane. 57. x = u cos y, y = u sin y, 0 … y … 2p
z = u,
0 … u … 2,
58. x = u cos y, y = u sin y, 0 … y … 2p
z = y,
0 … u … 2,
59. x = s2 + cos ud cos y, y = s2 + cos ud sin y, 0 … u … 2p, 0 … y … 2p 60. x = 2 cos u cos y, y = 2 cos u sin y, 0 … u … 2p, 0 … y … p
z = sin u,
z = 2 sin u,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
Limits and Continuity in Higher Dimensions
i
14.2
suf
976
You
This section treats limits and continuity for multivariable functions. The definition of the limit of a function of two or three variables is similar to the definition of the limit of a function of a single variable but with a crucial difference, as we now see.
Limits
nR
iaz
If the values of ƒ(x, y) lie arbitrarily close to a fixed real number L for all points (x, y) sufficiently close to a point sx0 , y0 d, we say that ƒ approaches the limit L as (x, y) approaches sx0 , y0 d. This is similar to the informal definition for the limit of a function of a single variable. Notice, however, that if sx0 , y0 d lies in the interior of ƒ’s domain, (x, y) can approach sx0 , y0 d from any direction. The direction of approach can be an issue, as in some of the examples that follow.
DEFINITION Limit of a Function of Two Variables We say that a function ƒ(x, y) approaches the limit L as (x, y) approaches sx0 , y0 d, and write lim
ass a
sx, yd:sx0, y0d
ƒsx, yd = L
if, for every number P 7 0, there exists a corresponding number d 7 0 such that for all (x, y) in the domain of ƒ, 0 6 2sx  x0 d2 + s y  y0 d2 6 d.
whenever
dH
ƒ ƒsx, yd  L ƒ 6 P
Mu
ham
ma
The definition of limit says that the distance between ƒ(x, y) and L becomes arbitrarily small whenever the distance from (x, y) to sx0 , y0 d is made sufficiently small (but not 0). The definition of limit applies to boundary points sx0 , y0 d as well as interior points of the domain of ƒ. The only requirement is that the point (x, y) remain in the domain at all times. It can be shown, as for functions of a single variable, that lim
x = x0
lim
y = y0
lim
k = k
sx, yd:sx0, y0d sx, yd:sx0, y0d sx, yd:sx0, y0d
sany number kd.
For example, in the first limit statement above, ƒsx, yd = x and L = x0. Using the definition of limit, suppose that P 7 0 is chosen. If we let d equal this P, we see that 0 6 2sx  x0 d2 + sy  y0 d2 6 d = P implies
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0 6 2sx  x0 d2 6 P ƒ x  x0 ƒ 6 P
2a 2 = ƒ a ƒ
ƒ ƒsx, yd  x0 ƒ 6 P
x = ƒsx, yd
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.2
Limits and Continuity in Higher Dimensions
977
0 6 2sx  x0 d2 + s y  y0 d2 6 d.
whenever
So lim
ƒsx, yd =
lim
sx, yd:sx0 , y0d
x = x0.
You
sx, yd:sx0 , y0d
suf
ƒ ƒsx, yd  x0 ƒ 6 P
i
That is,
It can also be shown that the limit of the sum of two functions is the sum of their limits (when they both exist), with similar results for the limits of the differences, products, constant multiples, quotients, and powers.
lim
ƒsx, yd = L
and
nR
sx, yd:sx0, y0d
iaz
THEOREM 1 Properties of Limits of Functions of Two Variables The following rules hold if L, M, and k are real numbers and
1.
Sum Rule:
2.
Difference Rule:
3.
Product Rule:
lim
sx, yd:sx0 , y0d
gsx, yd = M.
lim
(ƒsx, yd + gsx, ydd = L + M
lim
(ƒsx, yd  gsx, ydd = L  M
lim
sƒsx, yd # gsx, ydd = L # M
lim
skƒsx, ydd = kL
lim
ƒsx, yd L = M gsx, yd
sx, yd:sx0 , y0d
ass a
sx, yd:sx0 , y0d sx, yd:sx0, y0d
sany number kd
Constant Multiple Rule:
5.
Quotient Rule:
6.
Power Rule: If r and s are integers with no common factors, and s Z 0, then
dH
4.
sx, yd:sx0 , y0d
sx, yd:sx0 , y0d
lim
M Z 0
sƒsx, yddr>s = L r>s
sx, yd:sx0 , y0d
Mu
ham
ma
provided Lr>s is a real number. (If s is even, we assume that L 7 0.)
While we won’t prove Theorem 1 here, we give an informal discussion of why it’s true. If (x, y) is sufficiently close to sx0 , y0 d, then ƒ(x, y) is close to L and g(x, y) is close to M (from the informal interpretation of limits). It is then reasonable that ƒsx, yd + gsx, yd is close to L + M; ƒsx, yd  gsx, yd is close to L  M; ƒsx, ydgsx, yd is close to LM; kƒ(x, y) is close to kL; and that ƒ(x, y)> g(x, y) is close to L> M if M Z 0. When we apply Theorem 1 to polynomials and rational functions, we obtain the useful result that the limits of these functions as sx, yd : sx0 , y0 d can be calculated by evaluating the functions at sx0 , y0 d. The only requirement is that the rational functions be defined at sx0 , y0 d.
EXAMPLE 1 (a) (b)
lim
sx, yd:s0,1d
lim
Calculating Limits x  xy + 3 x 2y + 5xy  y 3
sx, yd:s3, 4d
To Read it Online & Download:
=
0  s0ds1d + 3 = 3 s0d2s1d + 5s0ds1d  s1d3
2x 2 + y 2 = 2s3d2 + s 4d2 = 225 = 5
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4100 AWL/Thomas_ch14p9651066 8/25/04 2:53 PM Page 978
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
Calculating Limits
i
EXAMPLE 2 Find
x 2  xy
lim
2x  2y
.
You
sx, yd:s0,0d
suf
978
Since the denominator 2x  2y approaches 0 as sx, yd : s0, 0d, we cannot use the Quotient Rule from Theorem 1. If we multiply numerator and denominator by 2x + 2y, however, we produce an equivalent fraction whose limit we can find:
Solution
x 2  xy 2x  2y
A 2x  2y B A 2x + 2y B
lim
=
sx, yd:s0,0d
x A x  y B A 2x + 2y B x  y sx, yd:s0,0d lim
x A 2x + 2y B
nR
=
iaz
lim
sx, yd:s0,0d
A x 2  xy B A 2x + 2y B
lim
=
sx, yd:s0,0d
Algebra
Cancel the nonzero factor sx  yd.
= 0 A 20 + 20 B = 0
ass a
We can cancel the factor sx  yd because the path y = x (along which x  y = 0) is not in the domain of the function x 2  xy
dH
2x  2y
EXAMPLE 3 Find
.
Applying the Limit Definition 4xy 2
lim
sx, yd:s0,0d
x2 + y2
if it exists.
We first observe that along the line x = 0, the function always has value 0 when y Z 0. Likewise, along the line y = 0, the function has value 0 provided x Z 0. So if the limit does exist as (x, y) approaches (0, 0), the value of the limit must be 0. To see if this is true, we apply the definition of limit. Let P 7 0 be given, but arbitrary. We want to find a d 7 0 such that
Mu
ham
ma
Solution
`
4xy 2 x2 + y2
 0` 6 P
whenever
0 6 2x 2 + y 2 6 d
or 4ƒ x ƒ y 2 x2 + y2
whenever
6 P
0 6 2x 2 + y 2 6 d.
Since y 2 … x 2 + y 2 we have that
To Read it Online & Download:
4ƒ x ƒ y 2 x2 + y2
… 4 ƒ x ƒ = 42x 2 … 42x 2 + y 2 .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.2
Limits and Continuity in Higher Dimensions
979
4xy 2
P  0 ` … 42x 2 + y 2 6 4d = 4 a b = P. 4 x + y 2
suf
`
2
It follows from the definition that sx, yd:s0,0d
You
4xy 2
lim
x2 + y2
Continuity
i
So if we choose d = P>4 and let 0 6 2x 2 + y 2 6 d, we get
= 0.
As with functions of a single variable, continuity is defined in terms of limits.
iaz
z
1. 2.
x –y
ƒ is defined at sx0 , y0 d, lim ƒsx, yd exists,
sx, yd:sx0, y0d
lim
sx, yd:sx0, y0d
ƒsx, yd = ƒsx0 , y0 d.
ass a
3.
nR
DEFINITION Continuous Function of Two Variables A function ƒ(x, y) is continuous at the point (x 0, y0) if
A function is continuous if it is continuous at every point of its domain. (a)
0
–0.8
0.8 1
–1
0.8
– 0.8
x
ma
0
As with the definition of limit, the definition of continuity applies at boundary points as well as interior points of the domain of ƒ. The only requirement is that the point (x, y) remain in the domain at all times. As you may have guessed, one of the consequences of Theorem 1 is that algebraic combinations of continuous functions are continuous at every point at which all the functions involved are defined. This means that sums, differences, products, constant multiples, quotients, and powers of continuous functions are continuous where defined. In particular, polynomials and rational functions of two variables are continuous at every point at which they are defined.
dH
y
0.8
– 0.8
–1
ham
1 0.8
–0.8
0
EXAMPLE 4 Show that
2xy
x + y2 ƒsx, yd = L 0,
(b)
FIGURE 14.11
2xy
x + y 0,
2
Mu
ƒsx, yd = L
2
(a) The graph of
2
,
A Function with a Single Point of Discontinuity
,
sx, yd Z s0, 0d sx, yd = s0, 0d
is continuous at every point except the origin (Figure 14.11).
sx, yd Z s0, 0d sx, yd = s0, 0d.
The function is continuous at every point except the origin. (b) The level curves of ƒ (Example 4).
The function ƒ is continuous at any point sx, yd Z s0, 0d because its values are then given by a rational function of x and y. At (0, 0), the value of ƒ is defined, but ƒ, we claim, has no limit as sx, yd : s0, 0d. The reason is that different paths of approach to the origin can lead to different results, as we now see.
Solution
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 980
Chapter 14: Partial Derivatives
ƒsx, yd `
2xy = y = mx
x2 + y2
`
= y = mx
suf
i
For every value of m, the function ƒ has a constant value on the “punctured” line y = mx, x Z 0, because 2xsmxd 2mx 2 2m = = . 2 2 2 2 2 x + smxd x + m x 1 + m2
Therefore, ƒ has this number as its limit as (x, y) approaches (0, 0) along the line: ƒsx, yd =
lim
sx, yd:s0,0d
cƒsx, yd `
d =
You
lim
sx, yd:s0,0d along y = mx
y = mx
2m . 1 + m2
iaz
This limit changes with m. There is therefore no single number we may call the limit of ƒ as (x, y) approaches the origin. The limit fails to exist, and the function is not continuous. Example 4 illustrates an important point about limits of functions of two variables (or even more variables, for that matter). For a limit to exist at a point, the limit must be the same along every approach path. This result is analogous to the singlevariable case where both the left and rightsided limits had to have the same value; therefore, for functions of two or more variables, if we ever find paths with different limits, we know the function has no limit at the point they approach.
nR
z
ass a
y
TwoPath Test for Nonexistence of a Limit If a function ƒ(x, y) has different limits along two different paths as (x, y) approaches sx0 , y0 d, then limsx, yd:sx0, y0d ƒsx, yd does not exist.
Show that the function
(a) y 0
ƒsx, yd =
2x 2 y x4 + y 2
(Figure 14.12) has no limit as (x, y) approaches (0, 0).
1
ma
1
Applying the TwoPath Test
dH
EXAMPLE 5 x
The limit cannot be found by direct substitution, which gives the form 0> 0. We examine the values of ƒ along curves that end at (0, 0). Along the curve y = kx 2, x Z 0, the function has the constant value Solution
0
–1
x
ham
0
–1
0
Mu
(b)
FIGURE 14.12 (a) The graph of ƒsx, yd = 2x 2y>sx 4 + y 2 d. As the graph suggests and the levelcurve values in part (b) confirm, limsx, yd:s0,0d ƒsx, yd does not exist (Example 5).
ƒsx, yd `
2x 2y = y = kx 2
x4 + y 2
`
= y = kx 2
2x 2skx 2 d 2kx 4 2k = 4 = . 4 2 2 x + skx d x + k 2x 4 1 + k2
Therefore, lim
sx, yd:s0,0d along y = kx2
ƒsx, yd =
lim
sx, yd:s0,0d
cƒsx, yd `
y=kx
2
d =
2k . 1 + k2
This limit varies with the path of approach. If (x, y) approaches (0, 0) along the parabola y = x 2, for instance, k = 1 and the limit is 1. If (x, y) approaches (0, 0) along the xaxis, k = 0 and the limit is 0. By the twopath test, ƒ has no limit as (x, y) approaches (0, 0). The language here may seem contradictory. You might well ask, “What do you mean ƒ has no limit as (x, y) approaches the origin — it has lots of limits.” But that is
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.2
Limits and Continuity in Higher Dimensions
981
suf
i
the point. There is no single pathindependent limit, and therefore, by the definition, limsx, yd:s0,0d ƒsx, yd does not exist.
You
Compositions of continuous functions are also continuous. The proof, omitted here, is similar to that for functions of a single variable (Theorem 10 in Section 2.6). Continuity of Composites If ƒ is continuous at sx0 , y0 d and g is a singlevariable function continuous at ƒsx0 , y0 d, then the composite function h = g f defined by hsx, yd = gsƒsx, ydd is continuous at sx0 , y0 d.
e x  y,
iaz
For example, the composite functions
cos
xy
2
x + 1
ln s1 + x 2y 2 d
,
ass a
nR
are continuous at every point (x, y). As with functions of a single variable, the general rule is that composites of continuous functions are continuous. The only requirement is that each function be continuous where it is applied.
Functions of More Than Two Variables
dH
The definitions of limit and continuity for functions of two variables and the conclusions about limits and continuity for sums, products, quotients, powers, and composites all extend to functions of three or more variables. Functions like ln sx + y + zd
y sin z x  1
and
ma
are continuous throughout their domains, and limits like lim
P:s1,0,1d
e x+z z 2 + cos 2xy
=
e11 1 = , 2 s 1d + cos 0 2
where P denotes the point (x, y, z), may be found by direct substitution.
Mu
ham
Extreme Values of Continuous Functions on Closed, Bounded Sets
We have seen that a function of a single variable that is continuous throughout a closed, bounded interval [a, b] takes on an absolute maximum value and an absolute minimum value at least once in [a, b]. The same is true of a function z = ƒsx, yd that is continuous on a closed, bounded set R in the plane (like a line segment, a disk, or a filledin triangle). The function takes on an absolute maximum value at some point in R and an absolute minimum value at some point in R. Theorems similar to these and other theorems of this section hold for functions of three or more variables. A continuous function w = ƒsx, y, zd, for example, must take on absolute maximum and minimum values on any closed, bounded set (solid ball or cube, spherical shell, rectangular solid) on which it is defined. We learn how to find these extreme values in Section 14.7, but first we need to study derivatives in higher dimensions. That is the topic of the next section.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 982
Chapter 14: Partial Derivatives
suf
i
EXERCISES 14.2 Continuity in the Plane
Find the limits in Exercises 1–12. 3x 2  y 2 + 5 lim 1. 2. sx, yd: s0,0d x 2 + y 2 + 2
At what points (x, y) in the plane are the functions in Exercises 27–30 continuous?
7. 9. 11.
4.
lim
sec x tan y
6.
lim
e xy
8.
sx, yd: s3,4d
sx, yd: s0,p>4d sx, yd: s0,ln 2d
e y sin x x sx, yd: s0,0d lim
10.
x sin y
lim
sx, yd: s1,0d
12.
2
x + 1
lim
sx, yd:s2, 3d
1 1 ax + y b
2
x2 + y3 x + y + 1
lim
cos
lim
ln ƒ 1 + x 2 y 2 ƒ
lim
3 cos 2 ƒ xy ƒ  1
sx, yd:s0,0d sx, yd:s1,1d
sx, yd:s1,1d
lim
sx, yd:sp>2,0d
cos y + 1 y  sin x
xZy
xy  y  2x + 2 15. lim x  1 sx, yd: s1,1d
dH
lim
2x  2y x + y  4
lim
sx, yd: s2,2d x+yZ4
lim
sx, yd: s4,3d xZy+1
x2 + y2
x 2  3x + 2
2x + y  2
32. a. ƒsx, y, zd = ln xyz
b. ƒsx, y, zd = e x + y cos z
1 33. a. hsx, y, zd = xy sin z
b. hsx, y, zd =
1 x2 + z2  1
b. hsx, y, zd =
1 ƒ xy ƒ + ƒ z ƒ
19.
lim
sx, yd:s2,0d 2x  y Z 4
1 ƒyƒ + ƒzƒ
No Limit at a Point By considering different paths of approach, show that the functions in Exercises 35–42 have no limit as sx, yd : s0, 0d. 35. ƒsx, yd = 
x 2x 2 + y 2
lim
24. 26.
lim
P:s3,3,0d
1 1 1 ax + y + z b
z
x y
P:s1>4,p>2,2d
lim
22.
lim
P:s0, 2,0d
tan1 xyz
25.
lim
P: sp,0,3d
x
y
2xy + yz
P: s1,1,1d
x2 + z2
37. ƒsx, yd =
ssin2 x + cos2 y + sec2 zd
lim
x4 x + y2 4
2x  2y + 1 x  y  1
Mu 23.
36. ƒsx, yd =
z
Find the limits in Exercises 21–26. P:s1,3,4d
1 x2  y
At what points (x, y, z) in space are the functions in Exercises 31–34 continuous?
22x  y  2 2x  y  4
Limits with Three Variables 21.
b. g sx, yd =
x + y 2 + cos x
Continuity in Space
ma
20.
x  y + 22x  2 2y
ham
18.
y + 4 x 2y  xy + 4x 2  4x
sx, yd: s0,0d xZy
30. a. g sx, yd =
34. a. hsx, y, zd =
xZ1
17.
b. g sx, yd =
ass a
xZy
lim
1 29. a. g sx, yd = sin xy
b. ƒsx, y, zd = 2x 2 + y 2  1
Find the limits in Exercises 13–20 by rewriting the fractions first. x 2  2xy + y 2 x2  y2 13. 14. lim lim x y sx, yd: s1,1d sx, yd:s1,1d x  y
sx, yd: s2, 4d y Z 4, x Z x2
b. ƒsx, yd = ln sx 2 + y 2 d y b. ƒsx, yd = 2 x + 1
31. a. ƒsx, y, zd = x 2 + y 2  2z 2
Limits of Quotients
16.
27. a. ƒsx, yd = sin sx + yd x + y 28. a. ƒsx, yd = x  y
iaz
5.
2x 2 + y 2  1
lim
2y
nR
3.
x
lim
sx, yd:s0,4d
You
Limits with Two Variables
ze2y cos 2x
ln 2x 2 + y 2 + z 2
To Read it Online & Download:
x4  y 2 4
x + y
2
38. ƒsx, yd =
xy ƒ xy ƒ
39. gsx, yd =
x  y x + y
x + y 40. g sx, yd = x  y
41. hsx, yd =
x2 + y y
42. hsx, yd =
x2 x  y 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.2 Limits and Continuity in Higher Dimensions
Theory and Examples 44. If ƒsx0 , y0 d = 3, what can you say about
i
b. Use the formula you obtained in part (a) to show that the limit of ƒ as sx, yd : s0, 0d along the line y = mx varies from 1 to 1 depending on the angle of approach. 50. Continuous extension
ƒsx, yd
if ƒ is continuous at sx0 , y0 d? If ƒ is not continuous at sx0 , y0 d? Give reasons for your answer.
sx, yd:sx0 , y0d
ƒsx, yd = L.
Use this result to support your answers to the questions in Exercises 45 – 48. 45. Does knowing that x 2y 2 tan1 xy 6 1 6 xy 3
tell you anything about tan1 xy lim xy ? sx, yd:s0,0d
tell you anything about lim
ƒrƒ 6 d
ma
4  4 cos 2ƒ xy ƒ ? ƒ xy ƒ
Give reasons for your answer.
47. Does knowing that ƒ sin s1>xd ƒ … 1 tell you anything about lim
1 y sin x ?
ham
sx, yd:s0,0d
Give reasons for your answer.
48. Does knowing that ƒ cos s1>yd ƒ … 1 tell you anything about lim
sx, yd:s0,0d
1 x cos y ?
Mu
Give reasons for your answer.
49. (Continuation of Example 4.) a. Reread Example 4. Then substitute m = tan u into the formula ƒsx, yd `
= y = mx
Q
ƒ ƒsr, ud  L ƒ 6 P.
(1)
If such an L exists, then
dH
x 2y 2 6 4  4 cos 2ƒ xy ƒ 6 2 ƒ xy ƒ 6
sx, yd:s0,0d
If you cannot make any headway with limsx, yd:s0,0d ƒsx, yd in rectangular coordinates, try changing to polar coordinates. Substitute x = r cos u, y = r sin u, and investigate the limit of the resulting expression as r : 0. In other words, try to decide whether there exists a number L satisfying the following criterion: Given P 7 0, there exists a d 7 0 such that for all r and u,
lim
sx, yd: s0,0d
2m 1 + m2
To Read it Online & Download:
ƒsx, yd = lim ƒsr, ud = L. r:0
For instance,
sx, yd: s0,0d
46. Does knowing that 2 ƒ xy ƒ 
Changing to Polar Coordinates
lim
Give reasons for your answer.
x2 + y2
to be continuous at the origin.
ass a
1 
x2  y2
iaz
lim
ƒsx, yd = xy
nR
The Sandwich Theorem for functions of two variables states that if gsx, yd … ƒsx, yd … hsx, yd for all sx, yd Z sx0 , y0 d in a disk centered at sx0 , y0 d and if g and h have the same finite limit L as sx, yd : sx0 , y0 d, then
Define ƒ(0, 0) in a way that extends
You
lim
sx, yd:sx0 , y0d
and simplify the result to show how the value of ƒ varies with the line’s angle of inclination.
suf
43. If limsx, yd:sx0 , y0d ƒsx, yd = L, must ƒ be defined at sx0 , y0 d? Give reasons for your answer.
983
r 3 cos3 u x3 = lim = lim r cos3 u = 0. 2 r: 0 r:0 x + y r2 2
To verify the last of these equalities, we need to show that Equation (1) is satisfied with ƒsr, ud = r cos3 u and L = 0. That is, we need to show that given any P 7 0 there exists a d 7 0 such that for all r and u, ƒrƒ 6 d
Q
ƒ r cos3 u  0 ƒ 6 P.
Since 3 3 ƒ r cos u ƒ = ƒ r ƒ ƒ cos u ƒ … ƒ r ƒ # 1 = ƒ r ƒ ,
the implication holds for all r and u if we take d = P. In contrast, r 2 cos2 u x2 = = cos2 u 2 x + y r2 2
takes on all values from 0 to 1 regardless of how small ƒ r ƒ is, so that limsx, yd:s0,0d x 2>sx 2 + y 2 d does not exist. In each of these instances, the existence or nonexistence of the limit as r : 0 is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) u = constant and yet fail to exist in the broader sense. Example 4 illustrates this point. In polar coordinates, ƒsx, yd = s2x 2yd>sx 4 + y 2 d becomes ƒsr cos u, r sin ud =
r cos u sin 2u r 2 cos4 u + sin2 u
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
r sin u 2r cos2 u sin u = 2 = 1. 2r 2 cos4 u r cos2 u
In Exercises 51–56, find the limit of ƒ as sx, yd : s0, 0d or show that the limit does not exist. x 3  xy 2 x3  y3 51. ƒsx, yd = 2 52. ƒsx, yd = cos a 2 b 2 x + y x + y2 53. ƒsx, yd =
y2 x2 + y2
55. ƒsx, yd = tan1 a 56. ƒsx, yd =
2x 54. ƒsx, yd = 2 x + x + y2
ƒxƒ + ƒyƒ 2
x + y
2
i
2x 2 + y 2 6 d 59. ƒsx, yd = x 2 + y 2, 2
60. ƒsx, yd = y>sx + 1d,
P = 0.05
61. ƒsx, yd = sx + yd>sx 2 + 1d,
Q
65. ƒsx, y, zd =
ƒ ƒsx, y, zd  ƒs0, 0, 0d ƒ 6 P.
P = 0.015
P = 0.008 x + y + z
, P = 0.015 x2 + y2 + z2 + 1 66. ƒsx, y, zd = tan2 x + tan2 y + tan2 z, P = 0.03
ass a
In Exercises 57 and 58, define ƒ(0, 0) in a way that extends ƒ to be continuous at the origin. 3x 2  x 2y 2 + 3y 2 b 57. ƒsx, yd = ln a x2 + y2 3x 2y 58. ƒsx, yd = 2 x + y2
P = 0.02
Each of Exercises 63–66 gives a function ƒ(x, y, z) and a positive number P. In each exercise, show that there exists a d 7 0 such that for all (x, y, z),
64. ƒsx, y, zd = xyz,
x2 + y2
P = 0.01
62. ƒsx, yd = sx + yd>s2 + cos xd,
63. ƒsx, y, zd = x 2 + y 2 + z 2,
x2  y2
ƒ ƒsx, yd  ƒs0, 0d ƒ 6 P.
P = 0.01
2x 2 + y 2 + z 2 6 d
b
Q
suf
=
Each of Exercises 59–62 gives a function ƒ(x, y) and a positive number P. In each exercise, show that there exists a d 7 0 such that for all (x, y),
You
r cos u sin 2u ƒsr cos u, r sin ud = 2 r cos4 u + sr cos2 ud2
Using the dP Definition
iaz
for r Z 0. If we hold u constant and let r : 0, the limit is 0. On the path y = x 2, however, we have r sin u = r 2 cos2 u and
nR
984
67. Show that ƒsx, y, zd = x + y  z is continuous at every point sx0 , y0 , z0 d.
Mu
ham
ma
dH
68. Show that ƒsx, y, zd = x 2 + y 2 + z 2 is continuous at the origin.
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i f su
Chapter 14: Partial Derivatives
14.3
u o zY
Partial Derivatives
a i R
The calculus of several variables is basically singlevariable calculus applied to several variables one at a time. When we hold all but one of the independent variables of a function constant and differentiate with respect to that one variable, we get a “partial” derivative. This section shows how partial derivatives are defined and interpreted geometrically, and how to calculate them by applying the rules for differentiating functions of a single variable.
n a s s a
H d a m am
Partial Derivatives of a Function of Two Variables
h u M
If sx0 , y0 d is a point in the domain of a function ƒ(x, y), the vertical plane y = y0 will cut the surface z = ƒsx, yd in the curve z = ƒsx, y0 d (Figure 14.13). This curve is the graph of the function z = ƒsx, y0 d in the plane y = y0. The horizontal coordinate in this plane is x; the vertical coordinate is z. The yvalue is held constant at y0 , so y is not a variable. We define the partial derivative of ƒ with respect to x at the point sx0 , y0 d as the ordinary derivative of ƒsx, y0 d with respect to x at the point x = x0. To distinguish partial derivatives from ordinary derivatives we use the symbol 0 rather than the d previously used.
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Partial Derivatives
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Vertical axis in the plane y y 0
i
z
P(x 0 , y 0 , f (x 0, y 0))
Tangent line x0
You
z f (x, y)
The curve z f (x, y 0 ) in the plane y y 0
0
x
iaz
y0
(x 0 h, y 0 )
(x 0, y 0 )
y
nR
Horizontal axis in the plane y y 0
FIGURE 14.13 The intersection of the plane y = y0 with the surface z = ƒsx, yd, viewed from above the first quadrant of the xyplane.
ass a
DEFINITION Partial Derivative with Respect to x The partial derivative of ƒ(x, y) with respect to x at the point sx0 , y0 d is 0ƒ ƒsx0 + h, y0 d  ƒsx0 , y0 d ` = lim , 0x sx0, y0d h:0 h
dH
provided the limit exists.
An equivalent expression for the partial derivative is
Mu
ham
ma
d ƒ(x, y0) ` . dx x = x0
The slope of the curve z = ƒsx, y0 d at the point Psx0 , y0, ƒsx0 , y0 dd in the plane y = y0 is the value of the partial derivative of ƒ with respect to x at sx0 , y0 d. The tangent line to the curve at P is the line in the plane y = y0 that passes through P with this slope. The partial derivative 0ƒ>0x at sx0 , y0 d gives the rate of change of ƒ with respect to x when y is held fixed at the value y0 . This is the rate of change of ƒ in the direction of i at sx0 , y0 d. The notation for a partial derivative depends on what we want to emphasize: 0ƒ sx , y d or ƒxsx0 , y0 d “Partial derivative of ƒ with respect to x at sx0 , y0 d” or “ƒ sub 0x 0 0 x at sx0 , y0 d.” Convenient for stressing the point sx0 , y0 d. 0z ` 0x sx0, y0d
ƒx,
0ƒ 0z , z , or 0x x 0x
To Read it Online & Download:
“Partial derivative of z with respect to x at sx0 , y0 d.” Common in science and engineering when you are dealing with variables and do not mention the function explicitly. “Partial derivative of ƒ (or z) with respect to x.” Convenient when you regard the partial derivative as a function in its own right.
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Chapter 14: Partial Derivatives
The definition of the partial derivative of ƒ(x, y) with respect to y at a point sx0 , y0 d is similar to the definition of the partial derivative of ƒ with respect to x. We hold x fixed at the value x0 and take the ordinary derivative of ƒsx0 , yd with respect to y at y0 .
i
Vertical axis in the plane x x0
suf
z
Tangent line P(x 0 , y 0 , f (x 0, y 0 ))
You
DEFINITION Partial Derivative with Respect to y The partial derivative of ƒ(x, y) with respect to y at the point sx0 , y0 d is ƒsx0 , y0 + hd  ƒsx0 , y0 d 0ƒ d , ` = ƒsx0 , yd ` = lim 0y sx0, y0d dy h h:0 y = y0
z f (x, y)
provided the limit exists.
0
(x 0 , y 0 )
y
(x 0 , y 0 k) The curve z f (x 0 , y) in the plane x x0
Horizontal axis in the plane x x 0
0ƒ sx , y d, 0y 0 0
ƒysx0 , y0 d,
ma ham
ƒy .
z
This tangent line P(x 0 , y 0, f (x 0 , y 0 )) has slope fy(x 0 , y 0 ).
Mu
0ƒ , 0y
Notice that we now have two tangent lines associated with the surface z = ƒsx, yd at the point Psx0 , y0 , ƒsx0 , y0 dd (Figure 14.15). Is the plane they determine tangent to the surface at P? We will see that it is, but we have to learn more about partial derivatives before we can find out why.
dH
FIGURE 14.14 The intersection of the plane x = x0 with the surface z = ƒsx, yd, viewed from above the first quadrant of the xyplane.
The slope of the curve z = ƒsx0 , yd at the point Psx0 , y0 , ƒsx0 , y0 dd in the vertical plane x = x0 (Figure 14.14) is the partial derivative of ƒ with respect to y at sx0 , y0 d. The tangent line to the curve at P is the line in the plane x = x0 that passes through P with this slope. The partial derivative gives the rate of change of ƒ with respect to y at sx0 , y0 d when x is held fixed at the value x0. This is the rate of change of ƒ in the direction of j at sx0 , y0 d. The partial derivative with respect to y is denoted the same way as the partial derivative with respect to x:
nR
x
ass a
x0
iaz
y0
The curve z f (x 0, y) in the plane x x 0
This tangent line has slope fx (x 0 , y 0 ). The curve z f (x, y0) in the plane y y0 z f (x, y)
x y y0
(x 0 , y0 )
x x0
y
FIGURE 14.15 Figures 14.13 and 14.14 combined. The tangent lines at the point sx0 , y0 , ƒsx0 , y0 dd determine a plane that, in this picture at least, appears to be tangent to the surface.
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Partial Derivatives
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i
Calculations
EXAMPLE 1
You
suf
The definitions of 0ƒ>0x and 0ƒ>0y give us two different ways of differentiating ƒ at a point: with respect to x in the usual way while treating y as a constant and with respect to y in the usual way while treating x as constant. As the following examples show, the values of these partial derivatives are usually different at a given point sx0 , y0 d.
Finding Partial Derivatives at a Point
Find the values of 0ƒ>0x and 0ƒ>0y at the point s4, 5d if
ƒsx, yd = x 2 + 3xy + y  1.
To find 0ƒ>0x, we treat y as a constant and differentiate with respect to x:
iaz
Solution
0ƒ 0 2 = sx + 3xy + y  1d = 2x + 3 # 1 # y + 0  0 = 2x + 3y. 0x 0x
nR
The value of 0ƒ>0x at s4, 5d is 2s4d + 3s 5d = 7. To find 0ƒ>0y, we treat x as a constant and differentiate with respect to y: 0ƒ 0 2 = sx + 3xy + y  1d = 0 + 3 # x # 1 + 1  0 = 3x + 1. 0y 0y
ass a
The value of 0ƒ>0y at s4, 5d is 3s4d + 1 = 13.
EXAMPLE 2
Finding a Partial Derivative as a Function
Find 0ƒ>0y if ƒsx, yd = y sin xy.
We treat x as a constant and ƒ as a product of y and sin xy:
dH
Solution
ma
0ƒ 0 0 0 = s y sin xyd = y sin xy + ssin xyd s yd 0y 0y 0y 0y = s y cos xyd
Mu
ham
USING TECHNOLOGY
0 sxyd + sin xy = xy cos xy + sin xy. 0y
Partial Differentiation
A simple grapher can support your calculations even in multiple dimensions. If you specify the values of all but one independent variable, the grapher can calculate partial derivatives and can plot traces with respect to that remaining variable. Typically, a CAS can compute partial derivatives symbolically and numerically as easily as it can compute simple derivatives. Most systems use the same command to differentiate a function, regardless of the number of variables. (Simply specify the variable with which differentiation is to take place).
EXAMPLE 3
Partial Derivatives May Be Different Functions
Find ƒx and ƒy if
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ƒsx, yd =
2y . y + cos x
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Chapter 14: Partial Derivatives
We treat ƒ as a quotient. With y held constant, we get
i
Solution
s y + cos xd
0 0 s2yd  2y s y + cos xd 0x 0x s y + cos xd2
s y + cos xds0d  2ys sin xd
2y sin x =
2
s y + cos xd
s y + cos xd2
With x held constant, we get s y + cos xd
s y + cos xd2
s y + cos xds2d  2ys1d =
2
s y + cos xd
0 0 s2yd  2y s y + cos xd 0y dy
iaz
2y 0 ƒy = a b = 0y y + cos x
.
You
=
suf
2y 0 ƒx = a b = 0x y + cos x
=
2 cos x . s y + cos xd2
EXAMPLE 4
nR
Implicit differentiation works for partial derivatives the way it works for ordinary derivatives, as the next example illustrates.
Implicit Partial Differentiation
ass a
Find 0z>0x if the equation
yz  ln z = x + y
defines z as a function of the two independent variables x and y and the partial derivative exists.
dH
Solution We differentiate both sides of the equation with respect to x, holding y constant and treating z as a differentiable function of x:
0y 0x 0 0 s yzd ln z = + 0x 0x 0x 0x
z
Plane x1
2
1
y
Mu
x
x1
FIGURE 14.16 The tangent to the curve of intersection of the plane x = 1 and surface z = x 2 + y 2 at the point (1, 2, 5) (Example 5).
EXAMPLE 5
0z 1 0z  z = 1 + 0 0x 0x
With y constant, 0 0z s yzd = y . 0x 0x
1 0z ay  z b = 1 0x
Tangent line
ham
(1, 2, 5)
ma
Surface z x2 y2
y
0z z = . 0x yz  1
Finding the Slope of a Surface in the yDirection
The plane x = 1 intersects the paraboloid z = x 2 + y 2 in a parabola. Find the slope of the tangent to the parabola at (1, 2, 5) (Figure 14.16). Solution
The slope is the value of the partial derivative 0z>0y at (1, 2): 0z 0 2 ` = sx + y 2 d ` = 2y ` = 2s2d = 4. 0y s1,2d 0y s1,2d s1,2d
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Partial Derivatives
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i
As a check, we can treat the parabola as the graph of the singlevariable function z = s1d2 + y 2 = 1 + y 2 in the plane x = 1 and ask for the slope at y = 2. The slope, calculated now as an ordinary derivative, is
Functions of More Than Two Variables
You
dz d ` = s1 + y 2 d ` = 2y ` = 4. dy y = 2 dy y=2 y=2
EXAMPLE 6
iaz
The definitions of the partial derivatives of functions of more than two independent variables are like the definitions for functions of two variables. They are ordinary derivatives with respect to one variable, taken while the other independent variables are held constant.
A Function of Three Variables
nR
If x, y, and z are independent variables and
Ć’sx, y, zd = x sin s y + 3zd,
then
ass a
0Ć’ 0 0 [x sin s y + 3zd] = x sin s y + 3zd = 0z 0z 0z = x cos s y + 3zd
0 s y + 3zd = 3x cos s y + 3zd. 0z
EXAMPLE 7
R2
If resistors of R1, R2 , and R3 ohms are connected in parallel to make an Rohm resistor, the value of R can be found from the equation 1 1 1 1 + + = R R1 R2 R3
(Figure 14.17). Find the value of 0R>0R2 when R1 = 30, R2 = 45, and R3 = 90 ohms.
ma
R3
Electrical Resistors in Parallel
dH
R1
To find 0R>0R2, we treat R1 and R3 as constants and, using implicit differentiation, differentiate both sides of the equation with respect to R2 : Solution
ham
FIGURE 14.17 Resistors arranged this way are said to be connected in parallel (Example 7). Each resistor lets a portion of the current through. Their equivalent resistance R is calculated with the formula

1 1 1 1 + + . = R R1 R2 R3
Mu
0 0 1 1 1 1 a b = a + + b 0R2 R 0R2 R1 R2 R3 1 1 0R = 0  2 + 0 R 2 0R2 R2 2
0R R2 R = 2 = a b . 0R2 R2 R2 When R1 = 30, R2 = 45, and R3 = 90,
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3 + 2 + 1 6 1 1 1 1 1 + + = = = , = R 30 45 90 90 90 15
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Chapter 14: Partial Derivatives
2
15 0R 1 1 = a b = a b = . 0R2 45 3 9
Partial Derivatives and Continuity
suf
2
i
so R = 15 and
You
A function ƒ(x, y) can have partial derivatives with respect to both x and y at a point without the function being continuous there. This is different from functions of a single variable, where the existence of a derivative implies continuity. If the partial derivatives of ƒ(x, y) exist and are continuous throughout a disk centered at sx0 , y0 d, however, then ƒ is continuous at sx0, y0 d, as we see at the end of this section.
z
0, xy 0 1, xy 0
Partials Exist, But ƒ Discontinuous
iaz
EXAMPLE 8 Let z
ƒsx, yd = e
L1 1
L2 x
y
(a) Since ƒ(x, y) is constantly zero along the line y = x (except at the origin), we have
The graph of
ƒsx, yd = e
0, 1,
Solution
xy Z 0 xy = 0
ƒsx, yd `
= y=x
lim
0 = 0.
sx, yd:s0,0d
(b) Since ƒs0, 0d = 1, the limit in part (a) proves that ƒ is not continuous at (0, 0). (c) To find 0ƒ>0x at (0, 0), we hold y fixed at y = 0. Then ƒsx, yd = 1 for all x, and the graph of ƒ is the line L1 in Figure 14.18. The slope of this line at any x is 0ƒ>0x = 0. In particular, 0ƒ>0x = 0 at (0, 0). Similarly, 0ƒ>0y is the slope of line L2 at any y, so 0ƒ>0y = 0 at (0, 0).
Mu
ham
ma
consists of the lines L1 and L2 and the four open quadrants of the xyplane. The function has partial derivatives at the origin but is not continuous there (Example 8).
lim
sx, yd:s0,0d
dH
FIGURE 14.18
(a) Find the limit of ƒ as (x, y) approaches (0, 0) along the line y = x. (b) Prove that ƒ is not continuous at the origin. (c) Show that both partial derivatives 0ƒ>0x and 0ƒ>0y exist at the origin.
ass a
0
xy Z 0 xy = 0
nR
(Figure 14.18).
0, 1,
Example 8 notwithstanding, it is still true in higher dimensions that differentiability at a point implies continuity. What Example 8 suggests is that we need a stronger requirement for differentiability in higher dimensions than the mere existence of the partial derivatives. We define differentiability for functions of two variables at the end of this section and revisit the connection to continuity.
SecondOrder Partial Derivatives When we differentiate a function ƒ(x, y) twice, we produce its secondorder derivatives. These derivatives are usually denoted by 0 2ƒ 0x 2 0 2ƒ 0y 2
To Read it Online & Download:
“d squared ƒdx squared” “d squared ƒdy squared”
or
ƒxx
“ƒ sub xx”
or
ƒyy
“ƒ sub yy”
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or
ƒyx
0 2ƒ 0y0x
“d squared ƒdy dx”
or
ƒxy
0 2ƒ 0x 2
=
“ƒ sub xy”
You
The defining equations are
“ƒ sub yx”
i
“d squared ƒdx dy”
991
suf
0 2ƒ 0x0y
Partial Derivatives
0 2ƒ 0 0ƒ = a b, 0x0y 0x 0y
0 0ƒ a b, 0x 0x
and so on. Notice the order in which the derivatives are taken:
Differentiate first with respect to y, then with respect to x.
ƒyx = sƒy dx
Means the same thing.
iaz
0 2ƒ 0x0y
HISTORICAL BIOGRAPHY
EXAMPLE 9
PierreSimon Laplace (1749–1827)
If ƒsx, yd = x cos y + ye x, find
nR
Finding SecondOrder Partial Derivatives 0 2ƒ
0 2ƒ , 0y0x
ass a
0x
, 2
0 2ƒ
0y
, 2
and
0 2ƒ . 0x0y
Solution
0ƒ 0 sx cos y + ye x d = 0x 0x
0ƒ 0 sx cos y + ye x d = 0y 0y
= cos y + ye x
dH
So 0 2ƒ 0 0ƒ a b = sin y + e x = 0y0x 0y 0x 0 2ƒ
ma
0x
2
=
= x sin y + e x So 0 2ƒ 0 0ƒ a b = sin y + e x = 0x0y 0x 0y 0 2ƒ
0 0ƒ a b = ye x. 0x 0x
0y
2
=
0 0ƒ a b = x cos y. 0y 0y
Mu
ham
The Mixed Derivative Theorem You may have noticed that the “mixed” secondorder partial derivatives 0 2ƒ 0 2ƒ and 0y0x 0x0y in Example 9 were equal. This was not a coincidence. They must be equal whenever ƒ, ƒx , ƒy , ƒxy , and ƒyx are continuous, as stated in the following theorem.
THEOREM 2 The Mixed Derivative Theorem If ƒ(x, y) and its partial derivatives ƒx , ƒy , ƒxy , and ƒyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then
To Read it Online & Download:
ƒxysa, bd = ƒyxsa, bd.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
Theorem 2 is also known as Clairaut’s Theorem, named after the French mathematician Alexis Clairaut who discovered it. A proof is given in Appendix 7. Theorem 2 says that to calculate a mixed secondorder derivative, we may differentiate in either order, provided the continuity conditions are satisfied. This can work to our advantage.
Alexis Clairaut (1713–1765)
EXAMPLE 10
Choosing the Order of Differentiation
2
Find 0 w>0x0y if w = xy +
suf
i
HISTORICAL BIOGRAPHY
You
992
ey . y + 1 2
0 2w = 1. 0y0x
nR
0w = y 0x
iaz
The symbol 0 2w>0x0y tells us to differentiate first with respect to y and then with respect to x. If we postpone the differentiation with respect to y and differentiate first with respect to x, however, we get the answer more quickly. In two steps,
Solution
and
ass a
If we differentiate first with respect to y, we obtain 0 2w>0x0y = 1 as well.
Partial Derivatives of Still Higher Order
0 3ƒ 0x0y 2 0 4ƒ 0x 20y 2
= ƒyyx
= ƒyyxx ,
ma
dH
Although we will deal mostly with first and secondorder partial derivatives, because these appear the most frequently in applications, there is no theoretical limit to how many times we can differentiate a function as long as the derivatives involved exist. Thus, we get third and fourthorder derivatives denoted by symbols like
EXAMPLE 11
Calculating a Partial Derivative of FourthOrder
Find ƒyxyz if ƒsx, y, zd = 1  2xy 2z + x 2y. We first differentiate with respect to the variable y, then x, then y again, and finally with respect to z:
Solution
Mu
ham
and so on. As with secondorder derivatives, the order of differentiation is immaterial as long as all the derivatives through the order in question are continuous.
To Read it Online & Download:
ƒy = 4xyz + x 2 ƒyx = 4yz + 2x ƒyxy = 4z ƒyxyz = 4
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Partial Derivatives
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Differentiability
You
¢y = ƒ¿sx0 d¢x + P¢x
suf
The starting point for differentiability is not Fermat’s difference quotient but rather the idea of increment. You may recall from our work with functions of a single variable in Section 3.8 that if y = ƒsxd is differentiable at x = x0 , then the change in the value of ƒ that results from changing x from x0 to x0 + ¢x is given by an equation of the form
iaz
in which P : 0 as ¢x : 0. For functions of two variables, the analogous property becomes the definition of differentiability. The Increment Theorem (from advanced calculus) tells us when to expect the property to hold.
nR
THEOREM 3 The Increment Theorem for Functions of Two Variables Suppose that the first partial derivatives of ƒ(x, y) are defined throughout an open region R containing the point sx0 , y0 d and that ƒx and ƒy are continuous at sx0 , y0 d. Then the change ¢z = ƒsx0 + ¢x, y0 + ¢yd  ƒsx0 , y0 d
in the value of ƒ that results from moving from sx0 , y0 d to another point (x0 + ¢x, y0 + ¢yd in R satisfies an equation of the form
ass a
¢z = ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y + P1 ¢x + P2 ¢y,
in which each of P1, P2 : 0 as both ¢x, ¢y : 0.
dH
You can see where the epsilons come from in the proof in Appendix 7. You will also see that similar results hold for functions of more than two independent variables.
ma
DEFINITION Differentiable Function A function z = ƒsx, yd is differentiable at sx0 , y0 d if ƒxsx0 , y0 d and ƒysx0 , y0 d exist and ¢z satisfies an equation of the form ¢z = ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y + P1 ¢x + P2 ¢y,
Mu
ham
in which each of P1, P2 : 0 as both ¢x, ¢y : 0. We call ƒ differentiable if it is differentiable at every point in its domain.
In light of this definition, we have the immediate corollary of Theorem 3 that a function is differentiable if its first partial derivatives are continuous.
COROLLARY OF THEOREM 3
Continuity of Partial Derivatives Implies Differentiability If the partial derivatives ƒx and ƒy of a function ƒ(x, y) are continuous throughout an open region R, then ƒ is differentiable at every point of R.
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Chapter 14: Partial Derivatives
suf
i
If z = ƒsx, yd is differentiable, then the definition of differentiability assures that ¢z = ƒsx0 + ¢x, y0 + ¢yd  ƒsx0 , y0 d approaches 0 as ¢x and ¢y approach 0. This tells us that a function of two variables is continuous at every point where it is differentiable.
You
THEOREM 4 Differentiability Implies Continuity If a function ƒ(x, y) is differentiable at sx0 , y0 d, then ƒ is continuous at sx0 , y0 d.
Mu
ham
ma
dH
ass a
nR
iaz
As we can see from Theorems 3 and 4, a function ƒ(x, y) must be continuous at a point sx0 , y0 d if ƒx and ƒy are continuous throughout an open region containing sx0 , y0 d. Remember, however, that it is still possible for a function of two variables to be discontinuous at a point where its first partial derivatives exist, as we saw in Example 8. Existence alone of the partial derivative at a point is not enough.
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i f su
Chapter 14: Partial Derivatives
EXERCISES 14.3
u o Y z a i R
Calculating FirstOrder Partial Derivatives
27. ƒsx, y, zd = sin1 sxyzd
In Exercises 1–22, find 0ƒ>0x and 0ƒ>0y .
29. ƒsx, y, zd = ln sx + 2y + 3zd
1. ƒsx, yd = 2x 2  3y  4
2. ƒsx, yd = x 2  xy + y 2
3. ƒsx, yd = sx 2  1ds y + 2d
30. ƒsx, y, zd = yz ln sxyd
2
n a s s a H
6. ƒsx, yd = s2x  3yd
7. ƒsx, yd = 2x 2 + y 2
8. ƒsx, yd = sx 3 + s y>2dd2>3 10. ƒsx, yd = x>sx 2 + y 2 d
9. ƒsx, yd = 1>sx + yd
11. ƒsx, yd = sx + yd>sxy  1d 12. ƒsx, yd = tan1 s y>xd
In Exercises 35–40, find the partial derivative of the function with respect to each variable. 35. ƒst, ad = cos s2pt  ad
14. ƒsx, yd = ex sin sx + yd
37. hsr, f, ud = r sin f cos u
15. ƒsx, yd = ln sx + yd
16. ƒsx, yd = e xy ln y
39. Work done by the heart
17. ƒsx, yd = sin2 sx  3yd
18. ƒsx, yd = cos2 s3x  y 2 d
19. ƒsx, yd = x y
20. ƒsx, yd = logy x
ad
y
sg continuous for all td
m m
q
22. ƒsx, yd = a sxyd
n
s ƒ xy ƒ 6 1d
a h u
n=0
+ y2 + z2d
34. ƒsx, y, zd = sinh sxy  z 2 d
13. ƒsx, yd = e sx + y + 1d
g std dt
2
33. ƒsx, y, zd = tanh sx + 2y + 3zd 3
5. ƒsx, yd = sxy  1d
Lx
31. ƒsx, y, zd = e sx
32. ƒsx, y, zd = exyz
4. ƒsx, yd = 5xy  7x 2  y 2 + 3x  6y + 2
21. ƒsx, yd =
28. ƒsx, y, zd = sec1 sx + yzd
36. g su, yd = y2e s2u>yd 38. g sr, u, zd = r s1  cos ud  z (Section 3.8, Exercise 51)
WsP, V, d, y, gd = PV +
40. Wilson lot size formula
Vdy 2 2g
(Section 4.5, Exercise 45)
hq km Asc, h, k, m, qd = q + cm + 2
In Exercises 23–34, find ƒx , ƒy , and ƒz .
Calculating SecondOrder Partial Derivatives
23. ƒsx, y, zd = 1 + xy 2  2z 2 24. ƒsx, y, zd = xy + yz + xz
Find all the secondorder partial derivatives of the functions in Exercises 41–46.
25. ƒsx, y, zd = x  2y + z 2
M
2
26. ƒsx, y, zd = sx 2 + y 2 + z 2 d1>2
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41. ƒsx, yd = x + y + xy
42. ƒsx, yd = sin xy
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.3 Partial Derivatives
y
45. r sx, yd = ln sx + yd
44. hsx, yd = xe + y + 1
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58. Find the value of 0x>0z at the point s1, 1, 3d if the equation
s y>xd
xz + y ln x  x 2 + 4 = 0
Mixed Partial Derivatives In Exercises 47–50, verify that wxy = wyx . 47. w = ln s2x + 3yd
48. w = e x + x ln y + y ln x
49. w = xy 2 + x 2y 3 + x 3y 4
50. w = x sin y + y sin x + xy
51. Which order of differentiation will calculate fxy faster: x first or y first? Try to answer without writing anything down.
defines x as a function of the two independent variables y and z and the partial derivative exists.
You
46. ssx, yd = tan
1
defines z as a function of the two independent variables x and y and the partial derivative exists.
i
43. g sx, yd = x 2y + cos y + y sin x
Exercises 59 and 60 are about the triangle shown here.
a. ƒsx, yd = x sin y + e y
B
b. ƒsx, yd = 1>x
c
iaz
a
c. ƒsx, yd = y + sx>yd
C
e. ƒsx, yd = x 2 + 5xy + sin x + 7e x f. ƒsx, yd = x ln xy 5
2
3
2 4 x
a. ƒsx, yd = y x e + 2 b. ƒsx, yd = y2 + yssin x  x4 d c. ƒsx, yd = x 2 + 5xy + sin x + 7e x >2
2
dH
Using the Partial Derivative Definition
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In Exercises 53 and 54, use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. 0ƒ 0ƒ 53. ƒsx, yd = 1  x + y  3x 2y, and at s1, 2d 0x 0y 54. ƒsx, yd = 4 + 2x  3y  xy 2,
59. Express A implicitly as a function of a, b, and c and calculate 0A>0a and 0A>0b .
60. Express a implicitly as a function of A, b, and B and calculate 0a>0A and 0a>0B .
61. Two dependent variables Express yx in terms of u and y if the equations x = y ln u and y = u ln y define u and y as functions of the independent variables x and y, and if yx exists. (Hint: Differentiate both equations with respect to x and solve for yx by eliminating ux .)
ass a
52. The fifthorder partial derivative 0 ƒ>0x 0y is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: x or y? Try to answer without writing anything down.
A
b
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d. ƒsx, yd = y + x 2y + 4y 3  ln s y 2 + 1d
d. ƒsx, yd = xe y
995
0ƒ 0ƒ and at s 2, 1d 0x 0y
ham
55. Three variables Let w = ƒsx, y, zd be a function of three independent variables and write the formal definition of the partial derivative 0ƒ>0z at sx0 , y0 , z0 d. Use this definition to find 0ƒ>0z at (1, 2, 3) for ƒsx, y, zd = x 2yz 2.
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56. Three variables Let w = ƒsx, y, zd be a function of three independent variables and write the formal definition of the partial derivative 0ƒ>0y at sx0 , y0 , z0 d. Use this definition to find 0ƒ>0y at s 1, 0, 3d for ƒsx, y, zd = 2xy 2 + yz 2.
Differentiating Implicitly 57. Find the value of 0z>0x at the point (1, 1, 1) if the equation xy + z 3x  2yz = 0
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62. Two dependent variables Find 0x>0u and 0y>0u if the equations u = x 2  y 2 and y = x 2  y define x and y as functions of the independent variables u and y, and the partial derivatives exist. (See the hint in Exercise 61.) Then let s = x 2 + y 2 and find 0s>0u.
Laplace Equations The threedimensional Laplace equation 0 2ƒ 0x 2
0 2ƒ +
0 2ƒ
0y 2
+
0z 2
= 0
is satisfied by steadystate temperature distributions T = ƒsx, y, zd in space, by gravitational potentials, and by electrostatic potentials. The twodimensional Laplace equation 0 2ƒ 0x 2
0 2ƒ +
0y 2
= 0,
obtained by dropping the 0 2ƒ>0z 2 term from the previous equation, describes potentials and steadystate temperature distributions in a plane (see the accompanying figure). The plane (a) may be treated as a thin slice of the solid (b) perpendicular to the zaxis.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 996
Chapter 14: Partial Derivatives water as the waves go by. We see periodic vertical motion in time. In physics, this beautiful symmetry is expressed by the onedimensional wave equation
suf
i
∂ 2f ∂ 2f 20 ∂x 2 ∂y
0 2w 0 2w = c2 2 , 2 0t 0x
(a)
You
where w is the wave height, x is the distance variable, t is the time variable, and c is the velocity with which the waves are propagated.
∂ 2f ∂ 2f ∂ 2f 2 2 0 2 ∂x ∂y ∂z
(b)
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w
x
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x
In our example, x is the distance across the ocean’s surface, but in other applications, x might be the distance along a vibrating string, distance through air (sound waves), or distance through space (light waves). The number c varies with the medium and type of wave. Show that the functions in Exercises 69–75 are all solutions of the wave equation.
ass a
Boundary temperatures controlled
Show that each function in Exercises 63–68 satisfies a Laplace equation. 64. ƒsx, y, zd = 2z 3  3sx 2 + y 2 dz 65. ƒsx, yd = e2y cos 2x 66. ƒsx, yd = ln 2x 2 + y 2 67. ƒsx, y, zd = sx 2 + y 2 + z 2 d1>2
The Wave Equation
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68. ƒsx, y, zd = e 3x + 4y cos 5z
74. w = 5 cos s3x + 3ctd + e
73. w = tan s2x  2ctd x + ct
75. w = ƒsud, where ƒ is a differentiable function of u, and u = asx + ctd, where a is a constant
Continuous Partial Derivatives 76. Does a function ƒ(x, y) with continuous first partial derivatives throughout an open region R have to be continuous on R? Give reasons for your answer. 77. If a function ƒ(x, y) has continuous second partial derivatives throughout an open region R, must the firstorder partial derivatives of ƒ be continuous on R? Give reasons for your answer.
Mu
ham
If we stand on an ocean shore and take a snapshot of the waves, the picture shows a regular pattern of peaks and valleys in an instant of time. We see periodic vertical motion in space, with respect to distance. If we stand in the water, we can feel the rise and fall of the
70. w = cos s2x + 2ctd
71. w = sin sx + ctd + cos s2x + 2ctd 72. w = ln s2x + 2ctd
dH
63. ƒsx, y, zd = x 2 + y 2  2z 2
69. w = sin sx + ctd
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 996
Chapter 14: Partial Derivatives
14.4
z a i R n a
The Chain Rule
s s a dH
The Chain Rule for functions of a single variable studied in Section 3.5 said that when w = Ć’sxd was a differentiable function of x and x = gstd was a differentiable function of t, w became a differentiable function of t and dw> dt could be calculated with the formula
a h u M
a m m
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dw dx dw = . dt dx dt
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.4
The Chain Rule
997
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For functions of two or more variables the Chain Rule has several forms. The form depends on how many variables are involved but works like the Chain Rule in Section 3.5 once we account for the presence of additional variables.
Functions of Two Variables
You
The Chain Rule formula for a function w = ƒsx, yd when x = xstd and y = ystd are both differentiable functions of t is given in the following theorem.
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THEOREM 5 Chain Rule for Functions of Two Independent Variables If w = ƒsx, yd has continuous partial derivatives ƒx and ƒy and if x = xstd, y = ystd are differentiable functions of t, then the composite w = ƒsxstd, ystdd is a differentiable function of t and
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dƒ = ƒxsxstd, ystdd # x¿std + ƒysxstd, ystdd # y¿std, dt or
ass a
0ƒ dx 0ƒ dy dw = + . 0x dt 0y dt dt
Proof The proof consists of showing that if x and y are differentiable at t = t0 , then w is differentiable at t0 and
dH
a
Chain Rule
0w 0x
y Intermediate variables
dx dt
dy dt
Mu
x
0w 0y
t dw 0 w dx 0 w dy 0 x dt 0 y dt dt
¢w = a
0w 0w b ¢x + a b ¢y + P1 ¢x + P2 ¢y, 0x P0 0y P0
where P1, P2 : 0 as ¢x, ¢y : 0. To find dw> dt, we divide this equation through by ¢t and let ¢t approach zero. The division gives
Dependent variable
ham
w f (x, y)
where P0 = sxst0 d, yst0 dd. The subscripts indicate where each of the derivatives are to be evaluated. Let ¢x, ¢y, and ¢w be the increments that result from changing t from t0 to t0 + ¢t. Since ƒ is differentiable (see the definition in Section 14.3),
ma
To remember the Chain Rule picture the diagram below. To find dw> dt, start at w and read down each route to t, multiplying derivatives along the way. Then add the products.
dy dw 0w dx 0w b = a b a b + a b a b , 0x P0 dt t0 0y P0 dt t0 dt t0
¢y ¢y 0w ¢x 0w ¢x ¢w = a b + a b + P1 + P2 . 0x P0 ¢t 0y P0 ¢t ¢t ¢t ¢t
Letting ¢t approach zero gives a
dw ¢w b = lim dt t0 ¢t:0 ¢t = a
Independent variable
dy dy 0w dx 0w dx b a b + a b a b + 0# a b + 0# a b . 0x P0 dt t0 0y P0 dt t0 dt t0 dt t0
The tree diagram in the margin provides a convenient way to remember the Chain Rule. From the diagram, you see that when t = t0 , the derivatives dx> dt and dy> dt are
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Chapter 14: Partial Derivatives
You
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i
evaluated at t0. The value of t0 then determines the value x0 for the differentiable function x and the value y0 for the differentiable function y. The partial derivatives 0w>0x and 0w>0y (which are themselves functions of x and y) are evaluated at the point P0sx0, y0 d corresponding to t0. The “true” independent variable is t, whereas x and y are intermediate variables (controlled by t) and w is the dependent variable. A more precise notation for the Chain Rule shows how the various derivatives in Theorem 5 are evaluated: dy 0ƒ 0ƒ dw dx st d = sx , y d # st d + sx , y d # st d. 0x 0 0 dt 0 0y 0 0 dt 0 dt 0
EXAMPLE 1
Applying the Chain Rule
iaz
Use the Chain Rule to find the derivative of
w = xy
Solution
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with respect to t along the path x = cos t, y = sin t. What is the derivative’s value at t = p>2? We apply the Chain Rule to find dw> dt as follows:
ass a
dw 0w dx 0w dy = + 0x dt 0y dt dt =
0sxyd 0x
#
0sxyd d scos td + 0y dt
#
d ssin td dt
= s yds sin td + sxdscos td
dH
= ssin tds sin td + scos tdscos td = sin2 t + cos2 t = cos 2t.
ma
In this example, we can check the result with a more direct calculation. As a function of t, w = xy = cos t sin t =
1 sin 2t , 2
Mu
ham
so
dw d 1 1 = a sin 2tb = 2 dt dt 2
#
2 cos 2t = cos 2t .
In either case, at the given value of t, a
dw p b = cos a2 # b = cos p = 1 . 2 dt t = p>2
Functions of Three Variables You can probably predict the Chain Rule for functions of three variables, as it only involves adding the expected third term to the twovariable formula.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.4
The Chain Rule
999
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THEOREM 6 Chain Rule for Functions of Three Independent Variables If w = ƒsx, y, zd is differentiable and x, y, and z are differentiable functions of t, then w is a differentiable function of t and
w f (x, y, z)
0w 0z
0w 0y
x
y dx dt
dy dt
Changes in a Function’s Values Along a Helix
Find dw> dt if w = xy + z,
x = cos t,
y = sin t,
z = t.
In this example the values of w are changing along the path of a helix (Section 13.1). What is the derivative’s value at t = 0 ? z
Intermediate variables
dz dt
Independent t variable dw 0 w dx 0 w dy 0 w dz 0 x dt 0 y dt 0 z dt dt
Solution
ass a
0w 0x
Dependent variable
EXAMPLE 2
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Chain Rule
The proof is identical with the proof of Theorem 5 except that there are now three intermediate variables instead of two. The diagram we use for remembering the new equation is similar as well, with three routes from w to t.
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Here we have three routes from w to t instead of two, but finding dw> dt is still the same. Read down each route, multiplying derivatives along the way; then add.
You
0ƒ dx 0ƒ dy 0ƒ dz dw = + + . 0x dt 0y dt 0z dt dt
dw 0w dx 0w dy 0w dz = + + 0x dt 0y dt 0z dt dt = s yds sin td + sxdscos td + s1ds1d
dH
= ssin tds sin td + scos tdscos td + 1
dw = 1 + cos s0d = 2. b dt t = 0
Mu
ham
ma
a
= sin2 t + cos2 t + 1 = 1 + cos 2t.
Substitute for the intermediate variables.
Here is a physical interpretation of change along a curve. If w = Tsx, y, zd is the temperature at each point (x, y, z) along a curve C with parametric equations x = xstd, y = ystd, and z = zstd, then the composite function w = Tsxstd, ystd, zstdd represents the temperature relative to t along the curve. The derivative dw> dt is then the instantaneous rate of change of temperature along the curve, as calculated in Theorem 6.
Functions Defined on Surfaces If we are interested in the temperature w = ƒsx, y, zd at points (x, y, z) on a globe in space, we might prefer to think of x, y, and z as functions of the variables r and s that give the points’ longitudes and latitudes. If x = gsr, sd, y = hsr, sd, and z = ksr, sd, we could then express the temperature as a function of r and s with the composite function w = ƒsgsr, sd, hsr, sd, ksr, sdd. Under the right conditions, w would have partial derivatives with respect to both r and s that could be calculated in the following way.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1000
Chapter 14: Partial Derivatives
THEOREM 7
You
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i
Chain Rule for Two Independent Variables and Three Intermediate Variables Suppose that w = Ć’sx, y, zd, x = gsr, sd, y = hsr, sd, and z = ksr, sd. If all four functions are differentiable, then w has partial derivatives with respect to r and s, given by the formulas 0w 0x 0w 0y 0w 0w = + + 0r 0x 0r 0y 0r 0z 0w 0x 0w 0y 0w 0w = + + 0s 0x 0s 0y 0s 0z
0z 0r
0z . 0s
w
0w 0x
f x
y g
Independent variables
h
k
0y 0r
w f (x, y, z)
0x 0s
0z 0r
0w 0z
0w 0y
x
z
r 0w 0w 0x 0w 0y 0w 0z 0r 0x 0r 0y 0r 0z 0r
dH
w f ( g(r, s), h (r, s), k (r, s)) (a)
y
0x 0r
r, s
FIGURE 14.19
x
z
0w 0x
0w 0z
ass a
Intermediate variables
0w 0y
nR
w f (x, y, z)
Dependent variable
iaz
The first of these equations can be derived from the Chain Rule in Theorem 6 by holding s fixed and treating r as t. The second can be derived in the same way, holding r fixed and treating s as t. The tree diagrams for both equations are shown in Figure 14.19.
y
0y 0s
z
0z 0s
s 0w 0 w 0x 0w 0y 0w 0z 0s 0x 0s 0y 0s 0z 0s (c)
(b)
Composite function and tree diagrams for Theorem 7.
ma
EXAMPLE 3
Partial Derivatives Using Theorem 7
Mu
ham
Express 0w>0r and 0w>0s in terms of r and s if w = x + 2y + z 2,
r x = s,
y = r 2 + ln s,
z = 2r.
Solution
0w 0w 0x 0w 0y 0w 0z = + + 0r 0x 0r 0y 0r 0z 0r 1 = s1d a s b + s2ds2rd + s2zds2d 1 1 = s + 4r + s4rds2d = s + 12r
Substitute for intermediate variable z.
0w 0w 0x 0w 0y 0w 0z = + + 0s 0x 0s 0y 0s 0z 0s = s1d a
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r 1 2 r b + s2d a s b + s2zds0d = s  2 s2 s
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.4
i
0w 0y y
0w 0x 0w 0y 0w = + 0r 0x 0r 0y 0r
0y 0r
and
r
0w 0w 0x 0w 0y 0r 0 x 0r 0y 0r
0w 0x 0w 0y 0w = + . 0s 0x 0s 0y 0s
iaz
Figure 14.20 shows the tree diagram for the first of these equations. The diagram for the second equation is similar; just replace r with s.
Tree diagram for the
EXAMPLE 4
0w 0w 0x 0w 0y = + . 0r 0x 0r 0y 0r
More Partial Derivatives
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FIGURE 14.20 equation
You
If w = ƒsx, yd, x = gsr, sd, and y = hsr, sd, then x
suf
w f (x, y)
0x 0r
1001
If ƒ is a function of two variables instead of three, each equation in Theorem 7 becomes correspondingly one term shorter.
Chain Rule
0w 0x
The Chain Rule
Express 0w>0r and 0w>0s in terms of r and s if w = x2 + y2,
ass a
Solution
x = r  s,
0w 0w 0x 0w 0y = + 0r 0x 0r 0y 0r
y = r + s.
0w 0w 0x 0w 0y = + 0s 0x 0s 0y 0s = s2xds 1d + s2yds1d
= 2sr  sd + 2sr + sd
= 2sr  sd + 2sr + sd
= 4r
= 4s
dH
= s2xds1d + s2yds1d
Substitute for the intermediate variables.
If ƒ is a function of x alone, our equations become even simpler.
ma
Chain Rule w f (x)
ham
dw dx
If w = ƒsxd and x = gsr, sd, then 0w dw 0x = 0r dx 0r
and
0w dw 0x = . 0s dx 0s
x
0x 0r
In this case, we can use the ordinary (singlevariable) derivative, dw> dx. The tree diagram is shown in Figure 14.21.
s
0w dw 0 x 0r dx 0r
Implicit Differentiation Revisited
0w dw 0 x 0s dx 0s
The twovariable Chain Rule in Theorem 5 leads to a formula that takes most of the work out of implicit differentiation. Suppose that
Mu
r
0x 0s
FIGURE 14.21 Tree diagram for differentiating ƒ as a composite function of r and s with one intermediate variable.
1.
2.
The function F(x, y) is differentiable and The equation Fsx, yd = 0 defines y implicitly as a differentiable function of x, say y = hsxd.
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Chapter 14: Partial Derivatives
Since w = Fsx, yd = 0, the derivative dw> dx must be zero. Computing the derivative from the Chain Rule (tree diagram in Figure 14.22), we find 0 =
y h(x)
x
dy dw dx = Fx + Fy dx dx dx
= Fx # 1 + Fy # dx 1 dx
dy h'(x) dx
Theorem 5 with t = x and ƒ = F
dy . dx
You
Fy 0 w 0y
0w F x 0x
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w F(x, y)
If Fy = 0w>0y Z 0, we can solve this equation for dy> dx to get
x
dy Fx =  . Fy dx
dw dy Fx • 1 Fy • dx dx
iaz
This relationship gives a surprisingly simple shortcut to finding derivatives of implicitly defined functions, which we state here as a theorem.
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FIGURE 14.22 Tree diagram for differentiating w = Fsx, yd with respect to x. Setting dw>dx = 0 leads to a simple computational formula for implicit differentiation (Theorem 8).
THEOREM 8 A Formula for Implicit Differentiation Suppose that F(x, y) is differentiable and that the equation Fsx, yd = 0 defines y as a differentiable function of x. Then at any point where Fy Z 0,
ass a
dy Fx =  . Fy dx
EXAMPLE 5
Implicit Differentiation
dH
Use Theorem 8 to find dy> dx if y 2  x 2  sin xy = 0. Take Fsx, yd = y 2  x 2  sin xy. Then
Mu
ham
ma
Solution
dy 2x  y cos xy Fx = = Fy 2y  x cos xy dx =
2x + y cos xy . 2y  x cos xy
This calculation is significantly shorter than the singlevariable calculation with which we found dy> dx in Section 3.6, Example 3.
Functions of Many Variables We have seen several different forms of the Chain Rule in this section, but you do not have to memorize them all if you can see them as special cases of the same general formula. When solving particular problems, it may help to draw the appropriate tree diagram by placing the dependent variable on top, the intermediate variables in the middle, and the selected independent variable at the bottom. To find the derivative of the dependent variable with respect to the selected independent variable, start at the dependent variable and read down each route of the tree to the independent variable, calculating and multiplying the derivatives along each route. Then add the products you found for the different routes.
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1003
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In general, suppose that w = ƒsx, y, Á , yd is a differentiable function of the variables x, y, Á , y (a finite set) and the x, y, Á , y are differentiable functions of p, q, Á , t (another finite set). Then w is a differentiable function of the variables p through t and the partial derivatives of w with respect to these variables are given by equations of the form
You
0w 0w 0x 0w 0y Á 0w 0y = + + + . 0p 0x 0p 0y 0p 0y 0p
The other equations are obtained by replacing p by q, Á , t, one at a time. One way to remember this equation is to think of the righthand side as the dot product of two vectors with components 0w 0w 0w , ,Á, b 0x 0y 0y
and
a
0x 0y 0y , , Á , b. 0p 0p 0p
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('''')''''* Derivatives of the intermediate variables with respect to the selected independent variable
Mu
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ma
dH
ass a
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('''')''''* Derivatives of w with respect to the intermediate variables
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u
14.4 The Chain Rule
EXERCISES 14.4 Chain Rule: One Independent Variable
terms of u and y before differentiating. Then (b) evaluate 0w>0u and 0w>0y at the given point (u, y).
In Exercises 1–6, (a) express dw> dt as a function of t, both by using the Chain Rule and by expressing w in terms of t and differentiating directly with respect to t. Then (b) evaluate dw> dt at the given value of t. 2
2
1. w = x + y ,
x = cos t,
y = sin t;
5. w = 2ye x  ln z, t = 1
x = cos t,
6. w = z  sin xy,
x = ln st 2 + 1d, x = t,
y = ln t,
y = tan1 t, z = e
t1
;
t = 1
a H
d a
z = uy;
y = ue y cos u,
In Exercises 11 and 12, (a) express 0u>0x, 0u>0y, and 0u>0z as functions of x, y, and z both by using the Chain Rule and by expressing u directly in terms of x, y, and z before differentiating. Then (b) evaluate 0u>0x, 0u>0y, and 0u>0z at the given point (x, y, z). p  q 11. u = q  r , p = x + y + z, q = x  y + z,
n a ss
z = 42t ; z = e t;
y = u  y,
10. w = ln sx 2 + y 2 + z 2 d, x = ue y sin u, z = ue y; su, yd = s 2, 0d
t = p
y = sin t,
z a i R
x = u + y,
9. w = xy + yz + xz, su, yd = s1>2, 1d
2. w = x 2 + y 2, x = cos t + sin t, y = cos t  sin t; t = 0 y x 3. w = z + z , x = cos2 t, y = sin2 t, z = 1>t; t = 3 4. w = ln sx 2 + y 2 + z 2 d, t = 3
o Y
1003
r = x + y  z;
sx, y, zd =
A 23, 2, 1 B
12. u = e qr sin1 p, p = sin x, q = z 2 ln y, sx, y, zd = sp>4, 1>2, 1>2d
r = 1>z;
Chain Rule: Two and Three Independent Variables
In Exercises 7 and 8, (a) express 0z>0u and 0z>0y as functions of u and y both by using the Chain Rule and by expressing z directly in terms of u and y before differentiating. Then (b) evaluate 0z>0u and 0z>0y at the given point su, yd.
m m
7. z = 4e x ln y, x = ln su cos yd, su, yd = s2, p>4d
a h
8. z = tan1 sx>yd, x = u cos y, su, yd = s1.3, p>6d
u M
y = u sin y;
y = u sin y;
In Exercises 9 and 10, (a) express 0w>0u and 0w>0y as functions of u and y both by using the Chain Rule and by expressing w directly in
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Using a Tree Diagram In Exercises 13–24, draw a tree diagram and write a Chain Rule formula for each derivative. 13.
dz for z = ƒsx, yd, dt
14.
dz for z = ƒsu, y, wd, dt
15.
x = gstd,
y = hstd
u = gstd,
0w 0w and for w = hsx, y, zd, 0u 0y z = ksu, yd
y = hstd,
x = ƒsu, yd,
w = kstd y = gsu, yd,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
0w 0w and for w = ƒsr, s, td, 0x 0y t = ksx, yd
r = gsx, yd,
s = hsx, yd,
Finding Specified Partial Derivatives 33. Find 0w>0r when r = 1, s = 1 if w = sx + y + zd2, x = r  s, y = cos sr + sd, z = sin sr + sd.
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16.
Chapter 14: Partial Derivatives
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1004
34. Find 0w>0y when u = 1, y = 2 x = y 2>u, y = u + y, z = cos u.
if
w = xy + ln z,
17.
0w 0w and for w = gsx, yd, 0u 0y
x = hsu, yd,
y = ksu, yd
18.
0w 0w and for w = gsu, yd, 0x 0y
u = hsx, yd,
y = ksx, yd
19.
0z 0z and for z = ƒsx, yd, 0t 0s
20.
0y for y = ƒsud, 0r
21.
0w 0w and for w = gsud, 0s 0t
22.
0w for w = ƒsx, y, z, yd, x = gs p, qd, 0p z = js p, qd, y = ks p, qd
y = hs p, qd,
Theory and Examples
23.
0w 0w and for w = ƒsx, yd, 0r 0s
y = hssd
39. Changing voltage in a circuit The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wears out. At the same time, the resistance R is increasing as the resistor heats up. Use the equation
24.
0w for w = gsx, yd, 0s
y = hst, sd
s 1, 1d s1, 2d
y
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ThreeVariable Implicit Differentiation
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Theorem 8 can be generalized to functions of three variables and even more. The threevariable version goes like this: If the equation Fsx, y, zd = 0 determines z as a differentiable function of x and y, then, at points where Fz Z 0, Fy 0z =  . 0y Fz
and
Use these equations to find the values of 0z>0x and 0z>0y at the points in Exercises 29–32.
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29. z 3  xy + yz + y 3  2 = 0, 1 1 1 30. x + y + z  1 = 0,
to find how the current is changing at the instant when R = 600 ohms, I = 0.04 amp, dR>dt = 0.5 ohm>sec, and dV>dt = 0.01 volt>sec. V Battery I
s0, ln 2d
28. xe + sin xy + y  ln 2 = 0,
Fx 0z = 0x Fz
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ass a
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s1, 1d
27. x + xy + y  7 = 0,
z = sin xy + x sin y,
dV 0V dI 0V dR = + 0I dt 0R dt dt
y = ksr, s, td
Assuming that the equations in Exercises 25–28 define y as a differentiable function of x, use Theorem 8 to find the value of dy> dx at the given point.
2
if
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x = gsrd,
Implicit Differentiation
2
36. Find 0z>0u when u = 0, y = 1 x = u 2 + y 2, y = uy.
38. Find 0z>0u and 0z>0y when u = 1 and y = 2 if z = ln q and q = 1y + 3 tan1 u.
u = hss, td
x = hsr, s, td,
26. xy + y 2  3x  3 = 0,
w = x 2 + s y>xd,
37. Find 0z>0u and 0z>0y when u = ln 2, y = 1 if z = 5 tan1 x and x = e u + ln y.
u = gsr, sd
25. x 3  2y 2 + xy = 0,
if
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x = gst, sd,
35. Find 0w>0y when u = 0, y = 0 x = u  2y + 1, y = 2u + y  2.
s1, 1, 1d
R
40. Changing dimensions in a box The lengths a, b, and c of the edges of a rectangular box are changing with time. At the instant in question, a = 1 m, b = 2 m, c = 3 m, da>dt = db>dt = 1 m>sec, and dc>dt = 3 m>sec. At what rates are the box’s volume V and surface area S changing at that instant? Are the box’s interior diagonals increasing in length or decreasing? 41. If ƒ(u, y, w) is differentiable and u = x  y, y = y  z, and w = z  x, show that 0ƒ 0ƒ 0ƒ + + = 0. 0x 0y 0z
s2, 3, 6d
31. sin sx + yd + sin s y + zd + sin sx + zd = 0, 32. xe y + ye z + 2 ln x  2  3 ln 2 = 0,
sp, p, pd
s1, ln 2, ln 3d
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42. Polar coordinates Suppose that we substitute polar coordinates x = r cos u and y = r sin u in a differentiable function w = ƒsx, yd.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.4 The Chain Rule
b. Suppose that T = 4x 2  4xy + 4y 2. Find the maximum and minimum values of T on the circle.
0w = ƒx cos u + ƒy sin u 0r
48. Temperature on an ellipse Let T = g sx, yd be the temperature at the point (x, y) on the ellipse
1 0w r 0u = ƒx sin u + ƒy cos u . b. Solve the equations in part (a) to express ƒx and ƒy in terms of 0w>0r and 0w>0u.
and suppose that
2
a. Locate the maximum and minimum temperatures on the ellipse by examining dT> dt and d 2T>dt 2.
and y = x  iy and i = 2 1. Show that w satisfies the Laplace equation wxx + wyy = 0 if all the necessary functions are differentiable.
Changes in Functions Along Curves
Under mild continuity restrictions, it is true that if Fsxd =
b
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g st, xd dt,
g x st, xd dt. Using this fact and the Chain Rule, we La can find the derivative of
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then F¿sxd =
ƒsxd
ƒz = t 2 + t  2.
ƒy = sin t,
At what points on the curve, if any, can ƒ take on extreme values?
dH
46. A space curve Let w = x 2e 2y cos 3z . Find the value of dw> dt at the point (1, ln 2, 0) on the curve x = cos t, y = ln st + 2d, z = t .
47. Temperature on a circle Let T = ƒsx, yd be the temperature at the point (x, y) on the circle x = cos t, y = sin t, 0 … t … 2p and suppose that 0T = 8y  4x. 0y
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0T = 8x  4y, 0x
Differentiating Integrals
b
45. Extreme values on a helix Suppose that the partial derivatives of a function ƒ(x, y, z) at points on the helix x = cos t, y = sin t, z = t are ƒx = cos t,
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Let w = ƒsud + g syd, where u = x + iy
b. Suppose that T = xy  2. Find the maximum and minimum values of T on the ellipse.
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43. Laplace equations Show that if w = ƒsu, yd satisfies the Laplace equation ƒuu + ƒyy = 0 and if u = sx 2  y 2 d>2 and y = xy, then w satisfies the Laplace equation wxx + wyy = 0. 44. Laplace equations
0 … t … 2p,
0T = x. 0y
0T = y, 0x
c. Show that 0w 1 0w b + 2 a b. 0r 0u r
y = 22 sin t,
x = 2 22 cos t,
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and
2
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a. Show that
sƒx d2 + sƒy d2 = a
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g st, xd dt
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by letting
u
Gsu, xd =
La
g st, xd dt,
where u = ƒsxd. Find the derivatives of the functions in Exercises 49 and 50. x2
49. Fsxd =
L0
50. Fsxd =
Lx
2t 4 + x 3 dt 1
2t 3 + x 2 dt
2
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a. Find where the maximum and minimum temperatures on the circle occur by examining the derivatives dT> dt and d 2T>dt 2.
Fsxd =
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
14.5 Directional Derivatives and Gradient Vectors
14.5
z a i R n a
Directional Derivatives and Gradient Vectors
s s a dH
1005
If you look at the map (Figure 14.23) showing contours on the West Point Area along the Hudson River in New York, you will notice that the tributary streams flow perpendicular to the contours. The streams are following paths of steepest descent so the waters reach the Hudson as quickly as possible. Therefore, the instantaneous rate of change in a streamâ€™s
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a m m a h
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1006
Chapter 14: Partial Derivatives
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altitude above sea level has a particular direction. In this section, you see why this direction, called the “downhill” direction, is perpendicular to the contours.
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FIGURE 14.23 Contours of the West Point Area in New York show streams, which follow paths of steepest descent, running perpendicular to the contours.
Directional Derivatives in the Plane We know from Section 14.4 that if ƒ(x, y) is differentiable, then the rate at which ƒ changes with respect to t along a differentiable curve x = gstd, y = hstd is
y
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Line x x 0 su 1, y y 0 su 2
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u u1i u 2 j
Direction of increasing s R
Mu
P0(x 0, y0 )
0
x
FIGURE 14.24 The rate of change of ƒ in the direction of u at a point P0 is the rate at which ƒ changes along this line at P0 .
dƒ 0ƒ dx 0ƒ dy = + . 0x dt 0y dt dt
At any point P0sx0, y0 d = P0sgst0 d, hst0 dd, this equation gives the rate of change of ƒ with respect to increasing t and therefore depends, among other things, on the direction of motion along the curve. If the curve is a straight line and t is the arc length parameter along the line measured from P0 in the direction of a given unit vector u, then dƒ> dt is the rate of change of ƒ with respect to distance in its domain in the direction of u. By varying u, we find the rates at which ƒ changes with respect to distance as we move through P0 in different directions. We now define this idea more precisely. Suppose that the function ƒ(x, y) is defined throughout a region R in the xyplane, that P0sx0 , y0 d is a point in R, and that u = u1 i + u2 j is a unit vector. Then the equations x = x0 + su1,
y = y0 + su2
parametrize the line through P0 parallel to u. If the parameter s measures arc length from P0 in the direction of u, we find the rate of change of ƒ at P0 in the direction of u by calculating dƒ> ds at P0 (Figure 14.24).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.5
Directional Derivatives and Gradient Vectors
1007
dƒ ƒsx0 + su1, y0 + su2 d  ƒsx0 , y0 d b = lim , s ds u,P0 s:0
provided the limit exists.
The directional derivative is also denoted by
“The derivative of ƒ at P0 in the direction of u”
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sDu ƒdP0.
EXAMPLE 1
(1)
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DEFINITION Directional Derivative The derivative of ƒ at P0(x 0 , y0) in the direction of the unit vector u u 1i u 2 j is the number
Finding a Directional Derivative Using the Definition
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Find the derivative of
ƒsx, yd = x 2 + xy
at P0s1, 2d in the direction of the unit vector u = A 1> 22 B i + A 1> 22 B j.
¢
ass a
Solution
dƒ ƒsx0 + su1, y0 + su2 d  ƒsx0 , y0 d = lim ≤ s ds u,P0 s:0 ƒ¢1 + s #
dH
= lim
s:0
¢1 +
= lim
s 22
1 1 ,2 + s# ≤  ƒs1, 2d 22 22 s 2
≤ + ¢1 +
ma ham Mu
¢1 + = lim
2s 22
s:0
5s s:0
s 22
≤ ¢2 +
s 22
≤  s12 + 1 # 2d
s
s:0
= lim
Equation (1)
22
+
3s s2 s2 + ¢2 + + ≤  3 ≤ 2 2 22 s
+ s2 s
= lim ¢ s:0
5 22
+ s≤ = ¢
5 22
+ 0≤ =
5 22
.
The rate of change of ƒsx, yd = x2 + xy at P0s1, 2d in the direction u = A 1> 12 B i + A 1> 22 B j is 5> 12.
Interpretation of the Directional Derivative The equation z = ƒsx, yd represents a surface S in space. If z0 = ƒsx0 , y0 d, then the point Psx0 , y0 , z0 d lies on S. The vertical plane that passes through P and P0sx0 , y0 d parallel to u
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
s y
C x P0(x 0 , y0 )
(x 0 su1, y0 su 2 ) u u 1i u 2 j
FIGURE 14.25 The slope of curve C at P0 is lim slope (PQ); this is the Q:P
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Tangent line Q P(x 0 , y0 , z 0 )
intersects S in a curve C (Figure 14.25). The rate of change of ƒ in the direction of u is the slope of the tangent to C at P. When u = i, the directional derivative at P0 is 0ƒ>0x evaluated at sx0 , y0 d. When u = j, the directional derivative at P0 is 0ƒ>0y evaluated at sx0 , y0 d. The directional derivative generalizes the two partial derivatives. We can now ask for the rate of change of ƒ in any direction u, not just the directions i and j. Here’s a physical interpretation of the directional derivative. Suppose that T = ƒsx, yd is the temperature at each point (x, y) over a region in the plane. Then ƒsx0 , y0 d is the temperature at the point P0sx0 , y0 d and sDu ƒdP0 is the instantaneous rate of change of the temperature at P0 stepping off in the direction u.
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f (x 0 su1, y0 su 2 ) f (x 0 , y0 )
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z Surface S: z f (x, y)
Calculation and Gradients
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1008
We now develop an efficient formula to calculate the directional derivative for a differentiable function ƒ. We begin with the line
directional derivative
y = y0 + su2 ,
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x = x0 + su1 ,
dƒ a b = sDu ƒdP0. ds u,P0
(2)
through P0sx0 , y0 d , parametrized with the arc length parameter s increasing in the direction of the unit vector u = u1 i + u2 j. Then dƒ 0ƒ 0ƒ dy dx b = a b + a b 0x P0 ds 0y P0 ds ds u,P0
ass a
a
= a
0ƒ 0ƒ b # u + a b # u2 0x P0 1 0y P0
dH
= ca
Chain Rule for differentiable f
From Equations (2), dx>ds = u1 and dy>ds = u2
0ƒ 0ƒ b i + a b j d # cu1 i + u2 j d. 0x P0 0y P0
(3)
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(''''')'''''* ('')''* Gradient of ƒ at P0 Direction u
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DEFINITION Gradient Vector The gradient vector (gradient) of ƒ(x, y) at a point P0sx0 , y0 d is the vector §ƒ =
0ƒ 0ƒ i + j 0x 0y
obtained by evaluating the partial derivatives of ƒ at P0 .
The notation §ƒ is read “grad ƒ ” as well as “gradient of ƒ ” and “del ƒ.” The symbol § by itself is read “del.” Another notation for the gradient is grad ƒ, read the way it is written. Equation (3) says that the derivative of a differentiable function ƒ in the direction of u at P0 is the dot product of u with the gradient of ƒ at P0 .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.5
Directional Derivatives and Gradient Vectors
1009
a
dƒ b = s§ƒdP0 # u, ds u,P0
EXAMPLE 2
(4)
You
the dot product of the gradient ƒ at P0 and u.
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THEOREM 9 The Directional Derivative Is a Dot Product If ƒ(x, y) is differentiable in an open region containing P0sx0 , y0 d, then
Finding the Directional Derivative Using the Gradient
Solution
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Find the derivative of ƒsx, yd = xe y + cos sxyd at the point (2, 0) in the direction of v = 3i  4j. The direction of v is the unit vector obtained by dividing v by its length: v v 3 4 = = i  j. 5 5 5 ƒvƒ
nR
u =
The partial derivatives of ƒ are everywhere continuous and at (2, 0) are given by ƒxs2, 0d = se y  y sin sxydds2,0d = e 0  0 = 1
y 2
ass a
ƒys2, 0d = sxe y  x sin sxydds2,0d = 2e 0  2 # 0 = 2.
∇f i 2j
The gradient of ƒ at (2, 0) is
§ƒ ƒ s2,0d = ƒxs2, 0di + ƒys2, 0dj = i + 2j
1
–1
1 P0 (2, 0)
3
u 3i 4j 5 5
4
x
sDuƒd ƒ s2,0d = §ƒ ƒ s2,0d # u
dH
0
(Figure 14.26). The derivative of ƒ at (2, 0) in the direction of v is therefore Equation (4)
3 3 8 4 = si + 2jd # a i  jb =  = 1 . 5 5 5 5
Evaluating the dot product in the formula Du ƒ = §ƒ # u = ƒ §ƒ ƒ ƒ u ƒ cos u = ƒ §ƒ ƒ cos u,
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FIGURE 14.26 Picture §ƒ as a vector in the domain of ƒ. In the case of ƒsx, yd = xe y + cos sxyd, the domain is the entire plane. The rate at which ƒ changes at (2, 0) in the direction u = s3>5di  s4>5dj is §ƒ # u = 1 (Example 2).
where u is the angle between the vectors u and §ƒ, reveals the following properties.
Properties of the Directional Derivative Duƒ = §ƒ # u = ƒ §ƒ ƒ cos u 1.
2. 3.
The function ƒ increases most rapidly when cos u = 1 or when u is the direction of §ƒ. That is, at each point P in its domain, ƒ increases most rapidly in the direction of the gradient vector §ƒ at P. The derivative in this direction is Duƒ = ƒ §ƒ ƒ cos s0d = ƒ §ƒ ƒ . Similarly, ƒ decreases most rapidly in the direction of  §ƒ. The derivative in this direction is Duƒ = ƒ §ƒ ƒ cos spd =  ƒ §ƒ ƒ . Any direction u orthogonal to a gradient §f Z 0 is a direction of zero change in ƒ because u then equals p>2 and Duƒ = ƒ §ƒ ƒ cos sp>2d = ƒ §ƒ ƒ # 0 = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1010
Chapter 14: Partial Derivatives
Finding Directions of Maximal, Minimal, and Zero Change
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EXAMPLE 3
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As we discuss later, these properties hold in three dimensions as well as two. Find the directions in which ƒsx, yd = sx 2>2d + s y 2>2d
z f (x, y) x2 y2 2 2
Solution
(a) The function increases most rapidly in the direction of §ƒ at (1, 1). The gradient there is
iaz
z
You
(a) Increases most rapidly at the point (1, 1) (b) Decreases most rapidly at (1, 1). (c) What are the directions of zero change in ƒ at (1, 1)?
s§ƒds1,1d = sxi + yjds1,1d = i + j . (1, 1, 1)
1 (1, 1)
Most rapid decrease in f Most rapid increase in f
Zero change in f
∇f i j
(b) The function decreases most rapidly in the direction of  §ƒ at (1, 1), which is
ass a
x
–∇f
i + j i + j 1 1 = = i + j. 2 2 i + j ƒ ƒ 2s1d + s1d 22 22
u = 
FIGURE 14.27 The direction in which ƒsx, yd = sx 2>2d + s y 2>2d increases most rapidly at (1, 1) is the direction of §ƒ ƒ s1,1d = i + j. It corresponds to the direction of steepest ascent on the surface at (1, 1, 1) (Example 3).
1 1 i j. 22 22
(c) The directions of zero change at (1, 1) are the directions orthogonal to §ƒ: n = 
1 1 i + j 22 22
dH
1
u =
y
nR
Its direction is
and
n =
1 1 i j. 22 22
See Figure 14.27.
ma
Gradients and Tangents to Level Curves
Mu
ham
If a differentiable function ƒ(x, y) has a constant value c along a smooth curve r = gstdi + hstdj (making the curve a level curve of ƒ), then ƒsgstd, hstdd = c. Differentiating both sides of this equation with respect to t leads to the equations d d ƒsgstd, hstdd = scd dt dt 0ƒ dg 0ƒ dh + = 0 0x dt 0y dt a
Chain Rule
0ƒ 0ƒ dg dh i + jb # a i + jb = 0. 0x 0y dt dt
('')''* §ƒ
('')''* dr dt
(5)
Equation (5) says that §ƒ is normal to the tangent vector dr> dt, so it is normal to the curve.
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4100 AWL/Thomas_ch14p9651066 8/25/04 2:53 PM Page 1011
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.5
Directional Derivatives and Gradient Vectors
1011
The level curve f (x, y) f (x 0 , y 0 )
suf
i
At every point sx0 , y0 d in the domain of a differentiable function ƒ(x, y), the gradient of ƒ is normal to the level curve through sx0 , y0 d (Figure 14.28). (x 0 , y 0 )
Equation (5) validates our observation that streams flow perpendicular to the contours in topographical maps (see Figure 14.23). Since the downflowing stream will reach its destination in the fastest way, it must flow in the direction of the negative gradient vectors from Property 2 for the directional derivative. Equation (5) tells us these directions are perpendicular to the level curves. This observation also enables us to find equations for tangent lines to level curves. They are the lines normal to the gradients. The line through a point P0sx0 , y0 d normal to a vector N = Ai + Bj has the equation
iaz
FIGURE 14.28 The gradient of a differentiable function of two variables at a point is always normal to the function’s level curve through that point.
You
∇f (x 0 , y 0 )
nR
Asx  x0 d + Bs y  y0 d = 0 (Exercise 35). If N is the gradient s§ƒdsx0, y0d = ƒxsx0 , y0 di + ƒysx0 , y0 dj, the equation is the tangent line given by
ass a
ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 dsy  y0 d = 0.
EXAMPLE 4
(6)
Finding the Tangent Line to an Ellipse
Find an equation for the tangent to the ellipse y
x2 y2 2 4
兹2 (–2, 1) –1
1 0
1
2
x
Solution
The ellipse is a level curve of the function
2兹2
ham
FIGURE 14.29 We can find the tangent to the ellipse sx 2>4d + y 2 = 2 by treating the ellipse as a level curve of the function ƒsx, yd = sx 2>4d + y 2 (Example 4).
Mu
x2 + y2 = 2 4
(Figure 14.29) at the point s 2, 1d.
ma
–2
dH
x 2y –4
∇f (–2, 1) – i 2j
ƒsx, yd =
x2 + y 2. 4
The gradient of ƒ at s 2, 1d is x §ƒ ƒ s2,1d = a i + 2yjb = i + 2j. 2 s2,1d The tangent is the line s 1dsx + 2d + s2ds y  1d = 0
Equation (6)
x  2y = 4. If we know the gradients of two functions ƒ and g, we automatically know the gradients of their constant multiples, sum, difference, product, and quotient. You are asked to establish the following rules in Exercise 36. Notice that these rules have the same form as the corresponding rules for derivatives of singlevariable functions.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1012
Chapter 14: Partial Derivatives
5.
Quotient Rule:
§skƒd = k§ƒ sany number kd §sƒ + gd = §ƒ + §g §sƒ  gd = §ƒ  §g §sƒgd = ƒ§g + g§ƒ ƒ g§ƒ  ƒ§g § ag b = g2
Illustrating the Gradient Rules
iaz
EXAMPLE 5
suf
Constant Multiple Rule: Sum Rule: Difference Rule: Product Rule:
You
1. 2. 3. 4.
i
Algebra Rules for Gradients
We illustrate the rules with
gsx, yd = 3y §g = 3j.
nR
ƒsx, yd = x  y §ƒ = i  j
We have
§s2ƒd = §s2x  2yd = 2i  2j = 2§ƒ
2.
§sƒ + gd = §sx + 2yd = i + 2j = §ƒ + §g
3.
§sƒ  gd = §sx  4yd = i  4j = §ƒ  §g
4.
§sƒgd = §s3xy  3y 2 d = 3yi + s3x  6ydj
ass a
1.
= 3ysi  jd + 3yj + s3x  6ydj = 3ysi  jd + s3x  3ydj
dH
= 3ysi  jd + sx  yd3j = g§ƒ + ƒ§g
5.
ƒ x  y x 1 § ag b = § a b = §a  b 3y 3y 3
ma
=
x 1 i j 3y 3y 2
ham Mu
3ysi  jd  s3x  3ydj
3yi  3xj
= 9y
2
=
9y 2
3ysi  jd  sx  yd3j
= 9y
2
g§ƒ  ƒ§g =
g2
.
Functions of Three Variables For a differentiable function ƒ(x, y, z) and a unit vector u = u1 i + u2 j + u3 k in space, we have 0ƒ 0ƒ 0ƒ §ƒ = i + j + k 0x 0y 0z and 0ƒ 0ƒ 0ƒ Duƒ = §ƒ # u = u + u + u. 0x 1 0y 2 0z 3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.5 Directional Derivatives and Gradient Vectors
1013
suf
Du ƒ = §ƒ # u = ƒ §ƒ ƒ ƒ u ƒ cos u = ƒ §ƒ ƒ cos u,
i
The directional derivative can once again be written in the form
EXAMPLE 6
You
so the properties listed earlier for functions of two variables continue to hold. At any given point, ƒ increases most rapidly in the direction of §ƒ and decreases most rapidly in the direction of  §ƒ. In any direction orthogonal to §ƒ, the derivative is zero.
Finding Directions of Maximal, Minimal, and Zero Change
iaz
(a) Find the derivative of ƒsx, y, zd = x 3  xy 2  z at P0s1, 1, 0d in the direction of v = 2i  3j + 6k. (b) In what directions does ƒ change most rapidly at P0 , and what are the rates of change in these directions? Solution
(a) The direction of v is obtained by dividing v by its length:
nR
ƒ v ƒ = 2s2d2 + s 3d2 + s6d2 = 249 = 7 u =
v 3 6 2 = i  j + k. 7 7 7 ƒvƒ
ass a
The partial derivatives of ƒ at P0 are ƒx = s3x 2  y 2 ds1,1,0d = 2,
ƒy = 2xy ƒ s1,1,0d = 2,
ƒz = 1 ƒ s1,1,0d = 1.
The gradient of ƒ at P0 is
§ƒ ƒ s1,1,0d = 2i  2j  k.
dH
The derivative of ƒ at P0 in the direction of v is therefore
ma
3 6 2 sDuƒds1,1,0d = §ƒ ƒ s1,1,0d # u = s2i  2j  kd # a i  j + kb 7 7 7 =
6 6 4 4 +  = . 7 7 7 7
ƒ §ƒ ƒ = 2s2d2 + s 2d2 + s 1d2 = 29 = 3
and
 ƒ §ƒ ƒ = 3 .
Mu
ham
(b) The function increases most rapidly in the direction of §ƒ = 2i  2j  k and decreases most rapidly in the direction of  §ƒ. The rates of change in the directions are, respectively,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
14.5 Directional Derivatives and Gradient Vectors
EXERCISES 14.5
n a s as
Calculating Gradients at Points
H d a
In Exercises 1–4, find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point.
m m a h Mu
1. ƒsx, yd = y  x,
s2, 1d
2
2
2. ƒsx, yd = ln sx + y d,
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s1, 1d
Y z a i R
3. gsx, yd = y  x 2,
s 1, 0d
4. gsx, yd =
y2 x2  , 2 2
In Exercises 5–8, find §f at the given point. 5. ƒsx, y, zd = x 2 + y 2  2z 2 + z ln x,
s1, 1, 1d
6. ƒsx, y, zd = 2z 3  3sx 2 + y 2 dz + tan1 xz,
s1, 1, 1d
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1013
A 22, 1 B
4100 AWL/Thomas_ch14p9651066 8/25/04 2:53 PM Page 1014
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
i
s0, 0, p>6d
Finding Directional Derivatives
2
2
10. ƒsx, yd = 2x + y ,
P0s5, 5d,
A = 4i + 3j
P0s 1, 1d,
A = 3i  4j
11. gsx, yd = x  s y 2>xd + 23 sec1 s2xyd, A = 12i + 5j 12. hsx, yd = tan A = 3i  2j
1
sy>xd + 23 sin
sxy>2d,
P0s1, 1, 2d,
13. ƒsx, y, zd = xy + yz + zx,
14. ƒsx, y, zd = x 2 + 2y 2  3z 2, x
1
P0s1, 1, 1d,
P0s0, 0, 0d,
15. gsx, y, zd = 3e cos yz,
16. hsx, y, zd = cos xy + e yz + ln zx, A = i + 2j + 2k
32. The derivative of ƒ(x, y, z) at a point P is greatest in the direction of v = i + j  k. In this direction, the value of the derivative is 2 13.
P0s1, 1d,
a. What is §ƒ at P ? Give reasons for your answer. P0s1, 1d,
b. What is the derivative of ƒ at P in the direction of i + j ?
A = 3i + 6j  2k A = i + j + k
A = 2i + j  2k
P0s1, 0, 1>2d,
Directions of Most Rapid Increase and Decrease
P0s 1, 1d
18. ƒsx, yd = x 2y + e xy sin y, 19. ƒsx, y, zd = sx>yd  yz,
P0s1, 0d P0s4, 1, 1d
P0s1, ln 2, 1>2d
dH
20. gsx, y, zd = xe y + z 2,
P0s1, 1, 1d
21. ƒsx, y, zd = ln xy + ln yz + ln xz,
22. hsx, y, zd = ln sx 2 + y 2  1d + y + 6z,
Tangent Lines to Curves
P0s1, 1, 0d
A 22, 22 B 24. x 2  y = 1, A 22, 1 B
s2, 2d
2
2
26. x  xy + y = 7,
ham
25. xy = 4,
ma
In Exercises 23–26, sketch the curve ƒsx, yd = c together with §f and the tangent line at the given point. Then write an equation for the tangent line. 23. x 2 + y 2 = 4,
33. Directional derivatives and scalar components How is the derivative of a differentiable function ƒ(x, y, z) at a point P0 in the direction of a unit vector u related to the scalar component of s§ƒdP0 in the direction of u? Give reasons for your answer. 34. Directional derivatives and partial derivatives Assuming that the necessary derivatives of ƒ(x, y, z) are defined, how are Di ƒ, Dj ƒ, and D k ƒ related to ƒx , ƒy , and ƒz? Give reasons for your answer. 35. Lines in the xyplane Show that Asx  x0 d + Bsy  y0 d = 0 is an equation for the line in the xyplane through the point sx0 , y0 d normal to the vector N = Ai + Bj.
ass a
In Exercises 17–22, find the directions in which the functions increase and decrease most rapidly at P0. Then find the derivatives of the functions in these directions. 17. ƒsx, yd = x 2 + xy + y 2,
31. The derivative of ƒ(x, y) at P0s1, 2d in the direction of i + j is 212 and in the direction of 2j is 3 . What is the derivative of ƒ in the direction of i  2j ? Give reasons for your answer.
You
In Exercises 9–16, find the derivative of the function at P0 in the direction of A. 9. ƒsx, yd = 2xy  3y 2,
30. Changing temperature along a circle Is there a direction u in which the rate of change of the temperature function T sx, y, zd = 2xy  yz (temperature in degrees Celsius, distance in feet) at Ps1, 1, 1d is 3°C>ft ? Give reasons for your answer.
suf
8. ƒsx, y, zd = e x + y cos z + s y + 1d sin1 x,
s 1, 2, 2d
iaz
7. ƒsx, y, zd = sx 2 + y 2 + z 2 d1>2 + ln sxyzd,
nR
1014
s 1, 2d
36. The algebra rules for gradients gradients
Given a constant k and the
§ƒ =
0ƒ 0ƒ 0ƒ i + j + k 0x 0y 0z
§g =
0g 0g 0g i + j + k, 0x 0y 0z
and
use the scalar equations 0ƒ 0 skƒd = k , 0x 0x
0g 0ƒ 0 sƒgd = ƒ + g , 0x 0x 0x
0ƒ 0g 0 sƒ ; gd = , ; 0x 0x 0x 0ƒ 0g g  ƒ 0x 0x 0 ƒ a b = , 0x g g2
and so on, to establish the following rules.
27. Zero directional derivative In what direction is the derivative of ƒsx, yd = xy + y 2 at P(3, 2) equal to zero?
a. §skƒd = k§ƒ
28. Zero directional derivative In what directions is the derivative of ƒsx, yd = sx 2  y 2 d>sx 2 + y 2 d at P(1, 1) equal to zero?
c. §sƒ  gd = §ƒ  §g
Mu
Theory and Examples
29. Is there a direction u in which the rate of change of ƒsx, yd = x 2  3xy + 4y 2 at P(1, 2) equals 14? Give reasons for your answer.
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b. §sƒ + gd = §ƒ + §g d. §sƒgd = ƒ§g + g§ƒ ƒ g§ƒ  ƒ§g e. § a g b = g2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6
Tangent Planes and Differentials
suf
i
Tangent Planes and Differentials
14.6
1015
Tangent Planes and Normal Lines
You
In this section we define the tangent plane at a point on a smooth surface in space. We calculate an equation of the tangent plane from the partial derivatives of the function defining the surface. This idea is similar to the definition of the tangent line at a point on a curve in the coordinate plane for singlevariable functions (Section 2.7). We then study the total differential and linearization of functions of several variables.
iaz
If r = gstdi + hstdj + kstdk is a smooth curve on the level surface ƒsx, y, zd = c of a differentiable function ƒ, then ƒsgstd, hstd, kstdd = c. Differentiating both sides of this equation with respect to t leads to d d ƒsgstd, hstd, kstdd = scd dt dt
nR
∇f
0ƒ dg 0ƒ dh 0ƒ dk + + = 0 0x dt 0y dt 0z dt
v2 P0
v1
a
f (x, y, z) c
0ƒ 0ƒ 0ƒ dg dh dk i + j + kb # a i + j + kb = 0. 0x 0y 0z dt dt dt
(1)
('''')''''* dr>dt
ass a
('''')''''* §ƒ
At every point along the curve, §ƒ is orthogonal to the curve’s velocity vector. Now let us restrict our attention to the curves that pass through P0 (Figure 14.30). All the velocity vectors at P0 are orthogonal to §ƒ at P0, so the curves’ tangent lines all lie in the plane through P0 normal to §ƒ. We call this plane the tangent plane of the surface at P0. The line through P0 perpendicular to the plane is the surface’s normal line at P0.
dH
FIGURE 14.30 The gradient §ƒ is orthogonal to the velocity vector of every smooth curve in the surface through P0. The velocity vectors at P0 therefore lie in a common plane, which we call the tangent plane at P0.
Chain Rule
Tangent Plane, Normal Line
ma
DEFINITIONS
Mu
ham
The tangent plane at the point P0sx0 , y0 , z0 d on the level surface ƒsx, y, zd = c of a differentiable function ƒ is the plane through P0 normal to §ƒ ƒ P0. The normal line of the surface at P0 is the line through P0 parallel to §ƒ ƒ P0.
Thus, from Section 12.5, the tangent plane and normal line have the following equations:
Tangent Plane to ƒsx, y, zd = c at P0sx0 , y0 , z0 d ƒxsP0 dsx  x0 d + ƒy sP0 dsy  y0 d + ƒzsP0 dsz  z0 d = 0
(2)
Normal Line to ƒsx, y, zd = c at P0sx0 , y0 , z0 d
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x = x0 + ƒxsP0 dt,
y = y0 + ƒy sP0 dt,
z = z0 + ƒzsP0 dt
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4100 AWL/Thomas_ch14p9651066 8/25/04 2:53 PM Page 1016
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1016
Chapter 14: Partial Derivatives
P0 (1, 2, 4)
EXAMPLE 1
Finding the Tangent Plane and Normal Line
i
The surface x2 y2 z 9 0
Find the tangent plane and normal line of the surface
suf
z
ƒsx, y, zd = x 2 + y 2 + z  9 = 0 at the point P0s1, 2, 4d.
You
Normal line
The surface is shown in Figure 14.31. The tangent plane is the plane through P0 perpendicular to the gradient of ƒ at P0 . The gradient is Solution
Tangent plane
§ƒ ƒ P0 = s2xi + 2yj + kds1,2,4d = 2i + 4j + k.
2
The tangent plane is therefore the plane
iaz
y
2sx  1d + 4s y  2d + sz  4d = 0,
x
The line normal to the surface at P0 is x = 1 + 2t,
y = 2 + 4t,
or
2x + 4y + z = 14.
z = 4 + t.
To find an equation for the plane tangent to a smooth surface z = ƒsx, yd at a point P0sx0 , y0 , z0 d where z0 = ƒsx0 , y0 d, we first observe that the equation z = ƒsx, yd is equivalent to ƒsx, yd  z = 0. The surface z = ƒsx, yd is therefore the zero level surface of the function Fsx, y, zd = ƒsx, yd  z. The partial derivatives of F are
dH
ass a
FIGURE 14.31 The tangent plane and normal line to the surface x 2 + y 2 + z  9 = 0 at P0s1, 2, 4d (Example 1).
nR
1
A circular paraboloid
Fx =
0 sƒsx, yd  zd = fx  0 = fx 0x
Fy =
0 sƒsx, yd  zd = fy  0 = fy 0y
Fz =
0 sƒsx, yd  zd = 0  1 = 1. 0z
The formula
FxsP0 dsx  x0 d + FysP0 dsy  y0 d + FzsP0 dsz  z0 d = 0
Mu
ham
ma
for the plane tangent to the level surface at P0 therefore reduces to ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 dsy  y0 d  sz  z0 d = 0.
Plane Tangent to a Surface z = ƒsx, yd at sx0 , y0 , ƒsx0 , y0 dd The plane tangent to the surface z = ƒsx, yd of a differentiable function ƒ at the point P0sx0 , y0 , z0 d = sx0 , y0 , ƒsx0 , y0 dd is ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 dsy  y0 d  sz  z0 d = 0.
EXAMPLE 2
Finding a Plane Tangent to a Surface z = ƒsx, yd
Find the plane tangent to the surface z = x cos y  ye x at (0, 0, 0).
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4100 AWL/Thomas_ch14p9651066 8/25/04 2:53 PM Page 1017
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6
Tangent Planes and Differentials
1017
We calculate the partial derivatives of ƒsx, yd = x cos y  ye x and use Equation (4):
suf
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Solution
ƒxs0, 0d = scos y  ye x ds0,0d = 1  0 # 1 = 1
ƒys0, 0d = s x sin y  e x ds0,0d = 0  1 = 1.
You
The tangent plane is therefore
1 # sx  0d  1 # s y  0d  sz  0d = 0, or
Equation (4)
x  y  z = 0.
Tangent Line to the Curve of Intersection of Two Surfaces
iaz
EXAMPLE 3 The surfaces
nR
ƒsx, y, zd = x 2 + y 2  2 = 0 and
gsx, y, zd = x + z  4 = 0
The plane x z 4 0 g(x, y, z)
A plane
meet in an ellipse E (Figure 14.32). Find parametric equations for the line tangent to E at the point P0s1, 1, 3d .
ass a
z
A cylinder
The tangent line is orthogonal to both §ƒ and §g at P0, and therefore parallel to v = §ƒ * §g. The components of v and the coordinates of P0 give us equations for the line. We have Solution
dH
§ƒ ƒ s1,1,3d = s2xi + 2yjds1,1,3d = 2i + 2j
§g ƒ s1,1,3d = si + kds1,1,3d = i + k
∇g
i v = s2i + 2jd * si + kd = 3 2 1
ma
The ellipse E
∇f
(1, 1, 3)
ham
∇f ∇g
x
The cylinder x2 y2 2 0
Mu
k 0 3 = 2i  2j  2k. 1
The tangent line is x = 1 + 2t,
y = 1  2t,
z = 3  2t.
Estimating Change in a Specific Direction y
The directional derivative plays the role of an ordinary derivative when we want to estimate how much the value of a function ƒ changes if we move a small distance ds from a point P0 to another point nearby. If ƒ were a function of a single variable, we would have
f (x, y, z)
FIGURE 14.32 The cylinder ƒsx, y, zd = x 2 + y 2  2 = 0 and the plane gsx, y, zd = x + z  4 = 0 intersect in an ellipse E (Example 3).
j 2 0
dƒ = ƒ¿sP0 d ds.
Ordinary derivative * increment
For a function of two or more variables, we use the formula dƒ = s§ƒ ƒ P0 # ud ds,
Directional derivative * increment
where u is the direction of the motion away from P0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1018
Chapter 14: Partial Derivatives
dƒ = s§ƒ ƒ P0 # ud
ds
()*
Directional derivative
Distance increment
You
EXAMPLE 4
#
('')''*
suf
i
Estimating the Change in ƒ in a Direction u To estimate the change in the value of a differentiable function ƒ when we move a small distance ds from a point P0 in a particular direction u, use the formula
Estimating Change in the Value of ƒ(x, y, z)
iaz
Estimate how much the value of
ƒsx, y, zd = y sin x + 2yz
nR
will change if the point P(x, y, z) moves 0.1 unit from P0s0, 1, 0d straight toward P1s2, 2, 2d. 1 We first find the derivative of ƒ at P0 in the direction of the vector P0 P1 = 2i + j  2k. The direction of this vector is 1 1 P0 P1 P0 P1 2 1 2 u = 1 = = i + j  k. 3 3 3 3 P P ƒ 0 1ƒ
ass a
Solution
The gradient of ƒ at P0 is
§ƒ ƒ s0,1,0d = ss y cos xdi + ssin x + 2zdj + 2ykdds0,1,0d = i + 2k.
Therefore,
dH
2 1 2 2 4 2 §ƒ ƒ P0 # u = si + 2kd # a i + j  kb =  =  . 3 3 3 3 3 3
Mu
ham
ma
The change dƒ in ƒ that results from moving ds = 0.1 unit away from P0 in the direction of u is approximately 2 dƒ = s§ƒ ƒ P0 # udsdsd = a bs0.1d L 0.067 unit. 3
How to Linearize a Function of Two Variables
Functions of two variables can be complicated, and we sometimes need to replace them with simpler ones that give the accuracy required for specific applications without being so difficult to work with. We do this in a way that is similar to the way we find linear replacements for functions of a single variable (Section 3.8). Suppose the function we wish to replace is z = ƒsx, yd and that we want the replacement to be effective near a point sx0 , y0 d at which we know the values of ƒ, ƒx , and ƒy and at which ƒ is differentiable. If we move from sx0 , y0 d to any point (x, y) by increments ¢x = x  x0 and ¢y = y  y0 , then the definition of differentiability from Section 14.3 gives the change ƒsx, yd  ƒsx0 , y0 d = fxsx0 , y0 d¢x + ƒysx0 , y0 d¢y + P1 ¢x + P2 ¢y,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6
1019
i
where P1, P2 : 0 as ¢x, ¢y : 0. If the increments ¢x and ¢y are small, the products P1 ¢x and P2 ¢y will eventually be smaller still and we will have
(x, y)
suf
A point near (x 0 , y 0 )
Tangent Planes and Differentials
ƒsx, yd L ƒsx0 , y0 d + fxsx0 , y0 dsx  x0 d + ƒysx0 , y0 ds y  y0 d. (''''''''''''')'''''''''''''* Lsx, yd y y y0
You
In other words, as long as ¢x and ¢y are small, ƒ will have approximately the same value as the linear function L. If ƒ is hard to use, and our work can tolerate the error involved, we may approximate ƒ by L (Figure 14.33).
A point where f is differentiable (x 0 , y0 ) x x x 0
iaz
Lsx, yd = ƒsx0 , y0 d + ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 ds y  y0 d. The approximation
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FIGURE 14.33 If ƒ is differentiable at sx0 , y0 d, then the value of ƒ at any point (x, y) nearby is approximately ƒsx0 , y0 d + ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y.
DEFINITIONS Linearization, Standard Linear Approximation The linearization of a function ƒ(x, y) at a point sx0 , y0 d where ƒ is differentiable is the function (5)
ƒsx, yd L Lsx, yd
is the standard linear approximation of ƒ at sx0 , y0 d.
ass a
From Equation (4), we see that the plane z = Lsx, yd is tangent to the surface z = ƒsx, yd at the point sx0 , y0 d. Thus, the linearization of a function of two variables is a tangentplane approximation in the same way that the linearization of a function of a single variable is a tangentline approximation.
Finding a Linearization
dH
EXAMPLE 5
Find the linearization of ƒsx, yd = x 2  xy +
1 2 y + 3 2
ma
at the point (3, 2).
Mu
ham
Solution
We first evaluate ƒ, ƒx , and ƒy at the point sx0 , y0 d = s3, 2d: ƒs3, 2d = ax 2  xy +
1 2 y + 3b = 8 2 s3,2d
ƒxs3, 2d =
0 1 ax 2  xy + y 2 + 3b = s2x  yds3,2d = 4 0x 2 s3,2d
ƒys3, 2d =
0 1 ax 2  xy + y 2 + 3b = s x + yds3,2d = 1, 0y 2 s3,2d
giving Lsx, yd = ƒsx0 , y0 d + ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 ds y  y0 d = 8 + s4dsx  3d + s 1ds y  2d = 4x  y  2. The linearization of ƒ at (3, 2) is Lsx, yd = 4x  y  2.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1020
Chapter 14: Partial Derivatives
y
k h (x 0 , y0 )
You
suf
i
When approximating a differentiable function ƒ(x, y) by its linearization L(x, y) at sx0 , y0 d, an important question is how accurate the approximation might be. If we can find a common upper bound M for ƒ ƒxx ƒ , ƒ ƒyy ƒ , and ƒ ƒxy ƒ on a rectangle R centered at sx0 , y0 d (Figure 14.34), then we can bound the error E throughout R by using a simple formula (derived in Section 14.10). The error is defined by Esx, yd = ƒsx, yd  Lsx, yd.
R
0
FIGURE 14.34 The rectangular region R: ƒ x  x0 ƒ … h, ƒ y  y0 ƒ … k in the xyplane.
The Error in the Standard Linear Approximation If ƒ has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at sx0 , y0 d and if M is any upper bound for the values of ƒ ƒxx ƒ , ƒ ƒyy ƒ , and ƒ ƒxy ƒ on R, then the error E(x, y) incurred in replacing ƒ(x, y) on R by its linearization
iaz
x
Lsx, yd = ƒsx0 , y0 d + ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 dsy  y0 d satisfies the inequality
nR
1 2 ƒ Esx, yd ƒ … 2 Ms ƒ x  x0 ƒ + ƒ y  y0 ƒ d .
To make ƒ Esx, yd ƒ small for a given M, we just make ƒ x  x0 ƒ and ƒ y  y0 ƒ small.
Bounding the Error in Example 5
ass a
EXAMPLE 6
Find an upper bound for the error in the approximation ƒsx, yd L Lsx, yd in Example 5 over the rectangle R:
ƒ x  3 ƒ … 0.1,
ƒ y  2 ƒ … 0.1 .
dH
Express the upper bound as a percentage of ƒ(3, 2), the value of ƒ at the center of the rectangle. We use the inequality
ma
Solution
1 ƒ Esx, yd ƒ … 2 Ms ƒ x  x0 ƒ + ƒ y  y0 ƒ d2 .
Mu
ham
To find a suitable value for M, we calculate ƒxx, ƒxy , and ƒyy , finding, after a routine differentiation, that all three derivatives are constant, with values ƒ ƒxx ƒ = ƒ 2 ƒ = 2,
ƒ ƒxy ƒ = ƒ 1 ƒ = 1,
ƒ ƒyy ƒ = ƒ 1 ƒ = 1.
The largest of these is 2, so we may safely take M to be 2. With sx0 , y0 d = s3, 2d, we then know that, throughout R, 1 2 2 ƒ Esx, yd ƒ … 2 s2ds ƒ x  3 ƒ + ƒ y  2 ƒ d = s ƒ x  3 ƒ + ƒ y  2 ƒ d . Finally, since ƒ x  3 ƒ … 0.1 and ƒ y  2 ƒ … 0.1 on R, we have ƒ Esx, yd ƒ … s0.1 + 0.1d2 = 0.04. As a percentage of ƒs3, 2d = 8, the error is no greater than
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0.04 * 100 = 0.5% . 8
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6
Tangent Planes and Differentials
1021
i
Differentials
¢ƒ = ƒsa + ¢xd  ƒsad
You
and the differential of ƒ as
suf
Recall from Section 3.8 that for a function of a single variable, y = ƒsxd, we defined the change in ƒ as x changes from a to a + ¢x by
dƒ = ƒ¿sad¢x.
iaz
We now consider a function of two variables. Suppose a differentiable function ƒ(x, y) and its partial derivatives exist at a point sx0 , y0 d. If we move to a nearby point sx0 + ¢x, y0 + ¢yd, the change in ƒ is ¢ƒ = ƒsx0 + ¢x, y0 + ¢yd  ƒsx0 , y0 d.
nR
A straightforward calculation from the definition of L(x, y), using the notation x  x0 = ¢x and y  y0 = ¢y, shows that the corresponding change in L is ¢L = Lsx0 + ¢x, y0 + ¢yd  Lsx0 , y0 d = ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y.
ass a
The differentials dx and dy are independent variables, so they can be assigned any values. Often we take dx = ¢x = x  x0 , and dy = ¢y = y  y0 . We then have the following definition of the differential or total differential of ƒ.
dH
DEFINITION Total Differential If we move from sx0 , y0 d to a point sx0 + dx, y0 + dyd nearby, the resulting change dƒ = ƒxsx0 , y0 d dx + ƒysx0 , y0 d dy
ma
in the linearization of ƒ is called the total differential of ƒ.
Mu
ham
EXAMPLE 7
Estimating Change in Volume
Suppose that a cylindrical can is designed to have a radius of 1 in. and a height of 5 in., but that the radius and height are off by the amounts dr = +0.03 and dh = 0.1. Estimate the resulting absolute change in the volume of the can. Solution
To estimate the absolute change in V = pr 2h, we use ¢V L dV = Vr sr0 , h0 d dr + Vhsr0 , h0 d dh.
With Vr = 2prh and Vh = pr 2, we get dV = 2pr0 h0 dr + pr 02 dh = 2ps1ds5ds0.03d + ps1d2s 0.1d
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= 0.3p  0.1p = 0.2p L 0.63 in.3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1022
Chapter 14: Partial Derivatives
dƒ ƒsx0 , y0 d
and
dƒ * 100, ƒsx0 , y0 d
You
respectively. In Example 7, the relative change is estimated by
suf
i
Instead of absolute change in the value of a function ƒ(x, y), we can estimate relative change or percentage change by
dV 0.2p 0.2p = = 0.04, = Vsr0 , h0 d pr 02h0 ps1d2s5d giving 4% as an estimate of the percentage change.
Sensitivity to Change
iaz
EXAMPLE 8
Your company manufactures right circular cylindrical molasses storage tanks that are 25 ft high with a radius of 5 ft. How sensitive are the tanks’ volumes to small variations in height and radius? With V = pr 2h, we have the approximation for the change in volume as
nR
Solution
dV = Vr s5, 25d dr + Vhs5, 25d dh = s2prhds5,25d dr + spr 2 ds5,25d dh
h5 (a)
(b)
dV = s2prhds25,5d dr + spr 2 ds25,5d dh = 250p dr + 625p dh.
Now the volume is more sensitive to changes in h than to changes in r (Figure 14.35). The general rule is that functions are most sensitive to small changes in the variables that generate the largest partial derivatives.
ma
FIGURE 14.35 The volume of cylinder (a) is more sensitive to a small change in r than it is to an equally small change in h. The volume of cylinder (b) is more sensitive to small changes in h than it is to small changes in r (Example 8).
Thus, a 1unit change in r will change V by about 250p units. A 1unit change in h will change V by about 25p units. The tank’s volume is 10 times more sensitive to a small change in r than it is to a small change of equal size in h. As a quality control engineer concerned with being sure the tanks have the correct volume, you would want to pay special attention to their radii. In contrast, if the values of r and h are reversed to make r = 25 and h = 5, then the total differential in V becomes
dH
r 25
h 25
ass a
= 250p dr + 25p dh.
r5
Mu
ham
EXAMPLE 9
Estimating Percentage Error
The volume V = pr2h of a right circular cylinder is to be calculated from measured values of r and h. Suppose that r is measured with an error of no more than 2% and h with an error of no more than 0.5%. Estimate the resulting possible percentage error in the calculation of V. Solution
We are told that dr ` r * 100 ` … 2
and
`
dh * 100 ` … 0.5. h
Since
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dV 2 dr dh 2prh dr + pr 2 dh = r + , = V h pr 2h
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6
Tangent Planes and Differentials
1023
dV dh dr ` = `2 r + ` V h
suf
`
i
we have
You
dr dh … `2 r ` + ` ` h
… 2s0.02d + 0.005 = 0.045.
We estimate the error in the volume calculation to be at most 4.5%.
Functions of More Than Two Variables
1.
iaz
Analogous results hold for differentiable functions of more than two variables. The linearization of ƒ(x, y, z) at a point P0sx0 , y0 , z0 d is Lsx, y, zd = ƒsP0 d + ƒxsP0 dsx  x0 d + ƒy sP0 ds y  y0 d + ƒzsP0 dsz  z0 d.
nR
Suppose that R is a closed rectangular solid centered at P0 and lying in an open region on which the second partial derivatives of ƒ are continuous. Suppose also that ƒ ƒxx ƒ , ƒ ƒyy ƒ , ƒ ƒzz ƒ , ƒ ƒxy ƒ , ƒ ƒxz ƒ , and ƒ ƒyz ƒ are all less than or equal to M throughout R. Then the error Esx, y, zd = ƒsx, y, zd  Lsx, y, zd in the approximation of ƒ by L is bounded throughout R by the inequality
ass a
2.
1 ƒ E ƒ … 2 Ms ƒ x  x0 ƒ + ƒ y  y0 ƒ + ƒ z  z0 ƒ d2.
If the second partial derivatives of ƒ are continuous and if x, y, and z change from x0 , y0 , and z0 by small amounts dx, dy, and dz, the total differential
dH
3.
dƒ = ƒxsP0 d dx + ƒysP0 d dy + ƒzsP0 d dz
gives a good approximation of the resulting change in ƒ.
EXAMPLE 10
Finding a Linear Approximation in 3Space
ma
Find the linearization L(x, y, z) of ƒsx, y, zd = x2  xy + 3 sin z
Mu
ham
at the point sx0 , y0 , z0 d = s2, 1, 0d . Find an upper bound for the error incurred in replacing ƒ by L on the rectangle R: Solution
ƒ x  2 ƒ … 0.01,
ƒ y  1 ƒ … 0.02,
ƒ z ƒ … 0.01.
A routine evaluation gives
ƒs2, 1, 0d = 2,
ƒxs2, 1, 0d = 3,
ƒys2, 1, 0d = 2,
ƒzs2, 1, 0d = 3.
Thus, Lsx, y, zd = 2 + 3sx  2d + s 2ds y  1d + 3sz  0d = 3x  2y + 3z  2. Since
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ƒxx = 2,
ƒyy = 0,
ƒzz = 3 sin z,
ƒxy = 1,
ƒxz = 0,
ƒyz = 0,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1024
Chapter 14: Partial Derivatives
suf
i
we may safely take M to be max ƒ 3 sin z ƒ = 3. Hence, the error incurred by replacing ƒ by L on R satisfies 1 ƒ E ƒ … 2 s3ds0.01 + 0.02 + 0.01d2 = 0.0024 .
Mu
ham
ma
dH
ass a
nR
iaz
You
The error will be no greater than 0.0024.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
EXERCISES 14.6 Tangent Planes and Normal Lines to Surfaces
Estimating Change
In Exercises 1–8, find equations for the
19. By about how much will
(b) normal line at the point P0 on the given surface.
3. 2z  x = 0,
P0s3, 5, 4d
P0s2, 0, 2d
4. x 2 + 2xy  y 2 + z 2 = 7, 2
6. x  xy  y  z = 0,
ƒsx, y, zd = e x cos yz
P0s1, 1, 3d
5. cos px  x 2y + e xz + yz = 4, 2
20. By about how much will
iaz
2
change if the point P(x, y, z) moves from P0s3, 4, 12d a distance of ds = 0.1 unit in the direction of 3i + 6j  2k?
P0s1, 1, 1d
2. x 2 + y 2  z 2 = 18,
change as the point P(x, y, z) moves from the origin a distance of ds = 0.1 unit in the direction of 2i + 2j  2k?
P0s0, 1, 2d
P0s1, 1, 1d
21. By about how much will
P0s0, 1, 0d
7. x + y + z = 1,
8. x 2 + y 2  2xy  x + 3y  z = 4,
P0s2, 3, 18d
11. z = 2y  x,
s1, 0, 0d 10. z = e sx
2
+ y 2d
2
s1, 2, 1d
,
2
12. z = 4x + y ,
s0, 0, 1d
s1, 1, 5d
ad
In Exercises 13–18, find parametric equations for the line tangent to the curve of intersection of the surfaces at the given point. 13. Surfaces: x + y 2 + 2z = 4, Point:
x = 1
(1, 1, 1)
x 2 + 2y 2 + 3z 2 = 6
mm
Point:
15. Surfaces: x 2 + 2y + 2z = 4, Point:
(1, 1, 1> 2)
(1> 2, 1, 1> 2)
y = 1
Mu
17. Surfaces: x 3 + 3x 2y 2 + y 3 + 4xy  z 2 = 0, = 11 Point:
Point:
x2 + y2 + z2
(1, 1, 3)
18. Surfaces: x 2 + y 2 = 4,
A 22, 22, 4 B
hsx, y, zd = cos spxyd + xz 2
23. Temperature change along a circle Suppose that the Celsius temperature at the point (x, y) in the xyplane is T sx, yd = x sin 2y and that distance in the xyplane is measured in meters. A particle is moving clockwise around the circle of radius 1 m centered at the origin at the constant rate of 2 m> sec.
b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?
y = 1
ha
Point:
22. By about how much will
a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point P A 1>2, 23>2 B ?
(1, 1, 1)
16. Surfaces: x + y 2 + z = 2,
change if the point P(x, y, z) moves from P0s2, 1, 0d a distance of ds = 0.2 unit toward the point P1s0, 1, 2)?
change if the point P(x, y, z) moves from P0s 1, 1, 1d a distance of ds = 0.1 unit toward the origin?
Tangent Lines to Curves
14. Surfaces: xyz = 1,
gsx, y, zd = x + x cos z  y sin z + y
Ha ssa
In Exercises 9–12, find an equation for the plane that is tangent to the given surface at the given point. 9. z = ln sx 2 + y 2 d,
Yo
1. x 2 + y 2 + z 2 = 3,
ƒsx, y, zd = ln2x 2 + y 2 + z 2
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(a) tangent plane and
usu fi
1024
x2 + y2  z = 0
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24. Changing temperature along a space curve The Celsius temperature in a region in space is given by T sx, y, zd = 2x 2  xyz. A particle is moving in this region and its position at time t is given by x = 2t 2, y = 3t, z = t 2, where time is measured in seconds and distance in meters.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6 Tangent Planes and Differentials
b. How fast is the temperature experienced by the particle changing in degrees Celsius per second at P?
40. ƒsx, y, zd = ssin xyd>z at a. sp>2, 1, 1d
i
b. (2, 0, 1)
suf
a. How fast is the temperature experienced by the particle changing in degrees Celsius per meter when the particle is at the point Ps8, 6, 4d?
x
41. ƒsx, y, zd = e + cos s y + zd at b. a0,
a. (0, 0, 0)
c. a0,
p , 0b 2
In Exercises 25–30, find the linearization L(x, y) of the function at each point.
p p , b 4 4
You
42. ƒsx, y, zd = tan1 sxyzd at
Finding Linearizations
1025
a. (1, 0, 0)
b. (1, 1, 0)
c. (1, 1, 1)
25. ƒsx, yd = x 2 + y 2 + 1 at
a. (0, 0),
b. (1, 1)
26. ƒsx, yd = sx + y + 2d2 at
a. (0, 0),
b. (1, 2)
In Exercises 43–46, find the linearization L(x, y, z) of the function ƒ(x, y, z) at P0 . Then find an upper bound for the magnitude of the error E in the approximation ƒsx, y, zd L Lsx, y, zd over the region R.
27. ƒsx, yd = 3x  4y + 5 at
a. (0, 0),
b. (1, 1)
43. ƒsx, y, zd = xz  3yz + 2
28. ƒsx, yd = x y at
a. (1, 1),
b. (0, 0)
29. ƒsx, yd = e x cos y at
a. (0, 0),
b. s0, p>2d
30. ƒsx, yd = e 2y  x at
44. ƒsx, y, zd = x + xy + yz + s1>4dz
a. (0, 0),
b. (1, 2)
R: ƒ x  1 ƒ … 0.01, ƒ y  1 ƒ … 0.01, ƒ z  2 ƒ … 0.08 45. ƒsx, y, zd = xy + 2yz  3xz at P0s1, 1, 0d
Upper Bounds for Errors in Linear Approximations
31. ƒsx, yd = x 2  3xy + 5 at P0s2, 1d, R:
ƒ x  2 ƒ … 0.1, ƒ y  1 ƒ … 0.1 32. ƒsx, yd = s1>2dx 2 + xy + s1>4dy 2 + 3x  3y + 4 at P0s2, 2d, R:
ƒ x  2 ƒ … 0.1, ƒ y  2 ƒ … 0.1 33. ƒsx, yd = 1 + y + x cos y at P0s0, 0d, R:
dH
ƒ x ƒ … 0.2, ƒ y ƒ … 0.2 sUse ƒ cos y ƒ … 1 and ƒ sin y ƒ … 1 in estimating E.d 34. ƒsx, yd = xy 2 + y cos sx  1d at P0s1, 2d,
ma
R: ƒ x  1 ƒ … 0.1, ƒ y  2 ƒ … 0.1 35. ƒsx, yd = e x cos y at P0s0, 0d,
R: ƒ x ƒ … 0.1, ƒ y ƒ … 0.1 sUse e x … 1.11 and ƒ cos y ƒ … 1 in estimating E.d 36. ƒsx, yd = ln x + ln y at P0s1, 1d, ƒ x  1 ƒ … 0.2,
ƒ y  1 ƒ … 0.2
ham
R:
R:
ƒ x  1 ƒ … 0.01, ƒ x ƒ … 0.01,
37. ƒsx, y, zd = xy + yz + xz at
Mu
b. (0, 1, 0)
c. (1, 0, 0)
39. ƒsx, y, zd = 2x 2 + y 2 + z 2 at a. (1, 0, 0)
b. (1, 1, 0)
at
ƒ y ƒ … 0.01,
at
P0s1, 1, 2d
ƒ z ƒ … 0.01 P0s0, 0, p>4d
ƒ z  p>4 ƒ … 0.01
Estimating Error; Sensitivity to Change 47. Estimating maximum error Suppose that T is to be found from the formula T = x se y + e y d, where x and y are found to be 2 and ln 2 with maximum possible errors of ƒ dx ƒ = 0.1 and ƒ dy ƒ = 0.02. Estimate the maximum possible error in the computed value of T.
48. Estimating volume of a cylinder About how accurately may V = pr 2h be calculated from measurements of r and h that are in error by 1%? 49. Maximum percentage error If r = 5.0 cm and h = 12.0 cm to the nearest millimeter, what should we expect the maximum percentage error in calculating V = pr 2h to be? 50. Variation in electrical resistance The resistance R produced by wiring resistors of R1 and R2 ohms in parallel (see accompanying figure) can be calculated from the formula
a. Show that dR = a
2
2
R R b dR1 + a b dR2 . R1 R2
c. (0, 0, 0)
38. ƒsx, y, zd = x 2 + y 2 + z 2 at a. (1, 1, 1)
ƒ z  2 ƒ … 0.02
ƒ y  1 ƒ … 0.01,
46. ƒsx, y, zd = 22 cos x sin s y + zd R:
2
1 1 1 + . = R R1 R2
Find the linearizations L(x, y, z) of the functions in Exercises 37–42 at the given points. b. (1, 0, 0)
P0s1, 1, 2d
ƒ y  1 ƒ … 0.01,
2
Functions of Three Variables
a. (1, 1, 1)
at
iaz
ƒ x  1 ƒ … 0.01,
ass a
In Exercises 31–36, find the linearization L(x, y) of the function ƒ(x, y) at P0. Then find an upper bound for the magnitude ƒ E ƒ of the error in the approximation ƒsx, yd L Lsx, yd over the rectangle R.
R:
nR
3 4
c. (1, 2, 2)
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b. You have designed a tworesistor circuit like the one shown on the next page to have resistances of R1 = 100 ohms and R2 = 400 ohms, but there is always some variation in manufacturing and the resistors received by your firm will probably not have these exact values. Will the value of R be
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives 55. Value of a 2 : 2 determinant If ƒ a ƒ is much greater than ƒ b ƒ , ƒ c ƒ , and ƒ d ƒ , to which of a, b, c, and d is the value of the determinant
i
more sensitive to variation in R1 or to variation in R2? Give reasons for your answer.
ƒsa, b, c, dd = ` R2
most sensitive? Give reasons for your answer.
51. You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reasons for your answer. 52. a. Around the point (1, 0), is ƒsx, yd = x 2s y + 1d more sensitive to changes in x or to changes in y? Give reasons for your answer. b. What ratio of dx to dy will make dƒ equal zero at (1, 0)?
y P(3 ; 0.01, 4 ; 0.01)
4
0
3
58. Surveying a triangular field The area of a triangle is (1> 2)ab sin C, where a and b are the lengths of two sides of the triangle and C is the measure of the included angle. In surveying a triangular plot, you have measured a, b, and C to be 150 ft, 200 ft, and 60°, respectively. By about how much could your area calculation be in error if your values of a and b are off by half a foot each and your measurement of C is off by 2°? See the accompanying figure. Remember to use radians.
dH
r
57. The Wilson lot size formula The Wilson lot size formula in economics says that the most economical quantity Q of goods (radios, shoes, brooms, whatever) for a store to order is given by the formula Q = 22KM>h , where K is the cost of placing the order, M is the number of items sold per week, and h is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables K, M, and h is Q most sensitive near the point sK0 , M0 , h0 d = s2, 20, 0.05d? Give reasons for your answer.
ass a
53. Error carryover in coordinate changes
You
c. In another circuit like the one shown you plan to change R1 from 20 to 20.1 ohms and R2 from 25 to 24.9 ohms. By about what percentage will this change R?
56. Estimating maximum error Suppose that u = xe y + y sin z and that x, y, and z can be measured with maximum possible errors of ;0.2, ;0.6, and ;p>180, respectively. Estimate the maximum possible error in calculating u from the measured values x = 2, y = ln 3, z = p>2.
iaz
R1
b ` d
nR
V
a c
suf
1026
a 150 ;
x
ham
ma
a. If x = 3 ; 0.01 and y = 4 ; 0.01, as shown here, with approximately what accuracy can you calculate the polar coordinates r and u of the point P(x, y) from the formulas r 2 = x 2 + y 2 and u = tan1 s y>xd? Express your estimates as percentage changes of the values that r and u have at the point sx0 , y0 d = s3, 4d.
b. At the point sx0 , y0 d = s3, 4d, are the values of r and u more sensitive to changes in x or to changes in y? Give reasons for your answer. 54. Designing a soda can A standard 12fl oz can of soda is essentially a cylinder of radius r = 1 in. and height h = 5 in.
Mu
a. At these dimensions, how sensitive is the can’s volume to a small change in radius versus a small change in height? b. Could you design a soda can that appears to hold more soda but in fact holds the same 12fl oz? What might its dimensions be? (There is more than one correct answer.)
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1 ft 2
C 60° ; 2° b 200 ;
1 ft 2
Theory and Examples 59. The linearization of ƒ(x, y) is a tangentplane approximation Show that the tangent plane at the point P0sx0 , y0 d, ƒsx0 , y0 dd on the surface z = ƒsx, yd defined by a differentiable function ƒ is the plane ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 ds y  y0 d  sz  ƒsx0 , y0 dd = 0 or z = ƒsx0 , y0 d + ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 ds y  y0 d. Thus, the tangent plane at P0 is the graph of the linearization of ƒ at P0 (see accompanying figure).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.6 Tangent Planes and Differentials
1027
(x 0, y0, f (x 0, y0 ))
z
z L(x, y)
y
62. Normal curves A smooth curve is normal to a surface ƒsx, y, zd = c at a point of intersection if the curve’s velocity vector is a nonzero scalar multiple of §ƒ at the point. Show that the curve
You
z f (x, y)
suf
i
at the points where t = p>4, 0, and p>4. The function ƒ gives the square of the distance from a point P(x, y, z) on the helix to the origin. The derivatives calculated here give the rates at which the square of the distance is changing with respect to t as P moves through the points where t = p>4, 0, and p>4.
rstd = 2t i + 2t j (x 0 , y0 )
1 st + 3dk 4
rstd = scos t + t sin tdi + ssin t  t cos tdj,
t 7 0.
is tangent to the surface x 2 + y 2  z = 1 when t = 1.
Mu
ham
ma
dH
rstd = scos tdi + ssin tdj + t k
rstd = 2t i + 2t j + s2t  1dk
ass a
61. Change along a helix Find the derivative of ƒsx, y, zd = x 2 + y 2 + z 2 in the direction of the unit tangent vector of the helix
63. Tangent curves A smooth curve is tangent to the surface at a point of intersection if its velocity vector is orthogonal to §f there. Show that the curve
nR
60. Change along the involute of a circle Find the derivative of ƒsx, yd = x 2 + y 2 in the direction of the unit tangent vector of the curve
iaz
is normal to the surface x 2 + y 2  z = 3 when t = 1.
x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u
14.7 Extreme Values and Saddle Points
Extreme Values and Saddle Points
14.7 z
Continuous functions of two variables assume extreme values on closed, bounded domains (see Figures 14.36 and 14.37). We see in this section that we can narrow the search for these extreme values by examining the functions’ first partial derivatives. A function of two variables can assume extreme values only at domain boundary points or at interior domain points where both first partial derivatives are zero or where one or both of the first partial derivatives fails to exist. However, the vanishing of derivatives at an interior point (a, b) does not always signal the presence of an extreme value. The surface that is the graph of the function might be shaped like a saddle right above (a, b) and cross its tangent plane there.
y
x
z a i R
s u Yo
1027
d a m
n a ss
a H
Derivative Tests for Local Extreme Values
FIGURE 14.36
The function
z = scos xdscos yde
2x 2 + y 2
m a h u M
has a maximum value of 1 and a minimum value of about 0.067 on the square region ƒ x ƒ … 3p>2, ƒ y ƒ … 3p>2 .
To find the local extreme values of a function of a single variable, we look for points where the graph has a horizontal tangent line. At such points, we then look for local maxima, local minima, and points of inflection. For a function ƒ(x, y) of two variables, we look for points where the surface z = ƒsx, yd has a horizontal tangent plane. At such points, we then look for local maxima, local minima, and saddle points (more about saddle points in a moment).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1028
Chapter 14: Partial Derivatives
2. x
ƒ(a, b) is a local maximum value of ƒ if ƒsa, bd Ú ƒsx, yd for all domain points (x, y) in an open disk centered at (a, b). ƒ(a, b) is a local minimum value of ƒ if ƒsa, bd … ƒsx, yd for all domain points (x, y) in an open disk centered at (a, b).
You
1.
suf
DEFINITIONS Local Maximum, Local Minimum Let ƒ(x, y) be defined on a region R containing the point (a, b). Then
i
z
y
iaz
Local maxima correspond to mountain peaks on the surface z = ƒsx, yd and local minima correspond to valley bottoms (Figure 14.38). At such points the tangent planes, when they exist, are horizontal. Local extrema are also called relative extrema. As with functions of a single variable, the key to identifying the local extrema is a first derivative test. FIGURE 14.37 The “roof surface” 1 2
A ƒ ƒ x ƒ  ƒ yƒ ƒ  ƒ x ƒ  ƒ y ƒB
Local maxima (no greater value of f nearby)
nR
z =
ass a
viewed from the point (10, 15, 20). The defining function has a maximum value of 0 and a minimum value of a on the square region ƒ x ƒ … a, ƒ y ƒ … a .
Local minimum (no smaller value of f nearby)
FIGURE 14.38 A local maximum is a mountain peak and a local minimum is a valley low.
dH
HISTORICAL BIOGRAPHY SiméonDenis Poisson (1781–1840)
ma
THEOREM 10 First Derivative Test for Local Extreme Values If ƒ(x, y) has a local maximum or minimum value at an interior point (a, b) of its domain and if the first partial derivatives exist there, then ƒxsa, bd = 0 and ƒysa, bd = 0.
z
ham
0f 0 0x z f (x, y)
0f 0 0y
0 a
g(x) f (x, b) b
Proof If ƒ has a local extremum at (a, b), then the function gsxd = ƒsx, bd has a local extremum at x = a (Figure 14.39). Therefore, g¿sad = 0 (Chapter 4, Theorem 2). Now g¿sad = ƒxsa, bd, so ƒxsa, bd = 0. A similar argument with the function hsyd = ƒsa, yd shows that ƒysa, bd = 0.
h( y) f (a, y)
If we substitute the values ƒxsa, bd = 0 and ƒysa, bd = 0 into the equation
y
ƒxsa, bdsx  ad + ƒysa, bds y  bd  sz  ƒsa, bdd = 0
(a, b, 0)
Mu
x
FIGURE 14.39 If a local maximum of ƒ occurs at x = a, y = b , then the first partial derivatives ƒxsa, bd and ƒysa, bd are both zero.
for the tangent plane to the surface z = ƒsx, yd at (a, b), the equation reduces to 0 # sx  ad + 0 # s y  bd  z + ƒsa, bd = 0 or
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.7 z
Extreme Values and Saddle Points
1029
suf
i
Thus, Theorem 10 says that the surface does indeed have a horizontal tangent plane at a local extremum, provided there is a tangent plane there.
DEFINITION Critical Point An interior point of the domain of a function ƒ(x, y) where both ƒx and ƒy are zero or where one or both of ƒx and ƒy do not exist is a critical point of ƒ.
You
y x
z
xy (x 2 y 2 )
Theorem 10 says that the only points where a function ƒ(x, y) can assume extreme values are critical points and boundary points. As with differentiable functions of a single variable, not every critical point gives rise to a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point.
x2 y2
iaz
z
DEFINITION Saddle Point A differentiable function ƒ(x, y) has a saddle point at a critical point (a, b) if in every open disk centered at (a, b) there are domain points (x, y) where ƒsx, yd 7 ƒsa, bd and domain points sx, yd where ƒsx, yd 6 ƒsa, bd. The corresponding point (a, b, ƒ(a, b)) on the surface z = ƒsx, yd is called a saddle point of the surface (Figure 14.40).
y
ass a
nR
x
z y2 y4 x2
EXAMPLE 1 FIGURE 14.40 origin.
Saddle points at the
Finding Local Extreme Values
Find the local extreme values of ƒsx, yd = x 2 + y 2. The domain of ƒ is the entire plane (so there are no boundary points) and the partial derivatives ƒx = 2x and ƒy = 2y exist everywhere. Therefore, local extreme values can occur only where
dH
Solution
z x 2 y2
ƒx = 2x = 0
and
ƒy = 2y = 0.
The only possibility is the origin, where the value of ƒ is zero. Since ƒ is never negative, we see that the origin gives a local minimum (Figure 14.41).
ma
z
EXAMPLE 2
Identifying a Saddle Point
x
ham
Find the local extreme values (if any) of ƒsx, yd = y 2  x 2.
y
Mu
FIGURE 14.41 The graph of the function ƒsx, yd = x 2 + y 2 is the paraboloid z = x 2 + y 2. The function has a local minimum value of 0 at the origin (Example 1).
The domain of ƒ is the entire plane (so there are no boundary points) and the partial derivatives ƒx = 2x and ƒy = 2y exist everywhere. Therefore, local extrema can occur only at the origin (0, 0). Along the positive xaxis, however, ƒ has the value ƒsx, 0d = x 2 6 0; along the positive yaxis, ƒ has the value ƒs0, yd = y 2 7 0. Therefore, every open disk in the xyplane centered at (0, 0) contains points where the function is positive and points where it is negative. The function has a saddle point at the origin (Figure 14.42) instead of a local extreme value. We conclude that the function has no local extreme values. Solution
That ƒx = ƒy = 0 at an interior point (a, b) of R does not guarantee ƒ has a local extreme value there. If ƒ and its first and second partial derivatives are continuous on R, however, we may be able to learn more from the following theorem, proved in Section 14.10.
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Chapter 14: Partial Derivatives z
FIGURE 14.42 The origin is a saddle point of the function ƒsx, yd = y 2  x 2. There are no local extreme values (Example 2).
iaz
x
ƒ has a local maximum at (a, b) if ƒxx 6 0 and ƒxx ƒyy  ƒxy2 7 0 at (a, b). ƒ has a local minimum at (a, b) if ƒxx 7 0 and ƒxx ƒyy  ƒxy2 7 0 at (a, b). ƒ has a saddle point at (a, b) if ƒxx ƒyy  ƒxy2 6 0 at (a, b). The test is inconclusive at (a, b) if ƒxx ƒyy  ƒxy2 = 0 at (a, b). In this case, we must find some other way to determine the behavior of ƒ at (a, b).
You
i. ii. iii. iv.
y
suf
i
THEOREM 11 Second Derivative Test for Local Extreme Values Suppose that ƒ(x, y) and its first and second partial derivatives are continuous throughout a disk centered at (a, b) and that ƒxsa, bd = ƒysa, bd = 0 . Then
z y2 x2
The expression ƒxx ƒyy  ƒxy2 is called the discriminant or Hessian of ƒ. It is sometimes easier to remember it in determinant form, ƒxx ƒxy
nR
ƒxx ƒyy  ƒxy2 = p
ƒxy p. ƒyy
ass a
Theorem 11 says that if the discriminant is positive at the point (a, b), then the surface curves the same way in all directions: downward if ƒxx 6 0, giving rise to a local maximum, and upward if ƒxx 7 0, giving a local minimum. On the other hand, if the discriminant is negative at (a, b), then the surface curves up in some directions and down in others, so we have a saddle point.
EXAMPLE 3
Finding Local Extreme Values
Find the local extreme values of the function
dH
ƒsx, yd = xy  x 2  y 2  2x  2y + 4.
The function is defined and differentiable for all x and y and its domain has no boundary points. The function therefore has extreme values only at the points where ƒx and ƒy are simultaneously zero. This leads to
ma
Solution
ƒx = y  2x  2 = 0,
ƒy = x  2y  2 = 0,
Mu
ham
or
x = y = 2.
Therefore, the point s 2, 2d is the only point where ƒ may take on an extreme value. To see if it does so, we calculate ƒxx = 2,
ƒyy = 2,
ƒxy = 1.
The discriminant of ƒ at sa, bd = s 2, 2d is ƒxx ƒyy  ƒxy2 = s 2ds 2d  s1d2 = 4  1 = 3. The combination ƒxx 6 0
and
ƒxx ƒyy  ƒxy2 7 0
tells us that ƒ has a local maximum at s 2, 2d . The value of ƒ at this point is ƒs 2, 2d = 8.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.7
EXAMPLE 4
z
Extreme Values and Saddle Points
1031
Searching for Local Extreme Values
i
y
suf
Find the local extreme values of ƒsx, yd = xy.
Since ƒ is differentiable everywhere (Figure 14.43), it can assume extreme values only where
Solution
and
Thus, the origin is the only point where ƒ might have an extreme value. To see what happens there, we calculate ƒxx = 0,
ƒyy = 0,
ƒxy = 1.
iaz
The discriminant,
z xy
ƒy = x = 0.
You
ƒx = y = 0 x
ƒxx ƒyy  ƒxy2 = 1,
is negative. Therefore, the function has a saddle point at (0, 0). We conclude that ƒsx, yd = xy has no local extreme values.
nR
FIGURE 14.43 The surface z = xy has a saddle point at the origin (Example 4).
Absolute Maxima and Minima on Closed Bounded Regions
1. 2.
y
ham
9 2
y0
x
A(9, 0)
FIGURE 14.44 This triangular region is the domain of the function in Example 5.
Mu
ƒsx, yd = 2 + 2x + 2y  x 2  y 2
on the triangular region in the first quadrant bounded by the lines x = 0, y = 0, y = 9  x.
(1, 1)
O
Finding Absolute Extrema
Find the absolute maximum and minimum values of
y9x 9 , 2
ma
EXAMPLE 5
B(0, 9)
x0
List the interior points of R where ƒ may have local maxima and minima and evaluate ƒ at these points. These are the critical points of ƒ. List the boundary points of R where ƒ has local maxima and minima and evaluate ƒ at these points. We show how to do this shortly. Look through the lists for the maximum and minimum values of ƒ. These will be the absolute maximum and minimum values of ƒ on R. Since absolute maxima and minima are also local maxima and minima, the absolute maximum and minimum values of ƒ appear somewhere in the lists made in Steps 1 and 2.
dH
3.
ass a
We organize the search for the absolute extrema of a continuous function ƒ(x, y) on a closed and bounded region R into three steps.
Solution Since ƒ is differentiable, the only places where ƒ can assume these values are points inside the triangle (Figure 14.44) where ƒx = ƒy = 0 and points on the boundary.
(a) Interior points. For these we have fx = 2  2x = 0,
fy = 2  2y = 0,
yielding the single point sx, yd = s1, 1d. The value of ƒ there is
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Chapter 14: Partial Derivatives
suf
ƒsx, yd = ƒsx, 0d = 2 + 2x  x 2
i
(b) Boundary points. We take the triangle one side at a time: (i) On the segment OA, y = 0. The function
You
may now be regarded as a function of x defined on the closed interval 0 … x … 9. Its extreme values (we know from Chapter 4) may occur at the endpoints x = 0
where
ƒs0, 0d = 2
x = 9
where
ƒs9, 0d = 2 + 18  81 = 61
and at the interior points where ƒ¿sx, 0d = 2  2x = 0. The only interior point where ƒ¿sx, 0d = 0 is x = 1, where
(ii) On the segment OB, x = 0 and
iaz
ƒsx, 0d = ƒs1, 0d = 3.
nR
ƒsx, yd = ƒs0, yd = 2 + 2y  y 2.
We know from the symmetry of ƒ in x and y and from the analysis we just carried out that the candidates on this segment are ƒs0, 0d = 2,
ƒs0, 9d = 61,
ƒs0, 1d = 3.
ass a
(iii) We have already accounted for the values of ƒ at the endpoints of AB, so we need only look at the interior points of AB. With y = 9  x, we have ƒsx, yd = 2 + 2x + 2s9  xd  x 2  s9  xd2 = 61 + 18x  2x 2.
dH
Setting ƒ¿sx, 9  xd = 18  4x = 0 gives x =
9 18 = . 4 2
At this value of x,
9 9 = 2 2
and
9 9 41 ƒsx, yd = ƒ a , b =  . 2 2 2
ma
y = 9 
Mu
ham
Summary We list all the candidates: 4, 2, 61, 3, s41>2d. The maximum is 4, which ƒ assumes at (1, 1). The minimum is 61, which ƒ assumes at (0, 9) and (9, 0). Solving extreme value problems with algebraic constraints on the variables usually requires the method of Lagrange multipliers in the next section. But sometimes we can solve such problems directly, as in the next example.
EXAMPLE 6
Solving a Volume Problem with a Constraint
A delivery company accepts only rectangular boxes the sum of whose length and girth (perimeter of a crosssection) does not exceed 108 in. Find the dimensions of an acceptable box of largest volume. Let x, y, and z represent the length, width, and height of the rectangular box, respectively. Then the girth is 2y + 2z. We want to maximize the volume V = xyz of the
Solution
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.7
Extreme Values and Saddle Points
box (Figure 14.45) satisfying x + 2y + 2z = 108 (the largest box accepted by the delivery company). Thus, we can write the volume of the box as a function of two variables.
i
Girth distance around here
= 108yz  2y 2z  2yz 2
z
You
Setting the first partial derivatives equal to zero, y
FIGURE 14.45
suf
V = xyz and x = 108  2y  2z
Vs y, zd = s108  2y  2zdyz
x
1033
Vys y, zd = 108z  4yz  2z 2 = s108  4y  2zdz = 0
Vzs y, zd = 108y  2y 2  4yz = s108  2y  4zdy = 0,
The box in Example 6.
Vyy = 4z, Then
iaz
gives the critical points (0, 0), (0, 54), (54, 0), and (18, 18). The volume is zero at (0, 0), (0, 54), (54, 0), which are not maximum values. At the point (18, 18), we apply the Second Derivative Test (Theorem 11): Vzz = 4y,
Vyz = 108  4y  4z.
nR
Vyy Vzz  V yz2 = 16yz  16s27  y  zd2.
Thus,
Vyys18, 18d = 4s18d 6 0
and
ass a
C Vyy Vzz  V yz2 D s18,18d = 16s18ds18d  16s 9d2 7 0
imply that (18, 18) gives a maximum volume. The dimensions of the package are x = 108  2s18d  2s18d = 36 in., y = 18 in., and z = 18 in. The maximum volume is V = s36ds18ds18d = 11,664 in.3 , or 6.75 ft3.
dH
Despite the power of Theorem 10, we urge you to remember its limitations. It does not apply to boundary points of a function’s domain, where it is possible for a function to have extreme values along with nonzero derivatives. Also, it does not apply to points where either ƒx or ƒy fails to exist.
Mu
ham
ma
Summary of MaxMin Tests The extreme values of ƒ(x, y) can occur only at i. boundary points of the domain of ƒ ii. critical points (interior points where ƒx = ƒy = 0 or points where ƒx or ƒy fail to exist). If the first and secondorder partial derivatives of ƒ are continuous throughout a disk centered at a point (a, b) and ƒxsa, bd = ƒysa, bd = 0, the nature of ƒ(a, b) can be tested with the Second Derivative Test: i. ii. iii. iv.
ƒxx 6 0 and ƒxx ƒyy  ƒxy2 7 0 at sa, bd Q local maximum ƒxx 7 0 and ƒxx ƒyy  ƒxy2 7 0 at sa, bd Q local minimum ƒxx ƒyy  ƒxy2 6 0 at sa, bd Q saddle point ƒxx ƒyy  ƒxy2 = 0 at sa, bd Q test is inconclusive.
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Chapter 14: Partial Derivatives
Finding Local Extrema 2
2
1. ƒsx, yd = x + xy + y + 3x  3y + 4 3. ƒsx, yd = 2xy  5x 2  2y 2 + 4x + 4y  4
36. ƒsx, yd = 48xy  32x 3  24y 2 0 … x … 1, 0 … y … 1
4. ƒsx, yd = 2xy  5x 2  2y 2 + 4x  4 5. ƒsx, yd = x 2 + xy + 3x + 2y + 5 7. ƒsx, yd = 5xy  7x 2 + 3x  6y + 2 8. ƒsx, yd = 2xy  x 2  2y 2 + 3x + 4 9. ƒsx, yd = x 2  4xy + y 2 + 6y + 2
12. ƒsx, yd = 4x 2  6xy + 5y 2  20x + 26y 14. ƒsx, yd = x 2  2xy + 2y 2  2x + 2y + 1 15. ƒsx, yd = x 2 + 2xy 16. ƒsx, yd = 3 + 2x + 2y  2x 2  2xy  y 2 17. ƒsx, yd = x 3  y 3  2xy + 6
dH
18. ƒsx, yd = x 3 + 3xy + y 3
ass a
13. ƒsx, yd = x 2  y 2  2x + 4y + 6
22. ƒsx, yd = 8x 3 + y 3 + 6xy
ma
3
2
26. ƒsx, yd = x + y + 4xy 1 x + y2  1 29. ƒsx, yd = y sin x
1 1 28. ƒsx, yd = x + xy + y
ham
27. ƒsx, yd =
2
plate
z
38. ƒsx, yd = 4x  8xy + 2y + 1 on the triangular plate bounded by the lines x = 0, y = 0, x + y = 1 in the first quadrant 39. Find two numbers a and b with a … b such that b
La
2
25. ƒsx, yd = 4xy  x 4  y 4 4
rectangular
y
24. ƒsx, yd = 2x + 2y  9x + 3y  12y 4
the
x
23. ƒsx, yd = x 3 + y 3 + 3x 2  3y 2  8 3
plate
nR
11. ƒsx, yd = 2x 2 + 3xy + 4y 2  5x + 2y
21. ƒsx, yd = 9x + y >3  4xy
rectangular
z (4x x 2 ) cos y
10. ƒsx, yd = 3x 2 + 6xy + 7y 2  2x + 4y
3
the
on
iaz
6. ƒsx, yd = y + xy  2x  2y + 2
3
on
37. ƒsx, yd = s4x  x 2 d cos y on the rectangular plate 1 … x … 3, p>4 … y … p>4 (see accompanying figure).
2
20. ƒsx, yd = 3y 2  2y 3  3x 2 + 6xy
34. Tsx, yd = x 2 + xy + y 2  6x 0 … x … 5, 3 … y … 3
35. Tsx, yd = x 2 + xy + y 2  6x + 2 on the rectangular plate 0 … x … 5, 3 … y … 0
2. ƒsx, yd = x 2 + 3xy + 3y 2  6x + 3y  6
19. ƒsx, yd = 6x 2  2x 3 + 3y 2 + 6xy
33. ƒsx, yd = x 2 + y 2 on the closed triangular plate bounded by the lines x = 0, y = 0, y + 2x = 2 in the first quadrant
You
Find all the local maxima, local minima, and saddle points of the functions in Exercises 1–30.
suf
i
EXERCISES 14.7
30. ƒsx, yd = e 2x cos y
Finding Absolute Extrema
Mu
In Exercises 31–38, find the absolute maxima and minima of the functions on the given domains. 31. ƒsx, yd = 2x 2  4x + y 2  4y + 1 on the closed triangular plate bounded by the lines x = 0, y = 2, y = 2x in the first quadrant 32. Dsx, yd = x 2  xy + y 2 + 1 on the closed triangular plate in the first quadrant bounded by the lines x = 0, y = 4, y = x
To Read it Online & Download:
s6  x  x 2 d dx
has its largest value. 40. Find two numbers a and b with a … b such that b
La
s24  2x  x 2 d1>3 dx
has its largest value. 41. Temperatures The flat circular plate in Figure 14.46 has the shape of the region x 2 + y 2 … 1. The plate, including the boundary where x 2 + y 2 = 1 , is heated so that the temperature at the point (x, y) is Tsx, yd = x 2 + 2y 2  x. Find the temperatures at the hottest and coldest points on the plate.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.7 Extreme Values and Saddle Points y
1035
i
Theory and Examples
b. ƒx = 2x  2
x
0
c. ƒx = 9x 2  9
and and and
ƒy = 2y  4x ƒy = 2y  4
ƒy = 2y + 4
You
a. ƒx = 2x  4y
suf
43. Find the maxima, minima, and saddle points of ƒ(x, y), if any, given that
Describe your reasoning in each case.
44. The discriminant ƒxx ƒyy  ƒxy 2 is zero at the origin for each of the following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimum, or neither at the origin by imagining what the surface z = ƒsx, yd looks like. Describe your reasoning in each case.
ƒsx, yd = xy + 2x  ln x 2y
c. ƒsx, yd = xy 2
d. ƒsx, yd = x 3y 2
e. ƒsx, yd = x y
f. ƒsx, yd = x 4y 4
nR
45. Show that (0, 0) is a critical point of ƒsx, yd = x 2 + kxy + y 2 no matter what value the constant k has. (Hint: Consider two cases: k = 0 and k Z 0.)
dH
ma ham 0
b. ƒsx, yd = 1  x 2y 2
3 3
in the open first quadrant sx 7 0, y 7 0d and show that ƒ takes on a minimum there (Figure 14.47).
y
a. ƒsx, yd = x 2y 2
46. For what values of the constant k does the Second Derivative Test guarantee that ƒsx, yd = x 2 + kxy + y 2 will have a saddle point at (0, 0)? A local minimum at (0, 0)? For what values of k is the Second Derivative Test inconclusive? Give reasons for your answers.
ass a
42. Find the critical point of
iaz
FIGURE 14.46 Curves of constant temperature are called isotherms. The figure shows isotherms of the temperature function Tsx, yd = x 2 + 2y 2  x on the disk x 2 + y 2 … 1 in the xyplane. Exercise 41 asks you to locate the extreme temperatures.
x
Mu
FIGURE 14.47 The function ƒsx, yd = xy + 2x  ln x 2y (selected level curves shown here) takes on a minimum value somewhere in the open first quadrant x 7 0, y 7 0 (Exercise 42).
To Read it Online & Download:
47. If ƒxsa, bd = ƒysa, bd = 0, must ƒ have a local maximum or minimum value at (a, b)? Give reasons for your answer. 48. Can you conclude anything about ƒ(a, b) if ƒ and its first and second partial derivatives are continuous throughout a disk centered at (a, b) and ƒxxsa, bd and ƒyysa, bd differ in sign? Give reasons for your answer.
49. Among all the points on the graph of z = 10  x 2  y 2 that lie above the plane x + 2y + 3z = 0, find the point farthest from the plane. 50. Find the point on the graph of z = x 2 + y 2 + 10 nearest the plane x + 2y  z = 0. 51. The function ƒsx, yd = x + y fails to have an absolute maximum value in the closed first quadrant x Ú 0 and y Ú 0. Does this contradict the discussion on finding absolute extrema given in the text? Give reasons for your answer. 52. Consider the function ƒsx, yd = x 2 + y 2 + 2xy  x  y + 1 over the square 0 … x … 1 and 0 … y … 1. a. Show that ƒ has an absolute minimum along the line segment 2x + 2y = 1 in this square. What is the absolute minimum value? b. Find the absolute maximum value of ƒ over the square.
Extreme Values on Parametrized Curves To find the extreme values of a function ƒ(x, y) on a curve x = xstd, y = ystd, we treat ƒ as a function of the single variable t and
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
use the Chain Rule to find where dƒ> dt is zero. As in any other singlevariable case, the extreme values of ƒ are then found among the values at the
Find the absolute maximum and minimum values of the following functions on the given curves. 53. Functions: a. ƒsx, yd = x + y
b. gsx, yd = xy
c. hsx, yd = 2x 2 + y 2 Curves:
with all sums running from k = 1 to k = n . Many scientific calculators have these formulas built in, enabling you to find m and b with only a few key strokes after you have entered the data. The line y = mx + b determined by these values of m and b is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1.
summarize data with a simple expression,
2.
predict values of y for other, experimentally untried values of x,
3.
handle data analytically.
y Ú 0
ii. The quarter circle x 2 + y 2 = 4,
iaz
i. The semicircle x 2 + y 2 = 4,
suf
b. endpoints of the parameter domain.
y
x Ú 0,
y Ú 0
Use the parametric equations x = 2 cos t, y = 2 sin t .
P1(x1, y1 )
b. gsx, yd = xy
c. hsx, yd = x 2 + 3y 2 Curves:
ii. The quarter ellipse sx 2>9d + s y 2>4d = 1,
y Ú 0 x Ú 0,
y Ú 0
Use the parametric equations x = 3 cos t, y = 2 sin t . 55. Function: ƒsx, yd = xy Curves: i. The line x = 2t,
y = t + 1 y = t + 1,
1 … t … 0
iii. The line segment x = 2t,
y = t + 1,
0 … t … 1
dH
ii. The line segment x = 2t, 56. Functions: a. ƒsx, yd = x 2 + y 2
b. gsx, yd = 1>sx 2 + y 2 d
i. The line x = t,
ma
Curves: y = 2  2t
ii. The line segment x = t,
y = 2  2t,
0 … t … 1
ham
Least Squares and Regression Lines
When we try to fit a line y = mx + b to a set of numerical data points sx1, y1 d, sx2 , y2 d, Á , sxn , yn d (Figure 14.48), we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of m and b that minimize the value of the function
Mu
w = smx1 + b  y1 d2 + Á + smxn + b  yn d2 .
(1)
The values of m and b that do this are found with the First and Second Derivative Tests to be
m =
a a xk b a a yk b  n a xk yk 2
a a xk b  n a x k2
Pn(xn , yn ) y mx b
P2(x 2 , y2 ) x
0
ass a
i. The semiellipse sx 2>9d + s y 2>4d = 1,
nR
54. Functions: a. ƒsx, yd = 2x + 3y
(3)
You
a. critical points (points where dƒ> dt is zero or fails to exist), and
1 b = n a a yk  m a xk b,
i
1036
,
To Read it Online & Download:
(2)
FIGURE 14.48 To fit a line to noncollinear points, we choose the line that minimizes the sum of the squares of the deviations.
EXAMPLE
Find the least squares line for the points (0, 1), (1, 3), (2, 2), (3, 4), (4, 5). Solution
We organize the calculations in a table:
k
xk
yk
x k2
xk yk
1 2 3 4 5
0 1 2 3 4
1 3 2 4 5
0 1 4 9 16
0 3 4 12 20
g
10
15
30
39
Then we find m =
s10ds15d  5s39d 2
s10d  5s30d
= 0.9
Equation (2) with n = 5 and data from the table
and use the value of m to find b =
1 A 15  A 0.9 B A 10 B B = 1.2 . 5
Equation (3) with n = 5, m = 0.9
The least squares line is y = 0.9x + 1.2 (Figure 14.49).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
y
TABLE 14.2 Crater sizes on Mars
2
y 0.9x 1.2
32–45 45–64 64–90 90–128
0.001 0.0005 0.00024 0.000123
51 22 14 4
P3(2, 2)
1 P1(0, 1) 1
2
3
4
x
FIGURE 14.49 The least squares line for the data in the example.
In Exercises 57–60, use Equations (2) and (3) to find the least squares line for each set of data points. Then use the linear equation you obtain to predict the value of y that would correspond to x = 4. s0, 2d,
s2, 3d
59. (0, 0), (1, 2), (2, 3)
60. (0, 1), (2, 2), (3, 2)
TABLE 14.1 Growth of alfalfa
y (average alfalfa yield, tons/acre) 5.27 5.68 6.25 7.21 8.20 8.71
Mu
ham
ma
x (total seasonal depth of water applied, in.)
dH
T 61. Write a linear equation for the effect of irrigation on the yield of alfalfa by fitting a least squares line to the data in Table 14.1 (from the University of California Experimental Station, Bulletin No. 450, p. 8). Plot the data and draw the line.
12 18 24 30 36 42
a. Plot y vs. K to show that y is close to being a linear function of K. b. Find a least squares line y = mK + b for the data and add the line to your plot in part (a).
ass a
s3, 4d 58. s 2, 0d,
T 63. Köchel numbers In 1862, the German musicologist Ludwig von Köchel made a chronological list of the musical works of Wolfgang Amadeus Mozart. This list is the source of the Köchel numbers, or “K numbers,” that now accompany the titles of Mozart’s pieces (Sinfonia Concertante in Eflat major, K.364, for example). Table 14.3 gives the Köchel numbers and composition dates (y) of ten of Mozart’s works.
iaz
0
suf
3
s0, 1d,
Frequency, F
nR
P2(1, 3)
You
P4(3, 4)
4
57. s 1, 2d,
Diameter in km, D
1>D2 (for left value of class interval)
P5(4, 5)
5
1037
i
14.7 Extreme Values and Saddle Points
T 62. Craters of Mars One theory of crater formation suggests that the frequency of large craters should fall off as the square of the diameter (Marcus, Science, June 21, 1968, p. 1334). Pictures from Mariner IV show the frequencies listed in Table 14.2. Fit a line of the form F = ms1>D 2 d + b to the data. Plot the data and draw the line.
To Read it Online & Download:
c. K.364 was composed in 1779. What date is predicted by the least squares line?
TABLE 14.3 Compositions by Mozart
Köchel number, K
Year composed, y
1 75 155 219 271 351 425 503 575 626
1761 1771 1772 1775 1777 1780 1783 1786 1789 1791
T 64. Submarine sinkings The data in Table 14.4 show the results of a historical study of German submarines sunk by the U.S. Navy during 16 consecutive months of World War II. The data given for each month are the number of reported sinkings and the number of actual sinkings. The number of submarines sunk was slightly greater than the Navy’s reports implied. Find a least squares line for estimating the number of actual sinkings from the number of reported sinkings.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
during 16 consecutive months of WWII
Month
Guesses by U.S. (reported sinkings) x
Actual number y
COMPUTER EXPLORATIONS
i
TABLE 14.4 Sinkings of German submarines by U.S.
Exploring Local Extrema at Critical Points
suf
1038
In Exercises 65–70, you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle.
140
c. Calculate the function’s first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer.
iaz
d. Calculate the function’s second partial derivatives and find the discriminant ƒxx ƒyy  ƒxy 2. e. Using the maxmin tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? 65. ƒsx, yd = x 2 + y 3  3xy,
5 … x … 5,
66. ƒsx, yd = x 3  3xy 2 + y 2, 4
2
2 … x … 2,
2
67. ƒsx, yd = x + y  8x  6y + 16,
5 … y … 5 2 … y … 2
3 … x … 3,
6 … y … 6
68. ƒsx, yd = 2x 4 + y 4  2x 2  2y 2 + 3,
3>2 … x … 3>2,
3>2 … y … 3>2
69. ƒsx, yd = 5x 6 + 18x 5  30x 4 + 30xy 2  120x 3, 4 … x … 3,
2 … y … 2
x ln sx 2 + y 2 d, 0, 2 … x … 2, 2 … y … 2
70. ƒsx, yd = e
5
sx, yd Z s0, 0d , sx, yd = s0, 0d
Mu
ham
ma
dH
123
3 2 6 3 4 3 11 9 10 16 13 5 6 19 15 15
nR
3 2 4 2 5 5 9 12 8 13 14 3 4 13 10 16
ass a
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
You
b. Plot some level curves in the rectangle.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1038
i f su
Chapter 14: Partial Derivatives
14.8
u o zY
Lagrange Multipliers
HISTORICAL BIOGRAPHY
a i R
Sometimes we need to find the extreme values of a function whose domain is constrained to lie within some particular subset of the plane—a disk, for example, a closed triangular region, or along a curve. In this section, we explore a powerful method for finding extreme values of constrained functions: the method of Lagrange multipliers.
Joseph Louis Lagrange (1736–1813)
n a s s a
Constrained Maxima and Minima
H d a m am EXAMPLE 1
Finding a Minimum with Constraint
Find the point P(x, y, z) closest to the origin on the plane 2x + y  z  5 = 0. Solution
h u M
The problem asks us to find the minimum value of the function 1 2 2 2 ƒ OP ƒ = 2sx  0d + s y  0d + sz  0d
To Read it Online & Download:
= 2x 2 + y 2 + z 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8
Lagrange Multipliers
1039
ƒsx, y, zd = x 2 + y 2 + z 2
suf
2x + y  z  5 = 0. 1 Since ƒ OP ƒ has a minimum value wherever the function
i
subject to the constraint that
You
has a minimum value, we may solve the problem by finding the minimum value of ƒ(x, y, z) subject to the constraint 2x + y  z  5 = 0 (thus avoiding square roots). If we regard x and y as the independent variables in this equation and write z as z = 2x + y  5,
our problem reduces to one of finding the points (x, y) at which the function
iaz
hsx, yd = ƒsx, y, 2x + y  5d = x 2 + y 2 + s2x + y  5d2
nR
has its minimum value or values. Since the domain of h is the entire xyplane, the First Derivative Test of Section 14.7 tells us that any minima that h might have must occur at points where hx = 2x + 2s2x + y  5ds2d = 0, This leads to
ass a
10x + 4y = 20,
hy = 2y + 2s2x + y  5d = 0.
4x + 4y = 10,
and the solution
x =
5 , 3
y =
5 . 6
dH
We may apply a geometric argument together with the Second Derivative Test to show that these values minimize h. The zcoordinate of the corresponding point on the plane z = 2x + y  5 is
z
5 5 5 z = 2a b +  5 =  . 3 6 6
ma
Therefore, the point we seek is Closest point:
5 5 5 P a , ,  b. 3 6 6
ham
The distance from P to the origin is 5> 26 L 2.04.
(–1, 0, 0)
(1, 0, 0)
x
Mu
x 2 z2 1
FIGURE 14.50 The hyperbolic cylinder x 2  z 2  1 = 0 in Example 2.
y
Attempts to solve a constrained maximum or minimum problem by substitution, as we might call the method of Example 1, do not always go smoothly. This is one of the reasons for learning the new method of this section.
EXAMPLE 2
Finding a Minimum with Constraint
Find the points closest to the origin on the hyperbolic cylinder x 2  z 2  1 = 0. Solution 1 The cylinder is shown in Figure 14.50. We seek the points on the cylinder closest to the origin. These are the points whose coordinates minimize the value of the function
To Read it Online & Download:
ƒsx, y, zd = x 2 + y 2 + z 2
Square of the distance
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1040
Chapter 14: Partial Derivatives
suf
i
subject to the constraint that x 2  z 2  1 = 0 . If we regard x and y as independent variables in the constraint equation, then z2 = x2  1
and the values of ƒsx, y, zd = x 2 + y 2 + z 2 on the cylinder are given by the function
You
hsx, yd = x 2 + y 2 + sx 2  1d = 2x 2 + y 2  1.
To find the points on the cylinder whose coordinates minimize ƒ, we look for the points in the xyplane whose coordinates minimize h. The only extreme value of h occurs where hx = 4x = 0
x
On this part, x –兹z 2 1
1
–1
iaz
z
x 兹z 2 1
that is, at the point (0, 0). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened was that the First Derivative Test found (as it should have) the point in the domain of h where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire xyplane, the domain from which we can select the first two coordinates of the points (x, y, z) on the cylinder is restricted to the “shadow” of the cylinder on the xyplane; it does not include the band between the lines x = 1 and x = 1 (Figure 14.51). We can avoid this problem if we treat y and z as independent variables (instead of x and y) and express x in terms of y and z as
nR
On this part,
hy = 2y = 0,
ass a
The hyperbolic cylinder x 2 z 2 1
and
x 2 = z 2 + 1.
x1 y
x –1
With this substitution, ƒsx, y, zd = x 2 + y 2 + z 2 becomes ks y, zd = sz 2 + 1d + y 2 + z 2 = 1 + y 2 + 2z 2
dH
and we look for the points where k takes on its smallest value. The domain of k in the yzplane now matches the domain from which we select the y and zcoordinates of the points (x, y, z) on the cylinder. Hence, the points that minimize k in the plane will have corresponding points on the cylinder. The smallest values of k occur where ky = 2y = 0
and
kz = 4z = 0,
or where y = z = 0. This leads to
ma
FIGURE 14.51 The region in the xyplane from which the first two coordinates of the points (x, y, z) on the hyperbolic cylinder x 2  z 2 = 1 are selected excludes the band 1 6 x 6 1 in the xyplane (Example 2).
x 2 = z 2 + 1 = 1,
x = ;1.
Mu
ham
The corresponding points on the cylinder are s ;1, 0, 0d. We can see from the inequality ks y, zd = 1 + y 2 + 2z 2 Ú 1
that the points s ;1, 0, 0d give a minimum value for k. We can also see that the minimum distance from the origin to a point on the cylinder is 1 unit. Another way to find the points on the cylinder closest to the origin is to imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the cylinder (Figure 14.52). At each point of contact, the cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting
Solution 2
ƒsx, y, zd = x 2 + y 2 + z 2  a 2
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and
gsx, y, zd = x 2  z 2  1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8 Lagrange Multipliers
z
You
suf
x 2 y 2 z 2 a2 0
i
x2 z2 1 0
1041
x
iaz
y
ass a
nR
FIGURE 14.52 A sphere expanding like a soap bubble centered at the origin until it just touches the hyperbolic cylinder x 2  z 2  1 = 0 (Example 2).
equal to 0, then the gradients §ƒ and §g will be parallel where the surfaces touch. At any point of contact, we should therefore be able to find a scalar l (“lambda”) such that
dH
or
§ƒ = l§g,
2xi + 2yj + 2zk = ls2xi  2zkd.
Thus, the coordinates x, y, and z of any point of tangency will have to satisfy the three scalar equations 2x = 2lx,
2y = 0,
2z = 2lz.
Mu
ham
ma
For what values of l will a point (x, y, z) whose coordinates satisfy these scalar equations also lie on the surface x 2  z 2  1 = 0? To answer this question, we use our knowledge that no point on the surface has a zero xcoordinate to conclude that x Z 0. Hence, 2x = 2lx only if 2 = 2l,
or
l = 1.
For l = 1, the equation 2z = 2lz becomes 2z = 2z. If this equation is to be satisfied as well, z must be zero. Since y = 0 also (from the equation 2y = 0), we conclude that the points we seek all have coordinates of the form sx, 0, 0d. What points on the surface x 2  z 2 = 1 have coordinates of this form? The answer is the points (x, 0, 0) for which x 2  s0d2 = 1,
x 2 = 1,
or
x = ;1.
The points on the cylinder closest to the origin are the points s ;1, 0, 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1042
Chapter 14: Partial Derivatives
i
The Method of Lagrange Multipliers
suf
In Solution 2 of Example 2, we used the method of Lagrange multipliers. The method says that the extreme values of a function ƒ(x, y, z) whose variables are subject to a constraint gsx, y, zd = 0 are to be found on the surface g = 0 at the points where
You
§ƒ = l§g
for some scalar l (called a Lagrange multiplier). To explore the method further and see why it works, we first make the following observation, which we state as a theorem.
rstd = gstdi + hstdj + kstdk.
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C:
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THEOREM 12 The Orthogonal Gradient Theorem Suppose that ƒ(x, y, z) is differentiable in a region whose interior contains a smooth curve
ass a
If P0 is a point on C where ƒ has a local maximum or minimum relative to its values on C, then §ƒ is orthogonal to C at P0 .
Proof We show that §ƒ is orthogonal to the curve’s velocity vector at P0 . The values of ƒ on C are given by the composite ƒ(g(t), h(t), k(t)), whose derivative with respect to t is dƒ 0ƒ dg 0ƒ dh 0ƒ dk = + + = §ƒ # v. 0x dt 0y dt 0z dt dt
dH
At any point P0 where ƒ has a local maximum or minimum relative to its values on the curve, dƒ>dt = 0, so §ƒ # v = 0.
Mu
ham
ma
By dropping the zterms in Theorem 12, we obtain a similar result for functions of two variables.
COROLLARY OF THEOREM 12 At the points on a smooth curve rstd = gstdi + hstdj where a differentiable function ƒ(x, y) takes on its local maxima and minima relative to its values on the curve, §ƒ # v = 0, where v = dr>dt.
Theorem 12 is the key to the method of Lagrange multipliers. Suppose that ƒ(x, y, z) and g(x, y, z) are differentiable and that P0 is a point on the surface gsx, y, zd = 0 where ƒ has a local maximum or minimum value relative to its other values on the surface. Then ƒ takes on a local maximum or minimum at P0 relative to its values on every differentiable curve through P0 on the surface gsx, y, zd = 0. Therefore, §ƒ is orthogonal to the velocity vector of every such differentiable curve through P0 . So is §g , moreover (because §g is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefore, at P0, §ƒ is some scalar multiple l of §g.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8
Lagrange Multipliers
1043
suf
i
The Method of Lagrange Multipliers Suppose that ƒ(x, y, z) and g(x, y, z) are differentiable. To find the local maximum and minimum values of ƒ subject to the constraint gsx, y, zd = 0, find the values of x, y, z, and l that simultaneously satisfy the equations and
gsx, y, zd = 0.
(1)
You
§ƒ = l§g
For functions of two independent variables, the condition is similar, but without the variable z.
ƒsx, yd = xy
x
takes on the ellipse (Figure 14.53)
2兹2
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0
Find the greatest and smallest values that the function
x2 y2 1 8 2
y2 x2 + = 1. 8 2
FIGURE 14.53 Example 3 shows how to find the largest and smallest values of the product xy on this ellipse.
Solution
We want the extreme values of ƒsx, yd = xy subject to the constraint
ass a
兹2
Using the Method of Lagrange Multipliers
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EXAMPLE 3
y
gsx, yd =
y2 x2 +  1 = 0. 8 2
To do so, we first find the values of x, y, and l for which and
§ƒ = l§g
gsx, yd = 0.
dH
The gradient equation in Equations (1) gives yi + xj =
l xi + lyj, 4
ma
from which we find y =
l x, 4
x = ly,
and
y =
l l2 slyd = y, 4 4
Mu
ham
so that y = 0 or l = ;2. We now consider these two cases. Case 1: If y = 0 , then x = y = 0. But (0, 0) is not on the ellipse. Hence, y Z 0. Case 2: If y Z 0, then l = ;2 and x = ;2y. Substituting this in the equation gsx, yd = 0 gives s ;2yd2 y2 + = 1, 8 2
4y 2 + 4y 2 = 8
and
y = ;1.
The function ƒsx, yd = xy therefore takes on its extreme values on the ellipse at the four points s ;2, 1d, s ;2, 1d. The extreme values are xy = 2 and xy = 2. The Geometry of the Solution The level curves of the function ƒsx, yd = xy are the hyperbolas xy = c (Figure 14.54). The farther the hyperbolas lie from the origin, the larger the absolute value of ƒ. We want
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1044
Chapter 14: Partial Derivatives y xy 2
∇g 1 i j 2
1
xy 2
1
x
You
0
i
∇f i 2j
x2 y 2 10 2 8
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xy –2
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xy –2
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FIGURE 14.54 When subjected to the constraint gsx, yd = x 2>8 + y 2>2  1 = 0, the function ƒsx, yd = xy takes on extreme values at the four points s ;2, ;1d. These are the points on the ellipse when §ƒ (red) is a scalar multiple of §g (blue) (Example 3).
ass a
to find the extreme values of ƒ(x, y), given that the point (x, y) also lies on the ellipse x 2 + 4y 2 = 8. Which hyperbolas intersecting the ellipse lie farthest from the origin? The hyperbolas that just graze the ellipse, the ones that are tangent to it, are farthest. At these points, any vector normal to the hyperbola is normal to the ellipse, so §ƒ = yi + xj is a multiple sl = ;2d of §g = sx>4di + yj. At the point (2, 1), for example,
dH
§ƒ = i + 2j,
§g =
1 i + j, 2
and
§ƒ = 2§g.
§g = 
1 i + j, 2
and
§ƒ = 2§g.
At the point s 2, 1d,
ma
§ƒ = i  2j,
EXAMPLE 4
Finding Extreme Function Values on a Circle
Mu
ham
Find the maximum and minimum values of the function ƒsx, yd = 3x + 4y on the circle x 2 + y 2 = 1. Solution
We model this as a Lagrange multiplier problem with ƒsx, yd = 3x + 4y,
gsx, yd = x 2 + y 2  1
and look for the values of x, y, and l that satisfy the equations §ƒ = l§g:
3i + 4j = 2xli + 2ylj
gsx, yd = 0:
x 2 + y 2  1 = 0.
The gradient equation in Equations (1) implies that l Z 0 and gives
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x =
3 , 2l
y =
2 . l
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8 Lagrange Multipliers
1045
x2 y2 1
3 , 5
a
6i 8j 5 5
9 4 + 2 = 1, 4l2 l
x
9 + 16 = 4l2,
Thus,
3x 4y –5
x =
y =
and
5 l = ; . 2
4 2 = ; , 5 l
3 25 4 3 a b + 4 a b = = 5. 5 5 5
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3 25 4 3a b + 4a b = = 5 5 5 5
and
The Geometry of the Solution
ass a
The level curves of ƒsx, yd = 3x + 4y are the lines 3x + 4y = c (Figure 14.55). The farther the lines lie from the origin, the larger the absolute value of ƒ. We want to find the extreme values of ƒ(x, y) given that the point (x, y) also lies on the circle x 2 + y 2 = 1 . Which lines intersecting the circle lie farthest from the origin? The lines tangent to the circle are farthest. At the points of tangency, any vector normal to the line is normal to the circle, so the gradient §ƒ = 3i + 4j is a multiple sl = ;5>2d of the gradient §g = 2xi + 2yj. At the point (3> 5, 4> 5), for example, §ƒ = 3i + 4j,
g2 0
3 3 = ; , 5 2l
4l2 = 25,
and ƒsx, yd = 3x + 4y has extreme values at sx, yd = ;s3>5, 4>5d. By calculating the value of 3x + 4y at the points ;s3>5, 4>5d, we see that its maximum and minimum values on the circle x 2 + y 2 = 1 are
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FIGURE 14.55 The function ƒsx, yd = 3x + 4y takes on its largest value on the unit circle gsx, yd = x 2 + y 2  1 = 0 at the point (3> 5, 4> 5) and its smallest value at the point s 3>5, 4>5d (Example 4). At each of these points, §ƒ is a scalar multiple of §g . The figure shows the gradients at the first point but not the second.
2
so
4 5
3x 4y 5
2
3 2 b + a b  1 = 0, l 2l
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∇g
5 ∇g 2
You
∇f 3i 4j
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y
i
These equations tell us, among other things, that x and y have the same sign. With these values for x and y, the equation gsx, yd = 0 gives
§g =
6 8 i + j, 5 5
and
§ƒ =
5 §g. 2
ma
Lagrange Multipliers with Two Constraints
Many problems require us to find the extreme values of a differentiable function ƒ(x, y, z) whose variables are subject to two constraints. If the constraints are
∇g1
ham
∇f
∇g2
g1sx, y, zd = 0
and
g2sx, y, zd = 0
and g1 and g2 are differentiable, with §g1 not parallel to §g2 , we find the constrained local maxima and minima of ƒ by introducing two Lagrange multipliers l and m (mu, pronounced “mew”). That is, we locate the points P(x, y, z) where ƒ takes on its constrained extreme values by finding the values of x, y, z, l, and m that simultaneously satisfy the equations
g1 0
Mu
C
FIGURE 14.56 The vectors §g1 and §g2 lie in a plane perpendicular to the curve C because §g1 is normal to the surface g1 = 0 and §g2 is normal to the surface g2 = 0 .
§ƒ = l§g1 + m§g2 ,
g1sx, y, zd = 0,
g2sx, y, zd = 0
(2)
Equations (2) have a nice geometric interpretation. The surfaces g1 = 0 and g2 = 0 (usually) intersect in a smooth curve, say C (Figure 14.56). Along this curve we seek the points where ƒ has local maximum and minimum values relative to its other values on the curve.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1046
Chapter 14: Partial Derivatives
EXAMPLE 5
z Cylinder
x2
y2
You
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These are the points where §ƒ is normal to C, as we saw in Theorem 12. But §g1 and §g2 are also normal to C at these points because C lies in the surfaces g1 = 0 and g2 = 0. Therefore, §ƒ lies in the plane determined by §g1 and §g2 , which means that §ƒ = l§g1 + m§g2 for some l and m . Since the points we seek also lie in both surfaces, their coordinates must satisfy the equations g1sx, y, zd = 0 and g2sx, y, zd = 0, which are the remaining requirements in Equations (2).
Finding Extremes of Distance on an Ellipse
The plane x + y + z = 1 cuts the cylinder x 2 + y 2 = 1 in an ellipse (Figure 14.57). Find the points on the ellipse that lie closest to and farthest from the origin.
1
Solution
We find the extreme values of
iaz
P2
ƒsx, y, zd = x 2 + y 2 + z 2
(the square of the distance from (x, y, z) to the origin) subject to the constraints
(0, 1, 0) (1, 0, 0)
(3)
g2sx, y, zd = x + y + z  1 = 0.
(4)
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P1 Plane xy z 1
g1sx, y, zd = x 2 + y 2  1 = 0
The gradient equation in Equations (2) then gives
§ƒ = l§g1 + m§g2
ass a
x
y
2xi + 2yj + 2zk = ls2xi + 2yjd + msi + j + kd 2xi + 2yj + 2zk = s2lx + mdi + s2ly + mdj + mk
or
2x = 2lx + m,
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FIGURE 14.57 On the ellipse where the plane and cylinder meet, what are the points closest to and farthest from the origin? (Example 5)
2y = 2ly + m,
2z = m.
(5)
The scalar equations in Equations (5) yield 2x = 2lx + 2z Q s1  ldx = z,
Mu
ham
ma
2y = 2ly + 2z Q s1  ldy = z.
(6)
Equations (6) are satisfied simultaneously if either l = 1 and z = 0 or l Z 1 and x = y = z>s1  ld. If z = 0, then solving Equations (3) and (4) simultaneously to find the corresponding points on the ellipse gives the two points (1, 0, 0) and (0, 1, 0). This makes sense when you look at Figure 14.57. If x = y, then Equations (3) and (4) give x2 + x2  1 = 0
x + x + z  1 = 0
2x 2 = 1 x = ;
z = 1  2x 22 2
z = 1 < 22.
The corresponding points on the ellipse are P1 = a
22 22 , , 1  22b 2 2
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and
P2 = a
22 22 ,, 1 + 22b. 2 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8 Lagrange Multipliers
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Mu
ham
ma
dH
ass a
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iaz
You
suf
i
Here we need to be careful, however. Although P1 and P2 both give local maxima of Ć’ on the ellipse, P2 is farther from the origin than P1. The points on the ellipse closest to the origin are (1, 0, 0) and (0, 1, 0). The point on the ellipse farthest from the origin is P2.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8 Lagrange Multipliers
3. Maximum on a line Find the maximum value of ƒsx, yd = 49  x 2  y 2 on the line x + 3y = 10. 4. Extrema on a line Find the local extreme values of ƒsx, yd = x 2y on the line x + y = 3 . 5. Constrained minimum Find the points on the curve xy 2 = 54 nearest the origin. 6. Constrained minimum Find the points on the curve x 2y = 2 nearest the origin. 7. Use the method of Lagrange multipliers to find
Three Independent Variables with One Constraint 17. Minimum distance to a point Find the point on the plane x + 2y + 3z = 13 closest to the point (1, 1, 1). 18. Maximum distance to a point Find the point on the sphere x 2 + y 2 + z 2 = 4 farthest from the point s1, 1, 1d .
ssa
a. Minimum on a hyperbola The minimum value of x + y, subject to the constraints xy = 16, x 7 0, y 7 0
16. Cheapest storage tank Your firm has been asked to design a storage tank for liquid petroleum gas. The customer’s specifications call for a cylindrical tank with hemispherical ends, and the tank is to hold 8000 m3 of gas. The customer also wants to use the smallest amount of material possible in building the tank. What radius and height do you recommend for the cylindrical portion of the tank?
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2. Extrema on a circle Find the extreme values of ƒsx, yd = xy subject to the constraint gsx, yd = x 2 + y 2  10 = 0.
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1. Extrema on an ellipse Find the points on the ellipse x 2 + 2y 2 = 1 where ƒsx, yd = xy as its extreme values.
15. Ant on a metal plate The temperature at a point (x, y) on a metal plate is Tsx, yd = 4x 2  4xy + y 2. An ant on the plate walks around the circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant?
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Two Independent Variables with One Constraint
usu
fi
EXERCISES 14.8
1047
b. Maximum on a line The maximum value of xy, subject to the constraint x + y = 16. Comment on the geometry of each solution.
20. Minimum distance to the origin z = xy + 1 nearest the origin.
Ha
8. Extrema on a curve Find the points on the curve x 2 + xy + y 2 = 1 in the xyplane that are nearest to and farthest from the origin.
19. Minimum distance to the origin Find the minimum distance from the surface x 2 + y 2  z 2 = 1 to the origin.
mm ad
9. Minimum surface area with fixed volume Find the dimensions of the closed right circular cylindrical can of smallest surface area whose volume is 16p cm3 .
10. Cylinder in a sphere Find the radius and height of the open right circular cylinder of largest surface area that can be inscribed in a sphere of radius a. What is the largest surface area? 11. Rectangle of greatest area in an ellipse Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse x 2>16 + y 2>9 = 1 with sides parallel to the coordinate axes.
Mu
ha
12. Rectangle of longest perimeter in an ellipse Find the dimensions of the rectangle of largest perimeter that can be inscribed in the ellipse x 2>a 2 + y 2>b 2 = 1 with sides parallel to the coordinate axes. What is the largest perimeter?
13. Extrema on a circle Find the maximum and minimum values of x 2 + y 2 subject to the constraint x 2  2x + y 2  4y = 0. 14. Extrema on a circle Find the maximum and minimum values of 3x  y + 6 subject to the constraint x 2 + y 2 = 4.
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Find the point on the surface
21. Minimum distance to the origin Find the points on the surface z 2 = xy + 4 closest to the origin.
22. Minimum distance to the origin Find the point(s) on the surface xyz = 1 closest to the origin. 23. Extrema on a sphere
Find the maximum and minimum values of
ƒsx, y, zd = x  2y + 5z on the sphere x 2 + y 2 + z 2 = 30. 24. Extrema on a sphere Find the points on the sphere x 2 + y 2 + z 2 = 25 where ƒsx, y, zd = x + 2y + 3z has its maximum and minimum values. 25. Minimizing a sum of squares Find three real numbers whose sum is 9 and the sum of whose squares is as small as possible. 26. Maximizing a product Find the largest product the positive numbers x, y, and z can have if x + y + z 2 = 16. 27. Rectangular box of longest volume in a sphere Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
4x 2 + y 2 + 4z 2 = 16 enters Earth’s atmosphere and its surface begins to heat. After 1 hour, the temperature at the point (x, y, z) on the probe’s surface is Tsx, y, zd = 8x 2 + 4yz  16z + 600 .
i
suf
A space probe in the shape of
36. Maximum value on line of intersection Find the maximum value that ƒsx, y, zd = x 2 + 2y  z 2 can have on the line of intersection of the planes 2x  y = 0 and y + z = 0. 37. Extrema on a curve of intersection Find the extreme values of ƒsx, y, zd = x 2yz + 1 on the intersection of the plane z = 1 with the sphere x 2 + y 2 + z 2 = 10. 38. a. Maximum on line of intersection Find the maximum value of w = xyz on the line of intersection of the two planes x + y + z = 40 and x + y  z = 0. b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of w.
Find the hottest point on the probe’s surface. 30. Extreme temperatures on a sphere Suppose that the Celsius temperature at the point (x, y, z) on the sphere x 2 + y 2 + z 2 = 1 is T = 400xyz 2 . Locate the highest and lowest temperatures on the sphere.
39. Extrema on a circle of intersection Find the extreme values of the function ƒsx, y, zd = xy + z 2 on the circle in which the plane y  x = 0 intersects the sphere x 2 + y 2 + z 2 = 4. 40. Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane 2y + 4z = 5 and the cone z 2 = 4x 2 + 4y 2.
Theory and Examples 41. The condition §f = l§g is not sufficient Although §ƒ = l§g is a necessary condition for the occurrence of an extreme value of ƒ(x, y) subject to the condition gsx, yd = 0, it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of ƒsx, yd = x + y subject to the constraint that xy = 16. The method will identify the two points (4, 4) and s 4, 4d as candidates for the location of extreme values. Yet the sum sx + yd has no maximum value on the hyperbola xy = 16. The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum ƒsx, yd = x + y becomes.
dH
ass a
31. Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts x and y of two capital goods G1 and G2 is sometimes measured by a function U(x, y). For example, G1 and G2 might be two chemicals a pharmaceutical company needs to have on hand and U(x, y) the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If G1 costs a dollars per kilogram, G2 costs b dollars per kilogram, and the total amount allocated for the purchase of G1 and G2 together is c dollars, then the company’s managers want to maximize U(x, y) given that ax + by = c . Thus, they need to solve a typical Lagrange multiplier problem. Suppose that
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29. Hottest point on a space probe the ellipsoid
35. Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes y + 2z = 12 and x + y = 6.
You
28. Box with vertex on a plane Find the volume of the largest closed rectangular box in the first octant having three faces in the coordinate planes and a vertex on the plane x>a + y>b + z>c = 1, where a 7 0, b 7 0, and c 7 0.
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1048
Usx, yd = xy + 2x
and that the equation ax + by = c simplifies to
ma
2x + y = 30.
Find the maximum value of U and the corresponding values of x and y subject to this latter constraint.
Mu
ham
32. Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it where the magnetic field of the planet is weakest. The planet is spherical, with a radius of 6 units. Based on a coordinate system whose origin is at the center of the planet, the strength of the magnetic field is given by Msx, y, zd = 6x  y 2 + xz + 60. Where should you locate the radio telescope?
Extreme Values Subject to Two Constraints 33. Maximize the function ƒsx, y, zd = x 2 + 2y  z 2 subject to the constraints 2x  y = 0 and y + z = 0.
34. Minimize the function ƒsx, y, zd = x 2 + y 2 + z 2 subject to the constraints x + 2y + 3z = 6 and x + 3y + 9z = 9.
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42. A least squares plane The plane z = Ax + By + C is to be “fitted” to the following points sxk , yk , zk d: s0, 0, 0d,
s0, 1, 1d,
s1, 1, 1d,
s1, 0, 1d.
Find the values of A, B, and C that minimize 4 2 a sAxk + Byk + C  zk d ,
k=1
the sum of the squares of the deviations. 43. a. Maximum on a sphere Show that the maximum value of a 2b 2c 2 on a sphere of radius r centered at the origin of a Cartesian abccoordinate system is sr 2>3d3. b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers a, b, and c, sabcd1>3 …
a + b + c ; 3
that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.8 Lagrange Multipliers
COMPUTER EXPLORATIONS
i
d. Evaluate ƒ at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise.
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44. Sum of products Let a1, a2 , Á , an be n positive numbers. Find the maximum of © ni = 1 ai xi subject to the constraint © ni = 1 x i 2 = 1.
1049
45. Minimize ƒsx, y, zd = xy + yz subject to the constraints x 2 + y 2  2 = 0 and x 2 + z 2  2 = 0.
a. Form the function h = ƒ  l1 g1  l2 g2 , where ƒ is the function to optimize subject to the constraints g1 = 0 and g2 = 0. b. Determine all the first partial derivatives of h, including the partials with respect to l1 and l2 , and set them equal to 0.
to
the
48. Minimize ƒsx, y, zd = x 2 + y 2 + z 2 subject to the constraints x 2  xy + y 2  z 2  1 = 0 and x 2 + y 2  1 = 0. 49. Minimize ƒsx, y, z, wd = x 2 + y 2 + z 2 + w 2 subject to the constraints 2x  y + z  w  1 = 0 and x + y  z + w  1 = 0. 50. Determine the distance from the line y = x + 1 to the parabola y 2 = x. (Hint: Let (x, y) be a point on the line and (w, z) a point on the parabola. You want to minimize sx  wd2 + sy  zd2.)
ass a dH ma ham Mu To Read it Online & Download:
constraints
47. Maximize ƒsx, y, zd = x 2 + y 2 + z 2 subject to the constraints 2y + 4z  5 = 0 and 4x 2 + 4y 2  z 2 = 0.
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c. Solve the system of equations found in part (b) for all the unknowns, including l1 and l2.
46. Minimize ƒsx, y, zd = xyz subject x 2 + y 2  1 = 0 and x  z = 0.
You
In Exercises 45–50, use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema:
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Implementing the Method of Lagrange Multipliers
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i f su
14.9 Partial Derivatives with Constrained Variables
Partial Derivatives with Constrained Variables
14.9
1049
u Yo
In finding partial derivatives of functions like w = ƒsx, yd , we have assumed x and y to be independent. In many applications, however, this is not the case. For example, the internal energy U of a gas may be expressed as a function U = ƒsP, V, Td of pressure P, volume V, and temperature T. If the individual molecules of the gas do not interact, however, P, V, and T obey (and are constrained by) the ideal gas law
z ia
R n
PV = nRT
sn and R constantd,
and fail to be independent. In this section we learn how to find partial derivatives in situations like this, which you may encounter in studying economics, engineering, or physics.†
a s s a
Decide Which Variables Are Dependent and Which Are Independent
H d
If the variables in a function w = ƒsx, y, zd are constrained by a relation like the one imposed on x, y, and z by the equation z = x 2 + y 2, the geometric meanings and the numerical values of the partial derivatives of ƒ will depend on which variables are chosen to be dependent and which are chosen to be independent. To see how this choice can affect the outcome, we consider the calculation of 0w>0x when w = x 2 + y 2 + z 2 and z = x 2 + y 2.
a m m a h
EXAMPLE 1
Finding a Partial Derivative with Constrained Independent Variables
Find 0w>0x if w = x 2 + y 2 + z 2 and z = x 2 + y 2.
u M
†This section is based on notes written for MIT by Arthur P. Mattuck.
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Chapter 14: Partial Derivatives
We are given two equations in the four unknowns x, y, z, and w. Like many such systems, this one can be solved for two of the unknowns (the dependent variables) in terms of the others (the independent variables). In being asked for 0w>0x, we are told that w is to be a dependent variable and x an independent variable. The possible choices for the other variables come down to Independent x, y x, z
You
Dependent w, z w, y
suf
i
Solution
iaz
In either case, we can express w explicitly in terms of the selected independent variables. We do this by using the second equation z = x 2 + y 2 to eliminate the remaining dependent variable in the first equation. In the first case, the remaining dependent variable is z. We eliminate it from the first equation by replacing it by x 2 + y 2 . The resulting expression for w is w = x 2 + y 2 + z 2 = x 2 + y 2 + sx 2 + y 2 d2
nR
= x 2 + y 2 + x 4 + 2x 2y 2 + y 4 and
(1)
ass a
0w = 2x + 4x 3 + 4xy 2. 0x
This is the formula for 0w>0x when x and y are the independent variables. In the second case, where the independent variables are x and z and the remaining dependent variable is y, we eliminate the dependent variable y in the expression for w by replacing y 2 in the second equation by z  x 2 . This gives
z z x 2, y 0
z x2 y2 P
(0, 0, 1)
Circle x 2 y 2 1 in the plane z 1
0 (1, 0, 0)
y
ham
x
FIGURE 14.58 If P is constrained to lie on the paraboloid z = x 2 + y 2, the value of the partial derivative of w = x 2 + y 2 + z 2 with respect to x at P depends on the direction of motion (Example 1). (1) As x changes, with y = 0 , P moves up or down the surface on the parabola z = x 2 in the xzplane with 0w>0x = 2x + 4x 3. (2) As x changes, with z = 1 , P moves on the circle x 2 + y 2 = 1, z = 1 , and 0w>0x = 0.
Mu
and
0w = 0. 0x
(2)
This is the formula for 0w>0x when x and z are the independent variables. The formulas for 0w>0x in Equations (1) and (2) are genuinely different. We cannot change either formula into the other by using the relation z = x 2 + y 2. There is not just one 0w>0x , there are two, and we see that the original instruction to find 0w>0x was incomplete. Which 0w>0x? we ask. The geometric interpretations of Equations (1) and (2) help to explain why the equations differ. The function w = x 2 + y 2 + z 2 measures the square of the distance from the point (x, y, z) to the origin. The condition z = x 2 + y 2 says that the point (x, y, z) lies on the paraboloid of revolution shown in Figure 14.58. What does it mean to calculate 0w>0x at a point P(x, y, z) that can move only on this surface? What is the value of 0w>0x when the coordinates of P are, say, (1, 0, 1)? If we take x and y to be independent, then we find 0w>0x by holding y fixed (at y = 0 in this case) and letting x vary. Hence, P moves along the parabola z = x 2 in the xzplane. As P moves on this parabola, w, which is the square of the distance from P to the origin, changes. We calculate 0w>0x in this case (our first solution above) to be
ma
(1, 0, 1)
dH
w = x 2 + y 2 + z 2 = x 2 + sz  x 2 d + z 2 = z + z 2
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0w = 2x + 4x 3 + 4xy 2. 0x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.9
Partial Derivatives with Constrained Variables
1051
suf
0w = 2 + 4 + 0 = 6. 0x
i
At the point P(1, 0, 1), the value of this derivative is
You
If we take x and z to be independent, then we find 0w>0x by holding z fixed while x varies. Since the zcoordinate of P is 1, varying x moves P along a circle in the plane z = 1. As P moves along this circle, its distance from the origin remains constant, and w, being the square of this distance, does not change. That is, 0w = 0, 0x
iaz
as we found in our second solution.
How to Find w>x When the Variables in w = Ć’sx, y, zd Are Constrained by Another Equation
ass a
nR
As we saw in Example 1, a typical routine for finding 0w>0x when the variables in the function w = Ć’sx, y, zd are related by another equation has three steps. These steps apply to finding 0w>0y and 0w>0z as well.
1.
dH
2. 3.
Decide which variables are to be dependent and which are to be independent. (In practice, the decision is based on the physical or theoretical context of our work. In the exercises at the end of this section, we say which variables are which.) Eliminate the other dependent variable(s) in the expression for w. Differentiate as usual.
ma
If we cannot carry out Step 2 after deciding which variables are dependent, we differentiate the equations as they are and try to solve for 0w>0x afterward. The next example shows how this is done.
Mu
ham
EXAMPLE 2
Finding a Partial Derivative with Identified Constrained Independent Variables
Find 0w>0x at the point sx, y, zd = s2, 1, 1d if w = x 2 + y 2 + z 2,
z 3  xy + yz + y 3 = 1,
and x and y are the independent variables. It is not convenient to eliminate z in the expression for w. We therefore differentiate both equations implicitly with respect to x, treating x and y as independent variables and w and z as dependent variables. This gives Solution
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0w 0z = 2x + 2z 0x 0x
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(3)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1052
Chapter 14: Partial Derivatives
0z 0z  y + y + 0 = 0. 0x 0x
suf
3z 2
i
and (4)
y 0z = 0x y + 3z 2 and substitute into Equation (3) to get
You
These equations may now be combined to express 0w>0x in terms of x, y, and z. We solve Equation (4) for 0z>0x to get
iaz
2yz 0w = 2x + . 0x y + 3z 2 The value of this derivative at sx, y, zd = s2, 1, 1d is
2s 1ds1d 0w 2 = 2s2d + = 4 + = 3. b 2 0x s2,1,1d 2 1 + 3s1d
nR
a
Notation
Sonya Kovalevsky (1850–1891)
To show what variables are assumed to be independent in calculating a derivative, we can use the following notation:
ass a
HISTORICAL BIOGRAPHY
0w b 0x y
0w>0x with x and y independent
a
0ƒ b 0y x, t
0ƒ>0y with y, x and t independent
dH
a
Finding a Partial Derivative with Constrained Variables Notationally Identified
ma
EXAMPLE 3
Find s0w>0xdy, z if w = x 2 + y  z + sin t and x + y = t.
Mu
ham
Solution
With x, y, z independent, we have t = x + y, a
w = x 2 + y  z + sin sx + yd
0w 0 b = 2x + 0  0 + cos sx + yd sx + yd 0x y, z 0x = 2x + cos sx + yd.
Arrow Diagrams In solving problems like the one in Example 3, it often helps to start with an arrow diagram that shows how the variables and functions are related. If
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w = x 2 + y  z + sin t
and
x + y = t
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.9 Partial Derivatives with Constrained Variables
1053
Independent variables
You
suf
i
and we are asked to find 0w>0x when x, y, and z are independent, the appropriate diagram is one like this: x x y (5) £y≥ : § ¥ : w z z t Intermediate variables
Dependent variable
:
u y § ¥ s t
nR
x £y≥ z
iaz
To avoid confusion between the independent and intermediate variables with the same symbolic names in the diagram, it is helpful to rename the intermediate variables (so they are seen as functions of the independent variables). Thus, let u = x, y = y , and s = z denote the renamed intermediate variables. With this notation, the arrow diagram becomes
ass a
Independent variables
Intermediate variables and relations u = x y = y s = z t = x + y
:
w
(6)
Dependent variable
The diagram shows the independent variables on the left, the intermediate variables and their relation to the independent variables in the middle, and the dependent variable on the right. The function w now becomes
dH
w = u 2 + y  s + sin t,
where
0w 0w 0u 0w 0y 0w 0s 0w 0t = + + + 0x 0u 0x 0y 0x 0s 0x 0t 0x = s2uds1d + s1ds0d + s 1ds0d + scos tds1d = 2u + cos t = 2x + cos sx + yd.
Substituting the original independent variables u = x and t = x + y.
Mu
ham
ma
y = y, s = z, and t = x + y. u = x, To find 0w>0x, we apply the fourvariable form of the Chain Rule to w, guided by the arrow diagram in Equation (6):
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i f u s u o
14.9 Partial Derivatives with Constrained Variables
EXERCISES 14.9
n a s s a H
Finding Partial Derivatives with Constrained Variables In Exercises 1â€“3, begin by drawing a diagram that shows the relations among the variables.
d a mm
1. If w = x 2 + y 2 + z 2 and z = x 2 + y 2 , find a. a
0w b 0y z
b. a
0w b 0z x
a h Mu
c. a
0w b . 0z y
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Y z a i R
2. If w = x 2 + y  z + sin t and x + y = t , find a. a
0w b 0y x, z
b. a
0w b 0y z, t
d. a
0w b 0z y, t
e. a
0w b 0t x, z
c. a f. a
0w b 0z x, y
0w b . 0t y, z
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1053
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
a. a
0U b 0P V
b. a
0U b . 0T V
0w b 0x y
b. a
0w b 0z y
Partial Derivatives Without Specific Formulas
i
3. Let U = ƒsP, V, Td be the internal energy of a gas that obeys the ideal gas law PV = nRT (n and R constant). Find
9. Establish the fact, widely used in hydrodynamics, that if ƒsx, y, zd = 0, then a
4. Find
(Hint: Express all the derivatives in terms of the formal partial derivatives 0ƒ>0x, 0ƒ>0y, and 0ƒ>0z. )
at the point sx, y, zd = s0, 1, pd if w = x2 + y2 + z2
and
y sin z + z sin x = 0 .
10. If z = x + ƒsud, where u = xy, show that x
5. Find 0w b 0y z
11. Suppose that the equation gsx, y, zd = 0 determines z as a differentiable function of the independent variables x and y and that gz Z 0. Show that
at the point sw, x, y, zd = s4, 2, 1, 1d if w = x 2y 2 + yz  z 3
x2 + y2 + z2 = 6 .
and
6. Find s0u>0ydx at the point su, yd = y = uy.
A 22, 1 B , if x = u 2 + y 2 and
0x b 0r u
and
a
0r b. 0x y
0g>0y 0z . b = 0y x 0g>0z
12. Suppose that ƒsx, y, z, wd = 0 and g sx, y, z, wd = 0 determine z and w as differentiable functions of the independent variables x and y, and suppose that 0ƒ 0g 0ƒ 0g Z 0. 0z 0w 0w 0z
Show that
8. Suppose that w = x 2  y 2 + 4z + t
and
x + 2z + t = 25.
and
dH
Show that the equations 0w = 2x  1 0x
a
ass a
7. Suppose that x 2 + y 2 = r 2 and x = r cos u, as in polar coordinates. Find a
0z 0z  y = x. 0x 0y
iaz
b. a
0w b 0y x
nR
a. a
0y 0x 0z b a b a b = 1. 0y z 0z x 0x y
You
a. a
suf
1054
0w = 2x  2 0x
0g 0ƒ 0g 0w 0w 0x 0g 0ƒ 0g 0w 0w 0z
and 0ƒ 0g 0ƒ 0g 0z 0y 0y 0z 0w a b = . 0y x 0ƒ 0g 0ƒ 0g 0z 0w 0w 0z
Mu
ham
ma
each give 0w>0x, depending on which variables are chosen to be dependent and which variables are chosen to be independent. Identify the independent variables in each case.
0ƒ 0x 0z a b = 0x y 0ƒ 0z
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i f u s u o
Chapter 14: Partial Derivatives
14.10
Taylor’s Formula for Two Variables
n a s as
Y z a i R
This section uses Taylor’s formula to derive the Second Derivative Test for local extreme values (Section 14.7) and the error formula for linearizations of functions of two independent variables (Section 14.6). The use of Taylor’s formula in these derivations leads to an extension of the formula that provides polynomial approximations of all orders for functions of two independent variables.
H d a m
Derivation of the Second Derivative Test
M
m a uh
Let ƒ(x, y) have continuous partial derivatives in an open region R containing a point P(a, b) where ƒx = ƒy = 0 (Figure 14.59). Let h and k be increments small enough to put the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.10
i
point Ssa + h, b + kd and the line segment joining it to P inside R. We parametrize the segment PS as y = b + tk,
x = a + th,
Parametrized segment in R (a th, b tk), a typical point on the segment
0 … t … 1.
If Fstd = ƒsa + th, b + tkd , the Chain Rule gives F¿std = ƒx
suf
S(a h, b k)
1055
dy dx + ƒy = hƒx + kƒy . dt dt
You
t1
Taylor’s Formula for Two Variables
Since ƒx and ƒy are differentiable (they have continuous partial derivatives), F¿ is a differentiable function of t and t0
P(a, b)
F– =
= h 2ƒxx + 2hkƒxy + k 2ƒyy .
fxy = fyx
Since F and F¿ are continuous on [0, 1] and F¿ is differentiable on (0, 1), we can apply Taylor’s formula with n = 2 and a = 0 to obtain
nR
FIGURE 14.59 We begin the derivation of the second derivative test at P(a, b) by parametrizing a typical line segment from P to a point S nearby.
0F¿ dx 0 0F¿ dy 0 + = shƒx + kƒy d # h + shƒx + kƒy d # k 0x dt 0y dt 0x 0y
iaz
Part of open region R
Fs1d = Fs0d + F¿s0ds1  0d + F–scd
ass a
Fs1d = Fs0d + F¿s0d +
s1  0d2 2
1 F–scd 2
(1)
for some c between 0 and 1. Writing Equation (1) in terms of ƒ gives ƒsa + h, b + kd = ƒsa, bd + hƒxsa, bd + kƒysa, bd
dH
+
1 2 . A h ƒxx + 2hkƒxy + k 2ƒyy B ` 2 sa + ch, b + ckd
(2)
Since ƒxsa, bd = ƒysa, bd = 0, this reduces to ƒsa + h, b + kd  ƒsa, bd =
1 2 . A h ƒxx + 2hkƒxy + k 2ƒyy B ` 2 sa + ch, b + ckd
(3)
Mu
ham
ma
The presence of an extremum of ƒ at (a, b) is determined by the sign of ƒsa + h, b + kd  ƒsa, bd. By Equation (3), this is the same as the sign of Qscd = sh 2ƒxx + 2hkƒxy + k 2ƒyy d ƒ sa + ch,b + ckd .
Now, if Qs0d Z 0 , the sign of Q(c) will be the same as the sign of Q(0) for sufficiently small values of h and k. We can predict the sign of Qs0d = h 2ƒxxsa, bd + 2hkƒxysa, bd + k 2ƒyysa, bd
(4)
from the signs of ƒxx and ƒxx ƒyy  ƒxy2 at (a, b). Multiply both sides of Equation (4) by ƒxx and rearrange the righthand side to get ƒxx Qs0d = shƒxx + kƒxy d2 + sƒxx ƒyy  ƒxy2 dk 2.
(5)
From Equation (5) we see that 1. 2.
If ƒxx 6 0 and ƒxx ƒyy  ƒxy2 7 0 at (a, b), then Qs0d 6 0 for all sufficiently small nonzero values of h and k, and ƒ has a local maximum value at (a, b). If ƒxx 7 0 and ƒxx ƒyy  ƒxy2 7 0 at (a, b), then Qs0d 7 0 for all sufficiently small nonzero values of h and k and ƒ has a local minimum value at (a, b).
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Chapter 14: Partial Derivatives
If ƒxx ƒyy  ƒxy2 6 0 at (a, b), there are combinations of arbitrarily small nonzero values of h and k for which Qs0d 7 0, and other values for which Qs0d 6 0. Arbitrarily close to the point P0sa, b, ƒsa, bdd on the surface z = ƒsx, yd there are points above P0 and points below P0 , so ƒ has a saddle point at (a, b). If ƒxx ƒyy  ƒxy2 = 0, another test is needed. The possibility that Q(0) equals zero prevents us from drawing conclusions about the sign of Q(c).
The Error Formula for Linear Approximations
You
4.
suf
i
3.
We want to show that the difference E(x, y), between the values of a function ƒ(x, y), and its linearization L(x, y) at sx0 , y0 d satisfies the inequality
iaz
1 ƒ Esx, yd ƒ … 2 Ms ƒ x  x0 ƒ + ƒ y  y0 ƒ d2.
nR
The function ƒ is assumed to have continuous second partial derivatives throughout an open set containing a closed rectangular region R centered at sx0 , y0 d. The number M is an upper bound for ƒ ƒxx ƒ , ƒ ƒyy ƒ , and ƒ ƒxy ƒ on R. The inequality we want comes from Equation (2). We substitute x0 and y0 for a and b, and x  x0 and y  y0 for h and k, respectively, and rearrange the result as ƒsx, yd = ƒsx0 , y0 d + ƒxsx0 , y0 dsx  x0 d + ƒysx0 , y0 ds y  y0 d ('''''''')'''''''''''''*
ass a
linearization Lsx, yd
+ 1 A sx  x0 d2ƒxx + 2sx  x0 ds y  y0 dƒxy + s y  y0 d2ƒyy B ` sx0 + csx  x0d, y0 + cs y  y0dd. 2 ('''''''''''''''')'''''''''''''''''*
dH
error Esx, yd
This equation reveals that 1 ƒ E ƒ … 2 A ƒ x  x0 ƒ 2 ƒ ƒxx ƒ + 2 ƒ x  x0 ƒ ƒ y  y0 ƒ ƒ ƒxy ƒ + ƒ y  y0 ƒ 2 ƒ ƒyy ƒ B .
Mu
ham
ma
Hence, if M is an upper bound for the values of ƒ ƒxx ƒ , ƒ ƒxy ƒ , and ƒ ƒyy ƒ on R, 1 ƒ E ƒ … 2 A ƒ x  x0 ƒ 2 M + 2 ƒ x  x0 ƒ ƒ y  y0 ƒ M + ƒ y  y0 ƒ 2M B =
1 Ms ƒ x  x0 ƒ + ƒ y  y0 ƒ d2. 2
Taylor’s Formula for Functions of Two Variables The formulas derived earlier for F¿ and F– can be obtained by applying to ƒ(x, y) the operators ah
0 0 + k b 0x 0y
and
ah
2
0 0 02 02 02 + k b = h 2 2 + 2hk + k2 2 . 0x 0y 0x 0y 0x 0y
These are the first two instances of a more general formula,
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n
F sndstd =
dn 0 0 Fstd = ah + k b ƒsx, yd, 0x 0y dt n
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.10
Taylor’s Formula for Two Variables
1057
0 0 + k b 0x 0y
n
suf
ah
i
which says that applying d n>dt n to Fstd gives the same result as applying the operator
Fstd = Fs0d + F¿s0dt +
You
to ƒ(x, y) after expanding it by the Binomial Theorem. If partial derivatives of ƒ through order n + 1 are continuous throughout a rectangular region centered at (a, b), we may extend the Taylor formula for F(t) to F–s0d 2 F snds0d snd t + Á + t + remainder, 2! n!
and take t = 1 to obtain
F–s0d F snds0d + Á + + remainder. 2! n!
iaz
Fs1d = Fs0d + F¿s0d +
nR
When we replace the first n derivatives on the right of this last series by their equivalent expressions from Equation (6) evaluated at t = 0 and add the appropriate remainder term, we arrive at the following formula.
ass a
Taylor’s Formula for ƒ(x, y) at the Point (a, b) Suppose ƒ(x, y) and its partial derivatives through order n + 1 are continuous throughout an open rectangular region R centered at a point (a, b). Then, throughout R, 1 2 sh ƒxx + 2hkƒxy + k 2ƒyy d ƒ sa,bd 2! n 0 0 1 + 3hk 2ƒxyy + k 3ƒyyy d ƒ sa,bd + Á + ah + k b ƒ` 0x 0y n! sa,bd
ƒsa + h, b + kd = ƒsa, bd + shƒx + kƒy d ƒ sa,bd + 1 3 sh ƒxxx + 3h 2kƒxxy 3!
+
0 0 1 ah + k b 0x 0y sn + 1d!
n+1
ƒ`
dH
+
.
(7)
sa + ch,b + ckd
ham
ma
The first n derivative terms are evaluated at (a, b). The last term is evaluated at some point sa + ch, b + ckd on the line segment joining (a, b) and sa + h, b + kd. If sa, bd = s0, 0d and we treat h and k as independent variables (denoting them now by x and y), then Equation (7) assumes the following simpler form.
Taylor’s Formula for ƒ(x, y) at the Origin
Mu
ƒsx, yd = ƒs0, 0d + xƒx + yƒy +
1 2 sx ƒxx + 2xyƒxy + y 2ƒyy d 2! n
+
0 0 1 3 1 + y b ƒ sx ƒxxx + 3x 2yƒxxy + 3xy 2ƒxyy + y 3ƒyyy d + Á + ax 0x 0y 3! n!
+
0 0 1 + y b ax 0x 0y sn + 1d!
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n+1
ƒ`
(8) scx,cyd
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Chapter 14: Partial Derivatives
EXAMPLE 1
You
suf
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The first n derivative terms are evaluated at (0, 0). The last term is evaluated at a point on the line segment joining the origin and (x, y). Taylor’s formula provides polynomial approximations of twovariable functions. The first n derivative terms give the polynomial; the last term gives the approximation error. The first three terms of Taylor’s formula give the function’s linearization. To improve on the linearization, we add higher power terms.
Finding a Quadratic Approximation
Find a quadratic approximation to ƒsx, yd = sin x sin y near the origin. How accurate is the approximation if ƒ x ƒ … 0.1 and ƒ y ƒ … 0.1? We take n = 2 in Equation (8):
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Solution
ƒsx, yd = ƒs0, 0d + sxƒx + yƒy d +
1 3 sx ƒxxx + 3x 2yƒxxy + 3xy 2ƒxyy + y 3ƒyyy dscx,cyd 6
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+
1 2 sx ƒxx + 2xyƒxy + y 2ƒyy d 2
with
ƒxxs0, 0d = sin x sin y ƒ s0,0d = 0,
ƒxs0, 0d = cos x sin y ƒ s0,0d = 0,
ƒxys0, 0d = cos x cos y ƒ s0,0d = 1,
ƒys0, 0d = sin x cos y ƒ s0,0d = 0,
ƒyys0, 0d = sin x sin y ƒ s0,0d = 0,
ass a
we have
ƒs0, 0d = sin x sin y ƒ s0,0d = 0,
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sin x sin y L 0 + 0 + 0 +
1 2 sx s0d + 2xys1d + y 2s0dd, 2
sin x sin y L xy.
The error in the approximation is
ma
Esx, yd =
1 3 sx ƒxxx + 3x 2yƒxxy + 3xy 2ƒxyy + y 3ƒyyy d ƒ scx,cyd . 6
8 1 3 3 3 3 3 ƒ Esx, yd ƒ … 6 ss0.1d + 3s0.1d + 3s0.1d + s0.1d d = 6 s0.1d … 0.00134
(rounded up). The error will not exceed 0.00134 if ƒ x ƒ … 0.1 and ƒ y ƒ … 0.1.
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The third derivatives never exceed 1 in absolute value because they are products of sines and cosines. Also, ƒ x ƒ … 0.1 and ƒ y ƒ … 0.1. Hence
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1058
Chapter 14: Partial Derivatives
Y z a i R n a s as
EXERCISES 14.10 Finding Quadratic and Cubic Approximations
H d a
In Exercises 1–10, use Taylor’s formula for ƒ(x, y) at the origin to find quadratic and cubic approximations of ƒ near the origin.
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1. ƒsx, yd = xe y
M
2. ƒsx, yd = e x cos y
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3. ƒsx, yd = y sin x x
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4. ƒsx, yd = sin x cos y
5. ƒsx, yd = e ln s1 + yd
6. ƒsx, yd = ln s2x + y + 1d
7. ƒsx, yd = sin sx 2 + y 2 d
8. ƒsx, yd = cos sx 2 + y 2 d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 14.10 Taylor’s Formula for Two Variables
10. ƒsx, yd =
1 1  x  y + xy
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11. Use Taylor’s formula to find a quadratic approximation of ƒsx, yd = cos x cos y at the origin. Estimate the error in the approximation if ƒ x ƒ … 0.1 and ƒ y ƒ … 0.1.
12. Use Taylor’s formula to find a quadratic approximation of e x sin y at the origin. Estimate the error in the approximation if ƒ x ƒ … 0.1 and ƒ y ƒ … 0.1.
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1 1  x  y
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9. ƒsx, yd =
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14 Questions to Guide Your Review
3. How can you display the values of a function ƒ(x, y) of two independent variables graphically? How do you do the same for a function ƒ(x, y, z) of three independent variables? 4. What does it mean for a function ƒ(x, y) to have limit L as sx, yd : sx0 , y0 d ? What are the basic properties of limits of functions of two independent variables? 5. When is a function of two (three) independent variables continuous at a point in its domain? Give examples of functions that are continuous at some points but not others. 6. What can be said about algebraic combinations and composites of continuous functions? 7. Explain the twopath test for nonexistence of limits.
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15. What is the gradient vector of a differentiable function ƒ(x, y)? How is it related to the function’s directional derivatives? State the analogous results for functions of three independent variables. 16. How do you find the tangent line at a point on a level curve of a differentiable function ƒ(x, y)? How do you find the tangent plane and normal line at a point on a level surface of a differentiable function ƒ(x, y, z)? Give examples. 17. How can you use directional derivatives to estimate change? 18. How do you linearize a function ƒ(x, y) of two independent variables at a point sx0 , y0 d ? Why might you want to do this? How do you linearize a function of three independent variables? 19. What can you say about the accuracy of linear approximations of functions of two (three) independent variables? 20. If (x, y) moves from sx0 , y0 d to a point sx0 + dx, y0 + dyd nearby, how can you estimate the resulting change in the value of a differentiable function ƒ(x, y)? Give an example.
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8. How are the partial derivatives 0ƒ>0x and 0ƒ>0y of a function ƒ(x, y) defined? How are they interpreted and calculated?
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2. What does it mean for sets in the plane or in space to be open? Closed? Give examples. Give examples of sets that are neither open nor closed.
14. What is the derivative of a function ƒ(x, y) at a point P0 in the direction of a unit vector u? What rate does it describe? What geometric interpretation does it have? Give examples.
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1. What is a realvalued function of two independent variables? Three independent variables? Give examples.
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Questions to Guide Your Review
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Chapter 14
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9. How does the relation between first partial derivatives and continuity of functions of two independent variables differ from the relation between first derivatives and continuity for realvalued functions of a single independent variable? Give an example.
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10. What is the Mixed Derivative Theorem for mixed secondorder partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples. 11. What does it mean for a function ƒ(x, y) to be differentiable? What does the Increment Theorem say about differentiability?
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12. How can you sometimes decide from examining ƒx and ƒy that a function ƒ(x, y) is differentiable? What is the relation between the differentiability of ƒ and the continuity of ƒ at a point?
22. What derivative tests are available for determining the local extreme values of a function ƒ(x, y)? How do they enable you to narrow your search for these values? Give examples. 23. How do you find the extrema of a continuous function ƒ(x, y) on a closed bounded region of the xyplane? Give an example. 24. Describe the method of Lagrange multipliers and give examples. 25. If w = ƒsx, y, zd , where the variables x, y, and z are constrained by an equation gsx, y, zd = 0 , what is the meaning of the notation s0w>0xdy ? How can an arrow diagram help you calculate this partial derivative with constrained variables? Give examples.
26. How does Taylor’s formula for a function ƒ(x, y) generate polynomial approximations and error estimates?
Mu
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13. What is the Chain Rule? What form does it take for functions of two independent variables? Three independent variables? Functions defined on surfaces? How do you diagram these different forms? Give examples. What pattern enables one to remember all the different forms?
21. How do you define local maxima, local minima, and saddle points for a differentiable function ƒ(x, y)? Give examples.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1060
Chapter 14: Partial Derivatives
Chapter 14
Domain, Range, and Level Curves
23. Psn, R, T, Vd =
In Exercises 1–4, find the domain and range of the given function and identify its level curves. Sketch a typical level curve.
24. ƒsr, l, T, wd =
2. ƒsx, yd = e x + y
3. gsx, yd = 1>xy
4. gsx, yd = 2x 2  y
SecondOrder Partials
In Exercises 5–8, find the domain and range of the given function and identify its level surfaces. Sketch a typical level surface. 6. gsx, y, zd = x 2 + 4y 2 + 9z 2
1 7. hsx, y, zd = 2 x + y2 + z2
28. ƒsx, yd = y 2  3xy + cos y + 7e y
2
Chain Rule Calculations
29. Find dw> dt at t = 0 if w = sin sxy + pd, x = e t , and y = ln st + 1d.
Evaluating Limits
11. 13.
lim
sx,yd:sp, ln 2d
e y cos x
x  y
lim
sx,yd:s1,1d
lim
2 + y x + cos y sx,yd: s0,0d x 3y 3  1 lim 12. sx,yd: s1,1d xy  1 lim
10.
x2  y2 ln ƒ x + y + z ƒ
14.
P:s1, 1, ed
lim
P: s1,1,1d
tan1 sx + y + zd
xy Z 0
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By considering different paths of approach, show that the limits in Exercises 15 and 16 do not exist. y x2 + y2 15. sx,ydlim 16. sx,ydlim xy : s0,0d 2 : s0,0d x  y y Z x2
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17. Continuous extension Let ƒsx, yd = sx 2  y 2 d>sx 2 + y 2 d for sx, yd Z s0, 0d . Is it possible to define ƒ(0, 0) in a way that makes ƒ continuous at the origin? Why? 18. Continuous extension Let
ƒxƒ + ƒyƒ Z 0
sx, yd = s0, 0d.
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sin sx  yd , ƒxƒ + ƒyƒ ƒsx, yd = L 0,
Is ƒ continuous at the origin? Why?
Partial Derivatives
In Exercises 19–24, find the partial derivative of the function with respect to each variable.
Mu
19. gsr, ud = r cos u + r sin u 20. ƒsx, yd =
30. Find dw> dt at t = 1 if w = xe y + y sin z  cos z, x = 22t, y = t  1 + ln t , and z = pt.
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Find the limits in Exercises 9–14. 9.
26. gsx, yd = e x + y sin x
27. ƒsx, yd = x + xy  5x 3 + ln sx 2 + 1d
1 x + y + z2 + 1 2
x 25. gsx, yd = y + y
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8. ksx, y, zd =
Find the secondorder partial derivatives of the functions in Exercises 25–28.
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5. ƒsx, y, zd = x 2 + y 2  z
T 1 2rl A pw
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1. ƒsx, yd = 9x 2 + y 2
nRT (the ideal gas law) V
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Practice Exercises
y 1 ln sx 2 + y 2 d + tan1 x 2
1 1 1 + + 21. ƒsR1, R2 , R3 d = R1 R2 R3
31. Find 0w>0r and 0w>0s when r = p and s = 0 if w = sin s2x  yd, x = r + sin s, y = rs. 32. Find
and
0w>0u
ln 21 + x  tan 2
1
u = y = 0
when
0w>0y
w =
x and x = 2e cos y.
33. Find the value of the derivative of ƒsx, y, zd = xy + yz + xz with respect to t on the curve x = cos t, y = sin t, z = cos 2t at t = 1 . 34. Show that if w = ƒssd is any differentiable function of s and if s = y + 5x , then 0w 0w  5 = 0. 0x 0y
Implicit Differentiation Assuming that the equations in Exercises 35 and 36 define y as a differentiable function of x, find the value of dy> dx at point P.
35. 1  x  y 2  sin xy = 0, 36. 2xy + e
x+y
 2 = 0,
Ps0, 1d
Ps0, ln 2d
Directional Derivatives In Exercises 37–40, find the directions in which ƒ increases and decreases most rapidly at P0 and find the derivative of ƒ in each direction. Also, find the derivative of ƒ at P0 in the direction of the vector v. 37. ƒsx, yd = cos x cos y, 2 2y
38. ƒsx, yd = x e
,
P0sp>4, p>4d,
P0s1, 0d,
v = 3i + 4j
v = i + j
39. ƒsx, y, zd = ln s2x + 3y + 6zd,
P0s 1, 1, 1d,
v = 2i + 3j + 6k
22. hsx, y, zd = sin s2px + y  3zd
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14
Tangent Lines to Curves
P0s0, 0, 0d,
41. Derivative in velocity direction Find the derivative of ƒsx, y, zd = xyz in the direction of the velocity vector of the helix
b. Find the derivative of ƒ at (1, 2) in the direction toward the point (4, 6).
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In Exercises 45 and 46, sketch the surface ƒsx, y, zd = c together with §ƒ at the given points. s2, 0, ;2d
ma
2
2
P0s2, 1, 1d P0s1, 1, 2d
ham
48. x + y + z = 4,
In Exercises 49 and 50, find an equation for the plane tangent to the surface z = ƒsx, yd at the given point. 49. z = ln sx 2 + y 2 d,
s0, 1, 0d
50. z = 1>sx 2 + y 2 d,
s1, 1, 1>2d
Mu
In Exercises 51 and 52, find equations for the lines that are tangent and normal to the level curve ƒsx, yd = c at the point P0 . Then sketch the lines and level curve together with §ƒ at P0 . 51. y  sin x = 1,
P0sp, 1d
52.
y2 3 x2 = , 2 2 2
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`y 
56. ƒsx, yd = xy  3y 2 + 2, ƒ x  1 ƒ … 0.1,
p ` … 0.1 4
P0s1, 1d
ƒ y  1 ƒ … 0.2
57. ƒsx, y, zd = xy + 2yz  3xz at (1, 0, 0) and (1, 1, 0) 58. ƒsx, y, zd = 22 cos x sin s y + zd at s0, 0, p>4d and sp>4, p>4, 0d
P0s1, 2d
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59. Measuring the volume of a pipeline You plan to calculate the volume inside a stretch of pipeline that is about 36 in. in diameter and 1 mile long. With which measurement should you be more careful, the length or the diameter? Why? 60. Sensitivity to change Near the point (1, 2), is ƒsx, yd = x 2  xy + y 2  3 more sensitive to changes in x or to changes in y? How do you know?
s0, 0, 0d
In Exercises 47 and 48, find an equation for the plane tangent to the level surface ƒsx, y, zd = c at the point P0 . Also, find parametric equations for the line that is normal to the surface at P0 . 47. x 2  y  5z = 0,
P0sp>4, p>4d
Estimates and Sensitivity to Change
Gradients, Tangent Planes, and Normal Lines
s2, ;2, 0d,
p ` x  ` … 0.1, 4
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c. The directional derivative of ƒ at sx0 , y0 d has its greatest value in the direction of §ƒ .
46. y + z = 4;
55. ƒsx, yd = sin x cos y,
Find the linearizations of the functions in Exercises 57 and 58 at the given points.
b. The derivative of ƒ at sx0 , y0 d in the direction of u is a vector.
s0, 1, ;1d,
In Exercises 55 and 56, find the linearization L(x, y) of the function ƒ(x, y) at the point P0 . Then find an upper bound for the magnitude of the error E in the approximation ƒsx, yd L Lsx, yd over the rectangle R.
R:
a. If u is a unit vector, the derivative of ƒ at sx0 , y0 d in the direction of u is sƒxsx0 , y0 di + ƒysx0 , y0 djd # u.
(1> 2, 1, 1> 2)
Linearizations
R:
44. Which of the following statements are true if ƒ(x, y) is differentiable at sx0 , y0 d ? Give reasons for your answers.
y = 1
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43. Directional derivatives with given values At the point (1, 2), the function ƒ(x, y) has a derivative of 2 in the direction toward (2, 2) and a derivative of 2 in the direction toward (1, 1). a. Find ƒxs1, 2d and ƒys1, 2d .
Point:
y = 1
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42. Maximum directional derivative What is the largest value that the directional derivative of ƒsx, y, zd = xyz can have at the point (1, 1, 1)?
2
(1, 1, 1> 2)
54. Surfaces: x + y 2 + z = 2,
at t = p>3.
45. x 2 + y + z 2 = 0;
53. Surfaces: x 2 + 2y + 2z = 4, Point:
rstd = scos 3tdi + ssin 3tdj + 3t k
d. At sx0 , y0 d , vector §ƒ is normal to the curve ƒsx, yd = ƒsx0 , y0 d.
In Exercises 53 and 54, find parametric equations for the line that is tangent to the curve of intersection of the surfaces at the given point.
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v = i + j + k
2
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40. ƒsx, y, zd = x 2 + 3xy  z 2 + 2y + z + 4,
Practice Exercises
61. Change in an electrical circuit Suppose that the current I (amperes) in an electrical circuit is related to the voltage V (volts) and the resistance R (ohms) by the equation I = V>R . If the voltage drops from 24 to 23 volts and the resistance drops from 100 to 80 ohms, will I increase or decrease? By about how much? Is the change in I more sensitive to change in the voltage or to change in the resistance? How do you know? 62. Maximum error in estimating the area of an ellipse If a = 10 cm and b = 16 cm to the nearest millimeter, what should you expect the maximum percentage error to be in the calculated area A = pab of the ellipse x 2>a 2 + y 2>b 2 = 1 ?
63. Error in estimating a product Let y = uy and z = u + y, where u and y are positive independent variables. a. If u is measured with an error of 2% and y with an error of 3%, about what is the percentage error in the calculated value of y ?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
C =
cardiac output . body surface area
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64. Cardiac index To make different people comparable in studies of cardiac output (Section 3.7, Exercise 25), researchers divide the measured cardiac output by the body surface area to find the cardiac index C:
75. ƒsx, yd = x 2  y 2  2x + 4y R: The triangular region bounded below by the xaxis, above by the line y = x + 2, and on the right by the line x = 2 76. ƒsx, yd = 4xy  x 4  y 4 + 16
R: The triangular region bounded below by the line y = 2, above by the line y = x, and on the right by the line x = 2 77. ƒsx, yd = x 3 + y 3 + 3x 2  3y 2
R: The square region enclosed by the lines x = ;1 and y = ;1
The body surface area B of a person with weight w and height h is approximated by the formula
78. ƒsx, yd = x 3 + 3xy + y 3 + 1
R: The square region enclosed by the lines x = ;1 and y = ;1
B = 71.84w 0.425h 0.725 ,
79. Extrema on a circle Find the extreme values of ƒsx, yd = x 3 + y 2 on the circle x 2 + y 2 = 1.
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80. Extrema on a circle Find the extreme values of ƒsx, yd = xy on the circle x 2 + y 2 = 1.
7 L>min 70 kg 180 cm
81. Extrema in a disk Find the extreme values of ƒsx, yd = x 2 + 3y 2 + 2y on the unit disk x 2 + y 2 … 1.
Which will have a greater effect on the calculation, a 1kg error in measuring the weight or a 1cm error in measuring the height?
82. Extrema in a disk Find the extreme values of ƒsx, yd = x 2 + y 2  3x  xy on the disk x 2 + y 2 … 9.
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83. Extrema on a sphere Find the extreme values of ƒsx, y, zd = x  y + z on the unit sphere x 2 + y 2 + z 2 = 1.
Local Extrema
Test the functions in Exercises 65–70 for local maxima and minima and saddle points. Find each function’s value at these points. 66. ƒsx, yd = 5x 2 + 4xy  2y 2 + 4x  4y 67. ƒsx, yd = 2x 3 + 3xy + 2y 3 68. ƒsx, yd = x 3 + y 3  3xy + 15 69. ƒsx, yd = x 3 + y 3 + 3x 2  3y 2
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70. ƒsx, yd = x 4  8x 2 + 3y 2  6y
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65. ƒsx, yd = x 2  xy + y 2 + 2x + 2y  4
Absolute Extrema
Lagrange Multipliers
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which gives B in square centimeters when w is measured in kilograms and h in centimeters. You are about to calculate the cardiac index of a person with the following measurements: Cardiac output: Weight: Height:
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b. Show that the percentage error in the calculated value of z is less than the percentage error in the value of y.
You
1062
In Exercises 71–78, find the absolute maximum and minimum values of ƒ on the region R. 71. ƒsx, yd = x 2 + xy + y 2  3x + 3y
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R: The triangular region cut from the first quadrant by the line x + y = 4 72. ƒsx, yd = x 2  y 2  2x + 4y + 1
R: The rectangular region in the first quadrant bounded by the coordinate axes and the lines x = 4 and y = 2
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73. ƒsx, yd = y 2  xy  3y + 2x
R: The square region enclosed by the lines x = ;2 and y = ;2
74. ƒsx, yd = 2x + 2y  x 2  y 2 R: The square region bounded by the coordinate axes and the lines x = 2, y = 2 in the first quadrant
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84. Minimum distance to origin Find the points on the surface z 2  xy = 4 closest to the origin.
85. Minimizing cost of a box A closed rectangular box is to have volume V cm3. The cost of the material used in the box is a cents>cm2 for top and bottom, b cents>cm2 for front and back, and c cents>cm2 for the remaining sides. What dimensions minimize the total cost of materials?
86. Least volume Find the plane x>a + y>b + z>c = 1 that passes through the point (2, 1, 2) and cuts off the least volume from the first octant. 87. Extrema on curve of intersecting surfaces Find the extreme values of ƒsx, y, zd = xs y + zd on the curve of intersection of the right circular cylinder x 2 + y 2 = 1 and the hyperbolic cylinder xz = 1. 88. Minimum distance to origin on curve of intersecting plane and cone Find the point closest to the origin on the curve of intersection of the plane x + y + z = 1 and the cone z 2 = 2x 2 + 2y 2.
Partial Derivatives with Constrained Variables In Exercises 89 and 90, begin by drawing a diagram that shows the relations among the variables. 89. If w = x 2e yz and z = x 2  y 2 find a. a
0w b 0y z
b. a
0w b 0z x
c. a
0w b . 0z y
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
0U b . 0V T
s y + zd2 + sz  xd2 = 16
where the normal line is parallel to the yzplane.
98. Tangent plane parallel to xyplane Find the points on the surface
91. Let w = ƒsr, ud, r = 2x 2 + y 2 , and u = tan1 sy>xd . Find 0w>0x and 0w>0y and express your answers in terms of r and u . 92. Let z = ƒsu, yd, u = ax + by, and y = ax  by . Express zx and zy in terms of fu , fy , and the constants a and b. 93. If a and b are constants, w = u 3 + tanh u + cos u, and u = ax + by, show that 0w 0w a = b . 0y 0x
xy + yz + zx  x  z 2 = 0
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Theory and Examples
where the tangent plane is parallel to the xyplane. 99. When gradient is parallel to position vector Suppose that §ƒsx, y, zd is always parallel to the position vector xi + yj + zk. Show that ƒs0, 0, ad = ƒs0, 0, ad for any a. 100. Directional derivative in all directions, but no gradient Show that the directional derivative of
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b. a
0U b 0T P
Find the points on the surface
ƒsx, y, zd = 2x 2 + y 2 + z 2
94. Using the Chain Rule If w = ln sx 2 + y 2 + 2zd, x = r + s, y = r  s, and z = 2rs, find wr and ws by the Chain Rule. Then check your answer another way. 95. Angle between vectors The equations e u cos y  x = 0 and e u sin y  y = 0 define u and y as differentiable functions of x and y. Show that the angle between the vectors
at the origin equals 1 in any direction but that ƒ has no gradient vector at the origin.
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a. a
97. Normal line parallel to a plane
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90. Let U = ƒsP, V, T d be the internal energy of a gas that obeys the ideal gas law PV = nRT (n and R constant). Find
Practice Exercises
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Chapter 14
101. Normal line through origin Show that the line normal to the surface xy + z = 2 at the point (1, 1, 1) passes through the origin. 102. Tangent plane and normal line
and
0y 0y i + j 0x 0y
is constant.
ass a
a. Sketch the surface x 2  y 2 + z 2 = 4 .
0u 0u i + j 0x 0y
b. Find a vector normal to the surface at s2, 3, 3d. Add the vector to your sketch. c. Find equations for the tangent plane and normal line at s2, 3, 3d.
dH
96. Polar coordinates and second derivatives Introducing polar coordinates x = r cos u and y = r sin u changes ƒ(x, y) to gsr, ud. Find the value of 0 2g>0u2 at the point sr, ud = s2, p>2d, given that 0ƒ 0 2ƒ 0ƒ 0 2ƒ = = 2 = 2 = 1 0x 0y 0x 0y
Mu
ham
ma
at that point.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14
Chapter 14
Additional and Advanced Exercises
Partial Derivatives
2
ƒsx, yd = L
xy
x  y
2
d a m
x2 + y2
0,
m a h
Mu
,
sx, yd Z s0, 0d sx, yd = s0, 0d
a i R n
To Read it Online & Download:
u o zY
1063
(see the accompanying figure) is continuous at (0, 0). Find ƒxys0, 0d and ƒyxs0, 0d.
a s s a H
1. Function with saddle at the origin If you did Exercise 50 in Section 14.2, you know that the function
i f su
Additional and Advanced Exercises
z
y x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
dy du d ƒstd dt = ƒsysxdd  ƒsusxdd . dx Lusxd dx dx
y
u = usxd,
y = ysxd
i
8. Gradient orthogonal to tangent Suppose that a differentiable function ƒ(x, y) has the constant value c along the differentiable curve x = gstd, y = hstd; that is ƒsgstd, hstdd = c
and calculating dg> dx with the Chain Rule.
4. Finding a function with constrained second partials Suppose that ƒ is a twicedifferentiable function of r, that r = 2x 2 + y 2 + z 2 , and that ƒxx + ƒyy + ƒzz = 0.
iaz
ƒstd dt,
c. Find a function whose gradient equals r.
e. Show that §sA # rd = A for any constant vector A.
Prove the rule by setting Lu
b. Show that §sr n d = nr n  2r. d. Show that r # dr = r dr.
ysxd
gsu, yd =
7. Properties of position vectors Let r = xi + yj + zk and let r = ƒ rƒ. a. Show that §r = r>r.
suf
3. A proof of Leibniz’s Rule Leibniz’s Rule says that if ƒ is continuous on [a, b] and if u(x) and y(x) are differentiable functions of x whose values lie in [a, b], then
Gradients and Tangents
You
2. Finding a function from second partials Find a function w = ƒsx, yd whose first partial derivatives are 0w>0x = 1 + e x cos y and 0w>0y = 2y  e x sin y and whose value at the point (ln 2, 0) is ln 2.
for all values of t. Differentiate both sides of this equation with respect to t to show that §ƒ is orthogonal to the curve’s tangent vector at every point on the curve.
nR
1064
9. Curve tangent to a surface
Show that the curve
rstd = sln tdi + st ln tdj + t k
Show that for some constants a and b,
is tangent to the surface
a ƒsrd = r + b.
ass a
xz 2  yz + cos xy = 1
at (0, 0, 1).
0 2ƒ 0 2ƒ b + y 2 a 2 b = nsn  1dƒ. 0x0y 0x 0y 6. Surface in polar coordinates Let 2
b + 2xy a
ƒsr, ud = L
ma
b. x 2 a
0 2ƒ
dH
5. Homogeneous functions A function ƒ(x, y) is homogeneous of degree n (n a nonnegative integer) if ƒstx, tyd = t nƒsx, yd for all t, x, and y. For such a function (sufficiently differentiable), prove that 0ƒ 0ƒ a. x + y = nƒsx, yd 0x 0y
sin 6r , 6r
r Z 0
1,
r = 0,
ham
where r and u are polar coordinates. Find a. lim ƒsr, ud r:0
b. ƒrs0, 0d
c. ƒusr, ud,
r Z 0.
Mu
z f (r, )
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10. Curve tangent to a surface rstd = a
Show that the curve
t3 4  2bi + a t  3 bj + cos st  2dk 4
is tangent to the surface x 3 + y 3 + z 3  xyz = 0 at s0, 1, 1d.
Extreme Values 11. Extrema on a surface Show that the only possible maxima and minima of z on the surface z = x 3 + y 3  9xy + 27 occur at (0, 0) and (3, 3). Show that neither a maximum nor a minimum occurs at (0, 0). Determine whether z has a maximum or a minimum at (3, 3). 12. Maximum in closed first quadrant Find the maximum value of ƒsx, yd = 6xye s2x + 3yd in the closed first quadrant (includes the nonnegative axes). 13. Minimum volume cut from first octant Find the minimum volume for a region bounded by the planes x = 0, y = 0, z = 0 and a plane tangent to the ellipsoid y2 x2 z2 + 2 + 2 = 1 2 a b c at a point in the first octant.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14
16. Suppose that rstd = gstdi + hstdj + kstdk is a smooth curve in the domain of a differentiable function ƒ(x, y, z). Describe the relation between dƒ> dt, §ƒ , and v = dr>dt. What can be said about §ƒ and v at interior points of the curve where ƒ has extreme values relative to its other values on the curve? Give reasons for your answer. 17. Finding functions from partial derivatives Suppose that ƒ and g are functions of x and y such that 0ƒ 0g = , 0x 0y
and
and suppose that 0ƒ = 0, 0x
ƒs1, 2d = gs1, 2d = 5
Find ƒ(x, y) and g(x, y).
and
i
suf
The OneDimensional Heat Equation If w(x, t) represents the temperature at position x at time t in a uniform conducting rod with perfectly insulated sides (see the accompanying figure), then the partial derivatives wxx and wt satisfy a differential equation of the form 1 wxx = 2 wt . c This equation is called the onedimensional heat equation. The value of the positive constant c 2 is determined by the material from which the rod is made. It has been determined experimentally for a broad range of materials. For a given application, one finds the appropriate value in a table. For dry soil, for example, c 2 = 0.19 ft2>day.
ass a
0ƒ 0g = 0y 0x
22. Drilling another borehole On a flat surface of land, geologists drilled a borehole straight down and hit a mineral deposit at 1000 ft. They drilled a second borehole 100 ft to the north of the first and hit the mineral deposit at 950 ft. A third borehole 100 ft east of the first borehole struck the mineral deposit at 1025 ft. The geologists have reasons to believe that the mineral deposit is in the shape of a dome, and for the sake of economy, they would like to find where the deposit is closest to the surface. Assuming the surface to be the xyplane, in what direction from the first borehole would you suggest the geologists drill their fourth borehole?
You
15. Boundedness of first partials implies continuity Prove the following theorem: If ƒ(x, y) is defined in an open region R of the xyplane and if ƒx and ƒy are bounded on R, then ƒ(x, y) is continuous on R. (The assumption of boundedness is essential.)
b. Which direction tangential to S at the point (1, 1, 8) will make the rate of change of temperature a maximum?
iaz
Theory and Examples
ƒs0, 0d = 4 .
x0
dH
18. Rate of change of the rate of change We know that if ƒ(x, y) is a function of two variables and if u = ai + bj is a unit vector, then Du ƒsx, yd = ƒxsx, yda + ƒysx, ydb is the rate of change of ƒ(x, y) at (x, y) in the direction of u. Give a similar formula for the rate of change of the rate of change of ƒ(x, y) at (x, y) in the direction u.
ma
ham
20. Velocity after a ricochet A particle traveling in a straight line with constant velocity i + j  5k passes through the point (0, 0, 30) and hits the surface z = 2x 2 + 3y 2 . The particle ricochets off the surface, the angle of reflection being equal to the angle of incidence. Assuming no loss of speed, what is the velocity of the particle after the ricochet? Simplify your answer.
Mu
21. Directional derivatives tangent to a surface Let S be the surface that is the graph of ƒsx, yd = 10  x 2  y 2 . Suppose that the temperature in space at each point (x, y, z) is Tsx, y, zd = x 2y + y 2z + 4x + 14y + z.
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w(x, t) is the temperature here at time t.
x
19. Path of a heatseeking particle A heatseeking particle has the property that at any point (x, y) in the plane it moves in the direction of maximum temperature increase. If the temperature at (x, y) is Tsx, yd = e 2y cos x, find an equation y = ƒsxd for the path of a heatseeking particle at the point sp>4, 0d.
a. Among all the possible directions tangential to the surface S at the point (0, 0, 10), which direction will make the rate of change of temperature at (0, 0, 10) a maximum?
1065
nR
14. Minimum distance from line to parabola in xyplane By minimizing the function ƒsx, y, u, yd = sx  ud2 + sy  yd2 subject to the constraints y = x + 1 and u = y 2, find the minimum distance in the xyplane from the line y = x + 1 to the parabola y 2 = x.
Additional and Advanced Exercises
x
In chemistry and biochemistry, the heat equation is known as the diffusion equation. In this context, w(x, t) represents the concentration of a dissolved substance, a salt for instance, diffusing along a tube filled with liquid. The value of w(x, t) is the concentration at point x at time t. In other applications, w(x, t) represents the diffusion of a gas down a long, thin pipe. In electrical engineering, the heat equation appears in the forms yxx = RCyt and ixx = RCit .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
ysx, td = voltage at point x at time t
23. Find all solutions of the onedimensional heat equation of the form w = e rt sin px, where r is a constant.
i
These equations describe the voltage y and the flow of current i in a coaxial cable or in any other cable in which leakage and inductance are negligible. The functions and constants in these equations are
24. Find all solutions of the onedimensional heat equation that have the form w = e rt sin k x and satisfy the conditions that ws0, td = 0 and wsL, td = 0. What happens to these solutions as t : q ?
suf
1066
R = resistance per unit length
You
C = capacitance to ground per unit of cable length
Mu
ham
ma
dH
ass a
nR
iaz
isx, td = current at point x at time t.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 14: Partial Derivatives
Technology Application Projects
i
Chapter 14
suf
1066
Mathematica / Maple Module
You
Plotting Surfaces Efficiently generate plots of surfaces, contours, and level curves.
Mathematica / Maple Module
Exploring the Mathematics Behind Skateboarding: Analysis of the Directional Derivative The path of a skateboarder is introduced, first on a level plane, then on a ramp, and finally on a paraboloid. Compute, plot, and analyze the directional derivative in terms of the skateboarder.
Mathematica / Maple Module
iaz
Looking for Patterns and Applying the Method of Least Squares to Real Data Fit a line to a set of numerical data points by choosing the line that minimizes the sum of the squares of the vertical distances from the points to the line.
Mathematica / Maple Module
Mu
ham
ma
dH
ass a
nR
Lagrange Goes Skateboarding: How High Does He Go? Revisit and analyze the skateboardersâ€™ adventures for maximum and minimum heights from both a graphical and analytic perspective using Lagrange multipliers.
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suf
i
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
15
You
Chapter
MULTIPLE INTEGRALS
ass a
Double Integrals
In Chapter 5 we defined the definite integral of a continuous function ƒ(x) over an interval [a, b] as a limit of Riemann sums. In this section we extend this idea to define the integral of a continuous function of two variables ƒ(x, y) over a bounded region R in the plane. In both cases the integrals are limits of approximating Riemann sums. The Riemann sums for the integral of a singlevariable function ƒ(x) are obtained by partitioning a finite interval into thin subintervals, multiplying the width of each subinterval by the value of ƒ at a point ck inside that subinterval, and then adding together all the products. A similar method of partitioning, multiplying, and summing is used to construct double integrals. However, this time we pack a planar region R with small rectangles, rather than small subintervals. We then take the product of each small rectangle’s area with the value of ƒ at a point inside that rectangle, and finally sum together all these products. When ƒ is continuous, these sums converge to a single number as each of the small rectangles shrinks in both width and height. The limit is the double integral of ƒ over R. As with single integrals, we can evaluate multiple integrals via antiderivatives, which frees us from the formidable task of calculating a double integral directly from its definition as a limit of Riemann sums. The major practical problem that arises in evaluating multiple integrals lies in determining the limits of integration. While the integrals of Chapter 5 were evaluated over an interval, which is determined by its two endpoints, multiple integrals are evaluated over a region in the plane or in space. This gives rise to limits of integration which often involve variables, not just constants. Describing the regions of integration is the main new issue that arises in the calculation of multiple integrals.
Mu
ham
ma
dH
15.1
nR
iaz
OVERVIEW In this chapter we consider the integral of a function of two variables ƒ(x, y) over a region in the plane and the integral of a function of three variables ƒ(x, y, z) over a region in space. These integrals are called multiple integrals and are defined as the limit of approximating Riemann sums, much like the singlevariable integrals presented in Chapter 5. We can use multiple integrals to calculate quantities that vary over two or three dimensions, such as the total mass or the angular momentum of an object of varying density and the volumes of solids with general curved boundaries.
Double Integrals over Rectangles We begin our investigation of double integrals by considering the simplest type of planar region, a rectangle. We consider a function ƒ(x, y) defined on a rectangular region R, R:
a … x … b,
c … y … d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: Multiple Integrals
y
We subdivide R into small rectangles using a network of lines parallel to the x and yaxes (Figure 15.1). The lines divide R into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece gets small. These rectangles form a partition of R. A small rectangular piece of width ¢x and height ¢y has area ¢A = ¢x¢y. If we number the small pieces partitioning R in some order, then their areas are given by numbers ¢A1, ¢A2 , Á , ¢An , where ¢Ak is the area of the kth small rectangle. To form a Riemann sum over R, we choose a point sxk , yk d in the kth small rectangle, multiply the value of ƒ at that point by the area ¢Ak, and add together the products:
R
suf
i
d Ak yk
(xk , yk ) xk
c 0
a
b
x
You
1068
n
Sn = a ƒsxk , yk d ¢Ak . k=1
iaz
Depending on how we pick sxk , yk d in the kth small rectangle, we may get different values for Sn. We are interested in what happens to these Riemann sums as the widths and heights of all the small rectangles in the partition of R approach zero. The norm of a partition P, written 7P7, is the largest width or height of any rectangle in the partition. If 7P7 = 0.1 then all the rectangles in the partition of R have width at most 0.1 and height at most 0.1. Sometimes the Riemann sums converge as the norm of P goes to zero, written 7P7 : 0. The resulting limit is then written as
nR
FIGURE 15.1 Rectangular grid partitioning the region R into small rectangles of area ¢Ak = ¢xk ¢yk.
n
lim a ƒsxk , yk d ¢Ak . ƒ ƒ P ƒ ƒ :0 k=1
ass a
As 7P7 : 0 and the rectangles get narrow and short, their number n increases, so we can also write this limit as n
lim a ƒsxk , yk d ¢Ak . n: q k=1
Mu
ham
ma
dH
with the understanding that ¢Ak : 0 as n : q and 7P7 : 0. There are many choices involved in a limit of this kind. The collection of small rectangles is determined by the grid of vertical and horizontal lines that determine a rectangular partition of R. In each of the resulting small rectangles there is a choice of an arbitrary point sxk , yk d at which ƒ is evaluated. These choices together determine a single Riemann sum. To form a limit, we repeat the whole process again and again, choosing partitions whose rectangle widths and heights both go to zero and whose number goes to infinity. When a limit of the sums Sn exists, giving the same limiting value no matter what choices are made, then the function ƒ is said to be integrable and the limit is called the double integral of ƒ over R, written as 6
ƒsx, yd dA
R
or
6
ƒsx, yd dx dy.
R
It can be shown that if ƒ(x, y) is a continuous function throughout R, then ƒ is integrable, as in the singlevariable case discussed in Chapter 5. Many discontinuous functions are also integrable, including functions which are discontinuous only on a finite number of points or smooth curves. We leave the proof of these facts to a more advanced text.
Double Integrals as Volumes When ƒ(x, y) is a positive function over a rectangular region R in the xyplane, we may interpret the double integral of ƒ over R as the volume of the 3dimensional solid region over the xyplane bounded below by R and above by the surface z = ƒsx, yd (Figure 15.2). Each term ƒsxk , yk d¢Ak in the sum Sn = g ƒsxk , yk d¢Ak is the volume of a vertical
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1 Double Integrals z
rectangular box that approximates the volume of the portion of the solid that stands directly above the base ¢Ak. The sum Sn thus approximates what we want to call the total volume of the solid. We define this volume to be Volume = lim Sn = n: q
f (x k , yk )
∆ Ak
(x k , yk )
You
R
where ¢Ak : 0 as n : q . As you might expect, this more general method of calculating volume agrees with the methods in Chapter 6, but we do not prove this here. Figure 15.3 shows Riemann sum approximations to the volume becoming more accurate as the number n of boxes increases.
iaz
FIGURE 15.2 Approximating solids with rectangular boxes leads us to define the volumes of more general solids as double integrals. The volume of the solid shown here is the double integral of ƒ(x, y) over the base region R.
(b) n 64
(c) n 256
ass a
(a) n 16
nR
b
ƒsx, yd dA,
R
d y
x
6
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i
z f (x, y)
a
1069
FIGURE 15.3 As n increases, the Riemann sum approximations approach the total volume of the solid shown in Figure 15.2.
Fubini’s Theorem for Calculating Double Integrals
dH
Suppose that we wish to calculate the volume under the plane z = 4  x  y over the rectangular region R: 0 … x … 2, 0 … y … 1 in the xyplane. If we apply the method of slicing from Section 6.1, with slices perpendicular to the xaxis (Figure 15.4), then the volume is
z
(1)
where A(x) is the crosssectional area at x. For each value of x, we may calculate A(x) as the integral
ham
z4xy
1
x
y
Mu
2
x
Asxd dx,
Lx = 0
ma
4
x=2
A(x) ⌠ ⌡
y1
y0
y=1
Asxd =
s4  x  yd dy,
Ly = 0
(2)
which is the area under the curve z = 4  x  y in the plane of the crosssection at x. In calculating A(x), x is held fixed and the integration takes place with respect to y. Combining Equations (1) and (2), we see that the volume of the entire solid is x=2
Volume =
(4 x y) dy
FIGURE 15.4 To obtain the crosssectional area A(x), we hold x fixed and integrate with respect to y.
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Lx = 0 x=2
=
Lx = 0
x=2
Asxd dx =
Lx = 0
c4y  xy 
a
y=1
Ly = 0
s4  x  yddyb dx
x=2 y 2 y=1 7 dx = a  xb dx d 2 y=0 2 Lx = 0
2
7 x2 = c x d = 5. 2 2 0
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(3)
4100 AWL/Thomas_ch15p10671142 8/25/04 2:57 PM Page 1070
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1070
Chapter 15: Multiple Integrals z
4
2
Volume =
z4xy
1 y
x=2 x2
A(y) ⌠ (4 x y) dx ⌡x 0
Lx = 0
s4  x  yd dx = c4x 
x=2
x2  xy d = 6  2y. 2 x=0
(4)
The volume of the entire solid is therefore y=1
Volume =
Ly = 0
y=1
As yd dy =
Ly = 0
s6  2yd dy = C 6y  y 2 D 0 = 5, 1
in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing
ass a
FIGURE 15.5 To obtain the crosssectional area A(y), we hold y fixed and integrate with respect to x.
As yd =
iaz
You
The expression on the right, called an iterated or repeated integral, says that the volume is obtained by integrating 4  x  y with respect to y from y = 0 to y = 1, holding x fixed, and then integrating the resulting expression in x with respect to x from x = 0 to x = 2. The limits of integration 0 and 1 are associated with y, so they are placed on the integral closest to dy. The other limits of integration, 0 and 2, are associated with the variable x, so they are placed on the outside integral symbol that is paired with dx. What would have happened if we had calculated the volume by slicing with planes perpendicular to the yaxis (Figure 15.5)? As a function of y, the typical crosssectional area is
2 x
s4  x  yd dy dx.
L0 L0
nR
y
1
suf
i
If we just wanted to write a formula for the volume, without carrying out any of the integrations, we could write
1
2
s4  x  yd dx dy. L0 L0 The expression on the right says we can find the volume by integrating 4  x  y with respect to x from x = 0 to x = 2 as in Equation (4) and integrating the result with respect to y from y = 0 to y = 1. In this iterated integral, the order of integration is first x and then y, the reverse of the order in Equation (3). What do these two volume calculations with iterated integrals have to do with the double integral
dH
Volume =
Mu
ham
Guido Fubini (1879–1943)
R
ma
HISTORICAL BIOGRAPHY
s4  x  yd dA 6
over the rectangle R: 0 … x … 2, 0 … y … 1? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integral measures the volume of the same region as the two iterated integrals. A theorem published in 1907 by Guido Fubini says that the double integral of any continuous function over a rectangle can be calculated as an iterated integral in either order of integration. (Fubini proved his theorem in greater generality, but this is what it says in our setting.)
THEOREM 1 Fubini’s Theorem (First Form) If ƒ(x, y) is continuous throughout the rectangular region R: a … x … b, c … y … d, then d
6
ƒsx, yd dA =
R
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Lc La
b
b
ƒsx, yd dx dy =
La Lc
d
ƒsx, yd dy dx.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1
Double Integrals
1071
EXAMPLE 1
Evaluating a Double Integral
Calculate 4R ƒsx, yd dA for ƒsx, yd = 1  6x 2y
1
0 … x … 2,
2
L1L0
s1  6x 2yd dx dy =
nR
6
ƒsx, yd dA =
R:
iaz
and
By Fubini’s Theorem,
Solution
You
suf
i
Fubini’s Theorem says that double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini’s Theorem also says that we may calculate the double integral by integrating in either order, a genuine convenience, as we see in Example 3. When we calculate a volume by slicing, we may use either planes perpendicular to the xaxis or planes perpendicular to the yaxis.
R
1
L1
L1
C x  2x 3y D xx == 20 dy
s2  16yd dy = C 2y  8y 2 D 1 = 4.
1
=
1 … y … 1.
1
ass a
Reversing the order of integration gives the same answer: 2
1
dH
L0 L1
y=1 C y  3x 2y 2 D y = 1 dx
2
s1  6x 2yd dy dx =
L0 2
=
[s1  3x 2 d  s 1  3x 2 d] dx
L0 2
=
ma
USING TECHNOLOGY
L0
2 dx = 4.
Multiple Integration
Mu
ham
Most CAS can calculate both multiple and iterated integrals. The typical procedure is to apply the CAS integrate command in nested iterations according to the order of integration you specify. Integral
Typical CAS Formulation int sint sx ¿ 2 * y, xd, yd ;
x 2y dx dy 6 p>4
Lp>3L0
1
x cos y dx dy
int sint sx * cos s yd, x = 0 . . 1d, y = Pi>3 . . Pi>4d;
If a CAS cannot produce an exact value for a definite integral, it can usually find an approximate value numerically. Setting up a multiple integral for a CAS to solve can be a highly nontrivial task, and requires an understanding of how to describe the boundaries of the region and set up an appropriate integral.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1072
Chapter 15: Multiple Integrals
i
Double Integrals over Bounded Nonrectangular Regions
suf
To define the double integral of a function ƒ(x, y) over a bounded, nonrectangular region R, such as the one in Figure 15.6, we again begin by covering R with a grid of small rectangular cells whose union contains all points of R. This time, however, we cannot exactly fill R with a finite number of rectangles lying inside R, since its boundary is curved, and some of the small rectangles in the grid lie partly outside R. A partition of R is formed by taking the rectangles that lie completely inside it, not using any that are either partly or completely outside. For commonly arising regions, more and more of R is included as the norm of a partition (the largest width or height of any rectangle used) approaches zero. Once we have a partition of R, we number the rectangles in some order from 1 to n and let ¢Ak be the area of the kth rectangle. We then choose a point sxk , yk d in the kth rectangle and form the Riemann sum
Ak
R yk
(xk , yk )
You
xk
iaz
FIGURE 15.6 A rectangular grid partitioning a bounded nonrectangular region into rectangular cells.
n
Sn = a ƒsxk , yk d ¢Ak . k=1
nR
As the norm of the partition forming Sn goes to zero, 7P7 : 0, the width and height of each enclosed rectangle goes to zero and their number goes to infinity. If ƒ(x, y) is a continuous function, then these Riemann sums converge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of ƒ(x, y) over R: n
ass a
lim a ƒsxk , yk d ¢Ak = ƒ ƒ P ƒ ƒ :0 k=1
6
ƒsx, yd dA.
R
ma
dH
The nature of the boundary of R introduces issues not found in integrals over an interval. When R has a curved boundary, the n rectangles of a partition lie inside R but do not cover all of R. In order for a partition to approximate R well, the parts of R covered by small rectangles lying partly outside R must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, ellipses, and from continuous graphs over an interval, joined end to end. A curve with a “fractal” type of shape would be problematic, but such curves are not relevant for most applications. A careful discussion of which type of regions R can be used for computing double integrals is left to a more advanced text. Double integrals of continuous functions over nonrectangular regions have the same algebraic properties (summarized further on) as integrals over rectangular regions. The domain Additivity Property says that if R is decomposed into nonoverlapping regions R1 and R2 with boundaries that are again made of a finite number of line segments or smooth curves (see Figure 15.7 for an example), then
R1
ham
y
R2
x
Mu
0
R R1 ∪ R2
⌠⌠ f(x, y) dA ⌠⌠ f (x, y) dA ⌠⌠ f (x, y) dA ⌡⌡ ⌡⌡ ⌡⌡ R R2 R1
6 R
ƒsx, yd dA =
6
ƒsx, yd dA +
R1
6
ƒsx, yd dA.
R2
If ƒ(x, y) is positive and continuous over R we define the volume of the solid region between R and the surface z = ƒsx, yd to be 4R ƒsx, yd dA, as before (Figure 15.8). If R is a region like the one shown in the xyplane in Figure 15.9, bounded “above” and “below” by the curves y = g2sxd and y = g1sxd and on the sides by the lines x = a, x = b, we may again calculate the volume by the method of slicing. We first calculate the crosssectional area
FIGURE 15.7 The Additivity Property for rectangular regions holds for regions bounded by continuous curves.
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y = g2sxd
Asxd =
Ly = g1sxd
ƒsx, yd dy
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Page 1073
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1 Double Integrals
z f (x, y)
0
a
z f (x, y)
b
y
y g1(x)
x
y
A(x)
y g2(x) A k
d
FIGURE 15.9 The area of the vertical slice shown here is
Volume lim f (xk , yk ) Ak 冮冮 f (x, y) dA
g2sxd
R
FIGURE 15.8 We define the volumes of solids with curved bases the same way we define the volumes of solids with rectangular bases.
d
ƒsx, yd dy.
Lg1sxd
x h1( y)
FIGURE 15.10 The volume of the solid shown here is Lc
nR
Asxd =
y
x h 2( y)
iaz
(xk , yk )
y
x
R R
A( y)
c
You
x x
z
0
suf
Height f (xk , yk )
i
z
z f (x, y)
z
1073
As yd dy =
d
h2syd
Lc Lh1syd
ƒsx, yd dx dy.
To calculate the volume of the solid, we integrate this area from x = a to x = b.
ass a
and then integrate A(x) from x = a to x = b to get the volume as an iterated integral: b
g2sxd
b
(5) ƒsx, yd dy dx. La La Lg1sxd Similarly, if R is a region like the one shown in Figure 15.10, bounded by the curves x = h2s yd and x = h1s yd and the lines y = c and y = d, then the volume calculated by slicing is given by the iterated integral
dH
V =
Asxd dx =
d
h2s yd
Volume =
Mu
ham
ma
(6) ƒsx, yd dx dy. Lc Lh1s yd That the iterated integrals in Equations (5) and (6) both give the volume that we defined to be the double integral of ƒ over R is a consequence of the following stronger form of Fubini’s Theorem.
THEOREM 2 Fubini’s Theorem (Stronger Form) Let ƒ(x, y) be continuous on a region R. 1. If R is defined by a … x … b, g1sxd … y … g2sxd, with g1 and g2 continuous on [a, b], then b
6
ƒsx, yd dA =
R
2.
g2sxd
La Lg1sxd
ƒsx, yd dy dx.
If R is defined by c … y … d, h1syd … x … h2syd, with h1 and h2 continuous on [c, d], then
To Read it Online & Download:
d
6 R
ƒsx, yd dA =
h2s yd
Lc Lh1s yd
ƒsx, yd dx dy.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1074
Chapter 15: Multiple Integrals
Finding Volume
i
EXAMPLE 2
suf
Find the volume of the prism whose base is the triangle in the xyplane bounded by the xaxis and the lines y = x and x = 1 and whose top lies in the plane
You
z = Ć’sx, yd = 3  x  y.
See Figure 15.11 on page 1075. For any x between 0 and 1, y may vary from y = 0 to y = x (Figure 15.11b). Hence,
Solution
x
L0 L0 1
=
L0
1
s3  x  yd dy dx =
a3x 
L0
y 2 y=x d dx 2 y=0
c3y  xy x=1
3x 2 3x 2 x3 b dx = c d = 1. 2 2 2 x=0
iaz
1
V =
When the order of integration is reversed (Figure 15.11c), the integral for the volume is 1
L0 Ly 1
a3 
L0
c3x 
L0
x=1
x2 dy  xy d 2 x=y
y2 1  y  3y + + y 2 b dy 2 2
y3 y=1 5 5 3 = 1.  4y + y 2 b dy = c y  2y 2 + d 2 2 2 2 y=0
ass a
=
1
s3  x  yd dx dy =
nR
1
V =
1
=
a
L0
The two integrals are equal, as they should be.
dH
Although Fubiniâ€™s Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen.
EXAMPLE 3
Evaluating a Double Integral
Mu
ham
ma
Calculate
6
sin x x dA,
R
where R is the triangle in the xyplane bounded by the xaxis, the line y = x , and the line x = 1. The region of integration is shown in Figure 15.12. If we integrate first with respect to y and then with respect to x, we find
Solution
1
L0
a
x
L0
sin x x dyb dx =
1
L0
sin x ay x d
y=x y=0
b dx =
1
L0
sin x dx
= cos s1d + 1 L 0.46. If we reverse the order of integration and attempt to calculate
To Read it Online & Download:
1
L0 Ly
1
sin x x dx dy,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1
1075
Double Integrals
y
i
z
suf
x1
(0, 0, 3)
yx
yx
You
z f (x, y) 3xy
R
y0 1
0
(1, 0, 2)
x
iaz
(b) y
nR
(1, 1, 1)
y
yx
xy
x1
(1, 1, 0)
(1, 0, 0)
x1
ass a
R
x
x1
0
yx
(a)
R 1
x
(c)
ma
dH
FIGURE 15.11 (a) Prism with a triangular base in the xyplane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may integrate first with respect to y and then with respect to x, or the other way around (Example 2). (b) Integration limits of x=1
y=x
Lx = 0 Ly = 0
ƒsx, yd dy dx.
y
x1
yx
Mu
1
ham
If we integrate first with respect to y, we integrate along a vertical line through R and then integrate from left to right to include all the vertical lines in R. (c) Integration limits of
R
0
1
FIGURE 15.12 in Example 3.
x
The region of integration
y=1
x=1
Ly = 0 Lx = y
ƒsx, yd dx dy.
If we integrate first with respect to x, we integrate along a horizontal line through R and then integrate from bottom to top to include all the horizontal lines in R.
we run into a problem, because 1 sssin xd>xd dx cannot be expressed in terms of elementary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you first choose doesn’t work, try the other. Sometimes neither order will work, and then we need to use numerical approximations.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1076
Chapter 15: Multiple Integrals
i
Finding Limits of Integration
1.
You
suf
We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works. When faced with evaluating 4R ƒsx, yddA, integrating first with respect to y and then with respect to x, do the following: Sketch. Sketch the region of integration and label the bounding curves. y x2 y 2 1
1
iaz
R
xy1
2.
1
x
nR
0
Find the ylimits of integration. Imagine a vertical line L cutting through R in the direction of increasing y. Mark the yvalues where L enters and leaves. These are the ylimits of integration and are usually functions of x (instead of constants).
ass a
y
Leaves at y 兹1 x 2
1
R
Enters at y1x
dH
L
x
1
x
Find the xlimits of integration. Choose xlimits that include all the vertical lines through R. The integral shown here is
Mu
ham
ma
3.
0
To Read it Online & Download:
6
ƒsx, yd dA =
R
x=1
y = 21  x2
ƒsx, yd dy dx.
Lx = 0 Ly = 1  x y
Leaves at y 兹1 x 2
1 R
Enters at y1x
L 0 Smallest x is x 0
x
1
x
Largest x is x 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1
Double Integrals
1077
21  y 2
1
6
ƒsx, yd dA =
R
You
Enters at x1y R
y
Leaves at x 兹1 y2 x
iaz
Smallest y is y 0 0
EXAMPLE 4
ƒsx, yd dx dy.
L0 L1  y
Largest y y is y 1 1
suf
i
To evaluate the same double integral as an iterated integral with the order of integration reversed, use horizontal lines instead of vertical lines in Steps 2 and 3. The integral is
1
Reversing the Order of Integration
nR
Sketch the region of integration for the integral 2
2x
L0 Lx2
s4x + 2d dy dx
ass a
and write an equivalent integral with the order of integration reversed. The region of integration is given by the inequalities x 2 … y … 2x and 0 … x … 2. It is therefore the region bounded by the curves y = x 2 and y = 2x between x = 0 and x = 2 (Figure 15.13a). Solution
dH
y
Mu
ham
ma
4
0
y 4
(2, 4)
(2, 4)
y 2x y x2
x
x
2
y 2
0
(a)
x 兹y
2 (b)
x
FIGURE 15.13 Region of integration for Example 4.
To find limits for integrating in the reverse order, we imagine a horizontal line passing from left to right through the region. It enters at x = y>2 and leaves at x = 2y. To include all such lines, we let y run from y = 0 to y = 4 (Figure 15.13b). The integral is 4
2y
L0 Ly>2
s4x + 2d dx dy.
The common value of these integrals is 8.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1078
Chapter 15: Multiple Integrals
i
Properties of Double Integrals
You
suf
Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications.
Properties of Double Integrals If ƒ(x, y) and g(x, y) are continuous, then 1.
Constant Multiple:
6 R
R
Sum and Difference:
iaz
2.
sany number cd
cƒsx, yd dA = c f (x, yd dA 6
sƒsx, yd ; gsx, ydd dA = ƒsx, yd dA ; gsx, yd dA 6 6 6 3.
R
Domination: (a)
6
ƒsx, yd dA Ú 0
6
ƒsx, yd dA Ú
R
4.
if
ƒsx, yd Ú 0 on R
ass a
R
(b)
Additivity:
R
nR
R
6
gsx, yd dA
if
ƒsx, yd Ú gsx, yd on R
R
6
ƒsx, yd dA =
ƒsx, yd dA +
R1
dH
R
6
6
ƒsx, yd dA
R2
if R is the union of two nonoverlapping regions R1 and R2 (Figure 15.7).
Mu
ham
ma
The idea behind these properties is that integrals behave like sums. If the function ƒ(x, y) is replaced by its constant multiple cƒ(x, y), then a Riemann sum for ƒ n
Sn = a ƒsxk , yk d ¢Ak k=1
is replaced by a Riemann sum for cƒ n
n
a cƒsxk , yk d ¢Ak = c a ƒsxk , yk d ¢Ak = cSn .
k=1
k=1
Taking limits as n : q shows that c limn: q Sn = c 4R f dA and limn: q cSn = 4R cf dA are equal. It follows that the constant multiple property carries over from sums to double integrals. The other properties are also easy to verify for Riemann sums, and carry over to double integrals for the same reason. While this discussion gives the idea, an actual proof that these properties hold requires a more careful analysis of how Riemann sums converge.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1 Double Integrals
21.
In Exercises 1–10, sketch the region of integration and evaluate the integral. L0 L0 0
3.
L1 L1
L0 L0
6.
Lp L0
2
e x + y dx dy
8.
4
3y 3e xy dx dy
10.
29.
3 y>2x e dy dx 2
1 … x … 2,
22ln 3
35. 37.
14. Rectangle ƒsx, yd = y cos xy over the rectangle 0 … x … p, 0 … y … 1
38.
dH
13. Triangle ƒsx, yd = x 2 + y 2 over the triangular region with vertices (0, 0), (1, 0), and (0, 1)
15. Triangle ƒsu, yd = y  2u over the triangular region cut from the first quadrant of the uyplane by the line u + y = 1
ma
16. Curved region ƒss, td = e s ln t over the region in the first quadrant of the stplane that lies above the curve s = ln t from t = 1 to t = 2 Each of Exercises 17–20 gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. 17.
L2 Ly 1
18.
L0 L0 p>3
19.
2 dp dy
8t dt ds
sthe stplaned
sec t
4  2u
3 cos t du dt
4  2u dy du y2
Mu
L0 L1
sthe pyplaned
21  s2
Lp>3L0 3
20.
y
ham
0
3y dx dy
30.
dx dy
4  y2
y dx dy
L0 L0 2
24  x 2
L0 L24  x 2
6x dy dx
In Exercises 31–40, sketch the region of integration, reverse the order of integration, and evaluate the integral. p p sin y 2 2 2y 2 sin xy dy dx 31. 32. y dy dx L0 Lx L0 Lx 1 1 2 4  x2 xe 2y x 2e xy dx dy dy dx 33. 34. 4  y L0 L0 L0 Ly
ass a
11. Quadrilateral ƒsx, yd = x>y over the region in the first quadrant bounded by the lines y = x, y = 2x, x = 1, x = 2 square
28.
2
L0 L21  y 2
Le
2
16x dy dx
21  y
L0
2
x
Evaluating Double Integrals
dx dy
the
26.
9  4x 2
L0 L0
dy dx
L0 L1  x ln 2
y
2x
24.
dy dx
2
L1 L0
over
27.
1
y dy dx
In Exercises 11–16, integrate ƒ over the given region.
12. Square ƒsx, yd = 1>sxyd 1 … y … 2
e
L0 L1
sin x
L1 Ly
y2
dx dy
dx dy
1  x2
1
x
3>2
ssin x + cos yd dx dy
L0 L0
ln y
25.
p
p
x sin y dy dx
L0 1
9.
4.
1
sx 2y  2xyd dy dx
L0 L2
x
L0 L0 L1
0
2p
sx + y + 1d dx dy
ln 8
7.
2.
1
p
5.
3
s4  y 2 d dy dx
2y
L0 Ly
0
L0 Ly  2
iaz
2
22.
nR
3
1.
2
dy dx
L0 L2 1
23.
4  2x
You
1
Finding Regions of Integration and Double Integrals
suf
i
EXERCISES 15.1
1079
sthe tuplaned sthe uyplaned
Reversing the Order of Integration In Exercises 21–30, sketch the region of integration and write an equivalent double integral with the order of integration reversed.
To Read it Online & Download:
L0
2ln 3
Ly>2
1>16
36.
1
L0 L2x>3
3
e y dy dx
1>2
Ly1>4
L0
3
2
e x dx dy
8
2
5
cos s16px d dx dy dy dx
4 L0 L2 x y + 1 3
39. Square region 4R s y  2x 2 d dA where R is the region bounded by the square ƒ x ƒ + ƒ y ƒ = 1 40. Triangular region 4R xy dA where R is the region bounded by the lines y = x, y = 2x, and x + y = 2
Volume Beneath a Surface z = ƒsx, yd 41. Find the volume of the region bounded by the paraboloid z = x 2 + y 2 and below by the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xyplane. 42. Find the volume of the solid that is bounded above by the cylinder z = x 2 and below by the region enclosed by the parabola y = 2  x 2 and the line y = x in the xyplane. 43. Find the volume of the solid whose base is the region in the xyplane that is bounded by the parabola y = 4  x 2 and the line y = 3x, while the top of the solid is bounded by the plane z = x + 4. 44. Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x 2 + y 2 = 4, and the plane z + y = 3.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: Multiple Integrals
47. Find the volume of the wedge cut from the first octant by the cylinder z = 12  3y 2 and the plane x + y = 2. 48. Find the volume of the solid cut from the square column ƒ x ƒ + ƒ y ƒ … 1 by the planes z = 0 and 3x + z = 3. 49. Find the volume of the solid that is bounded on the front and back by the planes x = 2 and x = 1, on the sides by the cylinders y = ;1>x, and above and below by the planes z = x + 1 and z = 0. 50. Find the volume of the solid bounded on the front and back by the planes x = ;p>3, on the sides by the cylinders y = ;sec x , above by the cylinder z = 1 + y 2, and below by the xyplane.
i
57. Circular sector Integrate ƒsx, yd = 24  x 2 over the smaller sector cut from the disk x 2 + y 2 … 4 by the rays u = p>6 and u = p>2.
suf
46. Find the volume of the solid cut from the first octant by the surface z = 4  x 2  y.
Theory and Examples
58. Unbounded region Integrate ƒsx, yd = 1>[sx 2  xdsy  1d2>3] over the infinite rectangle 2 … x 6 q , 0 … y … 2.
You
45. Find the volume of the solid in the first octant bounded by the coordinate planes, the plane x = 3, and the parabolic cylinder z = 4  y 2.
59. Noncircular cylinder A solid right (noncircular) cylinder has its base R in the xyplane and is bounded above by the paraboloid z = x 2 + y 2 . The cylinder’s volume is 1
y
2
2y
sx 2 + y 2 d dx dy +
sx 2 + y 2 d dx dy. L0 L0 L1 L0 Sketch the base region R and express the cylinder’s volume as a single iterated integral with the order of integration reversed. Then evaluate the integral to find the volume. V =
iaz
1080
60. Converting to a double integral
Evaluate the integral
2
q
53.
L q L q q
54.
q
L0 L0
1
52.
1> 21  x 2
L1 L1> 21  x 2
1 dx dy sx 2 + 1ds y 2 + 1d
q
xe sx + 2yd dx dy
s2y + 1d dy dx
dH
1 51. dy dx 3 L1 Lex x y
ma
Approximating Double Integrals
ham
In Exercises 55 and 56, approximate the double integral of ƒ(x, y) over the region R partitioned by the given vertical lines x = a and horizontal lines y = c. In each subrectangle, use sxk , yk d as indicated for your approximation. n
6
ƒsx, yd dA L a ƒsxk , yk d ¢Ak
R
(Hint: Write the integrand as an integral.)
61. Maximizing a double integral maximizes the value of
ass a
Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section 8.8. Evaluate the improper integrals in Exercises 51–54 as iterated integrals. 1
nR
Integrals over Unbounded Regions
q
stan1px  tan1 xd dx.
L0
k=1
Mu
55. ƒsx, yd = x + y over the region R bounded above by the semicircle y = 11  x 2 and below by the xaxis, using the partition x = 1, 1>2 , 0, 1> 4, 1> 2, 1 and y = 0 , 1> 2, 1 with sxk , yk d the lower left corner in the kth subrectangle (provided the subrectangle lies within R) 56. ƒsx, yd = x + 2y over the region R inside the circle sx  2d2 + s y  3d2 = 1 using the partition x = 1 , 3> 2, 2, 5> 2, 3 and y = 2, 5> 2, 3, 7> 2, 4 with sxk , yk d the center (centroid) in the kth subrectangle (provided the subrectangle lies within R)
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What region R in the xyplane
s4  x 2  2y 2 d dA? 6 R
Give reasons for your answer.
62. Minimizing a double integral minimizes the value of
What region R in the xyplane
sx 2 + y 2  9d dA? 6 R
Give reasons for your answer. 63. Is it possible to evaluate the integral of a continuous function ƒ(x, y) over a rectangular region in the xyplane and get different answers depending on the order of integration? Give reasons for your answer. 64. How would you evaluate the double integral of a continuous function ƒ(x, y) over the region R in the xyplane enclosed by the triangle with vertices (0, 1), (2, 0), and (1, 2)? Give reasons for your answer. Prove that
65. Unbounded region q
b
b
b: qL b
Lb
q
L qL q
e x
2
 y2
dx dy = lim = 4a
66. Improper double integral 1
q 2
L0
e x
2
 y2
dx dy
2
e x dxb .
Evaluate the improper integral
3
x2 dy dx. L0 L0 s y  1d2>3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1 Double Integrals
1081
Use a CAS doubleintegral evaluator to estimate the values of the integrals in Exercises 67–70.
71.
1
68.
L0 L0
1
esx
2
+ y2d
dy dx
L0 L0
74.
21  x 2
x cos s y 2 d dy dx
sx 2y  xy 2 d dx dy
4  y2
e xy dx dy x2
1 dy dx L1 L0 x + y
2
76.
8
1 dx dy L1 Ly3 2x 2 + y 2
Mu
ham
ma
dH
ass a
nR
iaz
L1 L0
75.
9
L0 Lx
422y
L0 L0 2
3 21  x2  y2 dy dx
72.
2
L0 Ly3 2
tan1 xy dy dx
3
2
e x dx dy
L0 L2y 2
73.
1
1
70.
1 xy dy dx
4
suf
L1 L1 1
69.
x
1
You
3
67.
i
Evaluating Double Integrals Numerically
Use a CAS doubleintegral evaluator to find the integrals in Exercises 71–76. Then reverse the order of integration and evaluate, again with a CAS.
COMPUTER EXPLORATIONS
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
15.2 Area, Moments, and Centers of Mass
Area, Moments, and Centers of Mass
15.2
1081
u o Y z
In this section, we show how to use double integrals to calculate the areas of bounded regions in the plane and to find the average value of a function of two variables. Then we study the physical problem of finding the center of mass of a thin plate covering a region in the plane.
yk
iR a
Areas of Bounded Regions in the Plane
Ak
R
If we take ƒsx, yd = 1 in the definition of the double integral over a region R in the preceding section, the Riemann sums reduce to
(xk , yk ) xk
n
n
Sn = a ƒsxk , yk d ¢Ak = a ¢Ak .
n sa k=1
FIGURE 15.14 As the norm of a partition of the region R approaches zero, the sum of the areas ¢Ak gives the area of R defined by the double integral 4R dA.
h u
M
This is simply the sum of the areas of the small rectangles in the partition of R, and approximates what we would like to call the area of R. As the norm of a partition of R approaches zero, the height and width of all rectangles in the partition approach zero, and the coverage of R becomes increasingly complete (Figure 15.14). We define the area of R to be the limit
sa
H d
a m
m a
(1)
k=1
n
Area = lim a ¢Ak = ƒ ƒ P ƒ ƒ :0 k=1
6
dA
(2)
R
DEFINITION Area The area of a closed, bounded plane region R is A =
6
dA.
R
As with the other definitions in this chapter, the definition here applies to a greater variety of regions than does the earlier singlevariable definition of area, but it agrees with the earlier definition on regions to which they both apply. To evaluate the integral in the definition of area, we integrate the constant function ƒsx, yd = 1 over R.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1082
Chapter 15: Multiple Integrals
Finding Area
i
EXAMPLE 1
suf
Find the area of the region R bounded by y = x and y = x 2 in the first quadrant. y
We sketch the region (Figure 15.15), noting where the two curves intersect, and calculate the area as
Solution (1, 1)
yx
1
yx
y x2
A =
x
L0 Lx2
1
dy dx =
L0
1
y x2 x
L0
1
x2 x3 1  d = . 2 3 0 6
1
FIGURE 15.15 The region in Example 1.
Notice that the single integral 10 sx  x 2 d dx, obtained from evaluating the inside iterated integral, is the integral for the area between these two curves using the method of Section 5.5.
iaz
1
sx  x2 d dx = c
x2
nR
0
=
x
cy d dx
You
1
EXAMPLE 2
Finding Area
Find the area of the region R enclosed by the parabola y = x 2 and the line y = x + 2. If we divide R into the regions R1 and R2 shown in Figure 15.16a, we may calculate the area as
ass a
Solution
A =
6
dA +
R1
6
dA =
R2
2y
1
L0 L2y
4
dx dy +
2y
L1 Ly  2
dx dy.
Mu
ham
ma
dH
On the other hand, reversing the order of integration (Figure 15.16b) gives
y
2
A =
x+2
L1 Lx2
dy dx.
y
y x2 (2, 4)
yx2
y x2
yx2
(2, 4)
4
⌠ ⌠ 兹y dx dy ⌡1 ⌡y – 2 2
R2 (–1, 1)
1
(–1, 1)
兹y
⌠ ⌠ dx dy ⌡0 ⌡–兹y
R1 0 (a)
x
x2
⌠ ⌠ ⌡–1⌡x2
dy dx
0 (b)
FIGURE 15.16 Calculating this area takes (a) two double integrals if the first integration is with respect to x, but (b) only one if the first integration is with respect to y (Example 2).
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x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.2
Area, Moments, and Centers of Mass
1083
2
L1
cy d
x+2
2
dx = x2
L1
sx + 2  x 2 d dx = c
2
9 x2 x3 + 2x d = . 2 3 1 2
You
A =
suf
i
This second result, which requires only one integral, is simpler and is the only one we would bother to write down in practice. The area is
Average Value
nR
iaz
The average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defined on a bounded region in the plane, the average value is the integral over the region divided by the area of the region. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of R. We are led to define the average value of an integrable function ƒ over a region R to be
ass a
Average value of ƒ over R =
1 ƒ dA. area of R 6
(3)
R
dH
If ƒ is the temperature of a thin plate covering R, then the double integral of ƒ over R divided by the area of R is the plate’s average temperature. If ƒ(x, y) is the distance from the point (x, y) to a fixed point P, then the average value of ƒ over R is the average distance of points in R from P.
EXAMPLE 3
Finding Average Value
ma
Find the average value of ƒsx, yd = x cos xy over the rectangle R: 0 … x … p, 0 … y … 1. The value of the integral of ƒ over R is
Solution p
Mu
ham
L0 L0
1
p
x cos xy dy dx =
L0
csin xy d
y=1
dx y=0
L
x cos xy dy = sin xy + C
p
=
L0
ssin x  0d dx = cos x d
p
= 1 + 1 = 2. 0
The area of R is p. The average value of ƒ over R is 2>p.
Moments and Centers of Mass for Thin Flat Plates In Section 6.4 we introduced the concepts of moments and centers of mass, and we saw how to compute these quantities for thin rods or strips and for plates of constant density. Using multiple integrals we can extend these calculations to a great variety of shapes with varying density. We first consider the problem of finding the center of mass of a thin flat plate: a disk of aluminum, say, or a triangular sheet of metal. We assume the distribution of
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1084
Chapter 15: Multiple Integrals
You
suf
i
mass in such a plate to be continuous. A material’s density function, denoted by dsx, yd, is the mass per unit area. The mass of a plate is obtained by integrating the density function over the region R forming the plate. The first moment about an axis is calculated by integrating over R the distance from the axis times the density. The center of mass is found from the first moments. Table 15.1 gives the double integral formulas for mass, first moments, and center of mass.
TABLE 15.1 Mass and first moment formulas for thin plates covering a region R
in the xyplane dsx, yddA 6
dsx, yd is the density at (x, y)
iaz
M =
Mass:
R
Mx =
First moments:
6
ydsx, yddA,
6
xdsx, yddA
R
y =
Mx M
ass a
EXAMPLE 4
(1, 2)
Finding the Center of Mass of a Thin Plate of Variable Density
A thin plate covers the triangular region bounded by the xaxis and the lines x = 1 and y = 2x in the first quadrant. The plate’s density at the point (x, y) is dsx, yd = 6x + 6y + 6. Find the plate’s mass, first moments, and center of mass about the coordinate axes.
y 2x x1
dH
2
My , M
x =
Center of mass:
y
My =
nR
R
We sketch the plate and put in enough de