Bismillah hir Rehman nir Raheem Assalat o Wasalam o Allika Ya RasoolALLAH
Calculus (11
th
Ed. Text Book)
Thomas, Weir, Hass, Giordano
Ch Appendixes
Published By: Muhammad Hassan Riaz Yousufi
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suf
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
16
INTEGRATION IN VECTOR FIELDS
You
Chapter
z
tb
(x k , yk , z k )
ta
ham
x
y
FIGURE 16.1 The curve r(t) partitioned into small arcs from t = a to t = b. The length of a typical subarc is ¢sk.
Mu
ass a
ma
r(t)
s k
In Chapter 5 we defined the definite integral of a function over a finite closed interval [a, b] on the xaxis. We used definite integrals to find the mass of a thin straight rod, or the work done by a variable force directed along the xaxis. Now we would like to calculate the masses of thin rods or wires lying along a curve in the plane or space, or to find the work done by a variable force acting along such a curve. For these calculations we need a more general notion of a “line” integral than integrating over a line segment on the xaxis. Instead we need to integrate over a curve C in the plane or in space. These more general integrals are called line integrals, although “curve” integrals might be more descriptive. We make our definitions for space curves, remembering that curves in the xyplane are just a special case with zcoordinate identically zero. Suppose that ƒ(x, y, z) is a realvalued function we wish to integrate over the curve rstd = gstdi + hstdj + kstdk, a … t … b, lying within the domain of ƒ. The values of ƒ along the curve are given by the composite function ƒ(g(t), h(t), k(t)). We are going to integrate this composite with respect to arc length from t = a to t = b. To begin, we first partition the curve into a finite number n of subarcs (Figure 16.1). The typical subarc has length ¢sk. In each subarc we choose a point sxk, yk, zk d and form the sum
dH
Line Integrals
16.1
nR
iaz
OVERVIEW This chapter treats integration in vector fields. It is the mathematics that engineers and physicists use to describe fluid flow, design underwater transmission cables, explain the flow of heat in stars, and put satellites in orbit. In particular, we define line integrals, which are used to find the work done by a force field in moving an object along a path through the field. We also define surface integrals so we can find the rate that a fluid flows across a surface. Along the way we develop key concepts and results, such as conservative force fields and Green’s Theorem, to simplify our calculations of these new integrals by connecting them to the single, double, and triple integrals we have already studied.
n
Sn = a ƒsxk , yk , zk d ¢sk . k=1
If ƒ is continuous and the functions g, h, and k have continuous first derivatives, then these sums approach a limit as n increases and the lengths ¢sk approach zero. We call this limit the line integral of ƒ over the curve from a to b. If the curve is denoted by a single letter, C for example, the notation for the integral is LC
ƒsx, y, zd ds
“The integral of ƒ over C”
(1)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1144
Chapter 16: Integration in Vector Fields
suf
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If r(t) is smooth for a … t … b (v = dr>dt is continuous and never 0), we can use the equation b
sstd =
La
Equation (3) of Section 13.3 with t0 = a
ƒ vstd ƒ dt
You
to express ds in Equation (1) as ds = ƒ vstd ƒ dt. A theorem from advanced calculus says that we can then evaluate the integral of ƒ over C as b
LC
ƒsx, y, zd ds =
ƒsgstd, hstd, kstdd ƒ vstd ƒ dt.
La
nR
iaz
Notice that the integral on the right side of this last equation is just an ordinary (single) definite integral, as defined in Chapter 5, where we are integrating with respect to the parameter t. The formula evaluates the line integral on the left side correctly no matter what parametrization is used, as long as the parametrization is smooth.
How to Evaluate a Line Integral To integrate a continuous function ƒ(x, y, z) over a curve C: Find a smooth parametrization of C,
ass a
1.
rstd = gstdi + hstdj + kstdk,
2.
a … t … b
Evaluate the integral as
b
La
(2)
ƒsgstd, hstd, kstdd ƒ vstd ƒ dt.
dH
LC
ƒsx, y, zd ds =
If ƒ has the constant value 1, then the integral of ƒ over C gives the length of C.
EXAMPLE 1
z (1, 1, 1)
Integrate ƒsx, y, zd = x  3y 2 + z over the line segment C joining the origin to the point (1, 1, 1) (Figure 16.2).
ma
C
ham
y
x
Evaluating a Line Integral
(1, 1, 0)
We choose the simplest parametrization we can think of: rstd = ti + tj + tk,
0 … t … 1.
The components have continuous first derivatives and ƒ vstd ƒ = ƒ i + j + k ƒ = 212 + 12 + 12 = 23 is never 0, so the parametrization is smooth. The integral of ƒ over C is 1
LC
ƒsx, y, zd ds =
Mu
FIGURE 16.2 The integration path in Example 1.
Solution
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L0
ƒst, t, td A 23 B dt
Equation (2)
1
=
L0
st  3t 2 + td23 dt 1
= 23
L0
s2t  3t 2 d dt = 23 C t 2  t 3 D 0 = 0. 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.1
Line Integrals
1145
i
Additivity
EXAMPLE 2
(1, 1, 1)
C2
(0, 0, 0)
LC1
ƒ ds +
LC2
ƒ ds + Á +
LCn
ƒ ds.
(3)
Line Integral for Two Joined Paths
Figure 16.3 shows another path from the origin to (1, 1, 1), the union of line segments C1 and C2. Integrate ƒsx, y, zd = x  3y 2 + z over C1 ´ C2.
iaz
z
ƒ ds =
You
LC
suf
Line integrals have the useful property that if a curve C is made by joining a finite number of curves C1, C2 , Á , Cn end to end, then the integral of a function over C is the sum of the integrals over the curves that make it up:
y
We choose the simplest parametrizations for C1 and C2 we can think of, checking the lengths of the velocity vectors as we go along:
C1 (1, 1, 0)
x
C1:
rstd = ti + tj,
C2:
rstd = i + j + tk,
The path of integration in
0 … t … 1;
ƒ v ƒ = 212 + 12 = 22
0 … t … 1;
ƒ v ƒ = 20 2 + 0 2 + 12 = 1.
With these parametrizations we find that
ass a
FIGURE 16.3 Example 2.
nR
Solution
LC1 ´C2
ƒsx, y, zd ds =
LC1
ƒsx, y, zd ds +
Equation (3)
LC2
ƒsx, y, zd ds
1
Mu
ham
ma
dH
=
ƒst, t, 0d22 dt +
L0
Equation (2)
1
L0
ƒs1, 1, tds1d dt
1
=
L0
1
st  3t 2 + 0d22 dt +
= 22 c
L0
s1  3 + tds1d dt
1 1 22 3 t2 t2  t 3 d + c  2t d =  . 2 2 2 2 0 0
Notice three things about the integrations in Examples 1 and 2. First, as soon as the components of the appropriate curve were substituted into the formula for ƒ, the integration became a standard integration with respect to t. Second, the integral of ƒ over C1 ´ C2 was obtained by integrating ƒ over each section of the path and adding the results. Third, the integrals of ƒ over C and C1 ´ C2 had different values. For most functions, the value of the integral along a path joining two points changes if you change the path between them. For some functions, however, the value remains the same, as we will see in Section 16.3.
Mass and Moment Calculations We treat coil springs and wires like masses distributed along smooth curves in space. The distribution is described by a continuous density function dsx, y, zd (mass per unit length). The spring’s or wire’s mass, center of mass, and moments are then calculated with the formulas in Table 16.1. The formulas also apply to thin rods.
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Chapter 16: Integration in Vector Fields
along a smooth curve C in space dsx, y, zd ds sd = d(x, y, z) = density) LC First moments about the coordinate planes:
You
M =
Mass:
suf
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TABLE 16.1 Mass and moment formulas for coil springs, thin rods, and wires lying
x d ds, Mxz = y d ds, LC LC Coordinates of the center of mass: y = Mxz >M, x = Myz >M, Moments of inertia about axes and other lines: LC
Iz =
LC
s y 2 + z 2 d d ds,
Iy =
LC
sx 2 + y 2 d d ds,
IL =
LC
nR
Ix =
Mxy =
LC
z d ds
z = Mxy >M
iaz
Myz =
sx 2 + z 2 d d ds r 2 d ds
rsx, y, zd = distance from the point sx, y, zd to line L RL = 2IL >M
ass a
Radius of gyration about a line L:
EXAMPLE 3
Finding Mass, Center of Mass, Moment of Inertia, Radius of Gyration
A coil spring lies along the helix
rstd = scos 4tdi + ssin 4tdj + tk,
0 … t … 2p.
dH
The spring’s density is a constant, d = 1. Find the spring’s mass and center of mass, and its moment of inertia and radius of gyration about the zaxis.
z 2
We sketch the spring (Figure 16.4). Because of the symmetries involved, the center of mass lies at the point s0, 0, pd on the zaxis. For the remaining calculations, we first find ƒ vstd ƒ :
Solution
ma
(1, 0, 2)
ham
c.m. (0, 0, )
Mu
(1, 0, 0) x
FIGURE 16.4 The helical spring in Example 3.
y
ƒ vstd ƒ =
2 2 dy 2 dx dz b + a b + a b dt dt B dt
a
= 2s 4 sin 4td2 + s4 cos 4td2 + 1 = 217 .
We then evaluate the formulas from Table 16.1 using Equation (2): 2p
M =
3
Helix
d ds =
L0
s1d217 dt = 2p217 2p
Iz =
3
Helix
sx 2 + y 2 dd ds =
L0
scos2 4t + sin2 4tds1d217 dt
2p
=
L0
217 dt = 2p217
Rz = 2Iz >M = 22p117>s2p117d = 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.1 Line Integrals
1147
EXAMPLE 4
z
Finding an Arch’s Center of Mass
c.m. –1
You
A slender metal arch, denser at the bottom than top, lies along the semicircle y 2 + z 2 = 1, z Ú 0, in the yzplane (Figure 16.5). Find the center of the arch’s mass if the density at the point (x, y, z) on the arch is dsx, y, zd = 2  z.
1
We know that x = 0 and y = 0 because the arch lies in the yzplane with its mass distributed symmetrically about the zaxis. To find z , we parametrize the circle as
Solution
rstd = scos tdj + ssin tdk,
FIGURE 16.5 Example 4 shows how to find the center of mass of a circular arch of variable density.
For this parametrization, ƒ vstd ƒ =
0 … t … p.
iaz
1 y 2 z 2 1, z ⱖ 0
y
2 2 dy 2 dx dz b + a b + a b = 2s0d2 + s sin td2 + scos td2 = 1. dt dt B dt
a
nR
x
suf
i
Notice that the radius of gyration about the zaxis is the radius of the cylinder around which the helix winds.
The formulas in Table 16.1 then give M =
zd ds =
LC
s2  zd ds =
LC
LC
p
=
s2 sin t  sin2 td dt =
p
L0
zs2  zd ds =
ass a
Mxy =
LC
d ds =
s2  sin tds1d dt = 2p  2 p
L0
ssin tds2  sin td dt
8  p 2
dH
L0 Mxy 8  p 8  p# 1 = L 0.57. z = = M 2 2p  2 4p  4
Mu
ham
ma
With z to the nearest hundredth, the center of mass is (0, 0, 0.57).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
u o Y z a i R n a s s a
16.1 Line Integrals
EXERCISES 16.1 Graphs of Vector Equations
Match the vector equations in Exercises 1–8 with the graphs (a)–(h) given here. a. b. z
H ad
z 2
–1 1
m am y
x
h u M
x
1
c.
d.
z
z
1
x
1
y
2 x
y
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1148
Chapter 16: Integration in Vector Fields
e.
f.
z
i
z
z
suf
z
2
(0, 0, 1) (0, 0, 0)
1 1
(0, 0, 0)
y
y
x
C2 –2
(1, 1, –1)
x
C1
h.
iaz
z
FIGURE 16.6 The paths of integration for Exercises 15 and 16.
–2
2 y
2
2
y
x
1. rstd = ti + s1  tdj, 2. rstd = i + j + t k,
0 … t … 1 1 … t … 1
3. rstd = s2 cos tdi + s2 sin tdj,
0 … t … 2p
1 … t … 1
6. rstd = t j + s2  2tdk, 7. rstd = st 2  1dj + 2tk,
C1:
rstd = tk,
C2:
rstd = tj + k,
C3:
rstd = ti + j + k,
0 … t … 1 0 … t … 1 0 … t … 1
17. Integrate ƒsx, y, zd = sx + y + zd>sx 2 + y 2 + z 2 d over the path rstd = ti + tj + tk, 0 6 a … t … b. 18. Integrate ƒsx, y, zd =  2x 2 + z 2 over the circle
0 … t … 2
rstd = sa cos tdj + sa sin tdk,
0 … t … 1
dH
5. rstd = ti + tj + tk,
ass a
x
16. Integrate ƒsx, y, zd = x + 1y  z 2 over the path from (0, 0, 0) to (1, 1, 1) (Figure 16.6b) given by
nR
2
4. rstd = ti,
(b)
(a)
z
y
x
(1, 1, 0)
g.
C3
(1, 1, 1)
y
x
(0, 1, 1)
C1
(1, 1, 1)
–1
You
(1, 1, 1)
C2
1 … t … 1
8. rstd = s2 cos tdi + s2 sin tdk,
0 … t … p
Evaluating Line Integrals over Space Curves
ma
9. Evaluate 1C sx + yd ds where C is the straightline segment x = t, y = s1  td, z = 0, from (0, 1, 0) to (1, 0, 0).
0 … t … 2p.
Line Integrals over Plane Curves In Exercises 19–22, integrate ƒ over the given curve. 19. ƒsx, yd = x 3>y,
C:
y = x 2>2,
20. ƒsx, yd = sx + y d> 21 + x 2, (0, 0) 2
0 … x … 2 C:
y = x 2>2 from (1, 1> 2) to
21. ƒsx, yd = x + y, (2, 0) to (0, 2)
11. Evaluate 1C sxy + y + zd ds along the curve rstd = 2ti + t j + s2  2tdk, 0 … t … 1.
22. ƒsx, yd = x 2  y, C: (0, 2) to s 12, 12d
12. Evaluate 1C 2x 2 + y 2 ds along the curve rstd = s4 cos tdi + s4 sin tdj + 3tk, 2p … t … 2p.
Mass and Moments
13. Find the line integral of ƒsx, y, zd = x + y + z over the straightline segment from (1, 2, 3) to s0, 1, 1d.
23. Mass of a wire Find the mass of a wire that lies along the curve rstd = st 2  1dj + 2tk, 0 … t … 1, if the density is d = s3>2dt.
14. Find the line integral of ƒsx, y, zd = 23>sx 2 + y 2 + z 2 d over the curve rstd = ti + tj + tk, 1 … t … q .
24. Center of mass of a curved wire A wire of density dsx, y, zd = 152y + 2 lies along the curve rstd = st 2  1dj + 2tk, 1 … t … 1. Find its center of mass. Then sketch the curve and center of mass together.
Mu
ham
10. Evaluate 1C sx  y + z  2d ds where C is the straightline segment x = t, y = s1  td, z = 1, from (0, 1, 1) to (1, 0, 1).
15. Integrate ƒsx, y, zd = x + 1y  z 2 over the path from (0, 0, 0) to (1, 1, 1) (Figure 16.6a) given by C1:
rstd = ti + t 2 j,
C2:
rstd = i + j + tk,
0 … t … 1 0 … t … 1
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C:
x 2 + y 2 = 4 in the first quadrant from x 2 + y 2 = 4 in the first quadrant from
25. Mass of wire with variable density Find the mass of a thin wire lying along the curve rstd = 22ti + 22tj + s4  t 2 dk, 0 … t … 1, if the density is (a) d = 3t and (b) d = 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.1 Line Integrals
29. Two springs of constant density lies along the helix
A spring of constant density d
rstd = scos tdi + ssin tdj + tk,
0 … t … 2p.
a. Find Iz and Rz. b. Suppose that you have another spring of constant density d that is twice as long as the spring in part (a) and lies along the helix for 0 … t … 4p. Do you expect Iz and Rz for the longer spring to be the same as those for the shorter one, or should they be different? Check your predictions by calculating Iz and Rz for the longer spring. A wire of constant density d = 1 lies
rstd = st cos tdi + st sin tdj + A 2 22>3 B t
Find z, Iz , and Rz.
k,
i
suf
if the density is d = 1>st + 1d COMPUTER EXPLORATIONS
Evaluating Line Integrals Numerically
In Exercises 33–36, use a CAS to perform the following steps to evaluate the line integrals. a. Find ds = ƒ vstd ƒ dt for the path rstd = gstdi + hstdj + kstdk. b. Express the integrand ƒsgstd, hstd, kstdd ƒ vstd ƒ as a function of the parameter t.
c. Evaluate 1C ƒ ds using Equation (2) in the text. 33. ƒsx, y, zd = 21 + 30x 2 + 10y ; rstd = ti + t 2j + 3t 2k, 0 … t … 2 1 34. ƒsx, y, zd = 21 + x 3 + 5y 3 ; rstd = ti + t 2j + 1tk, 3 0 … t … 2
0 … t … 1.
Find Ix and Rx for the arch in Example 4.
35. ƒsx, y, zd = x1y  3z 2 ; rstd = scos 2tdi + ssin 2tdj + 5tk, 0 … t … 2p 1>4 9 36. ƒsx, y, zd = a1 + z1>3 b ; rstd = scos 2tdi + ssin 2tdj + 4 t 5>2k, 0 … t … 2p
Mu
ham
ma
dH
31. The arch in Example 4
3>2
0 … t … 2,
ass a
30. Wire of constant density along the curve
222 3>2 t2 t j + k, 3 2
You
28. Inertia and radii of gyration of slender rod A slender rod of constant density lies along the line segment rstd = tj + s2  2tdk, 0 … t … 1, in the yzplane. Find the moments of inertia and radii of gyration of the rod about the three coordinate axes.
rstd = ti +
iaz
27. Moment of inertia and radius of gyration of wire hoop A circular wire hoop of constant density d lies along the circle x 2 + y 2 = a 2 in the xyplane. Find the hoop’s moment of inertia and radius of gyration about the zaxis.
32. Center of mass, moments of inertia, and radii of gyration for wire with variable density Find the center of mass, and the moments of inertia and radii of gyration about the coordinate axes of a thin wire lying along the curve
nR
26. Center of mass of wire with variable density Find the center of mass of a thin wire lying along the curve rstd = ti + 2tj + s2>3dt 3>2k, 0 … t … 2, if the density is d = 315 + t.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
16.2 Vector Fields, Work, Circulation, and Flux
16.2
u o zY
Vector Fields, Work, Circulation, and Flux
a i R an
1149
When we study physical phenomena that are represented by vectors, we replace integrals over closed intervals by integrals over paths through vector fields. We use such integrals to find the work done in moving an object along a path against a variable force (such as a vehicle sent into space against Earthâ€™s gravitational field) or to find the work done by a vector field in moving an object along a path through the field (such as the work done by an accelerator in raising the energy of a particle). We also use line integrals to find the rates at which fluids flow along and across curves.
d a m
m a h u M
s s Ha
Vector Fields
Suppose a region in the plane or in space is occupied by a moving fluid such as air or water. Imagine that the fluid is made up of a very large number of particles, and that at any instant of time a particle has a velocity v. If we take a picture of the velocities of some particles at
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields
different position points at the same instant, we would expect to find that these velocities vary from position to position. We can think of a velocity vector as being attached to each point of the fluid. Such a fluid flow exemplifies a vector field. For example, Figure 16.7 shows a velocity vector field obtained by attaching a velocity vector to each point of air flowing around an airfoil in a wind tunnel. Figure 16.8 shows another vector field of velocity vectors along the streamlines of water moving through a contracting channel. In addition to vector fields associated with fluid flows, there are vector force fields that are associated with gravitational attraction (Figure 16.9), magnetic force fields, electric fields, and even purely mathematical fields. Generally, a vector field on a domain in the plane or in space is a function that assigns a vector to each point in the domain. A field of threedimensional vectors might have a formula like
You
suf
i
1150
FIGURE 16.7 Velocity vectors of a flow around an airfoil in a wind tunnel. The streamlines were made visible by kerosene smoke.
iaz
Fsx, y, zd = Msx, y, zdi + Nsx, y, zdj + Psx, y, zdk.
nR
The field is continuous if the component functions M, N, and P are continuous, differentiable if M, N, and P are differentiable, and so on. A field of twodimensional vectors might have a formula like Fsx, yd = Msx, ydi + Nsx, ydj.
ass a
dH
y
ma
FIGURE 16.8 Streamlines in a contracting channel. The water speeds up as the channel narrows and the velocity vectors increase in length.
If we attach a projectile’s velocity vector to each point of the projectile’s trajectory in the plane of motion, we have a twodimensional field defined along the trajectory. If we attach the gradient vector of a scalar function to each point of a level surface of the function, we have a threedimensional field on the surface. If we attach the velocity vector to each point of a flowing fluid, we have a threedimensional field defined on a region in space. These and other fields are illustrated in Figures 16.10–16.15. Some of the illustrations give formulas for the fields as well. To sketch the fields that had formulas, we picked a representative selection of domain points and sketched the vectors attached to them. The arrows representing the vectors are drawn with their tails, not their heads, at the points where the vector functions are
ham
z
0 y
Mu
x
FIGURE 16.9 Vectors in a gravitational field point toward the center of mass that gives the source of the field.
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0
x
FIGURE 16.10 The velocity vectors v(t) of a projectile’s motion make a vector field along the trajectory.
f (x, y, z) c
FIGURE 16.11 The field of gradient vectors §ƒ on a surface ƒsx, y, zd = c.
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16.2 Vector Fields, Work, Circulation, and Flux
suf
y
x 2 y 2 ⱕ a2 x
F = s yi + xjd>sx 2 + y 2 d1>2
in the plane. The field is not defined at the origin.
Mu
ham
ma
dH
ass a
FIGURE 16.12 The flow of fluid in a long cylindrical pipe. The vectors v = sa 2  r 2 dk inside the cylinder that have their bases in the xyplane have their tips on the paraboloid z = a 2  r 2.
FIGURE 16.14 The circumferential or “spin” field of unit vectors
iaz
y
FIGURE 16.13 The radial field F = xi + yj of position vectors of points in the plane. Notice the convention that an arrow is drawn with its tail, not its head, at the point where F is evaluated.
nR
0
You
x
z a2 r 2
x
i
y
z
WIND SPEED, M/S 0
2
4
6
8
10 12 14
16+
FIGURE 16.15 NASA’s Seasat used radar to take 350,000 wind measurements over the world’s oceans. The arrows show wind direction; their length and the color contouring indicate speed. Notice the heavy storm south of Greenland.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1152
Chapter 16: Integration in Vector Fields
suf
i
evaluated. This is different from the way we draw position vectors of planets and projectiles, with their tails at the origin and their heads at the planet’s and projectile’s locations.
You
Gradient Fields
DEFINITION Gradient Field The gradient field of a differentiable function ƒ(x, y, z) is the field of gradient vectors
EXAMPLE 1
0ƒ 0ƒ 0ƒ i + j + k. 0x 0y 0z
iaz
§ƒ =
Finding a Gradient Field
nR
Find the gradient field of ƒsx, y, zd = xyz. B tb
Solution
The gradient field of ƒ is the field F = §ƒ = yzi + xzj + xyk.
ass a
As we will see in Section 16.3, gradient fields are of special importance in engineering, mathematics, and physics.
Work Done by a Force over a Curve in Space T
dH
Suppose that the vector field F = Msx, y, zdi + Nsx, y, zdj + Psx, y, zdk represents a force throughout a region in space (it might be the force of gravity or an electromagnetic force of some kind) and that
F
rstd = gstdi + hstdj + kstdk,
A ta
is a smooth curve in the region. Then the integral of F # T, the scalar component of F in the direction of the curve’s unit tangent vector, over the curve is called the work done by F over the curve from a to b (Figure 16.16).
Mu
ham
ma
FIGURE 16.16 The work done by a force F is the line integral of the scalar component F # T over the smooth curve from A to B.
a … t … b,
DEFINITION Work over a Smooth Curve The work done by a force F = Mi + Nj + Pk over a smooth curve r(t) from t = a to t = b is t=b
W =
Lt = a
F # T ds.
(1)
We motivate Equation (1) with the same kind of reasoning we used in Chapter 6 to b derive the formula W = 1a F(x) dx for the work done by a continuous force of magnitude F(x) directed along an interval of the xaxis. We divide the curve into short segments, apply the (constantforce) * (distance) formula for work to approximate the work over each curved segment, add the results to approximate the work over the entire curve, and calculate
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2
Vector Fields, Work, Circulation, and Flux
1153
ck a
tk
tk 1
b
You
suf
i
the work as the limit of the approximating sums as the segments become shorter and more numerous. To find exactly what the limiting integral should be, we partition the parameter interval [a, b] in the usual way and choose a point ck in each subinterval [tk, tk + 1]. The partition of [a, b] determines (“induces,” we say) a partition of the curve, with the point Pk being the tip of the position vector rstk d and ¢sk being the length of the curve segment Pk Pk + 1 (Figure 16.17).
iaz
tb
s k
ta
Pk 1(g(tk 1), h(tk 1), k(tk 1))
nR
t ck
Pk (g(tk ), h(tk ), k(tk ))
Fk
If Fk denotes the value of F at the point on the curve corresponding to t = ck and Tk denotes the curve’s unit tangent vector at this point, then Fk # Tk is the scalar component of F in the direction of T at t = ck (Figure 16.18). The work done by F along the curve segment Pk Pk + 1 is approximately
Tk Fk . Tk
dH
Pk1
t ck Pk
ass a
FIGURE 16.17 Each partition of [a, b] induces a partition of the curve rstd = gstdi + hstdj + kstdk.
Force component in distance a b * a b = Fk # Tk ¢sk. direction of motion applied
The work done by F along the curve from t = a to t = b is approximately n
# a Fk Tk ¢sk .
k=1
Mu
ham
ma
FIGURE 16.18 An enlarged view of the curve segment Pk Pk + 1 in Figure 16.17, showing the force and unit tangent vectors at the point on the curve where t = ck.
As the norm of the partition of [a, b] approaches zero, the norm of the induced partition of the curve approaches zero and these sums approach the line integral t=b
Lt = a
F # T ds.
The sign of the number we calculate with this integral depends on the direction in which the curve is traversed as t increases. If we reverse the direction of motion, we reverse the direction of T and change the sign of F # T and its integral. Table 16.2 shows six ways to write the work integral in Equation (1). Despite their variety, the formulas in Table 16.2 are all evaluated the same way. In the table, rstd = gstdi + hstdj + kstdk = xi + yj + zk is a smooth curve, and dr =
dr dt = dgi + dhj + dkk dt
is its differential.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1154
Chapter 16: Integration in Vector Fields
Lt = a
F # T ds
The definition
F # dr
Compact differential form
b
=
F#
La b
=
La b
=
La
dr dt dt
Expanded to include dt; emphasizes the parameter t and velocity vector dr>dt
aM
dg dh dk + N + P b dt dt dt dt
Emphasizes the component functions
aM
dy dx dz + N + P b dt dt dt dt
Abbreviates the components of r
b
M dx + N dy + P dz
dt’s canceled; the most common form
nR
=
iaz
Lt = a
You
t=b
=
suf
t=b
W =
i
TABLE 16.2 Six different ways to write the work integral
La
Evaluate F on the curve as a function of the parameter t. Find dr> dt Integrate F # dr>dt from t = a to t = b.
dH
1. 2. 3.
ass a
Evaluating a Work Integral To evaluate the work integral along a smooth curve r(t), take these steps:
EXAMPLE 2
z
Finding Work Done by a Variable Force over a Space Curve
ma
Find the work done by F = sy  x 2 di + sz  y 2 dj + sx  z 2 dk over the curve rstd = ti + t 2j + t 3k, 0 … t … 1, from (0, 0, 0) to (1, 1, 1) (Figure 16.19). Solution
(0, 0, 0)
First we evaluate F on the curve:
ham
(1, 1, 1)
F = s y  x 2 di + sz  y 2 dj + sx  z 2 dk 2 = st(')'*  t 2 di + st 3  t 4 dj + st  t 6 dk
y
x r(t) ti
t 2j
t 3k
Then we find dr> dt,
dr d = sti + t 2j + t 3kd = i + 2tj + 3t 2k dt dt
(1, 1, 0)
Mu
FIGURE 16.19 The curve in Example 2.
0
Finally, we find F # dr>dt and integrate from t = 0 to t = 1 : F#
dr = [st 3  t 4 dj + st  t 6 dk] # si + 2tj + 3t 2kd dt = st 3  t 4 ds2td + st  t 6 ds3t 2 d = 2t 4  2t 5 + 3t 3  3t 8,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2
Vector Fields, Work, Circulation, and Flux
1155
1
L0
s2t 4  2t 5 + 3t 3  3t 8 d dt
suf
Work =
i
so
1
You
3 3 29 2 2 = c t5  t6 + t4  t9d = . 5 6 4 9 60 0
Flow Integrals and Circulation for Velocity Fields
nR
iaz
Instead of being a force field, suppose that F represents the velocity field of a fluid flowing through a region in space (a tidal basin or the turbine chamber of a hydroelectric generator, for example). Under these circumstances, the integral of F # T along a curve in the region gives the fluid’s flow along the curve.
ass a
DEFINITIONS Flow Integral, Circulation If r(t) is a smooth curve in the domain of a continuous velocity field F, the flow along the curve from t = a to t = b is Flow =
b
La
F # T ds.
(2)
dH
The integral in this case is called a flow integral. If the curve is a closed loop, the flow is called the circulation around the curve.
We evaluate flow integrals the same way we evaluate work integrals.
ma
EXAMPLE 3
Finding Flow Along a Helix
A fluid’s velocity field is F = xi + zj + yk. Find the flow along the helix rstd = scos tdi + ssin tdj + tk, 0 … t … p>2.
Mu
ham
Solution
We evaluate F on the curve, F = xi + zj + yk = scos tdi + tj + ssin tdk
and then find dr> dt: dr = s sin tdi + scos tdj + k. dt Then we integrate F # sdr>dtd from t = 0 to t =
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F#
p : 2
dr = scos tds sin td + stdscos td + ssin tds1d dt = sin t cos t + t cos t + sin t
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1156
Chapter 16: Integration in Vector Fields
Lt = a
EXAMPLE 4
dr dt = dt L0 p>2
cos2 t + t sin t d 2 0
p>2
s sin t cos t + t cos t + sin td dt = a0 +
p p 1 1 b  a + 0b =  . 2 2 2 2
You
= c
F#
suf
t=b
Flow =
i
so,
Finding Circulation Around a Circle
Find the circulation of the field F = sx  ydi + xj around the circle rstd = scos tdi + ssin tdj, 0 … t … 2p. On the circle, F = sx  ydi + xj = scos t  sin tdi + scos tdj , and
iaz
Solution
dr = s sin tdi + scos tdj. dt
nR
Then
F#
ass a
gives
dr sin2 t + cos2 t = sin t cos t + (''')'''* dt
2p
Circulation =
L0
dr dt = dt L0
2p
s1  sin t cos td dt
2p
sin2 t d = 2p. 2 0
dH
= ct 
F#
1
Flux Across a Plane Curve
Mu
ham
ma
To find the rate at which a fluid is entering or leaving a region enclosed by a smooth curve C in the xyplane, we calculate the line integral over C of F # n , the scalar component of the fluid’s velocity field in the direction of the curve’s outwardpointing normal vector. The value of this integral is the flux of F across C. Flux is Latin for flow, but many flux calculations involve no motion at all. If F were an electric field or a magnetic field, for instance, the integral of F # n would still be called the flux of the field across C.
DEFINITION Flux Across a Closed Curve in the Plane If C is a smooth closed curve in the domain of a continuous vector field F = Msx, ydi + Nsx, ydj in the plane and if n is the outwardpointing unit normal vector on C, the flux of F across C is Flux of F across C =
LC
F # n ds.
(3)
Notice the difference between flux and circulation. The flux of F across C is the line integral with respect to arc length of F # n, the scalar component of F in the direction of the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2 Vector Fields, Work, Circulation, and Flux z
1157
For clockwise motion, k T points outward. y
kT z For counterclockwise motion, T k points outward. y k x
a … t … b,
You
C
T
y = hstd,
that traces the curve C exactly once as t increases from a to b. We can find the outward unit normal vector n by crossing the curve’s unit tangent vector T with the vector k. But which order do we choose, T * k or k * T? Which one points outward? It depends on which way C is traversed as t increases. If the motion is clockwise, k * T points outward; if the motion is counterclockwise, T * k points outward (Figure 16.20). The usual choice is n = T * k, the choice that assumes counterclockwise motion. Thus, although the value of the arc length integral in the definition of flux in Equation (3) does not depend on which way C is traversed, the formulas we are about to derive for evaluating the integral in Equation (3) will assume counterclockwise motion. In terms of components,
iaz
x
x = gstd,
nR
k
suf
i
outward normal. The circulation of F around C is the line integral with respect to arc length of F # T, the scalar component of F in the direction of the unit tangent vector. Flux is the integral of the normal component of F; circulation is the integral of the tangential component of F. To evaluate the integral in Equation (3), we begin with a smooth parametrization
n = T * k = a
C
Tk T
dy dy dx dx i + jb * k = i j. ds ds ds ds
ass a
If F = Msx, ydi + Nsx, ydj, then
F # n = Msx, yd
dy dx  Nsx, yd . ds ds
Hence,
dH
LC
F # n ds =
LC
aM
dy dx M dy  N dx.  N b ds = ds ds FC
FIGURE 16.20 To find an outward unit normal vector for a smooth curve C in the xyplane that is traversed counterclockwise as t increases, we take n = T * k. For clockwise motion, we take n = k * T.
Mu
ham
ma
We put a directed circle ~ on the last integral as a reminder that the integration around the closed curve C is to be in the counterclockwise direction. To evaluate this integral, we express M, dy, N, and dx in terms of t and integrate from t = a to t = b . We do not need to know either n or ds to find the flux.
Calculating Flux Across a Smooth Closed Plane Curve sFlux of F = Mi + Nj across Cd =
FC
M dy  N dx
(4)
The integral can be evaluated from any smooth parametrization x = gstd, y = hstd, a … t … b, that traces C counterclockwise exactly once.
EXAMPLE 5
Finding Flux Across a Circle
Find the flux of F = sx  ydi + xj across the circle x 2 + y 2 = 1 in the xyplane.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1158
Chapter 16: Integration in Vector Fields
The parametrization rstd = scos tdi + ssin tdj, 0 … t … 2p , traces the circle counterclockwise exactly once. We can therefore use this parametrization in Equation (4). With
suf
i
Solution
M = x  y = cos t  sin t,
dy = dssin td = cos t dt
N = x = cos t,
dx = dscos td = sin t dt,
You
We find 2p
Flux =
LC
M dy  N dx = 2p
=
L0
L0 2p
cos2 t dt =
L0
scos2 t  sin t cos t + cos t sin td dt
Equation (4)
2p
1 + cos 2t sin 2t t d = p. dt = c + 2 2 4 0
Mu
ham
ma
dH
ass a
nR
iaz
The flux of F across the circle is p. Since the answer is positive, the net flow across the curve is outward. A net inward flow would have given a negative flux.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1158
i f u s u
Chapter 16: Integration in Vector Fields
EXERCISES 16.2 Vector and Gradient Fields
z
Find the gradient fields of the functions in Exercises 1–4. 1. ƒsx, y, zd = sx 2 + y 2 + z 2 d1>2 2. ƒsx, y, zd = ln 2x + y + z 2
2
3. gsx, y, zd = e z  ln sx 2 + y 2 d
6. Give a formula F = Msx, ydi + Nsx, ydj for the vector field in the plane that has the properties that F = 0 at (0, 0) and that at any other point (a, b), F is tangent to the circle x 2 + y 2 = a 2 + b 2 and points in the clockwise direction with magnitude ƒ F ƒ = 2a 2 + b 2.
d a
Work
m m
a. The straightline path C1: rstd = ti + tj + tk, b. The curved path C2: rstd = ti + t 2j + t 4 k,
0 … t … 1
0 … t … 1
c. The path C3 ´ C4 consisting of the line segment from (0, 0, 0) to (1, 1, 0) followed by the segment from (1, 1, 0) to (1, 1, 1)
u M
9. F = 1zi  2xj + 1yk
n a ss x
a H
In Exercises 7 – 12, find the work done by force F from (0, 0, 0) to (1, 1, 1) over each of the following paths (Figure 16.21):
8. F = [1>sx 2 + 1d]j
10. F = xyi + yzj + xzk
11. F = s3x 2  3xdi + 3zj + k 12. F = s y + zdi + sz + xdj + sx + ydk
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o Y (1, 1, 1)
C4
y
C3
5. Give a formula F = Msx, ydi + Nsx, ydj for the vector field in the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from (x, y) to the origin. (The field is not defined at (0, 0).)
a h
C1
C2
4. gsx, y, zd = xy + yz + xz
7. F = 3yi + 2xj + 4zk
z a i R (0, 0, 0)
2
(1, 1, 0)
FIGURE 16.21 The paths from (0, 0, 0) to (1, 1, 1).
In Exercises 13–16, find the work done by F over the curve in the direction of increasing t. 13. F = xyi + yj  yzk rstd = ti + t 2j + tk, 0 … t … 1 14. F = 2yi + 3xj + sx + ydk rstd = scos tdi + ssin tdj + st>6dk, 0 … t … 2p 15. F = zi + xj + yk rstd = ssin tdi + scos tdj + tk, 0 … t … 2p 16. F = 6zi + y2j + 12xk rstd = ssin tdi + scos tdj + st>6dk,
0 … t … 2p
Line Integrals and Vector Fields in the Plane 17. Evaluate 1C xy dx + sx + yd dy along the curve y = x 2 from s 1, 1d to (2, 4). 18. Evaluate 1C sx  yd dx + sx + yd dy counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2 Vector Fields, Work, Circulation, and Flux
21. Work Find the work done by the force F = xyi + sy  xdj over the straight line from (1, 1) to (2, 3). 22. Work Find the work done by the gradient of ƒsx, yd = sx + yd2 counterclockwise around the circle x 2 + y 2 = 4 from (2, 0) to itself. Find the circulation and flux of the fields and
F2 = yi + xj
0 … t … 2p
b. The ellipse rstd = scos tdi + s4 sin tdj, 24. Flux across a circle
0 … t … 2p
Find the flux of the fields
F1 = 2xi  3yj
F2 = 2xi + sx  ydj
and
across the circle rstd = sa cos tdi + sa sin tdj,
dH
In Exercises 25–28, find the circulation and flux of the field F around and across the closed semicircular path that consists of the semicircular arch r1std = sa cos tdi + sa sin tdj, 0 … t … p, followed by the line segment r2std = ti, a … t … a. 25. F = xi + yj
26. F = x 2 i + y 2 j
27. F = yi + xj
28. F = y 2 i + x 2 j
ma
29. Flow integrals Find the flow of the velocity field F = sx + ydi  sx 2 + y 2 dj along each of the following paths from (1, 0) to s 1, 0d in the xyplane. a. The upper half of the circle x 2 + y 2 = 1 b. The line segment from (1, 0) to s 1, 0d
ham
c. The line segment from (1, 0) to s0, 1d followed by the line segment from s0, 1d to s 1, 0d. 30. Flux across a triangle Find the flux of the field F in Exercise 29 outward across the triangle with vertices (1, 0), (0, 1), s 1, 0d.
Sketching and Finding Fields in the Plane 31. Spin field Draw the spin field
Mu
i
magnitude 2a 2 + b 2 tangent to the circle x 2 + y 2 = a 2 + b 2 and pointing in the counterclockwise direction. (The field is undefined at (0, 0).) 34. A field of tangent vectors
a. Find a field G = Psx, ydi + Qsx, ydj in the xyplane with the property that at any point sa, bd Z s0, 0d, G is a unit vector tangent to the circle x 2 + y 2 = a 2 + b 2 and pointing in the clockwise direction. b. How is G related to the spin field F in Figure 16.14?
35. Unit vectors pointing toward the origin Find a field F = Msx, ydi + Nsx, ydj in the xyplane with the property that at each point sx, yd Z s0, 0d , F is a unit vector pointing toward the origin. (The field is undefined at (0, 0).)
ass a
0 … t … 2p.
Circulation and Flux
F = 
a. Find a field G = Psx, ydi + Qsx, ydj in the xyplane with the property that at any point sa, bd Z s0, 0d, G is a vector of
b. How is G related to the spin field F in Figure 16.14?
around and across each of the following curves. a. The circle rstd = scos tdi + ssin tdj,
33. A field of tangent vectors
iaz
F1 = xi + yj
(see Figure 16.13) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y 2 = 1.
nR
23. Circulation and flux
F = xi + yj
suf
20. Evaluate 1C F # dr for the vector field F = yi  xj counterclockwise along the unit circle x 2 + y 2 = 1 from (1, 0) to (0, 1).
32. Radial field Draw the radial field
You
19. Evaluate 1C F # T ds for the vector field F = x 2i  yj along the curve x = y 2 from (4, 2) to s1, 1d .
1159
y
2x + y 2 2
i +
x 2x 2 + y 2
j
(see Figure 16.14) along with its horizontal and vertical components at a representative assortment of points on the circle x 2 + y 2 = 4.
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36. Two “central” fields Find a field F = Msx, ydi + Nsx, ydj in the xyplane with the property that at each point sx, yd Z s0, 0d , F points toward the origin and ƒ F ƒ is (a) the distance from (x, y) to the origin, (b) inversely proportional to the distance from (x, y) to the origin. (The field is undefined at (0, 0).)
Flow Integrals in Space In Exercises 37–40, F is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing t. 37. F = 4xyi + 8yj + 2k rstd = ti + t2j + k,
0 … t … 2
38. F = x2i + yzj + y2k rstd = 3tj + 4tk,
0 … t … 1
39. F = sx  zdi + xk rstd = scos tdi + ssin tdk,
0 … t … p
40. F = yi + xj + 2k rstd = s 2 cos tdi + s2 sin tdj + 2tk,
0 … t … 2p
41. Circulation Find the circulation of F = 2xi + 2zj + 2yk around the closed path consisting of the following three curves traversed in the direction of increasing t: C1:
rstd = scos tdi + ssin tdj + tk,
C2:
rstd = j + sp>2ds1  tdk,
C 3:
rstd = ti + s1  tdj,
0 … t … p>2
0 … t … 1
0 … t … 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields 0, 1, 2
i
46. Work done by a radial force with constant magnitude A particle moves along the smooth curve y = ƒsxd from (a, ƒ(a)) to (b, ƒ(b)). The force moving the particle has constant magnitude k and always points away from the origin. Show that the work done by the force is
suf
C1
and the area of the region bounded by the taxis, the graph of ƒ, and the lines t = a and t = b? Give reasons for your answer.
C2 (0, 1, 0)
(1, 0, 0) C3
y
x
42. Zero circulation Let C be the ellipse in which the plane 2x + 3y  z = 0 meets the cylinder x 2 + y 2 = 12. Show, without evaluating either line integral directly, that the circulation of the field F = xi + yj + zk around C in either direction is zero. 43. Flow along a curve The field F = xyi + yj  yzk is the velocity field of a flow in space. Find the flow from (0, 0, 0) to (1, 1, 1) along the curve of intersection of the cylinder y = x 2 and the plane z = x . (Hint: Use t = x as the parameter.)
LC
You
z
F # T ds = k C sb 2 + sƒsbdd2 d1>2  sa 2 + sƒsadd2 d1>2 D .
COMPUTER EXPLORATIONS
Finding Work Numerically
iaz
1160
In Exercises 47–52, use a CAS to perform the following steps for finding the work done by force F over the given path: a. Find dr for the path rstd = gstdi + hstdj + kstdk.
nR
z
b. Evaluate the force F along the path.
F # dr. LC 47. F = xy 6 i + 3xsxy 5 + 2dj; 0 … t … 2p c. Evaluate
y
ass a
(1, 1, 1)
zx
3 2 i + j; 1 + x2 1 + y2 0 … t … p
48. F =
y x2 x
44. Flow of a gradient field Find the flow of the field F = §sxy 2z 3 d:
dH
a. Once around the curve C in Exercise 42, clockwise as viewed from above b. Along the line segment from (1, 1, 1) to s2, 1, 1d,
Theory and Examples
ma
45. Work and area Suppose that ƒ(t) is differentiable and positive for a … t … b . Let C be the path rstd = ti + ƒstdj, a … t … b, and F = yi. Is there any relation between the value of the work integral
rstd = scos tdi + ssin tdj,
49. F = s y + yz cos xyzdi + sx 2 + xz cos xyzdj + sz + xy cos xyzdk; rstd = (2 cos t)i + (3 sin t)j + k, 0 … t … 2p 50. F = 2xyi  y 2j + ze x k; 1 … t … 4
rstd = ti + 1tj + 3tk,
51. F = s2y + sin xdi + sz 2 + s1>3dcos ydj + x 4 k; rstd = ssin tdi + scos tdj + ssin 2tdk, p>2 … t … p>2 1 52. F = sx 2ydi + x 3j + xyk; rstd = scos tdi + ssin tdj + 3 s2 sin2 t  1dk, 0 … t … 2p
F # dr
Mu
ham
LC
rstd = s2 cos tdi + ssin tdj,
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1160
Chapter 16: Integration in Vector Fields
16.3
z a i R n a
Path Independence, Potential Functions, and Conservative Fields
Mu
s s a dH
In gravitational and electric fields, the amount of work it takes to move a mass or a charge from one point to another depends only on the objectâ€™s initial and final positions and not on the path taken in between. This section discusses the notion of path independence of work integrals and describes the properties of fields in which work integrals are path independent. Work integrals are often easier to evaluate if they are path independent.
a m m a h
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Path Independence, Potential Functions, and Conservative Fields
1161
i
Path Independence
suf
If A and B are two points in an open region D in space, the work 1 F # dr done in moving a particle from A to B by a field F defined on D usually depends on the path taken. For some special fields, however, the integral’s value is the same for all paths from A to B.
iaz
You
DEFINITIONS Path Independence, Conservative Field Let F be a field defined on an open region D in space, and suppose that for any B two points A and B in D the work 1A F # dr done in moving from A to B is the same over all paths from A to B. Then the integral 1 F # dr is path independent in D and the field F is conservative on D.
nR
The word conservative comes from physics, where it refers to fields in which the principle of conservation of energy holds (it does, in conservative fields). Under differentiability conditions normally met in practice, a field F is conservative if and only if it is the gradient field of a scalar function ƒ; that is, if and only if F = §ƒ for some ƒ. The function ƒ then has a special name.
ass a
DEFINITION Potential Function If F is a field defined on D and F = §ƒ for some scalar function ƒ on D, then ƒ is called a potential function for F.
dH
An electric potential is a scalar function whose gradient field is an electric field. A gravitational potential is a scalar function whose gradient field is a gravitational field, and so on. As we will see, once we have found a potential function ƒ for a field F, we can evaluate all the work integrals in the domain of F over any path between A and B by B
LA
B
F # dr =
LA
§ƒ # dr = ƒsBd  ƒsAd.
(1)
Mu
ham
ma
If you think of §ƒ for functions of several variables as being something like the derivative ƒ¿ for functions of a single variable, then you see that Equation (1) is the vector calculus analogue of the Fundamental Theorem of Calculus formula b
La
ƒ¿sxd dx = ƒsbd  ƒsad.
Conservative fields have other remarkable properties we will study as we go along. For example, saying that F is conservative on D is equivalent to saying that the integral of F around every closed path in D is zero. Naturally, certain conditions on the curves, fields, and domains must be satisfied for Equation (1) to be valid. We discuss these conditions below.
Assumptions in Effect from Now On: Connectivity and Simple Connectivity We assume that all curves are piecewise smooth, that is, made up of finitely many smooth pieces connected end to end, as discussed in Section 13.1. We also assume that
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Chapter 16: Integration in Vector Fields
nR
iaz
You
suf
i
the components of F have continuous first partial derivatives. When F = §ƒ, this continuity requirement guarantees that the mixed second derivatives of the potential function ƒ are equal, a result we will find revealing in studying conservative fields F. We assume D to be an open region in space. This means that every point in D is the center of an open ball that lies entirely in D. We assume D to be connected, which in an open region means that every point can be connected to every other point by a smooth curve that lies in the region. Finally, we assume D is simply connected, which means every loop in D can be contracted to a point in D without ever leaving D. (If D consisted of space with a line segment removed, for example, D would not be simply connected. There would be no way to contract a loop around the line segment to a point without leaving D.) Connectivity and simple connectivity are not the same, and neither implies the other. Think of connected regions as being in “one piece” and simply connected regions as not having any “holes that catch loops.” All of space itself is both connected and simply connected. Some of the results in this chapter can fail to hold if applied to domains where these conditions do not hold. For example, the component test for conservative fields, given later in this section, is not valid on domains that are not simply connected.
Line Integrals in Conservative Fields
ass a
The following result provides a convenient way to evaluate a line integral in a conservative field. The result establishes that the value of the integral depends only on the endpoints and not on the specific path joining them.
THEOREM 1
Let F = Mi + Nj + Pk be a vector field whose components are continuous throughout an open connected region D in space. Then there exists a differentiable function ƒ such that
dH
1.
The Fundamental Theorem of Line Integrals
0ƒ 0ƒ 0ƒ i + j + k 0x 0y 0z B
if and only if for all points A and B in D the value of 1A F # dr is independent of the path joining A to B in D. If the integral is independent of the path from A to B, its value is
ma Mu
ham
2.
F = §ƒ =
B
LA
F # dr = ƒsBd  ƒsAd.
Proof that F = §ƒ Implies Path Independence of the Integral Suppose that A and B are two points in D and that C: rstd = gstdi + hstdj + kstdk, a … t … b, is a smooth curve in D joining A and B. Along the curve, ƒ is a differentiable function of t and dƒ 0ƒ dx 0ƒ dy 0ƒ dz = + + 0x dt 0y dt 0z dt dt = §ƒ # a
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dy dx dz dr dr = F# . i + j + kb = §ƒ # dt dt dt dt dt
Chain Rule with x = gstd, y = hstd, z = kstd
Because F = §ƒ
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.3
Path Independence, Potential Functions, and Conservative Fields
1163
LC
F#
Lt = a
b dƒ dr dt = dt dt La dt b
suf
t=b
F # dr =
i
Therefore,
= ƒsgstd, hstd, kstdd d = ƒsBd  ƒsAd.
You
a
Thus, the value of the work integral depends only on the values of ƒ at A and B and not on the path in between. This proves Part 2 as well as the forward implication in Part 1. We omit the more technical proof of the reverse implication.
Finding Work Done by a Conservative Field
iaz
EXAMPLE 1
Find the work done by the conservative field
F = yzi + xzj + xyk = §sxyzd
Solution
nR
along any smooth curve C joining the point As 1, 3, 9d to Bs1, 6, 4d. With ƒsx, y, zd = xyz, we have B
LA
§ƒ # dr
F = §ƒ
= ƒsBd  ƒsAd
Fundamental Theorem, Part 2
= xyz ƒ s1,6, 4d  xyz ƒ s1,3,9d = s1ds6ds 4d  s 1ds3ds9d = 24 + 27 = 3.
dH
ass a
LA
B
F # dr =
ham
ma
THEOREM 2 ClosedLoop Property of Conservative Fields The following statements are equivalent.
B
C2
C1
Mu A
# 1 F dr = 0 around every closed loop in D.
2.
The field F is conservative on D.
B
–C 2
C1
1.
A
FIGURE 16.22 If we have two paths from A to B, one of them can be reversed to make a loop.
Proof that Part 1 Q Part 2 We want to show that for any two points A and B in D, the integral of F # dr has the same value over any two paths C1 and C2 from A to B. We reverse the direction on C2 to make a path C2 from B to A (Figure 16.22). Together, C1 and C2 make a closed loop C, and LC1
F # dr 
LC2
F # dr =
LC1
F # dr +
LC2
F # dr =
LC
F # dr = 0.
Thus, the integrals over C1 and C2 give the same value. Note that the definition of line integral shows that changing the direction along a curve reverses the sign of the line integral.
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Chapter 16: Integration in Vector Fields B
Proof that Part 2 Q Part 1 We want to show that the integral of F # dr is zero over any closed loop C. We pick two points A and B on C and use them to break C into two pieces: C1 from A to B followed by C2 from B back to A (Figure 16.23). Then
–C 2
B
F # dr =
C1
A
F # dr =
B
F # dr = 0. FC LC1 LC2 LA LA The following diagram summarizes the results of Theorems 1 and 2.
C1
F # dr +
F # dr 
Theorem 1
A
F = §ƒ on D
3
Theorem 2
F conservative on D
F # dr = 0 FC over any closed path in D
3
iaz
FIGURE 16.23 If A and B lie on a loop, we can reverse part of the loop to make two paths from A to B.
You
C2
suf
i
B
Now that we see how convenient it is to evaluate line integrals in conservative fields, two questions remain. How do we know when a given field F is conservative? If F is in fact conservative, how do we find a potential function ƒ (so that F = §f )?
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1. 2.
Finding Potentials for Conservative Fields
ass a
The test for being conservative is the following. Keep in mind our assumption that the domain of F is connected and simply connected.
dH
Component Test for Conservative Fields Let F = Msx, y, zdi + Nsx, y, zdj + Psx, y, zdk be a field whose component functions have continuous first partial derivatives. Then, F is conservative if and only if 0N 0P = , 0y 0z
0P 0M = , 0z 0x
and
0N 0M = . 0x 0y
(2)
ma
Proof that Equations (2) hold if F is conservative There is a potential function ƒ such that F = Mi + Nj + Pk =
0ƒ 0ƒ 0ƒ i + j + k. 0x 0y 0z
Mu
ham
Hence,
0 2ƒ 0P 0 0ƒ = a b = 0y 0y 0z 0y 0z =
0 2ƒ 0z 0y
=
0N 0 0ƒ a b = . 0z 0y 0z
Continuity implies that the mixed partial derivatives are equal.
The others in Equations (2) are proved similarly. The second half of the proof, that Equations (2) imply that F is conservative, is a consequence of Stokes’ Theorem, taken up in Section 16.7, and requires our assumption that the domain of F be simply connected.
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Path Independence, Potential Functions, and Conservative Fields
1165
suf
i
Once we know that F is conservative, we usually want to find a potential function for F. This requires solving the equation §ƒ = F or 0ƒ 0ƒ 0ƒ i + j + k = Mi + Nj + Pk 0x 0y 0z
0ƒ = M, 0x
You
for ƒ. We accomplish this by integrating the three equations 0ƒ = N, 0y
as illustrated in the next example.
EXAMPLE 2
0ƒ = P, 0z
Finding a Potential Function
We apply the test in Equations (2) to
nR
Solution
iaz
Show that F = se x cos y + yzdi + sxz  e x sin ydj + sxy + zdk is conservative and find a potential function for it.
M = e x cos y + yz, and calculate
0N 0P = x = , 0y 0z
N = xz  e x sin y,
ass a
0M 0P = y = , 0z 0x
P = xy + z
0N 0M = e x sin y + z = . 0x 0y
Together, these equalities tell us that there is a function ƒ with §ƒ = F. We find ƒ by integrating the equations
dH
0ƒ = e x cos y + yz, 0x
0ƒ = xz  e x sin y, 0y
0ƒ = xy + z. 0z
(3)
We integrate the first equation with respect to x, holding y and z fixed, to get ƒsx, y, zd = e x cos y + xyz + gsy, zd.
Mu
ham
ma
We write the constant of integration as a function of y and z because its value may change if y and z change. We then calculate 0ƒ>0y from this equation and match it with the expression for 0ƒ>0y in Equations (3). This gives e x sin y + xz +
0g = xz  e x sin y, 0y
so 0g>0y = 0. Therefore, g is a function of z alone, and ƒsx, y, zd = e x cos y + xyz + hszd. We now calculate 0ƒ>0z from this equation and match it to the formula for 0ƒ>0z in Equations (3). This gives xy +
dh = xy + z, dz
or
dh = z, dz
so
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hszd =
z2 + C. 2
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Chapter 16: Integration in Vector Fields
z2 + C. 2
suf
ƒsx, y, zd = e x cos y + xyz +
i
Hence,
We have infinitely many potential functions of F, one for each value of C.
Showing That a Field Is Not Conservative
You
EXAMPLE 3
Show that F = s2x  3di  zj + scos zdk is not conservative. Solution
We apply the component test in Equations (2) and find immediately that 0N 0 = s zd = 1. 0z 0z
iaz
0 0P = scos zd = 0, 0y 0y
The two are unequal, so F is not conservative. No further testing is required.
nR
Exact Differential Forms
As we see in the next section and again later on, it is often convenient to express work and circulation integrals in the “differential” form B
M dx + N dy + P dz
ass a
LA
mentioned in Section 16.2. Such integrals are relatively easy to evaluate if M dx + N dy + P dz is the total differential of a function ƒ. For then B
dH
LA
B
M dx + N dy + P dz =
0ƒ 0ƒ 0ƒ dx + dy + dz 0x 0y 0z LA B
§ƒ # dr LA = ƒsBd  ƒsAd. =
Theorem 1
Thus,
B
Mu
ham
ma
df = ƒsBd  ƒsAd, LA just as with differentiable functions of a single variable.
DEFINITIONS Exact Differential Form Any expression Msx, y, zd dx + Nsx, y, zd dy + Psx, y, zd dz is a differential form. A differential form is exact on a domain D in space if M dx + N dy + P dz =
0ƒ 0ƒ 0f dx + dy + dz = dƒ 0x 0y 0z
for some scalar function ƒ throughout D. Notice that if M dx + N dy + P dz = dƒ on D, then F = Mi + Nj + Pk is the gradient field of ƒ on D. Conversely, if F = §ƒ, then the form M dx + N dy + P dz is exact. The test for the form’s being exact is therefore the same as the test for F’s being conservative.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Path Independence, Potential Functions, and Conservative Fields
0N 0P = , 0y 0z
0M 0P = , 0z 0x
suf
Component Test for Exactness of M dx + N dy + P dz The differential form M dx + N dy + P dz is exact if and only if
1167
i
16.3
0N 0M = . 0x 0y
and
EXAMPLE 4
You
This is equivalent to saying that the field F = Mi + Nj + Pk is conservative.
Showing That a Differential Form Is Exact
Show that y dx + x dy + 4 dz is exact and evaluate the integral
iaz
s2,3, 1d
Ls1,1,1d
y dx + x dy + 4 dz
Solution
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over the line segment from (1, 1, 1) to s2, 3, 1d .
We let M = y, N = x, P = 4 and apply the Test for Exactness: 0N 0P = 0 = , 0y 0z
0N 0M = 1 = . 0x 0y
0M 0P = 0 = , 0z 0x
ass a
These equalities tell us that y dx + x dy + 4 dz is exact, so y dx + x dy + 4 dz = dƒ
for some function ƒ, and the integral’s value is ƒs2, 3, 1d  ƒs1, 1, 1d. We find ƒ up to a constant by integrating the equations
dH
0ƒ = y, 0x
0ƒ = x, 0y
0ƒ = 4. 0z
From the first equation we get ƒsx, y, zd = xy + gsy, zd.
ma
The second equation tells us that 0ƒ 0g = x + = x, 0y 0y
0g = 0. 0y
or
Mu
ham
Hence, g is a function of z alone, and ƒsx, y, zd = xy + hszd. The third of Equations (4) tells us that 0ƒ dh = 4, = 0 + 0z dz
or
hszd = 4z + C.
Therefore, ƒsx, y, zd = xy + 4z + C. The value of the integral is
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Chapter 16: Integration in Vector Fields
s1,2,3d
Which fields in Exercises 1–6 are conservative, and which are not?
19.
Ls1,1,1d s2,1,1d
1. F = yzi + xzj + xyk
20.
2. F = s y sin zdi + sx sin zdj + sxy cos zdk
Ls1,2,1d
21.
4. F = yi + xj 5. F = sz + ydi + zj + s y + xdk
22.
y x 1 1 y dx + a z  y 2 b dy  z 2 dz Ls1,1,1d s2,2,2d 2x dx + 2y dy + 2z dz
iaz
x2 + y2 + z2 Ls1, 1, 1d 23. Revisiting Example 4 Evaluate the integral
6. F = se x cos ydi  se x sin ydj + zk
Finding Potential Functions
s2,3, 1d
In Exercises 7–12, find a potential function ƒ for the field F.
Ls1,1,1d
7. F = 2xi + 3yj + 4zk
10. F = s y sin zdi + sx sin zdj + sxy cos zdk 11. F = sln x + sec2sx + yddi +
24. Evaluate
ass a
z asec2sx + yd + 2 bj + 2 k y + z2 y + z2 y
x z + bj + 1 + x2 y2 21  y2 z2 y 1 a + z bk 21  y2 z2
Evaluating Line Integrals
dH
i + a
In Exercises 13–17, show that the differential forms in the integrals are exact. Then evaluate the integrals. s2,3, 6d
2x dx + 2y dy + 2z dz
Ls0,0,0d
ma
13.
s3,5,0d
14.
yz dx + xz dy + xy dz
Ls1,1,2d s1,2,3d
Ls0,0,0d
2xy dx + sx 2  z 2 d dy  2yz dz
ham
15.
s3,3,1d
16.
Ls0,0,0d
2x dx  y 2 dy 
4 dz 1 + z2
s0,1,1d
17.
Ls1,0,0d
sin y cos x dx + cos y sin x dy + dz 3
Mu
Although they are not defined on all of space R , the fields associated with Exercises 18–22 are simply connected and the Component Test can be used to show they are conservative. Find a potential function for each field and evaluate the integrals as in Example 4. s1,p>2,2d
18.
Ls0,2,1d
nR
9. F = e y + 2zsi + xj + 2xkd
y
y dx + x dy + 4 dz
from Example 4 by finding parametric equations for the line segment from (1, 1, 1) to s2, 3, 1d and evaluating the line integral of F = yi + xj + 4k along the segment. Since F is conservative, the integral is independent of the path.
8. F = s y + zdi + sx + zdj + sx + ydk
1 + x2 y2
x2 s2x ln y  yzd dx + a y  xz b dy  xy dz
s2,2,2d
3. F = yi + sx + zdj  yk
12. F =
z2 3x 2 dx + y dy + 2z ln y dz
You
Testing for Conservative Fields
suf
i
EXERCISES 16.3
1 1 2 cos y dx + a y  2x sin yb dy + z dz
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LC
x 2 dx + yz dy + s y 2>2d dz
along the line segment C joining (0, 0, 0) to (0, 3, 4).
Theory, Applications, and Examples Independence of path Show that the values of the integrals in Exercises 25 and 26 do not depend on the path taken from A to B. B
25.
z 2 dx + 2y dy + 2xz dz
LA B
26.
x dx + y dy + z dz
LA 2x 2 + y 2 + z 2 In Exercises 27 and 28, find a potential function for F. 1  x2 2x bj 27. F = y i + a y2 ex 28. F = se x lnydi + a y + sin zbj + s y cos zdk 29. Work along different paths Find the work done by F = sx 2 + ydi + s y 2 + xdj + ze z k over the following paths from (1, 0, 0) to (1, 0, 1). a. The line segment x = 1, y = 0, 0 … z … 1 b. The helix rstd = scos tdi + ssin tdj + st>2pdk, 0 … t … 2p c. The xaxis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = x 2, y = 0 from (0, 0, 0) to (1, 0, 1) 30. Work along different paths Find the work done by F = e yz i + sxze yz + z cos ydj + sxye yz + sin ydk over the following paths from (1, 0, 1) to s1, p>2, 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.3 Path Independence, Potential Functions, and Conservative Fields
31. Evaluating a work integral two ways Let F = §sx 3y 2 d and let C be the path in the xyplane from s 1, 1d to (1, 1) that consists of the line segment from s 1, 1d to (0, 0) followed by the line segment from (0, 0) to (1, 1). Evaluate 1C F # dr in two ways. a. Find parametrizations for the segments that make up C and evaluate the integral. b. Using ƒsx, yd = x 3y 2 as a potential function for F. 32. Integral along different paths Evaluate 1C 2x cos y dx  x 2 sin y dy along the following paths C in the xyplane. a. The parabola y = sx  1d2 from (1, 0) to (0, 1)
d. The astroid rstd = scos3 tdi + ssin3 tdj, 0 … t … 2p, counterclockwise from (1, 0) back to (1, 0)
i
Show that §g = F.
F # dr.
35. Path of least work You have been asked to find the path along which a force field F will perform the least work in moving a particle between two locations. A quick calculation on your part shows F to be conservative. How should you respond? Give reasons for your answer.
36. A revealing experiment By experiment, you find that a force field F performs only half as much work in moving an object along path C1 from A to B as it does in moving the object along path C2 from A to B. What can you conclude about F? Give reasons for your answer.
38. Gravitational field
a. Find a potential function for the gravitational field
ass a
33. a. Exact differential form How are the constants a, b, and c related if the following differential form is exact? say 2 + 2czxd dx + ysbx + czd dy + say 2 + cx 2 d dz
dH
F = s y 2 + 2czxdi + ysbx + czdj + s y 2 + cx 2 dk
F = GmM
xi + yj + zk 2
sx + y 2 + z 2 d3>2
sG, m, and M are constantsd.
b. Let P1 and P2 be points at distance s1 and s2 from the origin. Show that the work done by the gravitational field in part (a) in moving a particle from P1 to P2 is 1 1 GmM a s2  s1 b.
Mu
ham
ma
be a gradient field?
Ls0,0,0d
nR
c. The xaxis from s 1, 0d to (1, 0)
For what values of b and c will
sx,y,zd
gsx, y, zd =
37. Work by a constant force Show that the work done by a constant force field F = ai + bj + ck in moving a particle along any 1 path from A to B is W = F # AB.
b. The line segment from s 1, pd to (1, 0)
b. Gradient field
suf
c. The line segment from (1, 0, 1) to (1, 0, 0), followed by the xaxis from (1, 0, 0) to the origin, followed by the parabola y = px 2>2, z = 0 from there to s1, p>2, 0d
You
b. The line segment from (1, 0, 1) to the origin followed by the line segment from the origin to s1, p>2, 0d
34. Gradient of a line integral Suppose that F = §ƒ is a conservative vector field and
iaz
a. The line segment x = 1, y = pt>2, z = 1  t, 0 … t … 1
1169
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
u o zY
16.4 Green’s Theorem in the Plane
16.4
Green’s Theorem in the Plane
From Table 16.2 in Section 16.2, we know that every line integral 1C M dx + N dy can be b written as a flow integral 1a F # T ds. If the integral is independent of path, so the field F is conservative (over a domain satisfying the basic assumptions), we can evaluate the integral easily from a potential function for the field. In this section we consider how to evaluate the integral if it is not associated with a conservative vector field, but is a flow or flux integral across a closed curve in the xyplane. The means for doing so is a result known as Green’s Theorem, which converts the line integral into a double integral over the region enclosed by the path. We frame our discussion in terms of velocity fields of fluid flows because they are easy to picture. However, Green’s Theorem applies to any vector field satisfying certain mathematical conditions. It does not depend for its validity on the field’s having a particular physical interpretation.
d a m
m a h u M
a i R an
1169
s s Ha
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1170
,x
(x , x, y , y)
Divergence
i
(x, y , y)
Chapter 16: Integration in Vector Fields
A (x, y)
,x
(x , x, y)
FIGURE 16.24 The rectangle for defining the divergence (flux density) of a vector field at a point (x, y).
suf
,y
You
,y
We need two new ideas for Green’s Theorem. The first is the idea of the divergence of a vector field at a point, sometimes called the flux density of the vector field by physicists and engineers. We obtain it in the following way. Suppose that Fsx, yd = Msx, ydi + Nsx, ydj is the velocity field of a fluid flow in the plane and that the first partial derivatives of M and N are continuous at each point of a region R. Let (x, y) be a point in R and let A be a small rectangle with one corner at (x, y) that, along with its interior, lies entirely in R (Figure 16.24). The sides of the rectangle, parallel to the coordinate axes, have lengths of ¢x and ¢y. The rate at which fluid leaves the rectangle across the bottom edge is approximately Fsx, yd # s jd ¢x = Nsx, yd¢x.
nR
iaz
This is the scalar component of the velocity at (x, y) in the direction of the outward normal times the length of the segment. If the velocity is in meters per second, for example, the exit rate will be in meters per second times meters or square meters per second. The rates at which the fluid crosses the other three sides in the directions of their outward normals can be estimated in a similar way. All told, we have Top: Bottom: Right: Left:
Fsx, y + ¢yd # j ¢x Fsx, yd # s jd ¢x = Fsx + ¢x, yd # i ¢y Fsx, yd # s id ¢y =
ass a
Exit Rates:
= Nsx, y + ¢yd¢x Nsx, yd¢x = Msx + ¢x, yd¢y Msx, yd¢y.
Combining opposite pairs gives
sNsx, y + ¢yd  Nsx, ydd¢x L a
Right and left:
sMsx + ¢x, yd  Msx, ydd¢y L a
dH
Top and bottom:
0N ¢yb ¢x 0y 0M ¢xb ¢y. 0x
Adding these last two equations gives
ma
Flux across rectangle boundary L a
0N 0M + b ¢x¢y. 0x 0y
Mu
ham
We now divide by ¢x¢y to estimate the total flux per unit area or flux density for the rectangle: Flux across rectangle boundary 0N 0M L a + b. 0x 0y rectangle area
Finally, we let ¢x and ¢y approach zero to define what we call the flux density of F at the point (x, y). In mathematics, we call the flux density the divergence of F. The symbol for it is div F, pronounced “divergence of F” or “div F.”
DEFINITION Divergence (Flux Density) The divergence (flux density) of a vector field F = Mi + Nj at the point (x, y) is
To Read it Online & Download:
div F =
0N 0M + . 0x 0y
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4
A gas expanding at the point (x 0 , y0 ).
Intuitively, if a gas is expanding at the point sx0 , y0), the lines of flow would diverge there (hence the name) and, since the gas would be flowing out of a small rectangle about sx0, y0 d the divergence of F at sx0, y0 d would be positive. If the gas were compressing instead of expanding, the divergence would be negative (see Figure 16.25).
i
div F (x 0 , y0 ) 0
EXAMPLE 1
Finding Divergence
Solution
You
Find the divergence of Fsx, yd = sx 2  ydi + sxy  y 2 dj. We use the formula in Equation (1):
Sink:
div F =
div F (x 0 , y0 ) 0
0N 0M 0 2 0 + = sx  yd + sxy  y 2 d 0x 0y 0x 0y
= 2x + x  2y = 3x  2y.
iaz
A gas compressing at the point (x 0 , y0 ).
1171
suf
Source:
Green’s Theorem in the Plane
Spin Around an Axis: The kComponent of Curl
nR
Fsx, yd = Msx, ydi + Nsx, ydj
ass a
FIGURE 16.25 If a gas is expanding at a point sx0, y0 d, the lines of flow have positive divergence; if the gas is compressing, the divergence is negative.
The second idea we need for Green’s Theorem has to do with measuring how a paddle wheel spins at a point in a fluid flowing in a plane region. This idea gives some sense of how the fluid is circulating around axes located at different points and perpendicular to the region. Physicists sometimes refer to this as the circulation density of a vector field F at a point. To obtain it, we return to the velocity field
and the rectangle A. The rectangle is redrawn here as Figure 16.26. The counterclockwise circulation of F around the boundary of A is the sum of flow rates along the sides. For the bottom edge, the flow rate is approximately
dH
Fsx, yd # i ¢x = Msx, yd¢x.
This is the scalar component of the velocity F(x, y) in the direction of the tangent vector i times the length of the segment. The rates of flow along the other sides in the counterclockwise direction are expressed in a similar way. In all, we have
ham
ma
Top: Fsx, y + ¢yd # s id ¢x = Msx, y + ¢yd¢x Bottom: Fsx, yd # i ¢x = Msx, yd¢x Right: Fsx + ¢x, yd # j ¢y = Nsx + ¢x, yd¢y Left: Fsx, yd # s jd ¢y = Nsx, yd¢y. We add opposite pairs to get
(x, y , y)
(x, y)
,y
,x
sMsx, y + ¢yd  Msx, ydd¢x L  a
(x , x, y , y)
Mu
,y
,x
Top and bottom:
Right and left: sNsx + ¢x, yd  Nsx, ydd¢y L a
A
(x , x, y)
0M ¢yb ¢x 0y
0N ¢xb ¢y. 0x
Adding these last two equations and dividing by ¢x¢y gives an estimate of the circulation density for the rectangle:
FIGURE 16.26 The rectangle for defining the curl (circulation density) of a vector field at a point (x, y).
To Read it Online & Download:
Circulation around rectangle 0N 0M L . 0x 0y rectangle area
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1172
Chapter 16: Integration in Vector Fields
We let ¢x and ¢y approach zero to define what we call the circulation density of F at the point (x, y). The positive orientation of the circulation density for the plane is the counterclockwise rotation around the vertical axis, looking downward on the xyplane from the tip of the (vertical) unit vector k (Figure 16.27). The circulation value is actually the kcomponent of a more general circulation vector we define in Section 16.7, called the curl of the vector field F. For Green’s Theorem, we need only this kcomponent.
i
Vertical axis
(x 0 , y 0 )
DEFINITION kComponent of Curl (Circulation Density) The kcomponent of the curl (circulation density) of a vector field F = Mi + Nj at the point (x, y) is the scalar
iaz
Curl F (x 0 , y 0 ) . k 0 Counterclockwise circulation
You
suf
k
Vertical axis
scurl Fd # k =
(2)
nR
k
0N 0M . 0x 0y
(x 0 , y 0 )
Curl F (x 0 , y 0 ) . k 0 Clockwise circulation
ass a
If water is moving about a region in the xyplane in a thin layer, then the kcomponent of the circulation, or curl, at a point sx0, y0 d gives a way to measure how fast and in what direction a small paddle wheel will spin if it is put into the water at sx0, y0 d with its axis perpendicular to the plane, parallel to k (Figure 16.27).
EXAMPLE 2
Finding the kComponent of the Curl
Find the kcomponent of the curl for the vector field Fsx, yd = sx2  ydi + sxy  y2 dj.
We use the formula in Equation (2):
dH
Solution
scurl Fd # k =
0N 0 2 0M 0 sxy  y2 d sx  yd = y + 1. = 0x 0y 0x 0y
ma
FIGURE 16.27 In the flow of an incompressible fluid over a plane region, the kcomponent of the curl measures the rate of the fluid’s rotation at a point. The kcomponent of the curl is positive at points where the rotation is counterclockwise and negative where the rotation is clockwise.
Two Forms for Green’s Theorem
ham
Simple
Simple
Mu
Not simple
FIGURE 16.28 In proving Green’s Theorem, we distinguish between two kinds of closed curves, simple and not simple. Simple curves do not cross themselves. A circle is simple but a figure 8 is not.
In one form, Green’s Theorem says that under suitable conditions the outward flux of a vector field across a simple closed curve in the plane (Figure 16.28) equals the double integral of the divergence of the field over the region enclosed by the curve. Recall the formulas for flux in Equations (3) and (4) in Section 16.2.
THEOREM 3 Green’s Theorem (FluxDivergence or Normal Form) The outward flux of a field F = Mi + Nj across a simple closed curve C equals the double integral of div F over the region R enclosed by C. F
F # n ds =
C
C
Outward flux
To Read it Online & Download:
F
M dy  N dx =
6
a
0N 0M + b dx dy 0x 0y
R
Divergence integral
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4
Green’s Theorem in the Plane
1173
You
suf
i
In another form, Green’s Theorem says that the counterclockwise circulation of a vector field around a simple closed curve is the double integral of the kcomponent of the curl of the field over the region enclosed by the curve. Recall the defining Equation (2) for circulation in Section 16.2.
THEOREM 4 Green’s Theorem (CirculationCurl or Tangential Form) The counterclockwise circulation of a field F = Mi + Nj around a simple closed curve C in the plane equals the double integral of scurl Fd # k over the region R enclosed by C. F # T ds =
C
F
C
M dx + N dy =
6
a
0N 0M b dx dy 0x 0y
iaz
F
(4)
R
Curl integral
nR
Counterclockwise circulation
ass a
The two forms of Green’s Theorem are equivalent. Applying Equation (3) to the field G1 = Ni  Mj gives Equation (4), and applying Equation (4) to G2 = Ni + Mj gives Equation (3).
Mathematical Assumptions
ma
dH
We need two kinds of assumptions for Green’s Theorem to hold. First, we need conditions on M and N to ensure the existence of the integrals. The usual assumptions are that M, N, and their first partial derivatives are continuous at every point of some open region containing C and R. Second, we need geometric conditions on the curve C. It must be simple, closed, and made up of pieces along which we can integrate M and N. The usual assumptions are that C is piecewise smooth. The proof we give for Green’s Theorem, however, assumes things about the shape of R as well. You can find proofs that are less restrictive in more advanced texts. First let’s look at examples.
EXAMPLE 3
Supporting Green’s Theorem Fsx, yd = sx  ydi + xj
and the region R bounded by the unit circle C: Solution
rstd = scos tdi + ssin tdj,
0 … t … 2p.
We have
Mu
ham
Verify both forms of Green’s Theorem for the field
To Read it Online & Download:
M = cos t  sin t, N = cos t, 0M = 1, 0x
0M = 1, 0y
dx = dscos td = sin t dt, dy = dssin td = cos t dt, 0N = 1, 0x
0N = 0. 0y
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1174
Chapter 16: Integration in Vector Fields
i
The two sides of Equation (3) are t = 2p
C
scos t  sin tdscos t dtd  scos tds sin t dtd
Lt = 0
suf
F
M dy  N dx =
2p
cos2 t dt = p
L0
0N 0M + b dx dy = s1 + 0d dx dy 0x 0y 6 6 a
R
=
6
dx dy = area inside the unit circle = p.
R
The two sides of Equation (4) are t = 2p
scos t  sin tds sin t dtd + scos tdscos t dtd
Lt = 0
nR
F
M dx + N dy =
iaz
R
You
=
C
2p
=
a
s sin t cos t + 1d dt = 2p
L0
R
ass a
0N 0M s1  s 1dd dx dy = 2 dx dy = 2p. b dx dy = 0x dy 6 6 6 R
R
Using Green’s Theorem to Evaluate Line Integrals
dH
If we construct a closed curve C by piecing a number of different curves end to end, the process of evaluating a line integral over C can be lengthy because there are so many different integrals to evaluate. If C bounds a region R to which Green’s Theorem applies, however, we can use Green’s Theorem to change the line integral around C into one double integral over R.
EXAMPLE 4
Evaluating a Line Integral Using Green’s Theorem
Mu
ham
ma
Evaluate the integral
F
xy dy  y 2 dx,
C
where C is the square cut from the first quadrant by the lines x = 1 and y = 1. We can use either form of Green’s Theorem to change the line integral into a double integral over the square.
Solution 1.
With the Normal Form Equation (3): Taking M = xy, N = y 2, and C and R as the square’s boundary and interior gives 1
F
xy dy  y2 dx =
C
To Read it Online & Download:
6
sy + 2yd dx dy =
R
1
=
L0
c3xy d
x=1
3y dx dy 1
1
dy = x=0
L0 L0
1
L0
3y dy =
3 2 3 y d = . 2 2 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4
1175
With the Tangential Form Equation (4): Taking M = y 2 and N = xy gives the same result:
F
 y 2 dx + xy dy =
EXAMPLE 5
6
s y  s 2ydd dx dy =
R
3 . 2
You
C
suf
i
2.
Green’s Theorem in the Plane
Finding Outward Flux
Calculate the outward flux of the field Fsx, yd = xi + y 2j across the square bounded by the lines x = ;1 and y = ;1 . Calculating the flux with a line integral would take four integrations, one for each side of the square. With Green’s Theorem, we can change the line integral to one double integral. With M = x, N = y 2 , C the square, and R the square’s interior, we have
nR
iaz
Solution
Flux =
F
F # n ds =
C
M dy  N dx
C
a
0N 0M + b dx dy 0x 0y 6
ass a
=
F
Green’s Theorem
R
1
=
L1L1
dH
1
=
1
L1
1
s1 + 2yd dx dy =
L1
cx + 2xy d
s2 + 4yd dy = c2y + 2y 2 d
x=1
dy x = 1
1
= 4. 1
Proof of Green’s Theorem for Special Regions
C 2: y f2 (x)
ham
P2 (x, f2 (x))
ma
y
Let C be a smooth simple closed curve in the xyplane with the property that lines parallel to the axes cut it in no more than two points. Let R be the region enclosed by C and suppose that M, N, and their first partial derivatives are continuous at every point of some open region containing C and R. We want to prove the circulationcurl form of Green’s Theorem, F
C
R
Mu 0
a
x
b
6
a
0N 0M b dx dy. 0x 0y
(5)
R
Figure 16.29 shows C made up of two directed parts:
P1(x, f1(x))
C1: y f1(x)
M dx + N dy =
C1: x
y = ƒ1sxd,
a … x … b,
C2:
y = ƒ2sxd,
b Ú x Ú a.
For any x between a and b, we can integrate 0M>0y with respect to y from y = ƒ1sxd to y = ƒ2sxd and obtain
FIGURE 16.29 The boundary curve C is made up of C1 , the graph of y = ƒ1sxd, and C2 , the graph of y = ƒ2sxd.
To Read it Online & Download:
ƒ2sxd
Lƒ1sxd
y = ƒ sxd
2 0M dy = Msx, yd d = Msx, ƒ2sxdd  Msx, ƒ1sxdd. 0y y = ƒ1sxd
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1176
Chapter 16: Integration in Vector Fields
ƒ2sxd
La Lƒ1sxd
b
0M dy dx = [Msx, ƒ2sxdd  Msx, ƒ1sxdd] dx 0y La a
= 
LC2
M dx 
= 
F
M dx.
R C 2' : x g2( y) x
0
M dx
iaz 6
a
0M b dx dy. 0y
(6)
R
Equation (6) is half the result we need for Equation (5). We derive the other half by integrating 0N>0x first with respect to x and then with respect to y, as suggested by Figure 16.30. This shows the curve C of Figure 16.29 decomposed into the two directed parts C 1œ : x = g1s yd, d Ú y Ú c and C 2œ : x = g2s yd, c … y … d. The result of this double integration is
ass a
Q 2(g2( y), y) Q1(g1( y), y)
c
M dx =
C
y
Msx, ƒ1sxdd dx
nR F
C 1' : x g1( y)
d
LC1
La
You
Lb
Msx, ƒ2sxdd dx 
Therefore y
b
= 
C
suf
b
i
We can then integrate this with respect to x from a to b:
F
0N dx dy. 6 0x
N dy =
C
(7)
R
dH
Summing Equations (6) and (7) gives Equation (5). This concludes the proof. FIGURE 16.30 The boundary curve C is made up of C 1œ , the graph of x = g1syd, and C 2œ , the graph of x = g2syd.
Extending the Proof to Other Regions
ma
The argument we just gave does not apply directly to the rectangular region in Figure 16.31 because the lines x = a, x = b, y = c , and y = d meet the region’s boundary in more than two points. If we divide the boundary C into four directed line segments, however,
y C3: y d
ham
d C4 xa
C2 xb
R
c
Mu
C1: y c
0
a
b
C1:
y = c,
a … x … b,
C2:
x = b,
c … y … d
C3:
y = d,
b Ú x Ú a,
C4:
x = a,
d Ú y Ú c,
we can modify the argument in the following way. Proceeding as in the proof of Equation (7), we have d
b
d
0N dx dy = sNsb, yd  Nsa, ydd dy Lc La 0x Lc d
x
FIGURE 16.31 To prove Green’s Theorem for a rectangle, we divide the boundary into four directed line segments.
To Read it Online & Download:
c
=
Lc
Nsb, yd dy +
=
LC2
N dy +
LC4
Ld
Nsa, yd dy
N dy.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4 y
Green’s Theorem in the Plane
1177
C
d
R
b
0N dx dy = N dy. Lc La 0x F x
0
Similarly, we can show that
(a)
b
(9)
You
C
suf
i
Because y is constant along C1 and C3, 1C1 N dy = 1C3 N dy = 0, so we can add 1C1 N dy = 1C3 N dy to the righthand side of Equation (8) without changing the equality. Doing so, we have
d
0M dy dx =  M dx. La Lc 0y F
y
(10)
C
Subtracting Equation (10) from Equation (9), we again arrive at
C R
a
F
iaz
b
M dx + N dy =
C
a
x
b
ass a
FIGURE 16.32 Other regions to which Green’s Theorem applies.
dH
R2
R1
a C2
–b C 2 –a
C1
C1
C1 0
ma
C2
LC1
M dx + N dy =
LC2
M dx + N dy =
a
0N 0M b dx dy 0x 0y 6 R1
a
0N 0M b dx dy. 0x 0y 6 R2
When we add these two equations, the line integral along the yaxis from b to a for C1 cancels the integral over the same segment but in the opposite direction for C2. Hence,
b C2
R
Regions like those in Figure 16.32 can be handled with no greater difficulty. Equation (5) still applies. It also applies to the horseshoeshaped region R shown in Figure 16.33, as we see by putting together the regions R1 and R2 and their boundaries. Green’s Theorem applies to C1, R1 and to C2, R2, yielding
(b)
y
0N 0M b dx dy. 0x 0y 6
nR
0
a
a
C1
b
x
Mu
ham
FIGURE 16.33 A region R that combines regions R1 and R2.
F
M dx + N dy =
C
a
0N 0M b dx dy, 0x 0y 6 R
where C consists of the two segments of the xaxis from b to a and from a to b and of the two semicircles, and where R is the region inside C. The device of adding line integrals over separate boundaries to build up an integral over a single boundary can be extended to any finite number of subregions. In Figure 16.34a let C1 be the boundary, oriented counterclockwise, of the region R1 in the first quadrant. Similarly, for the other three quadrants, Ci is the boundary of the region Ri , i = 2, 3, 4. By Green’s Theorem, F
M dx + N dy =
Ci
0N 0M b dx dy. 0x 0y 6 a
(11)
Ri
We sum Equation (11) over i = 1, 2, 3, 4, and get (Figure 16.34b): F
r=b
To Read it Online & Download:
sM dx + N dyd +
I
r=a
sM dx + N dyd =
0N 0M b dx dy. 0x 0y 6 hRi
a
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields y
R1 C1
C2
C2 C3
0
a
C3
EXAMPLE 6
C1 b
C4
C4
R3
x
R4 C4
C3
C3
i
C1
C1
C2 R2
suf
C2
Equation (12) says that the double integral of s0N>0xd  s0M>0yd over the annular ring R equals the line integral of M dx + N dy over the complete boundary of R in the direction that keeps R on our left as we progress (Figure 16.34b).
Verifying Green’s Theorem for an Annular Ring
Verify the circulation form of Green’s Theorem (Equation 4) on the annular ring R: h 2 … x 2 + y 2 … 1, 0 6 h 6 1 (Figure 16.35), if y M =
C4
,
2
x + y2
N =
(a)
x . x2 + y2
The boundary of R consists of the circle
Boundary of R
C1:
x = cos t,
iaz
Solution
y
You
1178
y = sin t,
0 … t … 2p,
traversed counterclockwise as t increases, and the circle
a
0
b
x
x = h cos u,
y = h sin u,
0 … u … 2p,
nR
Ch:
R
traversed clockwise as u increases. The functions M and N and their partial derivatives are continuous throughout R. Moreover,
ass a
sx2 + y2 ds 1d + ys2yd 0M = 0y sx2 + y2 d2 y2  x2
(b)
=
so
=
0N , 0x
a
0N 0M b dx dy = 0 dx dy = 0. 0x 0y 6 6
dH
FIGURE 16.34 The annular region R combines four smaller regions. In polar coordinates, r = a for the inner circle, r = b for the outer circle, and a … r … b for the region itself.
sx 2 + y 2 d2
R
R
The integral of M dx + N dy over the boundary of R is 3
C1
C
Ch R h
1
x
ham
0
M dx + N dy =
ma
y
Mu
FIGURE 16.35 Green’s Theorem may be applied to the annular region R by integrating along the boundaries as shown (Example 6).
x dy  y dx F
2
x + y
2
C1
x dy  y dx +
I
x2 + y2
Ch
2p
=
2p
scos2 t + sin2 td dt 
L0
L0
h 2scos2 u + sin2 ud du h2
= 2p  2p = 0. The functions M and N in Example 6 are discontinuous at (0, 0), so we cannot apply Green’s Theorem to the circle C1 and the region inside it. We must exclude the origin. We do so by excluding the points interior to Ch. We could replace the circle C1 in Example 6 by an ellipse or any other simple closed curve K surrounding Ch (Figure 16.36). The result would still be F
sM dx + N dyd +
K
To Read it Online & Download:
I
Ch
sM dx + N dyd =
0N 0M b dy dx = 0, 0x 0y 6 a
R
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4 Green’s Theorem in the Plane y
1179
sM dx + N dyd = 2p
for any such curve K. We can explain this result by changing to polar coordinates. With
x
x = r cos u,
Ch
y = r sin u,
dx = r sin u du + cos u dr,
K
we have x dy  y dx x2 + y2
dy = r cos u du + sin u dr,
r 2scos2 u + sin2 ud du = du, r2
iaz
and u increases by 2p as we traverse K once counterclockwise.
Mu
ham
ma
dH
ass a
nR
FIGURE 16.36 The region bounded by the circle Ch and the curve K.
=
You
0
suf
F
K
i
which leads to the conclusion that
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
16.4 Green’s Theorem in the Plane
EXERCISES 16.4 Verifying Green’s Theorem In Exercises 1–4, verify the conclusion of Green’s Theorem by evaluating both sides of Equations (3) and (4) for the field F = Mi + Nj. Take the domains of integration in each case to be the disk R: x 2 + y 2 … a 2 and its bounding circle C: r = sa cos tdi + sa sin tdj, 0 … t … 2p. 1. F = yi + xj
2. F = yi
3. F = 2xi  3yj
4. F = x2yi + xy2j
In Exercises 5–10, use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and curve C.
13. Find the outward flux of the field
2
6. F = sx + 4ydi + sx + y dj
H d
C: The triangle bounded by y = 0, x = 3 , and y = x
a m
8. F = sx + ydi  sx 2 + y 2 dj
n sa
In Exercises 15 and 16, find the work done by F in moving a particle once counterclockwise around the given curve. 15. F = 2xy 3i + 4x 2y 2j
2
C: The boundary of the “triangular” region in the first quadrant enclosed by the xaxis, the line x = 1 , and the curve y = x 3
16. F = s4x  2ydi + s2x  4ydj C: The circle sx  2d2 + s y  2d2 = 4
C: The triangle bounded by y = 0, x = 1 , and y = x 9. F = sx + e x sin ydi + sx + e x cos ydj
x bi + se x + tan1 ydj 1 + y2
14. Find the counterclockwise circulation of F = s y + e x ln ydi + se x>ydj around the boundary of the region that is bounded above by the curve y = 3  x 2 and below by the curve y = x 4 + 1.
as
C: The square bounded by x = 0, x = 1, y = 0, y = 1
7. F = s y 2  x 2 di + sx 2 + y 2 dj
z a i R
F = a3xy 
Work
5. F = sx  ydi + sy  xdj
C: The square bounded by x = 0, x = 1, y = 0, y = 1
u o Y
12. Find the counterclockwise circulation and the outward flux of the field F = s sin ydi + sx cos ydj around and over the square cut from the first quadrant by the lines x = p>2 and y = p>2 .
across the cardioid r = as1 + cos ud, a 7 0 .
Counterclockwise Circulation and Outward Flux
2
1179
Evaluating Line Integrals in the Plane
C: The righthand loop of the lemniscate r = cos 2u y 10. F = atan1 x bi + ln sx 2 + y 2 dj C: The boundary of the region defined by the polar coordinate inequalities 1 … r … 2, 0 … u … p
Apply Green’s Theorem to evaluate the integrals in Exercises 17 – 20.
11. Find the counterclockwise circulation and outward flux of the field F = xyi + y 2j around and over the boundary of the region enclosed by the curves y = x 2 and y = x in the first quadrant.
18.
h u
m a
M
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17.
F
s y2 dx + x2 dyd
C
C: The triangle bounded by x = 0, x + y = 1, y = 0 F
s3y dx + 2x dyd
C
C: The boundary of 0 … x … p, 0 … y … sin x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Theory and Examples
s6y + xd dx + s y + 2xd dy
i
F
25. Let C be the boundary of a region on which Green’s Theorem holds. Use Green’s Theorem to calculate
C
C: The circle sx  2d2 + sy  3d2 = 4 20.
F
a.
s2x + y 2 d dx + s2xy + 3yd dy
F
ƒsxd dx + gsyd dy
F
ky dx + hx dy
suf
19.
Chapter 16: Integration in Vector Fields
C
C
C: Any simple closed curve in the plane for which Green’s Theorem holds
b.
C
Calculating Area with Green’s Theorem
You
1180
sk and h constantsd.
Show that the value of
26. Integral dependent only on area
If a simple closed curve C in the plane and the region R it encloses satisfy the hypotheses of Green’s Theorem, the area of R is given by
xy 2 dx + sx 2y + 2xd dy
F
Green’s Theorem Area Formula
27. What is special about the integral
1 x dy  y dx 2F
(13)
C
nR
Area of R =
iaz
C
around any square depends only on the area of the square and not on its location in the plane.
F
4x 3y dx + x 4 dy?
C
Give reasons for your answer.
The reason is that by Equation (3), run backward,
28. What is special about the integral
R
ass a
1 1 dy dx = a + b dy dx Area of R = 2 6 2 6 R
1 1 x dy  y dx . = 2 2 F
dH
Use the Green’s Theorem area formula (Equation 13) to find the areas of the regions enclosed by the curves in Exercises 21–24.
y
ham t0
Mu –1
29. Area as a line integral Show that if R is a region in the plane bounded by a piecewisesmooth simple closed curve C, then Area of R =
 23 … t … 23 (see
2
4
x dy = 
F
y dx.
C
30. Definite integral as a line integral Suppose that a nonnegative function y = ƒsxd has a continuous first derivative on [a, b]. Let C be the boundary of the region in the xyplane that is bounded below by the xaxis, above by the graph of ƒ, and on the sides by the lines x = a and x = b. Show that b
La
ƒsxd dx = 
F
y dx.
C
31. Area and the centroid Let A be the area and x the xcoordinate of the centroid of a region R that is bounded by a piecewisesmooth simple closed curve C in the xyplane. Show that
t ;兹3
1
F
C
t0
1
0
0 … t … 2p
0 … t … 2p
ma
24. The curve rstd = t 2i + sst 3>3d  tdj, accompanying figure).
C
0 … t … 2p
22. The ellipse rstd = sa cos tdi + sb sin tdj, 23. The astroid rstd = scos3 tdi + ssin3 tdj,
 y 3 dy + x 3 dx?
Give reasons for your answer.
C
21. The circle rstd = sa cos tdi + sa sin tdj,
F
x
1 1 x 2 dy = xy dx = x 2 dy  xy dx = Ax. 2 F 3 F F C
C
C
32. Moment of inertia Let Iy be the moment of inertia about the yaxis of the region in Exercise 31. Show that t0
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1 1 x 3 dy = x 2y dx = x 3 dy  x 2y dx = Iy . 3 F 4 F F C
C
C
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4 Green’s Theorem in the Plane
0 2ƒ 0x 2
0 2ƒ +
0y 2
i
that does not pass through (0, 0). Use Green’s Theorem to show that
F
= 0,
suf
33. Green’s Theorem and Laplace’s equation Assuming that all the necessary derivatives exist and are continuous, show that if ƒ(x, y) satisfies the Laplace equation
1181
§ƒ # n ds
K
for all closed curves C to which Green’s Theorem applies. (The converse is also true: If the line integral is always zero, then ƒ satisfies the Laplace equation.) 34. Maximizing work Among all smooth simple closed curves in the plane, oriented counterclockwise, find the one along which the work done by 1 1 F = a x 2y + y 3 bi + xj 4 3
iaz
C
36. Bendixson’s criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid’s individual particles. The vectors F = Msx, ydi + Nsx, ydj of the flow’s velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region R (no holes or missing points) and that if Mx + Ny Z 0 throughout R, then none of the streamlines in R is closed. In other words, no particle of fluid ever has a closed trajectory in R. The criterion Mx + Ny Z 0 is called Bendixson’s criterion for the nonexistence of closed trajectories.
nR
0ƒ 0ƒ dx dy = 0 0y 0x F
You
has two possible values, depending on whether (0, 0) lies inside K or outside K.
then
37. Establish Equation (7) to finish the proof of the special case of Green’s Theorem. 38. Establish Equation (10) to complete the argument for the extension of Green’s Theorem.
is greatest. (Hint: Where is scurl Fd # k positive?)
39. Curl component of conservative fields Can anything be said about the curl component of a conservative twodimensional vector field? Give reasons for your answer.
ma
dH
ass a
35. Regions with many holes Green’s Theorem holds for a region R with any finite number of holes as long as the bounding curves are smooth, simple, and closed and we integrate over each component of the boundary in the direction that keeps R on our immediate left as we go along (Figure 16.37).
ham
FIGURE 16.37 Green’s Theorem holds for regions with more than one hole (Exercise 35).
Mu
a. Let ƒsx, yd = ln sx2 + y2 d and let C be the circle x 2 + y 2 = a 2 . Evaluate the flux integral F
§ƒ # n ds.
C
b. Let K be an arbitrary smooth simple closed curve in the plane
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40. Circulation of conservative fields Does Green’s Theorem give any information about the circulation of a conservative field? Does this agree with anything else you know? Give reasons for your answer. COMPUTER EXPLORATIONS
Finding Circulation In Exercises 41–44, use a CAS and Green’s Theorem to find the counterclockwise circulation of the field F around the simple closed curve C. Perform the following CAS steps. a. Plot C in the xyplane. b. Determine the integrand s0N>0xd  s0M>0yd for the curl form of Green’s Theorem. c. Determine the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. C: The ellipse x 2 + 4y 2 = 4 y2 x2 + = 1 42. F = s2x 3  y 3 di + sx 3 + y 3 dj, C: The ellipse 4 9 41. F = s2x  ydi + sx + 3ydj,
43. F = x 1e yi + se y ln x + 2xdj, C: The boundary of the region defined by y = 1 + x 4 (below) and y = 2 (above) 44. F = xe y i + 4x 2 ln y j, C: The triangle with vertices (0, 0), (2, 0), and (0, 4)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields
suf
Surface f (x, y, z) c
i
Surface Area and Surface Integrals
16.5
We know how to integrate a function over a flat region in a plane, but what if the function is defined over a curved surface? To evaluate one of these socalled surface integrals, we rewrite it as a double integral over a region in a coordinate plane beneath the surface (Figure 16.38). Surface integrals are used to compute quantities such as the flow of liquid across a membrane or the upward force on a falling parachute.
You
1182
S
Surface Area
iaz
f (x, y, z) c Tk (xk , yk , zk )
∆ Pk
ƒ suk * vk d # p ƒ = ¢Ak.
ma
∆k
nR
FIGURE 16.38 As we soon see, the integral of a function g(x, y, z) over a surface S in space can be calculated by evaluating a related double integral over the vertical projection or “shadow” of S on a coordinate plane.
ass a
The vertical projection or “shadow” of S on a coordinate plane
dH
R
S
ham
Ck R
Figure 16.39 shows a surface S lying above its “shadow” region R in a plane beneath it. The surface is defined by the equation ƒsx, y, zd = c. If the surface is smooth ( §ƒ is continuous and never vanishes on S), we can define and calculate its area as a double integral over R. We assume that this projection of the surface onto its shadow R is onetoone. That is, each point in R corresponds to exactly one point (x, y, z) satisfying ƒsx, y, zd = c. The first step in defining the area of S is to partition the region R into small rectangles ¢Ak of the kind we would use if we were defining an integral over R. Directly above each ¢Ak lies a patch of surface ¢sk that we may approximate by a parallelogram ¢Pk in the tangent plane to S at a point Tk sxk , yk , zk d in ¢sk. This parallelogram in the tangent plane projects directly onto ¢Ak . To be specific, we choose the point Tksxk , yk , zk d lying directly above the back corner Ck of ¢Ak , as shown in Figure 16.39. If the tangent plane is parallel to R, then ¢Pk will be congruent to ¢Ak . Otherwise, it will be a parallelogram whose area is somewhat larger than the area of ¢Ak . Figure 16.40 gives a magnified view of ¢sk and ¢Pk , showing the gradient vector §ƒsxk , yk , zk d at Tk and a unit vector p that is normal to R. The figure also shows the angle gk between §ƒ and p. The other vectors in the picture, uk and vk , lie along the edges of the patch ¢Pk in the tangent plane. Thus, both uk * vk and §ƒ are normal to the tangent plane. We now need to know from advanced vector geometry that ƒ suk * vk d # p ƒ is the area of the projection of the parallelogram determined by uk and vk onto any plane whose normal is p. (A proof is given in Appendix 8.) In our case, this translates into the statement
To simplify the notation in the derivation that follows, we are now denoting the area of the small rectangular region by ¢Ak as well. Likewise, ¢Pk will also denote the area of the portion of the tangent plane directly above this small region. Now, ƒ uk * vk ƒ itself is the area ¢Pk (standard fact about cross products) so this last equation becomes
Mu
FIGURE 16.39 A surface S and its vertical projection onto a plane beneath it. You can think of R as the shadow of S on the plane. The tangent plane ¢Pk approximates the surface patch ¢sk above ¢Ak.
ƒ cos sangle between u * v and pd ƒ = ¢Ak
ƒu * v ƒ
ƒpƒ
k k ('''''''')''''''''*
¢Pk
1
Same as ƒ cos gk ƒ because §ƒ and uk * vk are both normal to the tangent plane
k k ('')'* ()*
∆Ak
or ¢Pk ƒ cos gk ƒ = ¢Ak or
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¢Pk =
¢Ak , ƒ cos gk ƒ
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5
Surface Area and Surface Integrals
1183
provided cos gk Z 0. We will have cos gk Z 0 as long as §ƒ is not parallel to the ground plane and §ƒ # p Z 0. Since the patches ¢Pk approximate the surface patches ¢sk that fit together to make S, the sum
p
Tk
¢Ak a ¢Pk = a ƒ cos gk ƒ
vk
(1)
You
uk
suf
i
f (x k , yk , z k ) ␥k
∆ Pk
looks like an approximation of what we might like to call the surface area of S. It also looks as if the approximation would improve if we refined the partition of R. In fact, the sums on the righthand side of Equation (1) are approximating sums for the double integral
∆k
1 dA. gƒ cos ƒ 6
(2)
We therefore define the area of S to be the value of this integral whenever it exists. For any surface ƒsx, y, zd = c, we have ƒ §ƒ # p ƒ = ƒ §ƒ ƒ ƒ p ƒ ƒ cos g ƒ , so
p
ƒ §ƒ ƒ 1 = #pƒ . cos g §ƒ ƒ ƒ ƒ
nR
Ck
iaz
R
∆ Ak
This combines with Equation (2) to give a practical formula for surface area.
ass a
FIGURE 16.40 Magnified view from the preceding figure. The vector uk * vk (not shown) is parallel to the vector §ƒ because both vectors are normal to the plane of ¢Pk .
Formula for Surface Area The area of the surface ƒsx, y, zd = c over a closed and bounded plane region R is ƒ §ƒ ƒ # dA, 6 ƒ §ƒ p ƒ
(3)
R
dH
Surface area =
where p is a unit vector normal to R and §ƒ # p Z 0.
Mu
ham
ma
Thus, the area is the double integral over R of the magnitude of §ƒ divided by the magnitude of the scalar component of §ƒ normal to R. We reached Equation (3) under the assumption that §ƒ # p Z 0 throughout R and that §f is continuous. Whenever the integral exists, however, we define its value to be the area of the portion of the surface ƒsx, y, zd = c that lies over R. (Recall that the projection is assumed to be onetoone.) In the exercises (see Equation 11), we show how Equation (3) simplifies if the surface is defined by z = ƒsx, yd.
EXAMPLE 1
Finding Surface Area
Find the area of the surface cut from the bottom of the paraboloid x 2 + y 2  z = 0 by the plane z = 4. Solution We sketch the surface S and the region R below it in the xyplane (Figure 16.41). The surface S is part of the level surface ƒsx, y, zd = x 2 + y 2  z = 0, and R is the disk x 2 + y 2 … 4 in the xyplane. To get a unit vector normal to the plane of R, we can take p = k.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1184
Chapter 16: Integration in Vector Fields z
ƒsx, y, zd = x 2 + y 2  z §ƒ = 2xi + 2yj  k
4
ƒ §ƒ ƒ = 2s2xd2 + s2yd2 + s 1d2
S z
y2
ƒ §ƒ # p ƒ = ƒ §ƒ # k ƒ = ƒ 1 ƒ = 1. In the region R, dA = dx dy. Therefore,
R 0
Surface area =
y
x2 y2 4
ƒ §ƒ ƒ # p ƒ dA §ƒ 6 ƒ
= FIGURE 16.41 The area of this parabolic surface is calculated in Example 1.
L0 L0 2p
EXAMPLE 2
1
0 1
ham
Mu
FIGURE 16.42 The cap cut from the hemisphere by the cylinder projects vertically onto the disk R: x 2 + y 2 … 1 in the xyplane (Example 2).
2
1 s4r 2 + 1d3>2 d du 12 0
L0
p 1 s173>2  1d du = A 17217  1 B . 12 6
dH
Find the area of the cap cut from the hemisphere x 2 + y 2 + z 2 = 2, z Ú 0, by the cylinder x 2 + y 2 = 1 (Figure 16.42).
The cap S is part of the level surface ƒsx, y, zd = x 2 + y 2 + z 2 = 2. It projects onetoone onto the disk R: x 2 + y 2 … 1 in the xyplane. The unit vector p = k is normal to the plane of R. At any point on the surface, Solution
y
x
Polar coordinates
Finding Surface Area
ma
R
x2 y2 1
c
24r 2 + 1 r dr du
ass a
L0
2p
=
x2 y2 z2 2
2
nR
2p
=
z
24x 2 + 4y 2 + 1 dx dy
6 x2 + y2 … 4
=
Equation (3)
iaz
R
x
S
You
= 24x 2 + 4y 2 + 1 x2
suf
i
At any point (x, y, z) on the surface, we have
ƒsx, y, zd = x 2 + y 2 + z 2 §ƒ = 2xi + 2yj + 2zk ƒ §ƒ ƒ = 22x 2 + y 2 + z 2 = 222
Because x 2 + y 2 + z 2 = 2 at points of S
ƒ §ƒ # p ƒ = ƒ §ƒ # k ƒ = ƒ 2z ƒ = 2z . Therefore, Surface area =
ƒ §ƒ ƒ 222 dA dA = dA = 22 z . # 2z §ƒ p ƒ 6 ƒ 6 6 R
R
(4)
R
What do we do about the z? Since z is the zcoordinate of a point on the sphere, we can express it in terms of x and y as
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z = 22  x 2  y 2.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5 Surface Area and Surface Integrals
1185
6
dA z = 22
R
dA
6
x2 + y2 … 1
2p
suf
Surface area = 22
22  x 2  y 2
1
r dr du = 22 L0 L0 22  r 2 L0 L0
(xk , yk , zk )
∆ Pk
∆k
We now show how to integrate a function over a surface, using the ideas just developed for calculating surface area. Suppose, for example, that we have an electrical charge distributed over a surface ƒsx, y, zd = c like the one shown in Figure 16.43 and that the function g(x, y, z) gives the charge per unit area (charge density) at each point on S. Then we may calculate the total charge on S as an integral in the following way. We partition the shadow region R on the ground plane beneath the surface into small rectangles of the kind we would use if we were defining the surface area of S. Then directly above each ¢Ak lies a patch of surface ¢sk that we approximate with a parallelogramshaped portion of tangent plane, ¢Pk . (See Figure 16.43.) Up to this point the construction proceeds as in the definition of surface area, but now we take an additional step: We evaluate g at sxk , yk , zk d and approximate the total charge on the surface path ¢sk by the product gsxk , yk , zk d ¢Pk . The rationale is that when the partition of R is sufficiently fine, the value of g throughout ¢sk is nearly constant and ¢Pk is nearly the same as ¢sk . The total charge over S is then approximated by the sum
ma
∆Ak
ham
FIGURE 16.43 If we know how an electrical charge g(x, y, z) is distributed over a surface, we can find the total charge with a suitably modified surface integral.
Mu
A 22  1 B du = 2p A 2  22 B .
dH
S
R
du
r=0
ass a
f (x, y, z) c
r=1
nR
Surface Integrals
cs2  r 2 d1>2 d
iaz
2p
= 22
Polar coordinates
You
2p
= 22
i
We continue the work of Equation (4) with this substitution:
¢Ak Total charge L a g sxk , yk , zk d ¢Pk = a g sxk , yk , zk d . gk ƒ cos ƒ
If ƒ, the function defining the surface S, and its first partial derivatives are continuous, and if g is continuous over S, then the sums on the righthand side of the last equation approach the limit
6 R
gsx, y, zd
ƒ §ƒ ƒ dA = gsx, y, zd dA ƒ cos g ƒ ƒ §ƒ # p ƒ 6
(5)
R
as the partition of R is refined in the usual way. This limit is called the integral of g over the surface S and is calculated as a double integral over R. The value of the integral is the total charge on the surface S. As you might expect, the formula in Equation (5) defines the integral of any function g over the surface S as long as the integral exists.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1186
Chapter 16: Integration in Vector Fields
ƒ §ƒ ƒ dA, ƒ §ƒ # p ƒ
You
6
gsx, y, zd
suf
i
DEFINITION Surface Integral If R is the shadow region of a surface S defined by the equation ƒsx, y, zd = c, and g is a continuous function defined at the points of S, then the integral of g over S is the integral
R
(6)
where p is a unit vector normal to R and §ƒ # p Z 0. The integral itself is called a surface integral.
nR
iaz
The integral in Equation (6) takes on different meanings in different applications. If g has the constant value 1, the integral gives the area of S. If g gives the mass density of a thin shell of material modeled by S, the integral gives the mass of the shell. We can abbreviate the integral in Equation (6) by writing ds for s ƒ §ƒ ƒ > ƒ §ƒ # p ƒ d dA. The Surface Area Differential and the Differential Form for Surface Integrals ƒ §ƒ ƒ dA ƒ §ƒ # p ƒ
ass a
ds =
6
g ds
(7)
S
Surface area differential
Differential formula for surface integrals
dH
Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum of their integrals and so on. The domain Additivity Property takes the form 6
g ds =
ham
z
0
Mu
1
1
g ds + Á +
S2
6
g ds.
Sn
The idea is that if S is partitioned by smooth curves into a finite number of nonoverlapping smooth patches (i.e., if S is piecewise smooth), then the integral over S is the sum of the integrals over the patches. Thus, the integral of a function over the surface of a cube is the sum of the integrals over the faces of the cube. We integrate over a turtle shell of welded plates by integrating one plate at a time and adding the results.
y
Side C
EXAMPLE 3
The cube in Example 3.
Integrating Over a Surface
Integrate g sx, y, zd = xyz over the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1 (Figure 16.44). We integrate xyz over each of the six sides and add the results. Since xyz = 0 on the sides that lie in the coordinate planes, the integral over the surface of the cube reduces to
Solution
Side B
FIGURE 16.44
6
Side A
1
x
S1
g ds +
ma
S
6
6
xyz ds =
Cube surface
To Read it Online & Download:
6 Side A
xyz ds +
6 Side B
xyz ds +
6
xyz ds.
Side C
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5
Surface Area and Surface Integrals
1187
ƒ §ƒ # p ƒ = ƒ k # k ƒ = 1
ƒ §ƒ ƒ = 1,
§ƒ = k,
ƒ §ƒ ƒ 1 dA = dx dy = dx dy 1 ƒ §ƒ # p ƒ
ds =
You
p = k,
suf
i
Side A is the surface ƒsx, y, zd = z = 1 over the square region Rxy: 0 … x … 1, 0 … y … 1, in the xyplane. For this surface and region,
xyz = xys1d = xy and
1
6
Side A
Positive direction
6
Rxy
6
b
3 1 1 1 + + = . 4 4 4 4
We call a smooth surface S orientable or twosided if it is possible to define a field n of unit normal vectors on S that varies continuously with position. Any patch or subportion of an orientable surface is orientable. Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable. By convention, we choose n on a closed surface to point outward. Once n has been chosen, we say that we have oriented the surface, and we call the surface together with its normal field an oriented surface. The vector n at any point is called the positive direction at that point (Figure 16.45). The Möbius band in Figure 16.46 is not orientable. No matter where you start to construct a continuousunit normal field (shown as the shaft of a thumbtack in the figure), moving the vector continuously around the surface in the manner shown will return it to the starting point with a direction opposite to the one it had when it started out. The vector at that point cannot point both ways and yet it must if the field is to be continuous. We conclude that no such field exists.
ma
ac
xyz ds =
dH
Start Finish
y 1 dy = . 2 4 0 L
ass a
c
db
1
xy dx dy =
nR
Orientation
FIGURE 16.45 Smooth closed surfaces in space are orientable. The outward unit normal vector defines the positive direction at each point.
a
L0 L0
1
Symmetry tells us that the integrals of xyz over sides B and C are also 1> 4. Hence,
Cube surface
d
xy dx dy =
iaz
n
xyz ds =
Mu
ham
FIGURE 16.46 To make a Möbius band, take a rectangular strip of paper abcd, give the end bc a single twist, and paste the ends of the strip together to match a with c and b with d. The Möbius band is a nonorientable or onesided surface.
Surface Integral for Flux
Suppose that F is a continuous vector field defined over an oriented surface S and that n is the chosen unit normal field on the surface. We call the integral of F # n over S the flux of F across S in the positive direction. Thus, the flux is the integral over S of the scalar component of F in the direction of n.
DEFINITION Flux The flux of a threedimensional vector field F across an oriented surface S in the direction of n is
To Read it Online & Download:
Flux =
6
F # n ds.
S
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1188
Chapter 16: Integration in Vector Fields
3
suf
i
The definition is analogous to the flux of a twodimensional field F across a plane curve C. In the plane (Section 16.2), the flux is F # n ds,
C
You
the integral of the scalar component of F normal to the curve. If F is the velocity field of a threedimensional fluid flow, the flux of F across S is the net rate at which fluid is crossing S in the chosen positive direction. We discuss such flows in more detail in Section 16.7. If S is part of a level surface gsx, y, zd = c, then n may be taken to be one of the two fields §g , ƒ §g ƒ
iaz
n = ;
(9)
depending on which one gives the preferred direction. The corresponding flux is 6 S
=
6
aF #
; §g ƒ §g ƒ b # p ƒ dA §g §g ƒ ƒ ƒ
Equations (9) and (7)
ass a
R
F # n ds
nR
Flux =
=
6
F#
; §g dA. ƒ §g # p ƒ
(8)
(10)
R
EXAMPLE 4
Finding Flux
dH
Find the flux of F = yzj + z 2k outward through the surface S cut from the cylinder y 2 + z 2 = 1, z Ú 0, by the planes x = 0 and x = 1.
Solution The outward normal field on S (Figure 16.47) may be calculated from the gradient of gsx, y, zd = y 2 + z 2 to be §g 2yj + 2zk 2yj + 2zk n = + = = yj + zk. = ƒ §g ƒ 24y 2 + 4z 2 221
z
z2
1 n
ma
y2
With p = k, we also have
1
R xy
ham
(1, –1, 0)
(1, 1, 0)
x
ds =
y
We can drop the absolute value bars because z Ú 0 on S. The value of F # n on the surface is F # n = syzj + z2 kd # s yj + zkd
FIGURE 16.47 Calculating the flux of a vector field outward through this surface. The area of the shadow region Rxy is 2 (Example 4).
Mu
ƒ §g ƒ 2 1 dA = dA = z dA. ƒ §g # k ƒ ƒ 2z ƒ
= y 2z + z 3 = zs y 2 + z 2 d = z.
y2 + z2 = 1 on S
Therefore, the flux of F outward through S is 6
F # n ds =
S
To Read it Online & Download:
6 S
1 szd a z dAb =
6
dA = areasRxy d = 2.
Rxy
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5
Surface Area and Surface Integrals
1189
i
Moments and Masses of Thin Shells
suf
Thin shells of material like bowls, metal drums, and domes are modeled with surfaces. Their moments and masses are calculated with the formulas in Table 16.3.
Mass:
M =
6
dsx, y, zd ds
S
You
TABLE 16.3 Mass and moment formulas for very thin shells
sdsx, y, zd = density at sx, y, zd, mass per unit area)
First moments about the coordinate planes: 6
x d ds,
S
Mxz =
6
y d ds,
iaz
Myz =
Mxy =
S
6
z d ds
S
Coordinates of center of mass:
y = Mxz >M,
nR
x = Myz >M,
z = Mxy >M
Moments of inertia about coordinate axes: Ix =
6
s y 2 + z 2 d d ds,
Iy =
ass a
S
Iz =
6
sx 2 + y 2 d d ds,
6
sx 2 + z 2 d d ds,
S
IL =
S
6
r 2d ds,
S
rsx, y, zd = distance from point sx, y, zd to line L
dH
Radius of gyration about a line L: RL = 2IL >M
EXAMPLE 5
Finding Center of Mass
ma
Find the center of mass of a thin hemispherical shell of radius a and constant density d. Solution
z
y2
z2
c.m.
S
a
R
Mu
a
x
We model the shell with the hemisphere ƒsx, y, zd = x 2 + y 2 + z 2 = a 2,
a2
ham
0, 0, a 2
x2
(Figure 16.48). The symmetry of the surface about the zaxis tells us that x = y = 0. It remains only to find z from the formula z = Mxy >M. The mass of the shell is M =
6 S
y
x2 y2 a2
z Ú 0
d ds = d ds = sddsarea of Sd = 2pa 2d. 6 S
To evaluate the integral for Mxy, we take p = k and calculate
FIGURE 16.48 The center of mass of a thin hemispherical shell of constant density lies on the axis of symmetry halfway from the base to the top (Example 5).
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ƒ §ƒ ƒ = ƒ 2xi + 2yj + 2zk ƒ = 22x 2 + y 2 + z 2 = 2a ƒ §ƒ # p ƒ = ƒ §ƒ # k ƒ = ƒ 2z ƒ = 2z ds =
ƒ §ƒ ƒ a dA = z dA. ƒ §ƒ # p ƒ
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1190
Chapter 16: Integration in Vector Fields
S
z =
6
a z z dA = da
6
R
dA = daspa2 d = dpa3
R
Mxy pa 3d a = = . M 2 2pa 2d
Mu
ham
ma
dH
ass a
nR
iaz
The shellâ€™s center of mass is the point (0, 0, a>2).
suf
6
zd ds = d
You
Mxy =
i
Then
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Chapter 16: Integration in Vector Fields
EXERCISES 16.5 Surface Area 2
2
1. Find the area of the surface cut from the paraboloid x + y  z = 0 by the plane z = 2. 2. Find the area of the band cut from the paraboloid x 2 + y 2  z = 0 by the planes z = 2 and z = 6. 3. Find the area of the region cut from the plane x + 2y + 2z = 5 by the cylinder whose walls are x = y 2 and x = 2  y 2. 4. Find the area of the portion of the surface x 2  2z = 0 that lies above the triangle bounded by the lines x = 23, y = 0, and y = x in the xyplane. 2
2
2
2
6. Find the area of the cap cut from the sphere x + y + z = 2 by the cone z = 2x 2 + y 2.
7. Find the area of the ellipse cut from the plane z = cx (c a constant) by the cylinder x 2 + y 2 = 1. 2
2
8. Find the area of the upper portion of the cylinder x + z = 1 that lies between the planes x = ;1>2 and y = ;1>2.
a m m
9. Find the area of the portion of the paraboloid x = 4  y 2  z 2 that lies above the ring 1 … y 2 + z 2 … 4 in the yzplane. 2
2
10. Find the area of the surface cut from the paraboloid x + y + z = 2 by the plane y = 0.
11. Find the area of the surface x 2  2 ln x + 215y  z = 0 above the square R: 1 … x … 2, 0 … y … 1, in the xyplane.
a h u M
14. Integrate gsx, y, zd = y + z over the surface of the wedge in the first octant bounded by the coordinate planes and the planes x = 2 and y + z = 1.
Y z
15. Integrate gsx, y, zd = xyz over the surface of the rectangular solid cut from the first octant by the planes x = a, y = b, and z = c.
12. Find the area of the surface 2x 3>2 + 2y 3>2  3z = 0 above the square R: 0 … x … 1, 0 … y … 1, in the xyplane.
Surface Integrals
13. Integrate gsx, y, zd = x + y + z over the surface of the cube cut from the first octant by the planes x = a, y = a, z = a.
To Read it Online & Download:
iR a
16. Integrate gsx, y, zd = xyz over the surface of the rectangular solid bounded by the planes x = ;a, y = ;b, and z = ;c. 17. Integrate gsx, y, zd = x + y + z over the portion of the plane 2x + 2y + z = 2 that lies in the first octant.
n a sa s H d
5. Find the area of the surface x  2y  2z = 0 that lies above the triangle bounded by the lines x = 2, y = 0, and y = 3x in the xyplane.
i f u s u o
18. Integrate gsx, y, zd = x2y 2 + 4 over the surface cut from the parabolic cylinder y 2 + 4z = 16 by the planes x = 0, x = 1, and z = 0.
Flux Across a Surface
In Exercises 19 and 20, find the flux of the field F across the portion of the given surface in the specified direction. 19. Fsx, y, zd = i + 2j + 3k S: rectangular surface direction k
z = 0,
0 … x … 2,
0 … y … 3,
20. Fsx, y, zd = yx 2i  2j + xzk S: rectangular surface direction j
y = 0,
1 … x … 2,
2 … z … 7,
In Exercises 21–26, find the flux of the field F across the portion of the sphere x 2 + y 2 + z 2 = a 2 in the first octant in the direction away from the origin. 21. Fsx, y, zd = zk
22. Fsx, y, zd = yi + xj
23. Fsx, y, zd = yi  xj + k
24. Fsx, y, zd = zxi + zyj + z2k
25. Fsx, y, zd = xi + yj + zk xi + yj + zk 26. Fsx, y, zd = 2x 2 + y 2 + z 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5 Surface Area and Surface Integrals z
suf
i
27. Find the flux of the field Fsx, y, zd = z 2i + xj  3zk outward through the surface cut from the parabolic cylinder z = 4  y 2 by the planes x = 0, x = 1, and z = 0.
29. Let S be the portion of the cylinder y = e x in the first octant that projects parallel to the xaxis onto the rectangle Ryz: 1 … y … 2, 0 … z … 1 in the yzplane (see the accompanying figure). Let n be the unit vector normal to S that points away from the yzplane. Find the flux of the field Fsx, y, zd = 2i + 2yj + zk across S in the direction of n.
4x 2 4y2 z 2 0
zⱖ0
y
2
x2
y2
2x or r 2 cos
iaz
x
z
Ry z
nR
37. Spherical shells
a. Find the moment of inertia about a diameter of a thin spherical shell of radius a and constant density d . (Work with a hemispherical shell and double the result.)
1 ex
S
2
y
b. Use the Parallel Axis Theorem (Exercises 15.5) and the result in part (a) to find the moment of inertia about a line tangent to the shell.
ass a
x y
You
28. Find the flux of the field Fsx, y, zd = 4xi + 4yj + 2k outward (away from the zaxis) through the surface cut from the bottom of the paraboloid z = x 2 + y 2 by the plane z = 1.
1
1191
b. Use Pappus’s formula (Exercises 15.5) and the result in part (a) to find the centroid of the complete surface of a solid cone (side plus base).
dH
30. Let S be the portion of the cylinder y = ln x in the first octant whose projection parallel to the yaxis onto the xzplane is the rectangle Rxz: 1 … x … e, 0 … z … 1. Let n be the unit vector normal to S that points away from the xzplane. Find the flux of F = 2yj + zk through S in the direction of n.
38. a. Cones with and without ice cream Find the centroid of the lateral surface of a solid cone of base radius a and height h (cone surface minus the base).
c. A cone of radius a and height h is joined to a hemisphere of radius a to make a surface S that resembles an ice cream cone. Use Pappus’s formula and the results in part (a) and Example 5 to find the centroid of S. How high does the cone have to be to place the centroid in the plane shared by the bases of the hemisphere and cone?
31. Find the outward flux of the field F = 2xyi + 2yzj + 2xzk across the surface of the cube cut from the first octant by the planes x = a, y = a, z = a.
ma
32. Find the outward flux of the field F = xzi + yzj + k across the surface of the upper cap cut from the solid sphere x 2 + y 2 + z 2 … 25 by the plane z = 3.
Moments and Masses
ham
33. Centroid Find the centroid of the portion of the sphere x 2 + y 2 + z 2 = a 2 that lies in the first octant. 34. Centroid Find the centroid of the surface cut from the cylinder y 2 + z 2 = 9, z Ú 0, by the planes x = 0 and x = 3 (resembles the surface in Example 4).
Mu
35. Thin shell of constant density Find the center of mass and the moment of inertia and radius of gyration about the zaxis of a thin shell of constant density d cut from the cone x 2 + y 2  z 2 = 0 by the planes z = 1 and z = 2,
36. Conical surface of constant density Find the moment of inertia about the zaxis of a thin shell of constant density d cut from the cone 4x 2 + 4y 2  z 2 = 0, z Ú 0, by the circular cylinder x 2 + y 2 = 2x (see the accompanying figure).
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Special Formulas for Surface Area If S is the surface defined by a function z = ƒsx, yd that has continuous first partial derivatives throughout a region Rxy in the xyplane (Figure 16.49), then S is also the level surface Fsx, y, zd = 0 of the function Fsx, y, zd = ƒsx, yd  z . Taking the unit normal to Rxy to be p = k then gives 2 2 ƒ §F ƒ = ƒ ƒx i + ƒy j  k ƒ = 2ƒ x + ƒ y + 1
ƒ §F # p ƒ = ƒ sƒx i + ƒy j  kd # k ƒ = ƒ 1 ƒ = 1 and ƒ §F ƒ 2 2 # p ƒ dA = 6 2ƒx + ƒ y + 1 dx dy, §F 6 ƒ Rxy
Rxy
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1192
Chapter 16: Integration in Vector Fields Surface z f (x, y)
6
2ƒy2 + ƒz 2 + 1 dy dz,
(12) S
Ryz
6
2ƒ x2 + ƒ z 2 + 1 dx dz.
You
and the area of a smooth y = ƒsx, zd over a region Rxz in the xzplane is A =
suf
A =
i
Similarly, the area of a smooth surface x = ƒs y, zd over a region Ryz in the yzplane is
(13)
Rxz
Use Equations (11)–(13) to find the area of the surfaces in Exercises 39 – 44.
41. The portion of the cone z = 2x 2 + y 2 that lies over the region between the circle x 2 + y 2 = 1 and the ellipse 9x 2 + 4y 2 = 36 in the xyplane. (Hint: Use formulas from geometry to find the area of the region.)
A =
6
2ƒ x2 + ƒ y2 + 1 dx dy.
Rxy
43. The surface in the first octant cut from the cylinder y = s2>3dz 3>2 by the planes x = 1 and y = 16>3
ass a
42. The triangle cut from the plane 2x + 6y + 3z = 6 by the bounding planes of the first octant. Calculate the area three ways, once with each area formula
iaz
40. The surface cut from the “nose” of the paraboloid x = 1  y 2  z 2 by the yzplane
FIGURE 16.49 For a surface z = ƒsx, yd, the surface area formula in Equation (3) takes the form
nR
39. The surface cut from the bottom of the paraboloid z = x 2 + y 2 by the plane z = 3
Mu
ham
ma
dH
44. The portion of the plane y + z = 4 that lies above the region cut from the first quadrant of the xzplane by the parabola x = 4  z2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
1192
i f su
Chapter 16: Integration in Vector Fields
Parametrized Surfaces
16.6
z a i R
We have defined curves in the plane in three different ways: Explicit form: Implicit form: Parametric vector form:
y = ƒsxd Fsx, yd = 0 rstd = ƒstdi + gstdj,
n a ss
u o Y
a … t … b.
We have analogous definitions of surfaces in space:
a H
Explicit form: Implicit form:
z = ƒsx, yd Fsx, y, zd = 0.
There is also a parametric form that gives the position of a point on the surface as a vector function of two variables. The present section extends the investigation of surface area and surface integrals to surfaces described parametrically.
d a m m a uh
Parametrizations of Surfaces Let
M
rsu, yd = ƒsu, ydi + gsu, ydj + hsu, ydk
(1)
be a continuous vector function that is defined on a region R in the uyplane and onetoone on the interior of R (Figure 16.50). We call the range of r the surface S defined or traced by r. Equation (1) together with the domain R constitute a parametrization of the surface. The variables u and y are the parameters, and R is the parameter domain.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.6 Parametrized Surfaces u constant
To simplify our discussion, we take R to be a rectangle defined by inequalities of the form a … u … b, c … y … d. The requirement that r be onetoone on the interior of R ensures that S does not cross itself. Notice that Equation (1) is the vector equivalent of three parametric equations: x = ƒsu, yd,
R
y = gsu, yd,
u
0
EXAMPLE 1
Parametrizing a Cone
Find a parametrization of the cone
Parametrization
z = 2x 2 + y 2, z
P
Here, cylindrical coordinates provide everything we need. A typical point (x, y, z) on the cone (Figure 16.51) has x = r cos u, y = r sin u, and z = 2x 2 + y 2 = r, with 0 … r … 1 and 0 … u … 2p . Taking u = r and y = u in Equation (1) gives the parametrization
Solution
iaz
S
rsr, ud = sr cos udi + sr sin udj + rk,
Curve u constant x
r(u, y) f (u, y)i g(u, y)j h(u, y)k, Position vector to surface point
0 … z … 1.
EXAMPLE 2
0 … r … 1,
nR
Curve y constant
z = hsu, yd.
You
y constant (u, y)
suf
i
y
1193
0 … u … 2p.
Parametrizing a Sphere
Find a parametrization of the sphere x 2 + y 2 + z 2 = a 2.
ass a
y
Spherical coordinates provide what we need. A typical point (x, y, z) on the sphere (Figure 16.52) has x = a sin f cos u, y = a sin f sin u, and z = a cos f, 0 … f … p, 0 … u … 2p. Taking u = f and y = u in Equation (1) gives the parametrization Solution
FIGURE 16.50 A parametrized surface S expressed as a vector function of two variables defined on a region R.
Cone: z 兹x 2 y 2 r
dH
rsf, ud = sa sin f cos udi + sa sin f sin udj + sa cos fdk,
z
EXAMPLE 3
0 … f … p,
0 … u … 2p.
Parametrizing a Cylinder
Find a parametrization of the cylinder
ma
1
x 2 + sy  3d2 = 9,
0 … z … 5.
In cylindrical coordinates, a point (x, y, z) has x = r cos u, y = r sin u, and z = z. For points on the cylinder x 2 + s y  3d2 = 9 (Figure 16.53), the equation is the same as the polar equation for the cylinder’s base in the xyplane: Solution
r(r, ) (r cos )i (r sin )j rk
ham
(x, y, z) (r cos , r sin , r)
r
x
FIGURE 16.51 The cone in Example 1 can be parametrized using cylindrical coordinates.
Mu
x 2 + s y 2  6y + 9d = 9
y
r 2  6r sin u = 0 or r = 6 sin u,
0 … u … p.
A typical point on the cylinder therefore has
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x = r cos u = 6 sin u cos u = 3 sin 2u y = r sin u = 6 sin2 u z = z.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1194
Chapter 16: Integration in Vector Fields z
rsu, zd = s3 sin 2udi + s6 sin2 udj + zk, 0 … u … p,
0 … z … 5.
suf
(x, y, z) (a sin cos , a sin sin , a cos )
i
Taking u = u and y = z in Equation (1) gives the parametrization
a
Surface Area
rsu, yd = ƒsu, ydi + gsu, ydj + hsu, ydk, y
x
(x, y, z) (3 sin 2, 6 sin 2 , z) y
r 6 sin
ry =
0g 0ƒ 0h 0r i + j + k. = 0y 0y 0y 0y
nR
The condition that ru * ry is never the zero vector in the definition of smoothness means that the two vectors ru and ry are nonzero and never lie along the same line, so they always determine a plane tangent to the surface. Now consider a small rectangle ¢Auy in R with sides on the lines u = u0 , u = u0 + ¢u, y = y0 and y = y0 + ¢y (Figure 16.54). Each side of ¢Auy maps to a curve on the surface S, and together these four curves bound a “curved area element” ¢suy. In the notation of the figure, the side y = y0 maps to curve C1, the side u = u0 maps to C2 , and their common vertex su0 , y0 d maps to P0. z
Mu
ham
ma
FIGURE 16.53 The cylinder in Example 3 can be parametrized using cylindrical coordinates.
0ƒ 0g 0h 0r = i + j + k 0u 0u 0u 0u
dH
x
ru =
DEFINITION Smooth Parametrized Surface A parametrized surface rsu, yd = ƒsu, ydi + gsu, ydj + hsu, ydk is smooth if ru and ry are continuous and ru * ry is never zero on the parameter domain.
z r(, z)
c … y … d.
ass a
Cylinder: x 2 ( y 3)2 9 or r 6 sin
a … u … b,
We need S to be smooth for the construction we are about to carry out. The definition of smoothness involves the partial derivatives of r with respect to u and y:
FIGURE 16.52 The sphere in Example 2 can be parametrized using spherical coordinates.
z
You
H (, )
iaz
a
Our goal is to find a double integral for calculating the area of a curved surface S based on the parametrization
C1: y y0
P0
C 2: u u 0
Parametrization
y
S
d
∆uy
y0 ∆y
∆ A uy
y0
u u 0 ∆u
R
c 0
y y0 ∆y
x a
u0
u0 ∆u
b
u y
FIGURE 16.54 A rectangular area element ¢Auy in the uyplane maps onto a curved area element ¢suy on S.
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Parametrized Surfaces
1195
Figure 16.55 shows an enlarged view of ¢suy. The vector rusu0, y0 d is tangent to C1 at P0. Likewise, rysu0, y0 d is tangent to C2 at P0. The cross product ru * ry is normal to the surface at P0. (Here is where we begin to use the assumption that S is smooth. We want to be sure that ru * ry Z 0.) We next approximate the surface element ¢suy by the parallelogram on the tangent plane whose sides are determined by the vectors ¢uru and ¢yry (Figure 16.56). The area of this parallelogram is
P0 ru C1: y y0
ry C2: u u0
∆uy
You
z
suf
i
ru ry
y
x
ƒ ¢uru * ¢yry ƒ = ƒ ru * ry ƒ ¢u ¢y.
A partition of the region R in the uyplane by rectangular regions ¢Auy generates a partition of the surface S into surface area elements ¢suy. We approximate the area of each surface element ¢suy by the parallelogram area in Equation (2) and sum these areas together to obtain an approximation of the area of S:
iaz
FIGURE 16.55 A magnified view of a surface area element ¢suy.
a a ƒ ru * ry ƒ ¢u ¢y.
z P0
u
y
∆uy
C2
DEFINITION Area of a Smooth Surface The area of the smooth surface rsu, yd = ƒsu, ydi + gsu, ydj + hsu, ydk,
is
dH
FIGURE 16.56 The parallelogram determined by the vectors ¢uru and ¢yry approximates the surface area element ¢suy.
As ¢u and ¢y approach zero independently, the continuity of ru and ry guarantees that the d b sum in Equation (3) approaches the double integral 1c 1a ƒ ru * ry ƒ du dy. This double integral defines the area of the surface S and agrees with previous definitions of area, though it is more general.
nR
∆yry
C1
(3)
y
ass a
∆uru
x
(2)
A =
d
a … u … b,
c … y … d
b
Lc La
ƒ ru * ry ƒ du dy.
(4)
ma
As in Section 16.5, we can abbreviate the integral in Equation (4) by writing ds for ƒ ru * ry ƒ du dy.
Mu
ham
Surface Area Differential and Differential Formula for Surface Area ds = ƒ ru * ry ƒ du dy
6
ds
(5)
S
Surface area differential
EXAMPLE 4
Differential formula for surface area
Finding Surface Area (Cone)
Find the surface area of the cone in Example 1 (Figure 16.51). Solution
In Example 1, we found the parametrization rsr, ud = sr cos udi + sr sin udj + rk,
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0 … r … 1,
0 … u … 2p.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1196
Chapter 16: Integration in Vector Fields
j sin u r cos u
k 13 0
suf
i 3 rr * ru = cos u r sin u
i
To apply Equation (4), we first find rr * ru :
= sr cos udi  sr sin udj + sr cos2 u + r sin2 udk.
You
('''')''''* r
Thus, ƒ rr * ru ƒ = 2r 2 cos2 u + r 2 sin2 u + r 2 = 22r 2 = 22r. The area of the cone is 1
2p
=
ƒ rr * ru ƒ dr du
L0 L0
Equation (4) with u = r, y = u
1
L0 L0
L0
22 22 du = s2pd = p22 units squared. 2 2
Finding Surface Area (Sphere)
nR
EXAMPLE 5
2p
22 r dr du =
iaz
2p
A =
Find the surface area of a sphere of radius a. Solution
We use the parametrization from Example 2:
ass a
rsf, ud = sa sin f cos udi + sa sin f sin udj + sa cos fdk, 0 … f … p, 0 … u … 2p.
For rf * ru, we get
dH
i rf * ru = 3 a cos f cos u a sin f sin u
j a cos f sin u a sin f cos u
k  a sin f 3 0
= sa 2 sin2 f cos udi + sa 2 sin2 f sin udj + sa 2 sin f cos fdk.
Thus,
Mu
ham
ma
ƒ rf * ru ƒ = 2a 4 sin4 f cos2 u + a 4 sin4 f sin2 u + a 4 sin2 f cos2 f = 2a 4 sin4 f + a 4 sin2 f cos2 f = 2a 4 sin2 f ssin2 f + cos2 fd = a 2 2sin2 f = a 2 sin f,
since sin f Ú 0 for 0 … f … p . Therefore, the area of the sphere is 2p
A =
L0 2p
=
L0
p
L0
a 2 sin f df du 2p
p
ca 2 cos f d du = 0
L0
2a 2 du = 4pa 2 units squared.
This agrees with the wellknown formula for the surface area of a sphere.
Surface Integrals Having found a formula for calculating the area of a parametrized surface, we can now integrate a function over the surface using the parametrized form.
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Parametrized Surfaces
1197
6 S
EXAMPLE 6
b
Lc La
Gsƒsu, yd, gsu, yd, hsu, ydd ƒ ru * ry ƒ du dy.
You
d
Gsx, y, zd ds =
suf
i
DEFINITION Parametric Surface Integral If S is a smooth surface defined parametrically as rsu, yd = ƒsu, ydi + gsu, ydj + hsu, ydk, a … u … b, c … y … d, and G(x, y, z) is a continuous function defined on S, then the integral of G over S is
Integrating Over a Surface Defined Parametrically
Solution
Continuing the work in Examples 1 and 4, we have ƒ rr * ru ƒ = 22r and 2p
x 2 ds =
S
1
L0 L0
A r 2 cos2 u B A 22r B dr du
2p
= 22
1
r 3 cos2 u dr du
L0 L0
2p 22 2p 22 u p22 1 . cos2 u du = c + sin 2u d = 4 L0 4 2 4 4 0
ass a
=
EXAMPLE 7
z
Finding Flux
Find the flux of F = yzi + xj  z 2 k outward through the parabolic cylinder y = x 2, 0 … x … 1, 0 … z … 4 (Figure 16.57).
dH
4 (1, 0, 4)
x = r cos u
nR
6
iaz
Integrate Gsx, y, zd = x 2 over the cone z = 2x 2 + y 2, 0 … z … 1.
y x2
On the surface we have x = x, y = x 2, and z = z, so we automatically have the parametrization rsx, zd = xi + x 2 j + zk, 0 … x … 1, 0 … z … 4. The cross product of tangent vectors is
Solution
1 1 x
i 3 rx * rz = 1 0
ma
n y
j 2x 0
k 0 3 = 2xi  j. 1
The unit normal pointing outward from the surface is n =
2xi  j rx * rz = . ƒ rx * rz ƒ 24x 2 + 1
On the surface, y = x 2 , so the vector field there is F = yzi + xj  z 2 k = x 2zi + xj  z 2 k. Thus,
Mu
ham
FIGURE 16.57 Finding the flux through the surface of a parabolic cylinder (Example 7).
To Read it Online & Download:
F#n =
=
1 ssx 2zds2xd + sxds 1d + s z 2 ds0dd 24x 2 + 1 2x 3z  x 24x 2 + 1
.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1198
Chapter 16: Integration in Vector Fields
S
1
2x 3z  x
L0 L0 24x 2 + 1 4
=
ƒ rx * rz ƒ dx dz
L0 L0 24x 2 + 1 4
=
2x 3z  x
24x 2 + 1 dx dz
You
6
1
1
1 1 sz  1d dz = sz  1d2 d 2 4 0 L0
=
1 1 s9d  s1d = 2. 4 4
iaz
=
Finding a Center of Mass
nR
EXAMPLE 8
1 1 c x 4z  x 2 d dz 2 2 x=0 L0 4
4
z
x=1
4
s2x 3z  xd dx dz =
L0 L0
suf
4
F # n ds =
i
The flux of F outward through the surface is
Find the center of mass of a thin shell of constant density d cut from the cone z = 2x 2 + y 2 by the planes z = 1 and z = 2 (Figure 16.58). z 兹x 2 y 2
1
The symmetry of the surface about the zaxis tells us that x = y = 0 . We find z = Mxy >M . Working as in Examples 1 and 4, we have
Solution
ass a
2
rsr, ud = r cos ui + r sin uj + rk, and
Therefore,
M =
ham
2p
6
d ds = 2p
L0
= d22 c
c
2
L0 L1
S
= d22
ma
FIGURE 16.58 The cone frustum formed when the cone z = 2x 2 + y 2 is cut by the planes z = 1 and z = 2 (Example 8).
Mu
0 … u … 2p,
ƒ rr * ru ƒ = 22r.
y
dH
x
1 … r … 2,
2
r2 d du = d22 2 1 L0
6
L0 L1
S
2p
= d22
1 b du 2
2p
L0
2
dr22r dr du
2
L0 L1
z =
a2 
2p
dz ds =
= d22
2p
3u d = 3pd22 2 0 2p
Mxy =
d22r dr du
2p
r 2 dr du = d22
L0
c
2
r3 d du 3 1
7 14 du = pd22 3 3
Mxy 14pd22 14 = . = M 9 3 A 3pd22 B
The shell’s center of mass is the point (0, 0, 14> 9).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.6 Parametrized Surfaces
2
1. The paraboloid z = x + y , z … 4 2. The paraboloid z = 9  x 2  y 2, z Ú 0 3. Cone frustum The firstoctant portion of the cone z = 2x 2 + y 2>2 between the planes z = 0 and z = 3
4. Cone frustum The portion of the cone z = 22x 2 + y 2 between the planes z = 2 and z = 4 5. Spherical cap The cap cut from the sphere x 2 + y 2 + z 2 = 9 by the cone z = 2x 2 + y 2 6. Spherical cap The portion of the sphere x 2 + y 2 + z 2 = 4 in the first octant between the xyplane and the cone z = 2x 2 + y 2 7. Spherical band
The portion of the sphere x 2 + y 2 + z 2 = 3
between the planes z = 23>2 and z =  23>2
17. Titled plane inside cylinder The portion of the plane y + 2z = 2 inside the cylinder x 2 + y 2 = 1 18. Plane inside cylinder The portion of the plane z = x inside the cylinder x 2 + y 2 = 4 19. Cone frustum The portion of the cone z = 2 2x 2 + y 2 between the planes z = 2 and z = 6
20. Cone frustum The portion of the cone z = 2x 2 + y 2>3 between the planes z = 1 and z = 4>3 21. Circular cylinder band The portion of the cylinder x 2 + y 2 = 1 between the planes z = 1 and z = 4 22. Circular cylinder band The portion of the cylinder x 2 + z 2 = 10 between the planes y = 1 and y = 1 23. Parabolic cap The cap cut from the paraboloid z = 2  x 2  y 2 by the cone z = 2x 2 + y 2
24. Parabolic band The portion of the paraboloid z = x 2 + y 2 between the planes z = 1 and z = 4
ass a
8. Spherical cap The upper portion cut from the sphere x 2 + y 2 + z 2 = 8 by the plane z = 2
You
2
iaz
In Exercises 1–16, find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.)
many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.)
nR
Finding Parametrizations for Surfaces
suf
i
EXERCISES 16.6
1199
9. Parabolic cylinder between planes The surface cut from the parabolic cylinder z = 4  y 2 by the planes x = 0, x = 2 , and z = 0
dH
10. Parabolic cylinder between planes The surface cut from the parabolic cylinder y = x 2 by the planes z = 0, z = 3 and y = 2 11. Circular cylinder band The portion of the cylinder y 2 + z 2 = 9 between the planes x = 0 and x = 3 12. Circular cylinder band The portion of the cylinder x 2 + z 2 = 4 above the xyplane between the planes y = 2 and y = 2
ma
13. Tilted plane inside cylinder The portion of the plane x + y + z = 1 a. Inside the cylinder x 2 + y 2 = 9 b. Inside the cylinder y 2 + z 2 = 9
The portion of the plane
ham
14. Tilted plane inside cylinder x  y + 2z = 2
a. Inside the cylinder x 2 + z 2 = 3
b. Inside the cylinder y 2 + z 2 = 2
15. Circular cylinder band The portion of the cylinder sx  2d2 + z 2 = 4 between the planes y = 0 and y = 3
Mu
16. Circular cylinder band The portion of the cylinder y 2 + sz  5d2 = 25 between the planes x = 0 and x = 10
25. Sawedoff sphere The lower portion cut from the sphere x 2 + y 2 + z 2 = 2 by the cone z = 2x 2 + y 2 26. Spherical band The portion of the sphere x 2 + y 2 + z 2 = 4 between the planes z = 1 and z = 23
Integrals Over Parametrized Surfaces In Exercises 27–34, integrate the given function over the given surface. 27. Parabolic cylinder Gsx, y, zd = x, over the parabolic cylinder y = x 2, 0 … x … 2, 0 … z … 3 28. Circular cylinder Gsx, y, zd = z, over the cylindrical surface y 2 + z 2 = 4, z Ú 0, 1 … x … 4 29. Sphere Gsx, y, zd = x 2, over the unit sphere x 2 + y 2 + z 2 = 1 30. Hemisphere Gsx, y, zd = z 2, over the hemisphere x 2 + y 2 + z 2 = a 2, z Ú 0 31. Portion of plane Fsx, y, zd = z, over the portion of the plane x + y + z = 4 that lies above the square 0 … x … 1, 0 … y … 1, in the xyplane 32. Cone Fsx, y, zd = z  x, over the cone 0 … z … 1
z = 2x 2 + y 2,
Areas of Parametrized Surfaces
33. Parabolic dome Hsx, y, zd = x 2 25  4z, over the parabolic dome z = 1  x 2  y 2, z Ú 0
In Exercises 17–26, use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are
34. Spherical cap Hsx, y, zd = yz, over the part of the sphere x 2 + y 2 + z 2 = 4 that lies above the cone z = 2x 2 + y 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields
36. Parabolic cylinder F = x 2j  xzk outward (normal away from the yzplane) through the surface cut from the parabolic cylinder y = x 2, 1 … x … 1 , by the planes z = 0 and z = 2 37. Sphere F = zk across the portion of the sphere x 2 + y 2 + z 2 = a 2 in the first octant in the direction away from the origin 2
2
2
38. Sphere F = xi + yj + zk across the sphere x + y + z = a in the direction away from the origin
2
39. Plane F = 2xyi + 2yzj + 2xzk upward across the portion of the plane x + y + z = 2a that lies above the square 0 … x … a, 0 … y … a , in the xyplane 40. Cylinder F = xi + yj + zk outward through the portion of the cylinder x 2 + y 2 = 1 cut by the planes z = 0 and z = a 41. Cone F = xyi  zk outward (normal away from the zaxis) through the cone z = 2x 2 + y 2, 0 … z … 1 2
i
50. Hemisphere The hemisphere surface rsf, ud = s4 sin f cos udi + s4 sin f sin udj + s4 cos fdk, 0 … f … p>2, 0 … u … 2p, at the point P0 A 22, 22, 223 B corresponding to sf, ud = sp>6, p>4d 51. Circular cylinder The circular cylinder rsu, zd = s3 sin 2udi + s6 sin2 udj + zk, 0 … u … p, at the point P0 A 323>2, 9>2, 0 B corresponding to su, zd = sp>3, 0d (See Example 3.) 52. Parabolic cylinder The parabolic cylinder surface rsx, yd = xi + yj  x 2k,  q 6 x 6 q ,  q 6 y 6 q , at the point P0s1, 2, 1d corresponding to sx, yd = s1, 2d
Further Examples of Parametrizations 53. a. A torus of revolution (doughnut) is obtained by rotating a circle C in the xzplane about the zaxis in space. (See the accompanying figure.) If C has radius r 7 0 and center (R, 0, 0), show that a parametrization of the torus is rsu, yd = ssR + r cos udcos ydi + ssR + r cos udsin ydj + sr sin udk,
ass a
42. Cone F = y i + xzj  k outward (normal away from the zaxis) through the cone z = 2 2x 2 + y 2, 0 … z … 2
suf
35. Parabolic cylinder F = z 2i + xj  3zk outward (normal away from the xaxis) through the surface cut from the parabolic cylinder z = 4  y 2 by the planes x = 0, x = 1 , and z = 0
49. Cone The cone rsr, ud = sr cos udi + sr sin udj + rk, r Ú 0, 0 … u … 2p at the point P0 A 22, 22, 2 B corresponding to sr, ud = s2, p>4d
You
In Exercises 35–44, use a parametrization to find the flux 4S F # n ds across the surface in the given direction.
iaz
Flux Across Parametrized Surfaces
nR
1200
43. Cone frustum F = xi  yj + z 2k outward (normal away from the zaxis) through the portion of the cone z = 2x 2 + y 2 between the planes z = 1 and z = 2
where 0 … u … 2p and 0 … y … 2p are the angles in the figure.
b. Show that the surface area of the torus is A = 4p2Rr.
Moments and Masses
dH
44. Paraboloid F = 4xi + 4yj + 2k outward (normal way from the zaxis) through the surface cut from the bottom of the paraboloid z = x 2 + y 2 by the plane z = 1
z
C r
45. Find the centroid of the portion of the sphere x 2 + y 2 + z 2 = a 2 that lies in the first octant.
u x
ma
0 R
46. Find the center of mass and the moment of inertia and radius of gyration about the zaxis of a thin shell of constant density d cut from the cone x 2 + y 2  z 2 = 0 by the planes z = 1 and z = 2.
z
ham
47. Find the moment of inertia about the zaxis of a thin spherical shell x 2 + y 2 + z 2 = a 2 of constant density d. 48. Find the moment of inertia about the zaxis of a thin conical shell z = 2x 2 + y 2, 0 … z … 1, of constant density d.
Planes Tangent to Parametrized Surfaces
Mu
The tangent plane at a point P0sƒsu0 , y0 d, gsu0 , y0 d, hsu0 , y0 dd on a parametrized surface rsu, yd = ƒsu, ydi + gsu, ydj + hsu, ydk is the plane through P0 normal to the vector rusu0 , y0 d * rysu0 , y0 d, the cross product of the tangent vectors rusu0 , y0 d and rysu0 , y0 d at P0. In Exercises 49–52, find an equation for the plane tangent to the surface at P0. Then find a Cartesian equation for the surface and sketch the surface and tangent plane together.
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y r(u, y)
x
y u
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.6 Parametrized Surfaces
rsu, yd = ƒsudi + sgsudcos ydj + sgsudsin ydk is a parametrization of the resulting surface of revolution, where 0 … y … 2p is the angle from the xyplane to the point r(u, y) on the surface. (See the accompanying figure.) Notice that ƒ(u) measures distance along the axis of revolution and g(u) measures distance from the axis of revolution. y
i
55. a. Parametrization of an ellipsoid Recall the parametrization x = a cos u, y = b sin u, 0 … u … 2p for the ellipse sx2>a2 d + s y2>b2 d = 1 (Section 3.5, Example 13). Using the angles u and f in spherical coordinates, show that
suf
a. Show that
b. Find a parametrization for the surface obtained by revolving the curve x = y 2, y Ú 0 , about the xaxis.
rsu, fd = sa cos u cos fdi + sb sin u cos fdj + sc sin fdk
You
54. Parametrization of a surface of revolution Suppose that the parametrized curve C: (ƒ(u), g(u)) is revolved about the xaxis, where gsud 7 0 for a … u … b .
1201
is a parametrization of the ellipsoid sx 2>a 2 d + s y 2>b 2 d + sz 2>c 2 d = 1.
b. Write an integral for the surface area of the ellipsoid, but do not evaluate the integral.
iaz
56. Hyperboloid of one sheet a. Find a parametrization for the hyperboloid of one sheet x 2 + y 2  z 2 = 1 in terms of the angle u associated with the circle x 2 + y 2 = r 2 and the hyperbolic parameter u associated with the hyperbolic function r 2  z 2 = 1. (See Section 7.8, Exercise 84.)
( f (u), g(u), 0) r(u, y)
nR
C y
b. Generalize the result in part (a) to the hyperboloid sx2>a2 d + s y2>b2 d  sz2>c2 d = 1.
g(u) z
f (u)
57. (Continuation of Exercise 56.) Find a Cartesian equation for the plane tangent to the hyperboloid x 2 + y 2  z 2 = 25 at the point sx0 , y0 , 0d, where x02 + y02 = 25.
ass a
x
Mu
ham
ma
dH
58. Hyperboloid of two sheets Find a parametrization of the hyperboloid of two sheets sz2>c2 d  sx2>a2 d  s y2>b2 d = 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
16.7 Stokes’ Theorem
Stokes’ Theorem
16.7 Curl F
P
FIGURE 16.59 The circulation vector at a point P in a plane in a threedimensional fluid flow. Notice its righthand relation to the circulation line.
u o Y z a i R
As we saw in Section 16.4, the circulation density or curl component of a twodimensional field F = Mi + Nj at a point (x, y) is described by the scalar quantity s0N>0x  0M>0yd. In three dimensions, the circulation around a point P in a plane is described with a vector. This vector is normal to the plane of the circulation (Figure 16.59) and points in the direction that gives it a righthand relation to the circulation line. The length of the vector gives the rate of the fluid’s rotation, which usually varies as the circulation plane is tilted about P. It turns out that the vector of greatest circulation in a flow with velocity field F = Mi + Nj + Pk is the curl vector
H d
n a s s a
curl F = a
a m m a h
u M
1201
0N 0N 0P 0M 0P 0M bi + a bj + a bk. 0y 0z 0z 0x 0x 0y
(1)
We get this information from Stokes’ Theorem, the generalization of the circulationcurl form of Green’s Theorem to space. Notice that scurl Fd # k = s0N>0x  0M>0yd is consistent with our definition in Section 16.4 when F = Msx, ydi + Nsx, ydj. The formula for curl F in Equation (1) is often written using the symbolic operator
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0 0 0 + j + k . 0x 0y 0z
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(2)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1202
Chapter 16: Integration in Vector Fields
k
0 § * F = 4 0x
0 0y
0 4 0z
M
N
P
= a
suf
j
You
i
i
(The symbol § is pronounced “del.”) The curl of F is § * F :
0N 0N 0M 0P 0M 0P bi + a bj + a bk 0y 0z 0z 0x 0x 0y
= curl F.
EXAMPLE 1
iaz
curl F = § * F
Finding Curl F
(3)
nR
Find the curl of F = sx 2  ydi + 4zj + x 2k. Solution
curl F = § * F
Equation (3)
j
k
0 0x
0 0y
0 4 0z
x2  y
4z
x2
ass a
i
= 4
dH
= a
0 2 0 0 0 2 sx d s4zdbi  a sx 2 d sx  ydbj 0y 0z 0x 0z
ma
+ a
0 0 2 s4zd sx  ydbk 0x 0y
= s0  4di  s2x  0dj + s0 + 1dk = 4i  2xj + k
S
ham
As we will see, the operator § has a number of other applications. For instance, when applied to a scalar function ƒ(x, y, z), it gives the gradient of ƒ: §ƒ =
0ƒ 0ƒ 0ƒ i + j + k. 0x 0y 0z
n
Mu
C
FIGURE 16.60 The orientation of the bounding curve C gives it a righthanded relation to the normal field n.
This may now be read as “del ƒ” as well as “grad ƒ.”
Stokes’ Theorem Stokes’ Theorem says that, under conditions normally met in practice, the circulation of a vector field around the boundary of an oriented surface in space in the direction counterclockwise with respect to the surface’s unit normal vector field n (Figure 16.60) equals the integral of the normal component of the curl of the field over the surface.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.7 Stokes’ Theorem
1203
F # dr =
C
6 S
(4)
Curl integral
iaz
Counterclockwise circulation
§ * F # n ds
You
F
suf
i
THEOREM 5 Stokes’ Theorem The circulation of a vector field F = Mi + Nj + Pk around the boundary C of an oriented surface S in the direction counterclockwise with respect to the surface’s unit normal vector n equals the integral of § * F # n over S.
nR
Notice from Equation (4) that if two different oriented surfaces S1 and S2 have the same boundary C, their curl integrals are equal:
6
§ * F # n1 ds =
S1
R
dH
Stokes:
S2
ass a
k
Circulation
§ * F # n2 ds.
Both curl integrals equal the counterclockwise circulation integral on the left side of Equation (4) as long as the unit normal vectors n1 and n2 correctly orient the surfaces. Naturally, we need some mathematical restrictions on F, C, and S to ensure the existence of the integrals in Stokes’ equation. The usual restrictions are that all functions, vector fields, and their derivatives be continuous. If C is a curve in the xyplane, oriented counterclockwise, and R is the region in the xyplane bounded by C, then ds = dx dy and
Green:
Curl
6
s§ * Fd # n = s§ * Fd # k = a
0N 0M b. 0x 0y
Under these conditions, Stokes’ equation becomes
Cur
F
S
l
ion u l at
ham
C irc
ma
n
Mu
FIGURE 16.61 Comparison of Green’s Theorem and Stokes’ Theorem.
F # dr =
C
6
a
0N 0M b dx dy, 0x 0y
R
which is the circulationcurl form of the equation in Green’s Theorem. Conversely, by reversing these steps we can rewrite the circulationcurl form of Green’s Theorem for twodimensional fields in del notation as F
C
F # dr =
6
§ * F # k dA.
(5)
R
See Figure 16.61.
EXAMPLE 2
Verifying Stokes’ Equation for a Hemisphere
Evaluate Equation (4) for the hemisphere S: x 2 + y 2 + z 2 = 9, z Ú 0, its bounding circle C: x 2 + y 2 = 9, z = 0, and the field F = yi  xj.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1204
Chapter 16: Integration in Vector Fields
dr = s 3 sin u dudi + s3 cos u dudj F = yi  xj = s3 sin udi  s3 cos udj
suf
i
Solution We calculate the counterclockwise circulation around C (as viewed from above) using the parametrization rsud = s3 cos udi + s3 sin udj, 0 … u … 2p:
2p
F
F # dr =
C
9 du = 18p.
L0
For the curl integral of F, we have
0N 0N 0P 0M 0P 0M bi + a bj + a bk 0y 0z 0z 0x 0x 0y
iaz
§ * F = a
You
F # dr = 9 sin2 u du  9 cos2 u du = 9 du
= s0  0di + s0  0dj + s 1  1dk = 2k xi + yj + zk
2x + y + z 2
2
3 ds = z dA
and
2
=
Outer unit normal Section 16.5, Example 5, with a = 3
2z 3 dA = 2 dA 3 z
ass a
§ * F # n ds = 
xi + yj + zk 3
nR
n =
6
§ * F # n ds =
6 2
2 dA = 18p.
x +y …9
dH
S
2
The circulation around the circle equals the integral of the curl over the hemisphere, as it should.
EXAMPLE 3
Find the circulation of the field F = sx 2  ydi + 4zj + x 2k around the curve C in which the plane z = 2 meets the cone z = 2x 2 + y 2 , counterclockwise as viewed from above (Figure 16.62).
ma
z
Stokes’Theorem enables us to find the circulation by integrating over the surface of the cone. Traversing C in the counterclockwise direction viewed from above corresponds to taking the inner normal n to the cone, the normal with a positive zcomponent. We parametrize the cone as
Solution
rsr, ud = sr cos udi + sr sin udj + rk,
S: r(t) (r cos )i (r sin ) j rk
x
0 … r … 2,
0 … u … 2p.
We then have
Mu
ham
C: x 2 y 2 4, z 2 n
Finding Circulation
n =
y
FIGURE 16.62 The curve C and cone S in Example 3.
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=
sr cos udi  sr sin udj + rk rr * ru = ƒ rr * ru ƒ r22
Section 16.6, Example 4
1 Q scos udi  ssin udj + kR 22
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.7 Stokes’ Theorem
ds = r22 dr du § * F = 4i  2xj + k
1205
suf
i
Section 16.6, Example 4 Example 1
= 4i  2r cos uj + k.
x = r cos u
Accordingly,
1 a4 cos u + 2r cos u sin u + 1b 22 1 = a4 cos u + r sin 2u + 1b 22
You
§ * F#n =
F
F # dr =
C
6
§ * F # n ds
S
Stokes’ Theorem, Equation (4)
1 a4 cos u + r sin 2u + 1b A r22 dr du B = 4p. L0 L0 22 2
nR
2p
=
iaz
and the circulation is
Paddle Wheel Interpretation of § * F
ass a
Suppose that v(x, y, z) is the velocity of a moving fluid whose density at (x, y, z) is dsx, y, zd and let F = dv . Then F # dr
F
C
dH
is the circulation of the fluid around the closed curve C. By Stokes’ Theorem, the circulation is equal to the flux of § * F through a surface S spanning C: F
C
F # dr =
6
§ * F # n ds.
S
Mu
ham
ma
Suppose we fix a point Q in the domain of F and a direction u at Q. Let C be a circle of radius r , with center at Q, whose plane is normal to u. If § * F is continuous at Q, the average value of the ucomponent of § * F over the circular disk S bounded by C approaches the ucomponent of § * F at Q as r : 0: s§ * F # udQ = lim
p:0
1 § * F # u ds. pr2 6 S
If we replace the surface integral in this last equation by the circulation, we get s§ * F # udQ = lim
p:0
1 F # dr. pr2 F
(6)
C
The lefthand side of Equation (6) has its maximum value when u is the direction of § * F. When r is small, the limit on the righthand side of Equation (6) is approximately
To Read it Online & Download:
1 F # dr, pr2 F C
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields
Curl F
Q
EXAMPLE 4
Relating § * F to Circulation Density
suf
i
which is the circulation around C divided by the area of the disk (circulation density). Suppose that a small paddle wheel of radius r is introduced into the fluid at Q, with its axle directed along u. The circulation of the fluid around C will affect the rate of spin of the paddle wheel. The wheel will spin fastest when the circulation integral is maximized; therefore it will spin fastest when the axle of the paddle wheel points in the direction of § * F (Figure 16.63).
You
1206
A fluid of constant density rotates around the zaxis with velocity v = vs yi + xjd, where v is a positive constant called the angular velocity of the rotation (Figure 16.64). If F = v, find § * F and relate it to the circulation density. With F = v = vyi + vxj, § * F = a
iaz
Solution
FIGURE 16.63 The paddle wheel interpretation of curl F.
0N 0N 0P 0M 0P 0M bi + a bj + a bk 0y 0z 0z 0x 0x 0y
nR
= s0  0di + s0  0dj + sv  s vddk = 2vk.
z
By Stokes’ Theorem, the circulation of F around a circle C of radius r bounding a disk S in a plane normal to § * F , say the xyplane, is
F
P(x, y, z)
F # dr =
Thus,
6 S
6
s§ * Fd # k = 2v =
y
Applying Stokes’ Theorem
Use Stokes’ Theorem to evaluate 1C F # dr, if F = xzi + xyj + 3xzk and C is the boundary of the portion of the plane 2x + y + z = 2 in the first octant, traversed counterclockwise as viewed from above (Figure 16.65).
ma
x
ham
FIGURE 16.64 A steady rotational flow parallel to the xyplane, with constant angular velocity v in the positive (counterclockwise) direction (Example 4).
Mu
C
dH EXAMPLE 5
r P(x, y, 0)
1 F # dr, pr2 F
consistent with Equation (6) when u = k.
v (–yi xj)
O
2vk # k dx dy = s2vdspr2 d.
S
ass a
C
§ * F # n ds =
The plane is the level surface ƒsx, y, zd = 2 of the function ƒsx, y, zd = 2x + y + z. The unit normal vector
Solution
n =
§ƒ s2i + j + kd 1 = = a2i + j + kb §ƒ 2i + j + k ƒ ƒ ƒ ƒ 26
is consistent with the counterclockwise motion around C. To apply Stokes’Theorem, we find i
j
k
0 curl F = § * F = 4 0x
0 0y
0 4 = sx  3zdj + yk. 0z
xz
xy
3xz
On the plane, z equals 2  2x  y, so § * F = sx  3s2  2x  yddj + yk = s7x + 3y  6dj + yk
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.7 Stokes’ Theorem z
1 1 a7x + 3y  6 + yb = a7x + 4y  6b. 26 26
suf
(0, 0, 2)
i
and § * F#n =
The surface area element is ds = 2x y z 2
The circulation is
R
F
(0, 2, 0) y 2 2x
F # dr =
6
C
§ * F # n ds
S
y
1
FIGURE 16.65 Example 5.
=
The planar surface in
E
2  2x
L0 L0
s7x + 4y  6d dy dx = 1.
D
Let S be a polyhedral surface consisting of a finite number of plane regions. (See Figure 16.66 for an example.) We apply Green’s Theorem to each separate panel of S. There are two types of panels: 1.
The boundary ¢ of S consists of those edges of the type 2 panels that are not adjacent to other panels. In Figure 16.66, the triangles EAB, BCE, and CDE represent a part of S, with ABCD part of the boundary ¢. Applying Green’s Theorem to the three triangles in turn and adding the results, we get
ham
ma
Part of a polyhedral
Those that are surrounded on all sides by other panels Those that have one or more edges that are not adjacent to other panels.
dH
2. C
ass a
Proof of Stokes’ Theorem for Polyhedral Surfaces
A
Mu
1 a7x + 4y  6b 26 dy dx 26
nR
1
FIGURE 16.66 surface.
2  2x
L0 L0
=
B
Stokes’ Theorem, Equation (4)
iaz
x
You
ƒ §ƒ ƒ 26 dA = dx dy. 1 ƒ §ƒ # k ƒ
n C
(1, 0, 0)
1207
£
F
EAB
+
F
BCE
+
F
≥F # dr = £
CDE
+
6
EAB
6
BCE
+
6
≥ § * F # n ds.
(7)
CDE
The three line integrals on the lefthand side of Equation (7) combine into a single line integral taken around the periphery ABCDE because the integrals along interior segments cancel in pairs. For example, the integral along segment BE in triangle ABE is opposite in sign to the integral along the same segment in triangle EBC. The same holds for segment CE. Hence, Equation (7) reduces to F
F # dr =
ABCDE
6
§ * F # n ds.
ABCDE
When we apply Green’s Theorem to all the panels and add the results, we get
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F ¢
F # dr =
6
§ * F # n ds.
S
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1208
Chapter 16: Integration in Vector Fields
n
suf
i
This is Stokes’ Theorem for a polyhedral surface S. You can find proofs for more general surfaces in advanced calculus texts.
Stokes’ Theorem for Surfaces with Holes
FIGURE 16.67 Stokes’ Theorem also holds for oriented surfaces with holes.
An Important Identity
iaz
S
You
Stokes’ Theorem can be extended to an oriented surface S that has one or more holes (Figure 16.67), in a way analogous to the extension of Green’s Theorem: The surface integral over S of the normal component of § * F equals the sum of the line integrals around all the boundary curves of the tangential component of F, where the curves are to be traced in the direction induced by the orientation of S.
The following identity arises frequently in mathematics and the physical sciences.
or
§ * §f = 0
nR
curl grad ƒ = 0
(8)
ass a
This identity holds for any function ƒ(x, y, z) whose second partial derivatives are continuous. The proof goes like this: j
k
0 § * §ƒ = 5 0x
0 0y
0 5 = sƒzy  ƒyz di  sƒzx  ƒxz dj + sƒyx  ƒxy dk. 0z
0ƒ 0x
0ƒ 0y
0ƒ 0z
dH
i
If the second partial derivatives are continuous, the mixed second derivatives in parentheses are equal (Theorem 2, Section 14.3) and the vector is zero.
ma
Conservative Fields and Stokes’ Theorem
Mu
ham
In Section 16.3, we found that a field F is conservative in an open region D in space is equivalent to the integral of F around every closed loop in D being zero. This, in turn, is equivalent in simply connected open regions to saying that § * F = 0.
THEOREM 6 Curl F = 0 Related to the ClosedLoop Property If § * F = 0 at every point of a simply connected open region D in space, then on any piecewisesmooth closed path C in D, F
F # dr = 0.
C
Sketch of a Proof Theorem 6 is usually proved in two steps. The first step is for simple closed curves. A theorem from topology, a branch of advanced mathematics, states that
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.7 Stokes’ Theorem
1209
F
F # dr =
C
§ * F # n ds = 0.
S
You
The second step is for curves that cross themselves, like the one in Figure 16.68. The idea is to break these into simple loops spanned by orientable surfaces, apply Stokes’ Theorem one loop at a time, and add the results. The following diagram summarizes the results for conservative fields defined on connected, simply connected open regions.
ß ß
Ü
Vector identity (Eq. 8) (continuous second partial derivatives)
F 0 throughout D
Theorem 6 Domain's simple connectivity and Stokes' theorem
Mu
ham
ma
dH
ass a
F • dr 0 EC over any closed path in D
ß ß
F f on D
nR
ß
F conservative on D Theorem 2, Section 13.3
Theorem 1, Section 16.3
iaß z
FIGURE 16.68 In a simply connected open region in space, differentiable curves that cross themselves can be divided into loops to which Stokes’ Theorem applies.
6
suf
i
every differentiable simple closed curve C in a simply connected open region D is the boundary of a smooth twosided surface S that also lies in D. Hence, by Stokes’ Theorem,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
16.7 Stokes’ Theorem
EXERCISES 16.7
1. F = x 2i + 2xj + z 2k
3. F = yi + xzj + x 2k
a m
4. F = sy 2 + z 2 di + sx 2 + z 2 dj + sx 2 + y 2 dk
m a h u
C: The boundary of the triangle cut from the plane x + y + z = 1 by the first octant, counterclockwise when viewed from above 5. F = s y 2 + z 2 di + sx 2 + y 2 dj + sx 2 + y 2 dk C: The square bounded by the lines x = ;1 and y = ;1 in the xyplane, counterclockwise when viewed from above
M
7. Let n be the outer unit normal of the elliptical shell
a H d
C: The boundary of the triangle cut from the plane x + y + z = 1 by the first octant, counterclockwise when viewed from above
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n a ss
Flux of the Curl
C: The ellipse 4x 2 + y 2 = 4 in the xyplane, counterclockwise when viewed from above C: The circle x 2 + y 2 = 9 in the xyplane, counterclockwise when viewed from above
u o Y
C: The intersection of the cylinder x 2 + y 2 = 4 and the hemisphere x 2 + y 2 + z 2 = 16, z Ú 0, counterclockwise when viewed from above.
In Exercises 1–6, use the surface integral in Stokes’ Theorem to calculate the circulation of the field F around the curve C in the indicated direction.
2. F = 2yi + 3xj  z 2k
z a i R
6. F = x 2y 3i + j + zk
Using Stokes’ Theorem to Calculate Circulation
1209
S:
4x 2 + 9y 2 + 36z 2 = 36,
z Ú 0,
and let
F = yi + x 2j + sx 2 + y 4 d3>2 sin e 2xyz k.
Find the value of 6
§ * F # n ds.
S
(Hint: One parametrization of the ellipse at the base of the shell is x = 3 cos t, y = 2 sin t, 0 … t … 2p.) 8. Let n be the outer unit normal (normal away from the origin) of the parabolic shell S:
4x 2 + y + z 2 = 4,
y Ú 0,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1210
Chapter 16: Integration in Vector Fields 17. F = 3yi + s5  2xdj + sz 2  2dk
S: rsf, ud = A 23 sin f cos u B i + A 23 sin f sin u B j + A 23 cos f B k, 0 … f … p>2, 0 … u … 2p
i
1 1 b i + stan1 ydj + ax + bk. 2 + x 4 + z
18. F = y2i + z2j + xk
Find the value of
10. Evaluate 6
Theory and Examples
19. Zero circulation Use the identity § * §ƒ = 0 (Equation (8) in the text) and Stokes’ Theorem to show that the circulations of the following fields around the boundary of any smooth orientable surface in space are zero. a. F = 2xi + 2yj + 2zk
§ * s yid # n ds,
b. F = §sxy 2z 3 d
S
where S is the hemisphere x 2 + y 2 + z 2 = 1, z Ú 0.
c. F = § * sxi + yj + zkd d. F = §ƒ
11. Flux of curl F Show that 6
You
S
9. Let S be the cylinder x 2 + y 2 = a 2, 0 … z … h, together with its top, x 2 + y 2 … a 2, z = h. Let F = yi + xj + x 2k. Use Stokes’ Theorem to find the flux of § * F outward through S.
iaz
6
S: rsf, ud = s2 sin f cos udi + s2 sin f sin udj + s2 cos fdk, 0 … f … p>2, 0 … u … 2p
§ * F # n ds.
nR
F = az +
suf
and let
20. Zero circulation Let ƒsx, y, zd = sx 2 + y 2 + z 2 d1>2. Show that the clockwise circulation of the field F = §ƒ around the circle x 2 + y 2 = a 2 in the xyplane is zero
§ * F # n ds
S
a. by taking r = sa cos tdi + sa sin tdj, 0 … t … 2p, and integrating F # dr over the circle.
ass a
has the same value for all oriented surfaces S that span C and that induce the same positive direction on C.
b. by applying Stokes’ Theorem.
21. Let C be a simple closed smooth curve in the plane 2x + 2y + z = 2 , oriented as shown here. Show that F
§ * F # n ds?
6 S
Give reasons for your answer.
2y dx + 3z dy  x dz
C
dH
12. Let F be a differentiable vector field defined on a region containing a smooth closed oriented surface S and its interior. Let n be the unit normal vector field on S. Suppose that S is the union of two surfaces S1 and S2 joined along a smooth simple closed curve C. Can anything be said about
z 2x 2y z 2
2
Stokes’ Theorem for Parametrized Surfaces
ma
C
In Exercises 13–18, use the surface integral in Stokes’ Theorem to calculate the flux of the curl of the field F across the surface S in the direction of the outward unit normal n.
ham
13. F = 2zi + 3xj + 5yk
S: rsr, ud = sr cos udi + sr sin udj + s4  r 2 dk, 0 … r … 2, 0 … u … 2p 14. F = sy  zdi + sz  xdj + sx + zdk S: rsr, ud = sr cos udi + sr sin udj + s9  r 2 dk, 0 … r … 3, 0 … u … 2p
Mu
15. F = x2yi + 2y3zj + 3zk
S: rsr, ud = sr cos udi + sr sin udj + rk, 0 … r … 1, 0 … u … 2p
16. F = sx  ydi + s y  zdj + sz  xdk S: rsr, ud = sr cos udi + sr sin udj + s5  rdk, 0 … r … 5, 0 … u … 2p
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O
a1
y
1 x
depends only on the area of the region enclosed by C and not on the position or shape of C. 22. Show that if F = xi + yj + zk, then § * F = 0. 23. Find a vector field with twicedifferentiable components whose curl is xi + yj + zk or prove that no such field exists. 24. Does Stokes’ Theorem say anything special about circulation in a field whose curl is zero? Give reasons for your answer. 25. Let R be a region in the xyplane that is bounded by a piecewisesmooth simple closed curve C and suppose that the moments of
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.7 Stokesâ€™ Theorem
F
Â§sr 4 d # n ds,
C
C
where r = 2x 2 + y 2, in terms of Ix and Iy. 26. Zero curl, yet field not conservative Show that the curl of y x2 + y2
i +
x j + zk x2 + y2
i
is not zero if C is the circle x 2 + y 2 = 1 in the xyplane. (Theorem 6 does not apply here because the domain of F is not simply connected. The field F is not defined along the zaxis so there is no way to contract C to a point without leaving the domain of F.)
Mu
ham
ma
dH
ass a
nR
iaz
F =
F # dr
You
F
is zero but that
suf
inertia of R about the x and yaxes are known to be Ix and Iy. Evaluate the integral
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1211
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
16.8 The Divergence Theorem and a Unified Theory
16.8
The Divergence Theorem and a Unified Theory
1211
u o Y
The divergence form of Green’s Theorem in the plane states that the net outward flux of a vector field across a simple closed curve can be calculated by integrating the divergence of the field over the region enclosed by the curve. The corresponding theorem in three dimensions, called the Divergence Theorem, states that the net outward flux of a vector field across a closed surface in space can be calculated by integrating the divergence of the field over the region enclosed by the surface. In this section, we prove the Divergence Theorem and show how it simplifies the calculation of flux. We also derive Gauss’s law for flux in an electric field and the continuity equation of hydrodynamics. Finally, we unify the chapter’s vector integral theorems into a single fundamental theorem.
n sa
Divergence in Three Dimensions
z a i R
The divergence of a vector field F = Msx, y, zdi + Nsx, y, zdj + Psx, y, zdk is the scalar function
as
H d
h u
M
0N 0M 0P + + . 0x 0y 0z
(1)
The symbol “div F” is read as “divergence of F” or “div F.” The notation § # F is read “del dot F.” Div F has the same physical interpretation in three dimensions that it does in two. If F is the velocity field of a fluid flow, the value of div F at a point (x, y, z) is the rate at which fluid is being piped in or drained away at (x, y, z). The divergence is the flux per unit volume or flux density at the point.
a m
m a
div F = § # F =
EXAMPLE 1
Finding Divergence
Find the divergence of F = 2xzi  xyj  zk. Solution
The divergence of F is
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§#F =
0 0 0 s2xzd + s xyd + s zd = 2z  x  1. 0x 0y 0z
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1212
Chapter 16: Integration in Vector Fields
i
Divergence Theorem
suf
The Divergence Theorem says that under suitable conditions, the outward flux of a vector field across a closed surface (oriented outward) equals the triple integral of the divergence of the field over the region enclosed by the surface.
F # n ds =
S
§ # F dV.
(2)
D
Divergence integral
nR
Outward flux
EXAMPLE 2
9
iaz
6
You
THEOREM 7 Divergence Theorem The flux of a vector field F across a closed oriented surface S in the direction of the surface’s outward unit normal field n equals the integral of § # F over the region D enclosed by the surface:
Supporting the Divergence Theorem
ass a
Evaluate both sides of Equation (2) for the field F = xi + yj + zk over the sphere x 2 + y 2 + z 2 = a 2. The outer unit normal to S, calculated from the gradient of ƒsx, y, zd = x 2 + y + z  a 2, is
Solution 2
2
n =
24sx 2 + y 2 + z 2 d
dH
Hence,
2sxi + yj + zkd =
xi + yj + zk . a
x2 + y2 + z2 a2 ds = a a ds = a ds
F # n ds =
ma
because x 2 + y 2 + z 2 = a 2 on the surface. Therefore, 6
F # n ds =
S
6
a ds = a
6
S
ds = as4pa 2 d = 4pa 3.
S
Mu
ham
The divergence of F is §#F =
0 0 0 sxd + s yd + szd = 3, 0x 0y 0z
so 9
§ # F dV =
D
EXAMPLE 3
4 3 dV = 3 a pa 3 b = 4pa 3. 3 9 D
Finding Flux
Find the flux of F = xyi + yzj + xzk outward through the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.8
The Divergence Theorem and a Unified Theory
1213
Instead of calculating the flux as a sum of six separate integrals, one for each face of the cube, we can calculate the flux by integrating the divergence §#F =
suf
i
Solution
0 0 0 sxyd + s yzd + sxzd = y + z + x 0x 0y 0z
6
F # n ds =
Cube surface
1
=
R yz
1
1
L0 L0 L0
nR
y Rx y
ma
(n1, n 2, n 3)
ham
k
␣
␥
S1:
z = ƒ1sx, yd,
sx, yd in Rxy
S2:
z = ƒ2sx, yd,
sx, yd in Rxy,
with ƒ1 … ƒ2. We make similar assumptions about the projection of D onto the other coordinate planes. See Figure 16.69. The components of the unit normal vector n = n1i + n2j + n3k are the cosines of the angles a, b, and g that n makes with i, j, and k (Figure 16.70). This is true because all the vectors involved are unit vectors. We have
dH
FIGURE 16.69 We first prove the Divergence Theorem for the kind of threedimensional region shown here. We then extend the theorem to other regions.
n3
Routine integration
To prove the Divergence Theorem, we assume that the components of F have continuous first partial derivatives. We also assume that D is a convex region with no holes or bubbles, such as a solid sphere, cube, or ellipsoid, and that S is a piecewise smooth surface. In addition, we assume that any line perpendicular to the xyplane at an interior point of the region Rxy that is the projection of D on the xyplane intersects the surface S in exactly two points, producing surfaces
S2
z
3 . 2
Proof of the Divergence Theorem for Special Regions
S1
x
The Divergence Theorem
sx + y + zd dx dy dz =
ass a
D
§ # F dV
Cube interior
z
Rxz
9
iaz
Flux =
You
over the cube’s interior:
n1 = n # i = ƒ n ƒ ƒ i ƒ cos a = cos a n2 = n # j = ƒ n ƒ ƒ j ƒ cos b = cos b n3 = n # k = ƒ n ƒ ƒ k ƒ cos g = cos g
Thus,
n
n = scos adi + scos bdj + scos gdk

i
n2
Mu
x
n1
j
and
y
FIGURE 16.70 The scalar components of the unit normal vector n are the cosines of the angles a, b, and g that it makes with i, j, and k.
F # n = M cos a + N cos b + P cos g. In component form, the Divergence Theorem states that
6
sM cos a + N cos b + P cos gd ds =
S
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0N 0P 0M + + b dx dy dz. a 0y 0z 9 0x D
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1214
Chapter 16: Integration in Vector Fields z
d
z f1(x, y) d
6
P cos g ds =
D
0N dx dy dz 9 0y
S
D
n y
x
6
N cos b ds =
S
S1 O
6
0M dx dy dz 9 0x
S
Rxy
D
(3)
(4)
(5)
Proof of Equation (5) We prove Equation (5) by converting the surface integral on the left to a double integral over the projection Rxy of D on the xyplane (Figure 16.71). The surface S consists of an upper part S2 whose equation is z = ƒ2sx, yd and a lower part S1 whose equation is z = ƒ1sx, yd. On S2, the outer normal n has a positive kcomponent and
iaz
FIGURE 16.71 The threedimensional region D enclosed by the surfaces S1 and S2 shown here projects vertically onto a twodimensional region Rxy in the xyplane.
0P dx dy dz 9 0z
nR
dA dx dy
You
S2
M cos a ds =
suf
z f2(x, y)
i
We prove the theorem by proving the three following equalities:
n D
cos g ds = dx dy
because
ds =
dx dy dA = cos g . ƒ cos g ƒ
See Figure 16.72. On S1, the outer normal n has a negative kcomponent and ␥
cos g ds = dx dy.
n
Therefore, Here ␥ is acute, so d dx dy/cos ␥.
6
P cos g ds =
=
n
dy
Here ␥ is obtuse, so d –dx dy/cos ␥.
dx
Mu
ham
FIGURE 16.72 An enlarged view of the area patches in Figure 16.71. The relations ds = ;dx dy>cos g are derived in Section 16.5.
S1
6
Psx, y, ƒ2sx, ydd dx dy 
6
[Psx, y, ƒ2sx, ydd  Psx, y, ƒ1sx, ydd] dx dy
Rxy
=
ma
␥
P cos g ds + P cos g ds 6 6 S2
dH
S
k
ass a
k
6
Psx, y, ƒ1sx, ydd dx dy
Rxy
Rxy
=
c
ƒ2sx,yd
0P 0P dz d dx dy = dz dx dy. 6 Lƒ1sx,yd 0z 9 0z Rxy
D
This proves Equation (5). The proofs for Equations (3) and (4) follow the same pattern; or just permute x, y, z; M, N, P; a, b, g, in order, and get those results from Equation (5).
Divergence Theorem for Other Regions The Divergence Theorem can be extended to regions that can be partitioned into a finite number of simple regions of the type just discussed and to regions that can be defined as limits of simpler regions in certain ways. For example, suppose that D is the region between two concentric spheres and that F has continuously differentiable components throughout D and on the bounding surfaces. Split D by an equatorial plane and apply the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.8 z
The Divergence Theorem and a Unified Theory
1215
6
O
S1
y
§ # F dV1 .
D1
(6)
The unit normal n1 that points outward from D1 points away from the origin along the outer surface, equals k along the flat base, and points toward the origin along the inner surface. Next apply the Divergence Theorem to D2, and its surface S2 (Figure 16.74):
n1
x
FIGURE 16.73 The lower half of the solid region between two concentric spheres.
6
F # n2 ds2 =
S2
9
§ # F dV2 .
(7)
D2
nR
As we follow n2 over S2 , pointing outward from D2 , we see that n2 equals k along the washershaped base in the xyplane, points away from the origin on the outer sphere, and points toward the origin on the inner sphere. When we add Equations (6) and (7), the integrals over the flat base cancel because of the opposite signs of n1 and n2 . We thus arrive at the result
z
6
Finding Outward Flux
dH
xi + yj + zk
F =
r3
r = 2x 2 + y 2 + z 2
,
across the boundary of the region D: 0 6 a 2 … x 2 + y 2 + z 2 … b 2. Solution
ham
D
Find the net outward flux of the field
ma
FIGURE 16.74 The upper half of the solid region between two concentric spheres.
§ # F dV,
ass a
EXAMPLE 4 y –k
9
with D the region between the spheres, S the boundary of D consisting of two spheres, and n the unit normal to S directed outward from D.
n2
x
F # n ds =
S
D2
Mu
9
iaz
D1
F # n1 ds1 =
You
k
suf
i
Divergence Theorem to each half separately. The bottom half, D1, is shown in Figure 16.73. The surface S1 that bounds D1 consists of an outer hemisphere, a plane washershaped base, and an inner hemisphere. The Divergence Theorem says that
The flux can be calculated by integrating § # F over D. We have 0r x 1 = sx 2 + y 2 + z 2 d1>2s2xd = r 0x 2
and
0r 3x 2 0M 0 1 = = 3  5. sxr3 d = r3  3xr4 0x 0x 0x r r Similarly, 3y 2 0N 1 = 3  5 0y r r
and
3z 2 0P 1 = 3  5. 0z r r
Hence,
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div F =
3r2 3 2 3 3 2 2 sx + y + z d = = 0 5 r r5 r3 r3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1216
Chapter 16: Integration in Vector Fields
§ # F dV = 0.
D
suf
9
i
and
So the integral of § # F over D is zero and the net outward flux across the boundary of
Hence, on the sphere,
and
2x 2 + y 2 + z 2
=
xi + yj + zk . a
xi + yj + zk xi + yj + zk x2 + y2 + z2 a2 1 # = = = 2 a 3 4 4 a a a a
ass a
F#n =
xi + yj + zk
nR
n =
iaz
You
D is zero. There is more to learn from this example, though. The flux leaving D across the inner sphere Sa is the negative of the flux leaving D across the outer sphere Sb (because the sum of these fluxes is zero). Hence, the flux of F across Sa in the direction away from the origin equals the flux of F across Sb in the direction away from the origin. Thus, the flux of F across a sphere centered at the origin is independent of the radius of the sphere. What is this flux? To find it, we evaluate the flux integral directly. The outward unit normal on the sphere of radius a is
6
F # n ds =
Sa
1 1 ds = 2 s4pa 2 d = 4p. a 26 a Sa
dH
The outward flux of F across any sphere centered at the origin is 4p.
Gauss’s Law: One of the Four Great Laws of Electromagnetic Theory
ma
There is still more to be learned from Example 4. In electromagnetic theory, the electric field created by a point charge q located at the origin is
Mu
ham
Esx, y, zd =
q q xi + yj + zk q 1 r r a = , b = 4pP0 ƒ r ƒ 2 ƒ r ƒ 4pP0 ƒ r ƒ 3 4pP0 r3
where P0 is a physical constant, r is the position vector of the point (x, y, z), and r = ƒ r ƒ = 2x 2 + y 2 + z 2. In the notation of Example 4, E =
q F. 4pP0
The calculations in Example 4 show that the outward flux of E across any sphere centered at the origin is q>P0, but this result is not confined to spheres. The outward flux of E across any closed surface S that encloses the origin (and to which the Divergence Theorem applies) is also q>P0. To see why, we have only to imagine a large sphere Sa centered at the origin and enclosing the surface S. Since
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§#E = §#
q q F = §#F = 0 4pP0 4pP0
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The Divergence Theorem and a Unified Theory
1217
E # n ds = 0,
6
Boundary of D
suf
i
when r 7 0 , the integral of § # E over the region D between S and Sa is zero. Hence, by the Divergence Theorem,
6
q E # n ds = P0
iaz
Gauss’s law:
You
and the flux of E across S in the direction away from the origin must be the same as the flux of E across Sa in the direction away from the origin, which is q>P0. This statement, called Gauss’s Law, also applies to charge distributions that are more general than the one assumed here, as you will see in nearly any physics text.
S
n
v ∆t
h (v ∆ t) . n
nR
Continuity Equation of Hydrodynamics
Let D be a region in space bounded by a closed oriented surface S. If v(x, y, z) is the velocity field of a fluid flowing smoothly through D, d = dst, x, y, zd is the fluid’s density at (x, y, z) at time t, and F = dv, then the continuity equation of hydrodynamics states that
S
§#F +
0d = 0. 0t
If the functions involved have continuous first partial derivatives, the equation evolves naturally from the Divergence Theorem, as we now see. First, the integral
dH
FIGURE 16.75 The fluid that flows upward through the patch ¢s in a short time ¢t fills a “cylinder” whose volume is approximately base * height = v # n ¢s ¢t.
ass a
,
6
F # n ds
S
Mu
ham
ma
is the rate at which mass leaves D across S (leaves because n is the outer normal). To see why, consider a patch of area ¢s on the surface (Figure 16.75). In a short time interval ¢t, the volume ¢V of fluid that flows across the patch is approximately equal to the volume of a cylinder with base area ¢s and height sv¢td # n, where v is a velocity vector rooted at a point of the patch: ¢V L v # n ¢s ¢t.
The mass of this volume of fluid is about ¢m L dv # n ¢s ¢t, so the rate at which mass is flowing out of D across the patch is about ¢m L dv # n ¢s. ¢t This leads to the approximation
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a ¢m L a dv # n ¢s ¢t
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Chapter 16: Integration in Vector Fields
suf
i
as an estimate of the average rate at which mass flows across S. Finally, letting ¢s : 0 and ¢t : 0 gives the instantaneous rate at which mass leaves D across S as dm = dv # n ds, dt 6 S
You
which for our particular flow is dm = F # n ds. dt 6 S
Now let B be a solid sphere centered at a point Q in the flow. The average value of § # F over B is
iaz
1 § # F dV. volume of B 9 B
s§ # FdP =
nR
It is a consequence of the continuity of the divergence that § # F actually takes on this value at some point P in B. Thus, 6
F # n ds
S 1 § # F dV = volume of B 9 volume of B
ass a
B
=
rate at which mass leaves B across its surface S volume of B
(8)
dH
The fraction on the right describes decrease in mass per unit volume. Now let the radius of B approach zero while the center Q stays fixed. The left side of Equation (8) converges to s§ # FdQ, the right side to s 0d>0tdQ. The equality of these two limits is the continuity equation §#F = 
0d . 0t
Mu
ham
ma
The continuity equation “explains” § # F: The divergence of F at a point is the rate at which the density of the fluid is decreasing there. The Divergence Theorem 6
F # n ds =
S
9
§ # F dV
D
now says that the net decrease in density of the fluid in region D is accounted for by the mass transported across the surface S. So, the theorem is a statement about conservation of mass (Exercise 31).
Unifying the Integral Theorems If we think of a twodimensional field F = Msx, ydi + Nsx, ydj as a threedimensional field whose kcomponent is zero, then § # F = s0M>0xd + s0N>0yd and the normal form of Green’s Theorem can be written as F
C
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F # n ds =
a
0N 0M + b dx dy = § # F dA. 0x 0y 6 6 R
R
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.8
The Divergence Theorem and a Unified Theory
1219
F
F # dr =
C
6 R
a
suf
i
Similarly, § * F # k = s0N>0xd  s0M>0yd, so the tangential form of Green’s Theorem can be written as 0N 0M b dx dy = § * F # k dA. 0x 0y 6 R
You
With the equations of Green’s Theorem now in del notation, we can see their relationships to the equations in Stokes’ Theorem and the Divergence Theorem.
Green’s Theorem and Its Generalization to Three Dimensions F # n ds =
F
iaz
Normal form of Green’s Theorem:
C
Divergence Theorem:
6
6 R
F # n ds =
nR
S
Tangential form of Green’s Theorem:
F
F # dr =
C
ass a
Stokes’ Theorem:
F
§ # F dA
9
§ # F dV
D
6
§ * F # k dA
R
F # dr =
C
6
§ * F # n ds
S
n –i
ma
dH
Notice how Stokes’Theorem generalizes the tangential (curl) form of Green’s Theorem from a flat surface in the plane to a surface in threedimensional space. In each case, the integral of the normal component of curl F over the interior of the surface equals the circulation of F around the boundary. Likewise, the Divergence Theorem generalizes the normal (flux) form of Green’s Theorem from a twodimensional region in the plane to a threedimensional region in space. In each case, the integral of § # F over the interior of the region equals the total flux of the field across the boundary. There is still more to be learned here. All these results can be thought of as forms of a single fundamental theorem. Think back to the Fundamental Theorem of Calculus in Section 5.3. It says that if ƒ(x) is differentiable on (a, b) and continuous on [a, b], then
ni
a
b
x
Mu
ham
FIGURE 16.76 The outward unit normals at the boundary of [a, b] in onedimensional space.
b
dƒ dx = ƒsbd  ƒsad. La dx
If we let F = ƒsxdi throughout [a, b], then sdƒ>dxd = § # F. If we define the unit vector field n normal to the boundary of [a, b] to be i at b and i at a (Figure 16.76), then ƒsbd  ƒsad = ƒsbdi # sid + ƒsadi # s id = Fsbd # n + Fsad # n = total outward flux of F across the boundary of [a, b]. The Fundamental Theorem now says that
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Fsbd # n + Fsad # n =
3
§ # F dx.
[a,b]
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1220
Chapter 16: Integration in Vector Fields
You
suf
i
The Fundamental Theorem of Calculus, the normal form of Green’s Theorem, and the Divergence Theorem all say that the integral of the differential operator § # operating on a field F over a region equals the sum of the normal field components over the boundary of the region. (Here we are interpreting the line integral in Green’s Theorem and the surface integral in the Divergence Theorem as “sums” over the boundary.) Stokes’ Theorem and the tangential form of Green’s Theorem say that, when things are properly oriented, the integral of the normal component of the curl operating on a field equals the sum of the tangential field components on the boundary of the surface. The beauty of these interpretations is the observance of a single unifying principle, which we might state as follows.
Mu
ham
ma
dH
ass a
nR
iaz
The integral of a differential operator acting on a field over a region equals the sum of the field components appropriate to the operator over the boundary of the region.
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i f u s u
Chapter 16: Integration in Vector Fields
EXERCISES 16.8 8. Sphere F = x 2i + xzj + 3zk
Calculating Divergence
D: The region cut from the first octant by the sphere x 2 + y 2 + z2 = 4
2. The radial field in Figure 16.13. 3. The gravitational field in Figure 16.9.
10. Cylindrical can F = s6x 2 + 2xydi + s2y + x 2zdj + 4x 2y 3k
4. The velocity field in Figure 16.12.
H d
D: The cube bounded by the planes x = ;1, y = ;1 , and z = ;1 2
a m
6. F = x i + y j + z k
11. Wedge
as
In Exercises 5–16, use the Divergence Theorem to find the outward flux of F across the boundary of the region D. 5. Cube F = s y  xdi + sz  ydj + s y  xdk
as n
D: The region cut from the first octant by the cylinder x 2 + y 2 = 4 and the plane z = 3
Using the Divergence Theorem to Calculate Outward Flux
2
z a i R
9. Portion of sphere F = x 2i  2xyj + 3xzk
1. The spin field in Figure 16.14.
2
o Y
D: The solid sphere x 2 + y 2 + z 2 … 4
In Exercises 1–4, find the divergence of the field.
F = 2xzi  xyj  z 2k
D: The wedge cut from the first octant by the plane y + z = 4 and the elliptical cylinder 4x 2 + y 2 = 16
12. Sphere F = x 3i + y 3j + z 3k D: The solid sphere x 2 + y 2 + z 2 … a 2
13. Thick sphere F = 2x 2 + y 2 + z 2 sxi + yj + zkd D: The region 1 … x 2 + y 2 + z 2 … 2
a. Cube D: The cube cut from the first octant by the planes x = 1, y = 1 , and z = 1
14. Thick sphere F = sxi + yj + zkd> 2x 2 + y 2 + z 2
b. Cube D: The cube bounded by the planes x = ;1, y = ;1 , and z = ;1
15. Thick sphere F = s5x 3 + 12xy 2 di + s y 3 + e y sin zdj + s5z3 + e y cos zdk
c. Cylindrical can D: The region cut from the solid cylinder x 2 + y 2 … 4 by the planes z = 0 and z = 1
D: The solid region between the spheres x 2 + y 2 + z 2 = 1 and x2 + y2 + z2 = 2 y 2z 16. Thick cylinder F = ln sx 2 + y 2 di  a x tan1 x bj +
m a h
u M
7. Cylinder and paraboloid
F = yi + xyj  zk
D: The region inside the solid cylinder x 2 + y 2 … 4 between the plane z = 0 and the paraboloid z = x 2 + y 2
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D: The region 1 … x 2 + y 2 + z 2 … 4
z2x 2 + y 2 k D: The thickwalled cylinder 1 … x 2 + y 2 … 2,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.8 The Divergence Theorem and a Unified Theory
Properties of Curl and Divergence
i
z
17. div (curl G ) is zero
suf
Top
b. What, if anything, can you conclude about the flux of the field § * G across a closed surface? Give reasons for your answer.
1
18. Let F1 and F2 be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.
1
b. § * sgFd = g§ * F + §g * F
For differentiable vector fields F1 and F2 , verify the following identities.
b.
dH
a. § * sF1 * F2 d = sF2 # §dF1  sF1 # §dF2 + s§ # F2 dF1 s§ # F1 dF2 §sF1 # F2 d = sF1 # §dF2
+ sF2 # §dF1 + F1
F2 * s§ * F1 d
Theory and Examples
24. Maximum flux Among all rectangular solids defined by the inequalities 0 … x … a, 0 … y … b, 0 … z … 1 , find the one for which the total flux of F = s x 2  4xydi  6yzj + 12zk outward through the six sides is greatest. What is the greatest flux?
25. Volume of a solid region Let F = xi + yj + zk and suppose that the surface S and region D satisfy the hypotheses of the Divergence Theorem. Show that the volume of D is given by the formula Volume of D =
* s§ * F2 d +
ham
ma
21. Let F be a field whose components have continuous first partial derivatives throughout a portion of space containing a region D bounded by a smooth closed surface S. If ƒ F ƒ … 1 , can any bound be placed on the size of 9
b. Let n be the outward unit normal vector field on S. Show that it is not possible for F to be orthogonal to n at every point of S.
ass a
20. If F = Mi + Nj + Pk is a differentiable vector field, we define the notation F # § to mean
23. a. Show that the flux of the position vector field F = xi + yj + zk outward through a smooth closed surface S is three times the volume of the region enclosed by the surface.
iaz
a. § # sgFd = g§ # F + §g # F
(1, 1, 0)
nR
19. Let F be a differentiable vector field and let g(x, y, z) be a differentiable scalar function. Verify the following identities.
0 0 0 + N + P . 0x 0y 0z
Side A
x
c. § # sF1 * F2 d = F2 # § * F1  F1 # § * F2
y
Side B
a. § # saF1 + bF2 d = a§ # F1 + b§ # F2 b. § * saF1 + bF2 d = a§ * F1 + b§ * F2
You
a. Show that if the necessary partial derivatives of the components of the field G = Mi + Nj + Pk are continuous, then §# § * G = 0.
M
1221
§ # F dV ?
D
Give reasons for your answer.
Mu
22. The base of the closed cubelike surface shown here is the unit square in the xyplane. The four sides lie in the planes x = 0, x = 1, y = 0 , and y = 1 . The top is an arbitrary smooth surface whose identity is unknown. Let F = xi  2yj + sz + 3dk and suppose the outward flux of F through side A is 1 and through side B is 3 . Can you conclude anything about the outward flux through the top? Give reasons for your answer.
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1 F # n ds. 36 S
26. Flux of a constant field Show that the outward flux of a constant vector field F = C across any closed surface to which the Divergence Theorem applies is zero. 27. Harmonic functions A function ƒ(x, y, z) is said to be harmonic in a region D in space if it satisfies the Laplace equation 0 2ƒ 0 2ƒ 0 2ƒ §2ƒ = § # §ƒ = 2 + 2 + 2 = 0 0x 0y 0z throughout D. a. Suppose that ƒ is harmonic throughout a bounded region D enclosed by a smooth surface S and that n is the chosen unit normal vector on S. Show that the integral over S of §ƒ # n, the derivative of ƒ in the direction of n, is zero. b. Show that if ƒ is harmonic on D, then 6 S
ƒ §ƒ # n ds =
9
ƒ §ƒ ƒ 2 dV.
D
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Chapter 16: Integration in Vector Fields
i
b. Use the Divergence Theorem and Leibniz’s Rule,
§ƒ # n ds.
0p d pst, x, y, zd dV = dV, dt 9 9 0t
S
D
6
ƒ §g # n ds =
S
9
2
sƒ § g +
§ƒ # §gd dV.
(9)
D
Equation (9) is Green’s first formula. (Hint: Apply the Divergence Theorem to the field F = ƒ §g.) 30. Green’s second formula (Continuation of Exercise 29.) Interchange ƒ and g in Equation (9) to obtain a similar formula. Then subtract this formula from Equation (9) to show that sƒ §g  g§ƒd # n ds =
S
9
sƒ §2g  g§2ƒd dV.
D
This equation is Green’s second formula.
(10)
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31. Conservation of mass Let v(t, x, y, z) be a continuously differentiable vector field over the region D in space and let p(t, x, y, z) be a continuously differentiable scalar function. The variable t represents the time domain. The Law of Conservation of Mass asserts that d pst, x, y, zd dV = pv # n ds, dt 9 6 D
S
32. The heat diffusion equation Let T(t, x, y, z) be a function with continuous second derivatives giving the temperature at time t at the point (x, y, z) of a solid occupying a region D in space. If the solid’s heat capacity and mass density are denoted by the constants c and r , respectively, the quantity crT is called the solid’s heat energy per unit volume. a. Explain why  §T points in the direction of heat flow. b. Let k§T denote the energy flux vector. (Here the constant k is called the conductivity.) Assuming the Law of Conservation of Mass with k§T = v and crT = p in Exercise 31, derive the diffusion (heat) equation 0T = K§2T, 0t
where K = k>scrd 7 0 is the diffusivity constant. (Notice that if T(t, x) represents the temperature at time t at position x in a uniform conducting rod with perfectly insulated sides, then §2T = 0 2T>0x 2 and the diffusion equation reduces to the onedimensional heat equation in Chapter 14’s Additional Exercises.)
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where S is the surface enclosing D.
0p = 0. 0t
(In the first term § # pv, the variable t is held fixed, and in the second term 0p>0t , it is assumed that the point (x, y, z) in D is held fixed.)
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6
§ # pv +
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29. Green’s first formula Suppose that ƒ and g are scalar functions with continuous first and secondorder partial derivatives throughout a region D that is bounded by a closed piecewisesmooth surface S. Show that
D
to show that the Law of Conservation of Mass is equivalent to the continuity equation,
nR
( §ƒ # n is the derivative of ƒ in the direction of n.)
You
6
a. Give a physical interpretation of the conservation of mass law if v is a velocity flow field and p represents the density of the fluid at point (x, y, z) at time t.
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28. Flux of a gradient field Let S be the surface of the portion of the solid sphere x 2 + y 2 + z 2 … a 2 that lies in the first octant and let ƒsx, y, zd = ln2x 2 + y 2 + z 2. Calculate
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i f u s u o
Chapter 16: Integration in Vector Fields
Chapter 16
Questions to Guide Your Review
n a s as
1. What are line integrals? How are they evaluated? Give examples. 2. How can you use line integrals to find the centers of mass of springs? Explain.
H d a
3. What is a vector field? A gradient field? Give examples.
4. How do you calculate the work done by a force in moving a particle along a curve? Give an example.
m m a
5. What are flow, circulation, and flux?
6. What is special about path independent fields?
h u M
7. How can you tell when a field is conservative?
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Y z a i R
8. What is a potential function? Show by example how to find a potential function for a conservative field. 9. What is a differential form? What does it mean for such a form to be exact? How do you test for exactness? Give examples.
10. What is the divergence of a vector field? How can you interpret it? 11. What is the curl of a vector field? How can you interpret it? 12. What is Greenâ€™s theorem? How can you interpret it? 13. How do you calculate the area of a curved surface in space? Give an example.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Questions to Guide Your Review
Chapter 16
16. What is a parametrized surface? How do you find the area of such a surface? Give examples.
i
20. What is the Divergence Theorem? How can you interpret it?
21. How does the Divergence Theorem generalize Green’s Theorem? 22. How does Stokes’ Theorem generalize Green’s Theorem?
23. How can Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem be thought of as forms of a single fundamental theorem?
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dH
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nR
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17. How do you integrate a function over a parametrized surface? Give an example.
19. Summarize the chapter’s results on conservative fields.
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15. What are surface integrals? What can you calculate with them? Give an example.
18. What is Stokes’ Theorem? How can you interpret it?
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14. What is an oriented surface? How do you calculate the flux of a threedimensional vector field across an oriented surface? Give an example.
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i f u s
Chapter 16 Practice Exercises
Chapter 16
Practice Exercises
Evaluating Line Integrals
4. Integrate ƒsx, y, zd = 2x 2 + y 2 over the involute curve
1. The accompanying figure shows two polygonal paths in space joining the origin to the point (1, 1, 1). Integrate ƒsx, y, zd = 2x 3y 2  2z + 3 over each path. z (0, 0, 0)
z (0, 0, 0)
(1, 1, 1)
(1, 1, 1)
y
y
C3
y
2. The accompanying figure shows three polygonal paths joining the origin to the point (1, 1, 1). Integrate ƒsx, y, zd = x 2 + y  z over each path. z
(0, 0, 0)
H d a m (0, 0, 0)
(1, 1, 1)
C1 x
C2
C3
y
C4
(1, 1, 0)
m a
(0, 0, 1) (0, 0, 0)
h u M x
(1, 1, 0)
x
z C6
C5
(0, 1, 1)
C7 (1, 1, 1)
y
3. Integrate ƒsx, y, zd = 2x 2 + z 2 over the circle rstd = sa cos tdj + sa sin tdk,
8. Integrate F = 3x 2yi + sx 3 + 1dj + 9z 2k around the circle cut from the sphere x 2 + y 2 + z 2 = 9 by the plane x = 2.
Evaluate the integrals in Exercises 9 and 10.
z
(1, 0, 0)
y z dy  z dz y A A
Ls1,1,1d 7. Integrate F = s y sin zdi + sx sin zdj + sxy cos zdk around the circle cut from the sphere x 2 + y 2 + z 2 = 5 by the plane z = 1, clockwise as viewed from above.
(1, 1, 1)
Path 2
Path 1
iR a
dx 
n a s sa
(1, 1, 0)
x
0 … t … 23.
Evaluate the integrals in Exercises 5 and 6. s4,3,0d dx + dy + dz 5. Ls1,1,1d 2x + y + z s10,3,3d
(1, 1, 0)
Y z
rstd = scos t + t sin tdi + ssin t  t cos tdj,
6. x
u o
1223
0 … t … 2p.
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9.
10.
8x sin y dx  8y cos x dy LC C is the square cut from the first quadrant by the lines x = p>2 and y = p>2 .
y 2 dx + x 2 dy LC C is the circle x 2 + y 2 = 4.
Evaluating Surface Integrals 11. Area of an elliptical region Find the area of the elliptical region cut from the plane x + y + z = 1 by the cylinder x 2 + y 2 = 1. 12. Area of a parabolic cap Find the area of the cap cut from the paraboloid y 2 + z 2 = 3x by the plane x = 1. 13. Area of a spherical cap Find the area of the cap cut from the top of the sphere x 2 + y 2 + z 2 = 1 by the plane z = 22>2.
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b. Find the area of the portion of the cylinder that lies inside the hemisphere. (Hint: Project onto the xzplane. Or evaluate the integral 1 h ds, where h is the altitude of the cylinder and ds is the element of arc length on the circle x 2 + y 2 = 2x in the xyplane.) z
i
The cone z = 1 + 2x 2 + y 2, z … 3
22. Plane above square The portion of the plane 4x + 2y + 4z = 12 that lies above the square 0 … x … 2, 0 … y … 2 in the first quadrant 23. Portion of paraboloid The portion of the paraboloid y = 2sx2 + z2 d, y … 2, that lies above the xyplane 24. Portion of hemisphere The portion of the hemisphere x 2 + y 2 + z 2 = 10, y Ú 0, in the first octant 25. Surface area
Hemisphere z 兹4
21. Cone
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14. a. Hemisphere cut by cylinder Find the area of the surface cut from the hemisphere x 2 + y 2 + z 2 = 4, z Ú 0, by the cylinder x 2 + y 2 = 2x.
r2
You
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Find the area of the surface
rsu, yd = su + ydi + su  ydj + yk, 0 … u … 1, 0 … y … 1.
iaz
26. Surface integral Integrate ƒsx, y, zd = xy  z 2 over the surface in Exercise 25. 27. Area of a helicoid
Find the surface area of the helicoid
nR
rsr, ud = (r cos u)i + (r sin u)j + uk, 0
in the accompanying figure.
x
y
ass a
Cylinder r 2 cos
0 … u … 2p, 0 … r … 1,
dH
ma
17. Circular cylinder cut by planes Integrate gsx, y, zd = x 4ysy 2 + z 2 d over the portion of the cylinder y 2 + z 2 = 25 that lies in the first octant between the planes x = 0 and x = 1 and above the plane z = 3.
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18. Area of Wyoming The state of Wyoming is bounded by the meridians 111°3¿ and 104°3¿ west longitude and by the circles 41° and 45° north latitude. Assuming that Earth is a sphere of radius R = 3959 mi, find the area of Wyoming.
Parametrized Surfaces
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Find the parametrizations for the surfaces in Exercises 19–24. (There are many ways to do these, so your answers may not be the same as those in the back of the book.)
2
(1, 0, 2)
15. Area of a triangle Find the area of the triangle in which the plane sx>ad + s y>bd + sz>cd = 1 sa, b, c 7 0d intersects the first octant. Check your answer with an appropriate vector calculation. 16. Parabolic cylinder cut by planes Integrate yz z a. gsx, y, zd = b. gsx, y, zd = 2 24y + 1 24y 2 + 1 over the surface cut from the parabolic cylinder y 2  z = 1 by the planes x = 0, x = 3 , and z = 0.
z
(1, 0, 0) x
y
28. Surface integral Evaluate the integral 4S 2x 2 + y 2 + 1 ds, where S is the helicoid in Exercise 27.
Conservative Fields Which of the fields in Exercises 29–32 are conservative, and which are not? 29. F = xi + yj + zk 30. F = sxi + yj + zkd>sx 2 + y 2 + z 2 d3>2 31. F = xe yi + ye zj + ze xk 32. F = si + zj + ykd>sx + yzd Find potential functions for the fields in Exercises 33 and 34. 33. F = 2i + s2y + zdj + sy + 1dk 34. F = sz cos xzdi + e yj + sx cos xzdk
19. Spherical band The portion of the sphere x 2 + y 2 + z 2 = 36 between the planes z = 3 and z = 3 23
Work and Circulation
20. Parabolic cap The portion of the paraboloid z = sx 2 + y 2 d>2 above the plane z = 2
In Exercises 35 and 36, find the work done by each field along the paths from (0, 0, 0) to (1, 1, 1) in Exercise 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16 Practice Exercises 36. F = 2xyi + x 2j + k
F =
xi + yj 2
sx + y 2 d3>2
over the plane curve rstd = se t cos tdi + se t sin tdj from the point (1, 0) to the point se 2p, 0d in two ways: a. By using the parametrization of the curve to evaluate the work integral b. By evaluating a potential function for F.
a. Once around the ellipse C in which the plane x + y + z = 1 intersects the cylinder x 2 + z 2 = 25, clockwise as viewed from the positive yaxis
Flux Across a Plane Curve or Surface
Use Green’s Theorem to find the counterclockwise circulation and outward flux for the fields and curves in Exercises 49 and 50. C: The square bounded by x = 0, x = 1, y = 0, y = 1
In Exercises 39 and 40, use the surface integral in Stokes’ Theorem to find the circulation of the field F around the curve C in the indicated direction.
40. Circulation around a circle F = sx2 + ydi + sx + ydj + s4y2  zdk
F = s y  6x2 di + sx + y2 dj
C: The triangle made by the lines y = 0, y = x , and x = 1
51. Zero line integral Show that
ass a
F = y 2i  yj + 3z 2k
50. Triangle
C: The ellipse in which the plane 2x + 6y  3z = 6 meets the cylinder x 2 + y 2 = 1, counterclockwise as viewed from above
F
ln x sin y dy 
cos y x dx = 0
C
for any closed curve C to which Green’s Theorem applies.
dH
C: The circle in which the plane z = y meets the sphere x 2 + y 2 + z 2 = 4, counterclockwise as viewed from above
Mass and Moments
F = s2xy + xdi + sxy  ydj
49. Square
b. Along the curved boundary of the helicoid in Exercise 27 from (1, 0, 0) to s1, 0, 2pd.
39. Circulation around an ellipse
48. Moment of inertia of a cube Find the moment of inertia about the zaxis of the surface of the cube cut from the first octant by the planes x = 1, y = 1, and z = 1 if the density is d = 1.
iaz
38. Flow along different paths Find the flow of the field F = §sx 2ze y d
47. Inertia, radius of gyration, center of mass of a shell Find Iz, Rz, and the center of mass of a thin shell of density dsx, y, zd = z cut from the upper portion of the sphere x 2 + y 2 + z 2 = 25 by the plane z = 3.
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37. Finding work in two ways Find the work done by
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46. Helical wire with constant density Find the mass and center of mass of a wire of constant density d that lies along the helix rstd = s2 sin tdi + s2 cos tdj + 3tk, 0 … t … 2p.
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35. F = 2xyi + j + x 2k
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41. Wire with different densities Find the mass of a thin wire lying along the curve rstd = 22ti + 22tj + s4  t 2 dk, 0 … t … 1, if the density at t is (a) d = 3t and (b) d = 1.
52. a. Outward flux and area Show that the outward flux of the position vector field F = xi + yj across any closed curve to which Green’s Theorem applies is twice the area of the region enclosed by the curve. b. Let n be the outward unit normal vector to a closed curve to which Green’s Theorem applies. Show that it is not possible for F = xi + yj to be orthogonal to n at every point of C.
In Exercises 53 – 56, find the outward flux of F across the boundary of D.
43. Wire with variable density Find the center of mass and the moments of inertia and radii of gyration about the coordinate axes of a thin wire lying along the curve
D: The cube cut from the first octant by the planes x = 1, y = 1, z = 1
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42. Wire with variable density Find the center of mass of a thin wire lying along the curve rstd = ti + 2tj + s2>3dt 3>2k, 0 … t … 2 , if the density at t is d = 3 25 + t .
2 22 3>2 t2 rstd = ti + t j + k, 3 2
54. Spherical cap F = xzi + yzj + k
55. Spherical cap F = 2xi  3yj + zk
if the density at t is d = 1>st + 1d . 44. Center of mass of an arch A slender metal arch lies along the semicircle y = 2a 2  x 2 in the xyplane. The density at the point (x, y) on the arch is dsx, yd = 2a  y. Find the center of mass.
Mu
F = 2xyi + 2yzj + 2xzk
D: The entire surface of the upper cap cut from the solid sphere x 2 + y 2 + z 2 … 25 by the plane z = 3
0 … t … 2,
45. Wire with constant density A wire of constant density d = 1 lies along the curve rstd = se t cos tdi + se t sin tdj + e t k, 0 … t … ln 2. Find z, Iz, and Rz.
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53. Cube
D: The upper region cut from the solid sphere x 2 + y 2 + z 2 … 2 by the paraboloid z = x 2 + y 2 56. Cone and cylinder
F = s6x + ydi  sx + zdj + 4yzk
D: The region in the first octant bounded by the cone z = 2x 2 + y 2, the cylinder x 2 + y 2 = 1, and the coordinate planes
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60. Hemisphere Find the flux of F = s3z + 1dk upward across the hemisphere x 2 + y 2 + z 2 = a 2, z Ú 0 (a) with the Divergence Theorem and (b) by evaluating the flux integral directly.
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58. Cylinder and planes Find the outward flux of the field F = 3xz 2i + yj  z 3k across the surface of the solid in the first octant that is bounded by the cylinder x 2 + 4y 2 = 16 and the planes y = 2z, x = 0 , and z = 0.
59. Cylindrical can Use the Divergence Theorem to find the flux of F = xy 2i + x 2yj + yk outward through the surface of the region enclosed by the cylinder x 2 + y 2 = 1 and the planes z = 1 and z = 1.
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57. Hemisphere, cylinder, and plane Let S be the surface that is bounded on the left by the hemisphere x 2 + y 2 + z 2 = a 2, y … 0, in the middle by the cylinder x 2 + z 2 = a 2, 0 … y … a, and on the right by the plane y = a. Find the flux of F = yi + zj + xk outward across S.
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Chapter 16: Integration in Vector Fields
Chapter 16
Additional and Advanced Exercises
Finding Areas with Green’s Theorem
3. The eight curve x = s1>2d sin 2t, y = sin t, 0 … t … p (one loop)
x = 2 cos t  cos 2t,
1
y = 2 sin t  sin 2t,
y
0
x
1
Y z
y
Use the Green’s Theorem area formula, Equation (13) in Exercises 16.4, to find the areas of the regions enclosed by the curves in Exercises 1–4. 1. The limaçon 0 … t … 2p
i f u s u o
n a sa s H d
iR a
x
0
1
4. The teardrop x = 2a cos t  a sin 2t, y = b sin t, 0 … t … 2p
2. The deltoid 0 … t … 2p
a m m
x = 2 cos t + cos 2t,
y
a h u M 0
3
y b
y = 2 sin t  sin 2t,
x
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0
2a
x
Theory and Applications 5. a. Give an example of a vector field F (x, y, z) that has value 0 at only one point and such that curl F is nonzero everywhere. Be sure to identify the point and compute the curl.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16
6. Find all points (a, b, c) on the sphere x 2 + y 2 + z 2 = R 2 where the vector field F = yz 2i + xz 2j + 2xyzk is normal to the surface and Fsa, b, cd Z 0. 7. Find the mass of a spherical shell of radius R such that at each point (x, y, z) on the surface the mass density dsx, y, zd is its distance to some fixed point (a, b, c) of the surface.
i
c. Show that the total work done equals the work required to move the sheet’s center of mass sx, y, zd straight down to the xyplane.
13. Archimedes’ principle If an object such as a ball is placed in a liquid, it will either sink to the bottom, float, or sink a certain distance and remain suspended in the liquid. Suppose a fluid has constant weight density w and that the fluid’s surface coincides with the plane z = 4 . A spherical ball remains suspended in the fluid and occupies the region x 2 + y 2 + sz  2d2 … 1.
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c. Give an example of a vector field F (x, y, z) that has value 0 on a surface and such that curl F is nonzero everywhere. Be sure to identify the surface and compute the curl.
1227
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b. Give an example of a vector field F (x, y, z) that has value 0 on precisely one line and such that curl F is nonzero everywhere. Be sure to identify the line and compute the curl.
Additional and Advanced Exercises
a. Show that the surface integral giving the magnitude of the total force on the ball due to the fluid’s pressure is n
Force = lim a ws4  zk d ¢sk = n: q k=1
6
ws4  zd ds.
S
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8. Find the mass of a helicoid
0 … r … 1, 0 … u … 2p , if the density function is dsx, y, zd = 2 2x 2 + y 2. See Practice Exercise 27 for a figure. 9. Among all rectangular regions 0 … x … a, 0 … y … b , find the one for which the total outward flux of F = sx 2 + 4xydi  6yj across the four sides is least. What is the least flux?
Buoyant force =
6
wsz  4dk # n ds,
S
where n is the outer unit normal at (x, y, z). This illustrates Archimedes’ principle that the magnitude of the buoyant force on a submerged solid equals the weight of the displaced fluid.
ass a
10. Find an equation for the plane through the origin such that the circulation of the flow field F = zi + xj + yk around the circle of intersection of the plane with the sphere x 2 + y 2 + z 2 = 4 is a maximum.
nR
b. Since the ball is not moving, it is being held up by the buoyant force of the liquid. Show that the magnitude of the buoyant force on the sphere is
rsr, ud = sr cos udi + sr sin udj + uk,
11. A string lies along the circle x 2 + y 2 = 4 from (2, 0) to (0, 2) in the first quadrant. The density of the string is r sx, yd = xy
n
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a. Partition the string into a finite number of subarcs to show that the work done by gravity to move the string straight down to the xaxis is given by Work = lim a g xk yk2 ¢sk = n: q k=1
LC
14. Fluid force on a curved surface A cone in the shape of the surface z = 2x 2 + y 2, 0 … z … 2 is filled with a liquid of constant weight density w. Assuming the xyplane is “ground level,” show that the total force on the portion of the cone from z = 1 to z = 2 due to liquid pressure is the surface integral F =
g xy2 ds,
ws2  zd ds.
S
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where g is the gravitational constant.
c. Show that the total work done equals the work required to move the string’s center of mass sx, yd straight down to the xaxis.
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6
Evaluate the integral.
b. Find the total work done by evaluating the line integral in part (a).
12. A thin sheet lies along the portion of the plane x + y + z = 1 in the first octant. The density of the sheet is d sx, y, zd = xy . a. Partition the sheet into a finite number of subpieces to show that the work done by gravity to move the sheet straight down to the xyplane is given by
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c. Use the Divergence Theorem to find the magnitude of the buoyant force in part (b).
15. Faraday’s Law If E(t, x, y, z) and B(t, x, y, z) represent the electric and magnetic fields at point (x, y, z) at time t, a basic principle of electromagnetic theory says that § * E = 0B>0t. In this expression § * E is computed with t held fixed and 0B>0t is calculated with (x, y, z) fixed. Use Stokes’ Theorem to derive Faraday’s Law FC
E # dr = 
0 B # n ds, 0t 6 S
where C represents a wire loop through which current flows counterclockwise with respect to the surface’s unit normal n, giving rise to the voltage
n
Work = lim a g xk yk zk ¢sk = n: q k=1
6
g xyz ds,
S
where g is the gravitational constant.
b. Find the total work done by evaluating the surface integral in part (a).
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FC
E # dr
around C. The surface integral on the right side of the equation is called the magnetic flux, and S is any oriented surface with boundary C.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields 19. Prove or disprove that if § # F = 0 and § * F = 0, then F = 0.
be the gravitational force field defined for r Z 0. Use Gauss’s Law in Section 16.8 to show that there is no continuously differentiable vector field H satisfying F = § * H. 17. If ƒ(x, y, z) and g(x, y, z) are continuously differentiable scalar functions defined over the oriented surface S with boundary curve C, prove that 6
s§ƒ * §gd # n ds =
S
FC
20. Let S be an oriented surface parametrized by r(u, y). Define the notation dS = ru du * ry dy so that dS is a vector normal to the surface. Also, the magnitude ds = ƒ dS ƒ is the element of surface area (by Equation 5 in Section 16.6). Derive the identity
suf
GmM r ƒrƒ3
ds = sEG  F 2 d1>2 du dy where E = ƒ ru ƒ 2,
You
F = 
i
16. Let
F = ru # ry ,
and
G = ƒ ry ƒ 2.
21. Show that the volume V of a region D in space enclosed by the oriented surface S with outward normal n satisfies the identity
ƒ §g # dr.
V =
1 r # n ds, 36 S
where r is the position vector of the point (x, y, z) in D.
Mu
ham
ma
dH
ass a
nR
18. Suppose that § # F1 = § # F2 and § * F1 = § * F2 over a region D enclosed by the oriented surface S with outward unit normal n and that F1 # n = F2 # n on S. Prove that F1 = F2 throughout D.
iaz
1228
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: Integration in Vector Fields
Technology Application Projects
i
Chapter 16
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1228
Mathematica / Maple Module
Mathematica / Maple Module
You
Work in Conservative and Nonconservative Force Fields Explore integration over vector fields and experiment with conservative and nonconservative force functions along different paths in the field. How Can You Visualize Greenâ€™s Theorem? Explore integration over vector fields and use parametrizations to compute line integrals. Both forms of Greenâ€™s Theorem are explored.
Mathematica / Maple Module
Mu
ham
ma
dH
ass a
nR
iaz
Visualizing and Interpreting the Divergence Theorem Verify the Divergence Theorem by formulating and evaluating certain divergence and surface integrals.
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suf
i
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Chapter
You
15
FIRSTORDER DIFFERENTIAL EQUATIONS
Solutions, Slope Fields, and Picard’s Theorem
ass a
15.1
nR
iaz
OVERVIEW In Section 4.8 we introduced differential equations of the form dy>dx = ƒ(x), where ƒ is given and y is an unknown function of x. When ƒ is continuous over some interval, we found the general solution y(x) by integration, y = 1 ƒ(x) dx. In Section 6.5 we solved separable differential equations. Such equations arise when investigating exponential growth or decay, for example. In this chapter we study some other types of firstorder differential equations. They involve only first derivatives of the unknown function.
dH
We begin this section by defining general differential equations involving first derivatives. We then look at slope fields, which give a geometric picture of the solutions to such equations. Finally we present Picard’s Theorem, which gives conditions under which firstorder differential equations have exactly one solution.
General FirstOrder Differential Equations and Solutions
ma
A firstorder differential equation is an equation dy = ƒsx, yd dx
(1)
Mu
ham
in which ƒ(x, y) is a function of two variables defined on a region in the xyplane. The equation is of first order because it involves only the first derivative dy> dx (and not higherorder derivatives). We point out that the equations y¿ = ƒsx, yd
and
d y = ƒsx, yd, dx
are equivalent to Equation (1) and all three forms will be used interchangeably in the text. A solution of Equation (1) is a differentiable function y = ysxd defined on an interval I of xvalues (perhaps infinite) such that d ysxd = ƒsx, ysxdd dx on that interval. That is, when y(x) and its derivative y¿sxd are substituted into Equation (1), the resulting equation is true for all x over the interval I. The general solution to a firstorder differential equation is a solution that contains all possible solutions. The general
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 152
Chapter 15: FirstOrder Differential Equations
Show that every member of the family of functions C y = x + 2
You
EXAMPLE 1
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i
solution always contains an arbitrary constant, but having this property doesn’t mean a solution is the general solution. That is, a solution may contain an arbitrary constant without being the general solution. Establishing that a solution is the general solution may require deeper results from the theory of differential equations and is best studied in a more advanced course.
iaz
is a solution of the firstorder differential equation dy 1 = x s2  yd dx on the interval s0, q d, where C is any constant. Solution Differentiating y = C>x + 2 gives
nR
dy C d 1 a b + 0 =  2. = C dx dx x x
Thus we need only verify that for all x H s0, q d,
C C 1 = x c2  a x + 2b d . x2
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This last equation follows immediately by expanding the expression on the righthand side: C C C 1 1 x c2  a x + 2b d = x a x b =  x 2 .
dH
Therefore, for every value of C, the function y = C>x + 2 is a solution of the differential equation.
Mu
ham
ma
As was the case in finding antiderivatives, we often need a particular rather than the general solution to a firstorder differential equation y¿ = ƒsx, yd. The particular solution satisfying the initial condition ysx0 d = y0 is the solution y = ysxd whose value is y0 when x = x0. Thus the graph of the particular solution passes through the point sx0 , y0 d in the xyplane. A firstorder initial value problem is a differential equation y¿ = ƒsx, yd whose solution must satisfy an initial condition ysx0 d = y0.
EXAMPLE 2
Show that the function y = sx + 1d 
1 x e 3
is a solution to the firstorder initial value problem dy 2 = y  x, ys0d = . 3 dx Solution The equation
dy = y  x dx is a firstorder differential equation with ƒsx, yd = y  x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1
–2
–1
2
x
4
On the right side of the equation:
–2
y  x = sx + 1d 
–3 –4
i
dy d 1 1 ax + 1  e x b = 1  e x . = 3 3 dx dx
1
suf
–4
y (x 1) 1 e x 3
1 x 1 e  x = 1  ex. 3 3
You
0, 2 3
153
On the left side of the equation:
y 2
Solutions, Slope Fields, and Picard’s Theorem
The function satisfies the initial condition because ys0d = csx + 1d 
2 1 x 1 e d = 1  = . 3 3 3 x=0
The graph of the function is shown in Figure 15.1.
iaz
FIGURE 15.1 Graph of the solution 1 y = sx + 1d  e x to the differential 3 equation dy>dx = y  x , with initial 2 condition ys0d = (Example 2). 3
Slope Fields: Viewing Solution Curves
Mu
ham
ma
dH
ass a
nR
Each time we specify an initial condition ysx0 d = y0 for the solution of a differential equation y¿ = ƒsx, yd, the solution curve (graph of the solution) is required to pass through the point sx0 , y0 d and to have slope ƒsx0 , y0 d there. We can picture these slopes graphically by drawing short line segments of slope ƒ(x, y) at selected points (x, y) in the region of the xyplane that constitutes the domain of ƒ. Each segment has the same slope as the solution curve through (x, y) and so is tangent to the curve there. The resulting picture is called a slope field (or direction field) and gives a visualization of the general shape of the solution curves. Figure 15.2a shows a slope field, with a particular solution sketched into it in Figure 15.2b. We see how these line segments indicate the direction the solution curve takes at each point it passes through.
–4
y
–2
y
4
4
2
2
0
2
4
x
–4
–2
0
–2
–2
–4
–4
0, 2 3
2
4
x
(a)
(b)
FIGURE 15.2 (a) Slope field for
dy = y  x . (b) The particular solution dx
2 curve through the point a0, b (Example 2). 3
Figure 15.3 shows three slope fields and we see how the solution curves behave by following the tangent line segments in these fields.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
2 xy 1 x2
(c) y' (1 x)y x 2
iaz
(b) y' –
You
(a) y' y x 2
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i
154
nR
FIGURE 15.3 Slope fields (top row) and selected solution curves (bottom row). In computer renditions, slope segments are sometimes portrayed with arrows, as they are here. This is not to be taken as an indication that slopes have directions, however, for they do not.
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Constructing a slope field with pencil and paper can be quite tedious. All our examples were generated by a computer.
The Existence of Solutions
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A basic question in the study of firstorder initial value problems concerns whether a solution even exists. A second important question asks whether there can be more than one solution. Some conditions must be imposed to assure the existence of exactly one solution, as illustrated in the next example.
y
1
–1
(5, 1)
5
Mu
–5 (–5, –1)
(7, 5.4)
dy = y 4>5, dx
y(0) = 0
has more than one solution. One solution is the constant function y(x) = 0 for which the graph lies along the xaxis. A second solution is found by separating variables and integrating, as we did in Section 6.5. This leads to
ham
5
The initial value problem
ma
EXAMPLE 3
x
–5
5
x y = a b . 5 The two solutions y = 0 and y = (x>5) 5 both satisfy the initial condition y(0) = 0 (Figure 15.4). We have found a differential equation with multiple solutions satisfying the same initial condition. This differential equation has even more solutions. For instance, two additional solutions are
FIGURE 15.4 The graph of the solution y = (x>5) 5 to the initial value problem in Example 3. Another solution is y = 0.
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y = •
0,
for x … 0 5
x a b , 5
for x 7 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1
Solutions, Slope Fields, and Picard’s Theorem
155
5
x a b , y = • 5 0,
for x … 0
.
You
for x 7 0
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i
and
iaz
In many applications it is desirable to know that there is exactly one solution to an initial value problem. Such a solution is said to be unique. Picard’s Theorem gives conditions under which there is precisely one solution. It guarantees both the existence and uniqueness of a solution.
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THEOREM 1—Picard’s Theorem Suppose that both ƒ(x, y) and its partial derivative 0ƒ>0y are continuous on the interior of a rectangle R, and that (x0, y0) is an interior point of R. Then the initial value problem dy = ƒ(x, y), dx
y(x0) = y0
(2)
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has a unique solution y y(x) for x in some open interval containing x0.
ma
dH
The differential equation in Example 3 fails to satisfy the conditions of Picard’s Theorem. Although the function ƒ(x, y) = y 4>5 from Example 3 is continuous in the entire xyplane, the partial derivative 0ƒ>0y = (4>5) y 1>5 fails to be continuous at the point (0, 0) specified by the initial condition. Thus we found the possibility of more than one solution to the given initial value problem. Moreover, the partial derivative 0ƒ>0y is not even defined where y = 0. However, the initial value problem of Example 3 does have unique solutions whenever the initial condition y(x0) = y0 has y0 Z 0.
Mu
ham
Picard’s Iteration Scheme Picard’s Theorem is proved by applying Picard’s iteration scheme, which we now introduce. We begin by noticing that any solution to the initial value problem of Equations (2) must also satisfy the integral equation x
y(x) = y0 +
Lx0
ƒ(t, y(t)) dt
(3)
because x
dy dt = y(x)  y(x0). Lx0 dt The converse is also true: If y(x) satisfies Equation (3), then y¿ = ƒ(x, y(x)) and y(x0) = y0. So Equations (2) may be replaced by Equation (3). This sets the stage for Picard’s interation
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 156
Chapter 15: FirstOrder Differential Equations
x
y1(x) = y0 +
Lx0
ƒ(t, y0) dt.
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i
method: In the integrand in Equation (3), replace y(t) by the constant y0, then integrate and call the resulting righthand side of Equation (3) y1(x): (4)
You
This starts the process. To keep it going, we use the iterative formulas x
yn + 1(x) = y0 +
Lx0
ƒ(t, yn(t)) dt.
(5)
EXAMPLE 4
nR
iaz
The proof of Picard’s Theorem consists of showing that this process produces a sequence of functions {yn(x)} that converge to a function y(x) that satisfies Equations (2) and (3) for values of x sufficiently near x0. (The proof also shows that the solution is unique; that is, no other method will lead to a different solution.) The following examples illustrate the Picard iteration scheme, but in most practical cases the computations soon become too burdensome to continue. Illustrate the Picard iteration scheme for the initial value problem y¿ = x  y,
y(0) = 1.
ass a
Solution For the problem at hand, ƒ(x, y) = x  y, and Equation (4) becomes
y1(x) = 1 +
dH
= 1 +
x
L0
(t  1) dt
y0 = 1
x2  x. 2
If we now use Equation (5) with n = 1, we get
ma
y2(x) = 1 +
x
L0
at  1 
= 1  x + x2 
t2 + tb dt 2
Substitute y1 for y in ƒ(t, y).
x3 . 6
Mu
ham
The next iteration, with n = 2, gives x
y3(x) = 1 +
L0
at  1 + t  t 2 +
= 1  x + x2 
t3 b dt 6
Substitute y2 for y in ƒ(t, y).
x4 x3 + . 3 4!
In this example it is possible to find the exact solution because
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dy + y = x dx
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.1
Solutions, Slope Fields, and Picard’s Theorem
157
y = x  1 + Ce x
suf
i
is a firstorder differential equation that is linear in y. You will learn how to find the general solution
in the next section. The solution of the initial value problem is then
You
y = x  1 + 2e x.
If we substitute the Maclaurin series for e x in this particular solution, we get x2 x3 x4 +  Áb 2! 3! 4!
iaz
y = x  1 + 2 a1  x + = 1  x + x2 
x5 x3 x4 + Á b, + 2a 3 4! 5!
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and we see that the Picard scheme producing y3(x) has given us the first four terms of this expansion.
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In the next example we cannot find a solution in terms of elementary functions. The Picard scheme is one way we could get an idea of how the solution behaves near the initial point.
EXAMPLE 5
Find yn(x) for n = 0, 1, 2, and 3 for the initial value problem y¿ = x 2 + y 2,
y(0) = 0.
dH
Solution By definition, y0(x) = y(0) = 0. The other functions yn(x) are generated by the
integral representation x
yn + 1(x) = 0 +
L0
C t 2 + (yn(t)) 2 D dt
ma
x
=
x3 (yn(t)) 2 dt. + 3 L0
Mu
ham
We successively calculate y1(x) =
x3 , 3
y2(x) =
x7 x3 + , 3 63
y3(x) =
x3 x7 2x 11 x 15 + + + . 3 63 2079 59535
In Section 15.4 we introduce numerical methods for solving initial value problems like those in Examples 4 and 5.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 158
Chapter 15: FirstOrder Differential Equations
y
1. y¿ = x + y
2. y¿ = y + 1
x 3. y¿ =  y
4. y¿ = y 2  x 2
4
In Exercises 5 and 6, copy the slope fields and sketch in some of the solution curves.
2
5. y¿ = ( y + 2)( y  2) –4
You
In Exercises 1–4, match the differential equations with their slope fields, graphed here.
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EXERCISES 15.1
–2
2
4
x
y
4
iaz
–2
2
–4 (a)
4
–2
2
4
x
–2 –4
y 4
–4
–2
ma
2
2
4
ham Mu
–2
x
–4
In Exercises 7–10, write an equivalent firstorder differential equation and initial condition for y. x
L1
(t  y (t)) dt
x
1 dt L1 t x
9. y = 2 
2
2
–4
–2
8. y =
4
–2
4
–2
2
–4
7. y = 1 +
y
–4
2
x
y
–4
(c)
4
4
x
–2
2
6. y¿ = y( y + 1)( y  1)
dH
(b)
ass a
2
–4
–2
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–4
y
(1 + y (t)) sin t dt
L0 x
4
x
10. y = 1 +
L0
y (t) dt
–2
Use Picard’s iteration scheme to find yn(x) for n = 0, 1, 2, 3 in Exercises 11–16.
–4
11. y¿ = x,
y(1) = 2
12. y¿ = y,
y(0) = 1
(d)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.2
16. y¿ = 2x  y,
i
y(0) = 1
25. A logistic equation y¿ = ys2  yd, 0 … x … 4, 0 … y … 3
y(1) = 1
17. Show that the solution of the initial value problem y¿ = x + y,
26. y¿ = ssin xdssin yd,
y(x0) = y0
y = 1  x + (1 + x0 + y0) e
x  x0
.
18. What integral equation is equivalent to the initial value problem y¿ = ƒ(x), y(x0) = y0? COMPUTER EXPLORATIONS In Exercises 19–24, obtain a slope field and add to it graphs of the solution curves passing through the given points. 19. y¿ = y with
c. (0, 5)
b. (0, 2)
c. (0, 1>4) c. s0, 1d
23. y¿ = s y  1dsx + 2d with
d. s 1, 1d
0 … y … 5 ys0d = 1>3;
29. Use a CAS to find the solutions of y¿ + y = ƒsxd subject to the initial condition ys0d = 0 , if ƒ(x) is a. 2x
c. 3e x>2
b. sin 2x
d. 2e x>2 cos 2x .
c.
y¿ =
3x 2 + 4x + 2 2s y  1d
over the region 3 … x … 3 and 3 … y … 3 .
d. (0, 0)
b. Separate the variables and use a CAS integrator to find the general solution in implicit form.
d. (1, 1)
c. Using a CAS implicit function grapher, plot solution curves for the arbitrary constant values C = 6, 4, 2, 0, 2, 4, 6 .
dH
c. (0, 3)
d. Find and graph the solution that satisfies the initial condition ys0d = 1 .
A 2 23, 4 B
FirstOrder Linear Equations
ham Mu
ys0d = 2;
ma
a. s0, 1d b. (0, 1) xy 24. y¿ = 2 with x + 4 a. (0, 2) b. s0, 6d
15.2
0 … x … 5,
28. A Gompertz equation y¿ = ys1>2  ln yd, 0 … x … 4, 0 … y … 3
ass a
b. (0, 2)
22. y¿ = y 2 with a. (0, 1)
6 … y … 6
30. a. Use a CAS to plot the slope field of the differential equation
b. (0, 4)
21. y¿ = ysx + yd with a. (0, 1)
6 … x … 6,
Graph all four solutions over the interval 2 … x … 6 to compare the results.
c. s0, 1d
20. y¿ = 2s y  4d with a. (0, 1)
27. y¿ = cos s2x  yd,
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b. (0, 2)
ys0d = 2;
ys0d = 1>2;
Exercises 27 and 28 have no explicit solution in terms of elementary functions. Use a CAS to explore graphically each of the differential equations.
is
a. (0, 1)
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y(0) = 0
15. y¿ = x + y,
You
14. y¿ = x + y,
159
In Exercises 25 and 26, obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation.
y(1) = 1
iaz
13. y¿ = xy,
FirstOrder Linear Equations
A firstorder linear differential equation is one that can be written in the form dy + Psxdy = Qsxd, dx
(1)
where P and Q are continuous functions of x. Equation (1) is the linear equation’s standard form. Since the exponential growth> decay equation dy>dx = ky (Section 6.5) can be put in the standard form dy  ky = 0, dx we see it is a linear equation with Psxd = k and Qsxd = 0. Equation (1) is linear (in y) because y and its derivative dy> dx occur only to the first power, are not multiplied together, nor do they appear as the argument of a function A such as sin y, e y , or 2dy>dx B .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
Put the following equation in standard form: x
dy = x 2 + 3y, dx
i
EXAMPLE 1
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1510
x 7 0.
x
dy = x 2 + 3y dx dy 3 = x + xy dx
Divide by x
Standard form with Psxd = 3>x and Qsxd = x
iaz
dy 3  xy = x dx
You
Solution
Notice that P(x) is 3>x, not +3>x. The standard form is y¿ + Psxdy = Qsxd, so the minus sign is part of the formula for P(x).
We solve the equation
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Solving Linear Equations
dy + Psxdy = Qsxd dx
(2)
ass a
by multiplying both sides by a positive function y(x) that transforms the lefthand side into the derivative of the product ysxd # y. We will show how to find y in a moment, but first we want to show how, once found, it provides the solution we seek. Here is why multiplying by y(x) works: dy + Psxdy = Qsxd dx
dH
Original equation is in standard form.
dy + Psxdysxdy = ysxdQsxd dx d sysxd # yd = ysxdQsxd dx ysxd # y = y =
L
Multiply by positive y(x). ysxd is chosen to make dy d sy # yd . + Pyy = y dx dx Integrate with respect to x.
ysxdQsxd dx
1 ysxdQsxd dx ysxd L
(3)
Equation (3) expresses the solution of Equation (2) in terms of the function y(x) and Q(x). We call y(x) an integrating factor for Equation (2) because its presence makes the equation integrable. Why doesn’t the formula for P(x) appear in the solution as well? It does, but indirectly, in the construction of the positive function y(x). We have
Mu
ham
ma
ysxd
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dy d syyd = y + Pyy dx dx dy dy dy y + y = y + Pyy dx dx dx dy y = Pyy dx
Condition imposed on y Product Rule for derivatives The terms y
dy cancel. dx
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.2
FirstOrder Linear Equations
1511
i
This last equation will hold if
Variables separated, y 7 0
dy y =
ln y =
You
dy y = P dx L
P dx
Integrate both sides.
L
P dx
Since y 7 0 , we do not need absolute value signs in ln y.
e ln y = e 1 P dx
Exponentiate both sides to solve for y.
(4)
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y = e 1 P dx
iaz
L
suf
dy = Py dx
ass a
Thus a formula for the general solution to Equation (1) is given by Equation (3), where y(x) is given by Equation (4). However, rather than memorizing the formula, just remember how to find the integrating factor once you have the standard form so P(x) is correctly identified.
dH
To solve the linear equation y¿ + Psxdy = Qsxd, multiply both sides by the integrating factor ysxd = e 1 Psxd dx and integrate both sides.
When you integrate the lefthand side product in this procedure, you always obtain the product y(x)y of the integrating factor and solution function y because of the way y is defined. Solve the equation
ma
EXAMPLE 2
ham
HISTORICAL BIOGRAPHY
x
dy = x 2 + 3y, dx
x 7 0.
Solution First we put the equation in standard form (Example 1):
dy 3  x y = x, dx
Adrien Marie Legendre (1752–1833)
Mu
so Psxd = 3>x is identified. The integrating factor is
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ysxd = e 1 Psxd dx = e 1s3>xd dx = e 3 ln ƒ x ƒ
Constant of integration is 0, so y is as simple as possible.
= e 3 ln x
x 7 0
= e ln x
3
1 = 3. x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1512
Chapter 15: FirstOrder Differential Equations
i
Next we multiply both sides of the standard form by y(x) and integrate:
suf
3 1 # dy 1 a  x yb = 3 # x x 3 dx x
d 1 1 a yb = 2 dx x 3 x
You
3 1 dy 1  4y = 2 x 3 dx x x Lefthand side is
1 1 y = dx 2 x3 x L
d sy # yd. dx
Integrate both sides.
iaz
1 1 y =  x + C. x3
Solving this last equation for y gives the general solution:
EXAMPLE 3
nR
1 y = x 3 a x + Cb = x 2 + Cx 3,
x 7 0.
Find the particular solution of
ass a
3xy多  y = ln x + 1,
x 7 0,
satisfying ys1d = 2.
dH
Solution With x 7 0, we write the equation in standard form:
y多 
ln x + 1 1 y = . 3x 3x
Then the integrating factor is given by y = e 1  dx>3x = e s1>3dln x = x 1>3 .
x 7 0
ma
Thus
x 1>3y =
1 sln x + 1dx 4>3 dx. 3L
Lefthand side is yy.
Mu
ham
Integration by parts of the righthand side gives x 1>3y = x 1>3sln x + 1d +
L
x 4>3 dx + C.
Therefore x 1>3y = x 1>3sln x + 1d  3x 1>3 + C or, solving for y, y = sln x + 4d + Cx 1>3 . When x = 1 and y = 2 this last equation becomes
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2 = s0 + 4d + C,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.2
FirstOrder Linear Equations
1513
C = 2.
suf
i
so
Substitution into the equation for y gives the particular solution
You
y = 2x 1>3  ln x  4.
In solving the linear equation in Example 2, we integrated both sides of the equation after multiplying each side by the integrating factor. However, we can shorten the amount of work, as in Example 3, by remembering that the lefthand side always integrates into the product ysxd # y of the integrating factor times the solution function. From Equation (3) this means that ysxdy =
nR
iaz
ysxdQsxd dx. L We need only integrate the product of the integrating factor y(x) with the righthand side Q(x) of Equation (1) and then equate the result with y(x)y to obtain the general solution. Nevertheless, to emphasize the role of y(x) in the solution process, we sometimes follow the complete procedure as illustrated in Example 2. Observe that if the function Q(x) is identically zero in the standard form given by Equation (1), the linear equation is separable:
ass a
dy + Psxdy = Qsxd dx dy + Psxdy = 0 dx
Qsxd K 0
dy = Psxd dx
Separating the variables
dH
We now present two applied problems modeled by a firstorder linear differential equation.
RL Circuits
Switch a
b
ham
i
R
The diagram in Figure 15.5 represents an electrical circuit whose total resistance is a constant R ohms and whose selfinductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. Ohmâ€™s Law, V = RI, has to be modified for such a circuit. The modified form is
ma
V
L
Mu
FIGURE 15.5 The RL circuit in Example 4.
L
di + Ri = V, dt
(5)
where i is the intensity of the current in amperes and t is the time in seconds. By solving this equation, we can predict how the current will flow after the switch is closed.
EXAMPLE 4
The switch in the RL circuit in Figure 15.5 is closed at time t = 0. How will the current flow as a function of time?
Solution Equation (5) is a firstorder linear differential equation for i as a function of t. Its standard form is
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V di R + i = , L L dt
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(6)
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
I V R
I e
i =
i V (1 eRt/L) R
V V  e sR>Ldt R R
i
and the corresponding solution, given that i = 0 when t = 0, is
i
suf
1514
(7)
L R
2
L R
3
L R
4
L R
t
lim i = lim a
t: q
V V V V V  e sR>Ldt b =  #0 = . R R R R R
At any given time, the current is theoretically less than V> R, but as time passes, the current approaches the steadystate value V> R. According to the equation L
di + Ri = V, dt
I = V>R is the current that will flow in the circuit if either L = 0 (no inductance) or di>dt = 0 (steady current, i = constant) (Figure 15.6). Equation (7) expresses the solution of Equation (6) as the sum of two terms: a steadystate solution V> R and a transient solution sV>Rde sR>Ldt that tends to zero as t : q.
ass a
nR
FIGURE 15.6 The growth of the current in the RL circuit in Example 4. I is the current’s steadystate value. The number t = L>R is the time constant of the circuit. The current gets to within 5% of its steadystate value in 3 time constants (Exercise 31).
t: q
iaz
0
You
(Exercise 32). Since R and L are positive, sR>Ld is negative and e sR>Ldt : 0 as t : q. Thus,
Mixture Problems
dH
A chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas) with, possibly, a specified amount of the chemical dissolved as well. The mixture is kept uniform by stirring and flows out of the container at a known rate. In this process, it is often important to know the concentration of the chemical in the container at any given time. The differential equation describing the process is based on the formula
ma
Rate of change rate at which rate at which of amount = £ chemical ≥  £ chemical ≥ in container arrives departs.
(8)
Mu
ham
If y(t) is the amount of chemical in the container at time t and V(t) is the total volume of liquid in the container at time t, then the departure rate of the chemical at time t is Departure rate =
ystd # soutflow rated Vstd
concentration in = acontainer at time t b # soutflow rated.
(9)
Accordingly, Equation (8) becomes dy ystd # soutflow rated. = schemical’s arrival rated dt Vstd
(10)
If, say, y is measured in pounds, V in gallons, and t in minutes, the units in Equation (10) are
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pounds pounds gallons pounds # = . minutes minutes gallons minutes
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.2
FirstOrder Linear Equations
1515
You
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i
EXAMPLE 5 In an oil refinery, a storage tank contains 2000 gal of gasoline that initially has 100 lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 lb of additive per gallon is pumped into the tank at a rate of 40 gal> min. The wellmixed solution is pumped out at a rate of 45 gal> min. How much of the additive is in the tank 20 min after the pumping process begins (Figure 15.7)?
iaz
40 gal/min containing 2 lb/gal
45 gal/min containing y lb/gal V
nR
FIGURE 15.7 The storage tank in Example 5 mixes input liquid with stored liquid to produce an output liquid.
Solution Let y be the amount (in pounds) of additive in the tank at time t. We know that
ass a
y = 100 when t = 0. The number of gallons of gasoline and additive in solution in the tank at any time t is Vstd = 2000 gal + a40
gal gal  45 b st mind min min
dH
= s2000  5td gal.
Therefore,
Rate out =
= a
ma ham Mu
ystd # outflow rate Vstd
=
y b 45 2000  5t
Eq. (9) Outflow rate is 45 gal> min and y = 2000  5t .
45y lb . 2000  5t min
Also, Rate in = a2 = 80
gal lb b a40 b gal min lb . min
Eq. (10)
The differential equation modeling the mixture process is dy 45y = 80 2000  5t dt in pounds per minute.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1516
Chapter 15: FirstOrder Differential Equations
i
To solve this differential equation, we first write it in standard form:
suf
dy 45 + y = 80. 2000  5t dt
45
ystd = e 1 P dt = e 1 2000  5t dt = e 9 ln s2000  5td
2000  5t 7 0
iaz
= s2000  5td9 .
You
Thus, Pstd = 45>s2000  5td and Qstd = 80. The integrating factor is
Multiplying both sides of the standard equation by y(t) and integrating both sides gives
s2000  5td9
dy 45 yb = 80s2000  5td9 + 2000  5t dt
nR
s2000  5td9 # a
dy + 45s2000  5td10 y = 80s2000  5td9 dt
ass a
d C s2000  5td9y D = 80s2000  5td9 dt
dH
s2000  5td9y =
L
80s2000  5td9 dt
s2000  5td9y = 80 #
s2000  5td8 + C. s 8ds 5d
ma
The general solution is
y = 2s2000  5td + Cs2000  5td9 .
Mu
ham
Because y = 100 when t = 0, we can determine the value of C: 100 = 2s2000  0d + Cs2000  0d9 C = 
3900 . s2000d9
The particular solution of the initial value problem is y = 2s2000  5td 
3900 s2000  5td9 . s2000d9
The amount of additive 20 min after the pumping begins is ys20d = 2[2000  5s20d] 
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3900 [2000  5s20d]9 L 1342 lb. s2000d9
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.2
FirstOrder Linear Equations
a. as a firstorder linear equation.
Solve the differential equations in Exercises 1–14. dy dy 1. x 2. e x + y = e x, x 7 0 + 2e x y = 1 dx dx
b. as a separable equation.
You
23. Is either of the following equations correct? Give reasons for your answers.
sin x 3. xy¿ + 3y = 2 , x 7 0 x 4. y¿ + stan xdy = cos2 x, p>2 6 x 6 p>2 dy 1 + 2y = 1  x , x 7 0 5. x dx 6. s1 + xdy¿ + y = 2x
1 1 a. x x dx = x ln ƒ x ƒ + C b. x x dx = x ln ƒ x ƒ + Cx L L 24. Is either of the following equations correct? Give reasons for your answers. 1 a. cos x
7. 2y¿ = e x>2 + y
2x
8. e y¿ + 2e y = 2x 9. xy¿  y = 2x ln x dy cos x = x  2y, x 7 0 10. x dx
14. tan u
t 7 1
ds 1 , + 2s = 3st + 1d + dt st + 1d2
dr + scos udr = tan u, 13. sin u du dr + r = sin2 u, du
t 7 1
0 6 u 6 p>2
17. u
dy + y = sin u, du
18. u
dy  2y = u3 sec u tan u, du
t 7 0,
b. What is the volume of brine in the tank at time t ?
d. Write down and solve the initial value problem describing the mixing process.
dH
Solve the initial value problems in Exercises 15–20. dy + 2y = 3, ys0d = 1 15. dt dy + 2y = t 3, dt
a. At what rate (pounds per minute) does salt enter the tank at time t ? c. At what rate (pounds per minute) does salt leave the tank at time t ?
0 6 u 6 p>2
16. t
C cos x dx = tan x + cos x L 25. Salt mixture A tank initially contains 100 gal of brine in which 50 lb of salt are dissolved. A brine containing 2 lb> gal of salt runs into the tank at the rate of 5 gal> min. The mixture is kept uniform by stirring and flows out of the tank at the rate of 4 gal> min. 1 b. cos x
nR
12. st + 1d
ds + 4st  1d2s = t + 1, dt
e. Find the concentration of salt in the tank 25 min after the process starts. 26. Mixture problem A 200gal tank is half full of distilled water. At time t = 0 , a solution containing 0.5 lb> gal of concentrate enters the tank at the rate of 5 gal> min, and the wellstirred mixture is withdrawn at the rate of 3 gal> min.
ys2d = 1
ysp>2d = 1
ma
u 7 0,
u 7 0,
a. At what time will the tank be full?
ysp>3d = 2
b. At the time the tank is full, how many pounds of concentrate will it contain?
x2
20.
dy e ,  2sx 2 + xdy = x + 1 dx
x 7 1,
ys0d = 5
ham
19. sx + 1d
dy + xy = x, dx
ys0d = 6
21. Solve the exponential growth> decay initial value problem for y as a function of t thinking of the differential equation as a firstorder linear equation with Psxd = k and Qsxd = 0 :
Mu
dy = ky dt
sk constantd,
ys0d = y0
22. Solve the following initial value problem for u as a function of t: k du + mu = 0 dt
L
ass a
11. st  1d3
cos x dx = tan x + C
iaz
2x
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EXERCISES 15.2
1517
sk and m positive constantsd,
us0d = u0
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27. Fertilizer mixture A tank contains 100 gal of fresh water. A solution containing 1 lb> gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal> min, and the mixture is pumped out of the tank at the rate of 3 gal> min. Find the maximum amount of fertilizer in the tank and the time required to reach the maximum. 28. Carbon monoxide pollution An executive conference room of a corporation contains 4500 ft3 of air initially free of carbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the room at the rate of 0.3 ft3>min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft3>min. Find the time when the concentration of carbon monoxide in the room reaches 0.01%.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
L
i
V + Ce sR>Ldt . R
i =
suf
30. Current in an open RL circuit If the switch is thrown open after the current in an RL circuit has built up to its steadystate value I = V>R , the decaying current (graphed here) obeys the equation
is
b. Then use the initial condition is0d = 0 to determine the value of C. This will complete the derivation of Equation (7). c. Show that i = V>R is a solution of Equation (6) and that i = Ce sR>Ldt satisfies the equation
You
29. Current in a closed RL circuit How many seconds after the switch in an RL circuit is closed will it take the current i to reach half of its steadystate value? Notice that the time depends on R and L and not on how much voltage is applied.
di + Ri = 0 , dt
di R + i = 0. L dt
which is Equation (5) with V = 0 .
HISTORICAL BIOGRAPHY
a. Solve the equation to express i as a function of t. b. How long after the switch is thrown will it take the current to fall to half its original value?
A Bernoulli differential equation is of the form
nR
c. Show that the value of the current when t = L>R is I>e. (The significance of this time is explained in the next exercise.)
James Bernoulli (1654–1705)
iaz
1518
dy + Psxdy = Qsxdy n . dx
i V R
du + s1  ndPsxdu = s1  ndQsxd . dx
I e L R
2
L R
3
t
L R
For example, in the equation
dH
0
ass a
Observe that, if n = 0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u = y 1  n transforms the Bernoulli equation into the linear equation
ma
31. Time constants Engineers call the number L>R the time constant of the RL circuit in Figure 15.6. The significance of the time constant is that the current will reach 95% of its final value within 3 time constants of the time the switch is closed (Figure 15.6). Thus, the time constant gives a builtin measure of how rapidly an individual circuit will reach equilibrium. a. Find the value of i in Equation (7) that corresponds to t = 3L>R and show that it is about 95% of the steadystate value I = V>R .
ham
b. Approximately what percentage of the steadystate current will be flowing in the circuit 2 time constants after the switch is closed (i.e., when t = 2L>R)? 32. Derivation of Equation (7) in Example 4 a. Show that the solution of the equation
Mu
V di R + i = L L dt
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dy  y = e x y 2 dx
we have n = 2 , so that u = y 1  2 = y 1 and du>dx = y 2 dy>dx . Then dy>dx = y 2 du>dx = u 2 du>dx . Substitution into the original equation gives u 2
du  u 1 = e x u 2 dx
or, equivalently, du + u = e x . dx This last equation is linear in the (unknown) dependent variable u. Solve the differential equations in Exercises 33–36. 33. y¿  y = y 2
34. y¿  y = xy 2
35. xy¿ + y = y
36. x 2y¿ + 2xy = y 3
2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.3
i
Applications
1519
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15.3
Applications
You
We now look at three applications of firstorder differential equations. The first application analyzes an object moving along a straight line while subject to a force opposing its motion. The second is a model of population growth. The last application considers a curve or curves intersecting each curve in a second family of curves orthogonally (that is, at right angles).
Resistance Proportional to Velocity
iaz
In some cases it is reasonable to assume that the resistance encountered by a moving object, such as a car coasting to a stop, is proportional to the object’s velocity. The faster the object moves, the more its forward progress is resisted by the air through which it passes. Picture the object as a mass m moving along a coordinate line with position function s and velocity y at time t. From Newton’s second law of motion, the resisting force opposing the motion is
nR
Force = mass * acceleration = m
dy . dt
If the resisting force is proportional to velocity, we have dy = ky dt
dy k = my dt
or
ass a
m
sk 7 0d.
This is a separable differential equation representing exponential change. The solution to the equation with initial condition y = y0 at t = 0 is (Section 6.5) y = y0 e sk>mdt .
(1)
Mu
ham
ma
dH
What can we learn from Equation (1)? For one thing, we can see that if m is something large, like the mass of a 20,000ton ore boat in Lake Erie, it will take a long time for the velocity to approach zero (because t must be large in the exponent of the equation in order to make kt> m large enough for y to be small). We can learn even more if we integrate Equation (1) to find the position s as a function of time t. Suppose that a body is coasting to a stop and the only force acting on it is a resistance proportional to its speed. How far will it coast? To find out, we start with Equation (1) and solve the initial value problem ds = y0 e sk>mdt , dt
ss0d = 0.
Integrating with respect to t gives s = 
y0 m sk>mdt e + C. k
Substituting s = 0 when t = 0 gives 0 = 
y0 m + C k
and
C =
y0 m . k
The body’s position at time t is therefore
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sstd = 
y0 m sk>mdt y0 m y0 m e + = s1  e sk/mdt d. k k k
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(2)
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1520
Chapter 15: FirstOrder Differential Equations
y0 m s1  e sk>mdt d k y0 m y0 m = s1  0d = . k k
lim sstd = lim
t: q
Thus, Distance coasted =
You
t: q
suf
i
To find how far the body will coast, we find the limit of s(t) as t : q . Since sk>md 6 0, we know that e sk>mdt : 0 as t : q , so that
y0 m . k
(3)
Pounds = slugs * 32 ,
EXAMPLE 1 For a 192lb ice skater, the k in Equation (1) is about 1> 3 slug> sec and m = 192>32 = 6 slugs. How long will it take the skater to coast from 11 ft> sec (7.5 mph) to 1 ft> sec? How far will the skater coast before coming to a complete stop?
nR
In the English system, where weight is measured in pounds, mass is measured in slugs. Thus,
iaz
The number y0 m>k is only an upper bound (albeit a useful one). It is true to life in one respect, at least: if m is large, it will take a lot of energy to stop the body.
Solution We answer the first question by solving Equation (1) for t:
assuming the gravitational constant is 32 ft/sec2.
11e t>18 = 1
ass a
e t>18 = 1>11
Eq. (1) with k = 1>3, m = 6, y0 = 11, y = 1
t>18 = ln s1>11d = ln 11 t = 18 ln 11 L 43 sec.
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We answer the second question with Equation (3): Distance coasted =
y0 m 11 # 6 = k 1>3
= 198 ft.
ma
Modeling Population Growth
Mu
ham
In Section 6.5 we modeled population growth with the Law of Exponential Change: dP = kP, dt
Ps0d = P0
where P is the population at time t, k 7 0 is a constant growth rate, and P0 is the size of the population at time t = 0. In Section 6.5 we found the solution P = P0 e kt to this model. To assess the model, notice that the exponential growth differential equation says that dP>dt (4) = k P is constant. This rate is called the relative growth rate. Now, Table 15.1 gives the world population at midyear for the years 1980 to 1989. Taking dt = 1 and dP L Â˘P, we see from the table that the relative growth rate in Equation (4) is approximately the constant 0.017. Thus, based on the tabled data with t = 0 representing 1980, t = 1 representing 1981, and so forth, the world population could be modeled by the initial value problem,
To Read it Online & Download:
dP = 0.017P, dt
Ps0d = 4454.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
TABLE 15.1
6000
5000
4000
0
20
10
t
Population (millions)
1980 1981 1982 1983 1984 1985 1986 1987 1988 1989
4454 4530 4610 4690 4770 4851 4933 5018 5105 5190
≤P>P 76>4454 80>4530 80>4610 80>4690 81>4770 82>4851 85>4933 87>5018 85>5105
L L L L L L L L L
0.0171 0.0177 0.0174 0.0171 0.0170 0.0169 0.0172 0.0173 0.0167
nR
FIGURE 15.8 Notice that the value of the solution P = 4454e 0.017t is 6152.16 when t = 19 , which is slightly higher than the actual population in 1999.
Year
iaz
P
4454e 0.017t
World population (midyear)
suf
World population (1980–99)
1521
You
P
Applications
i
15.3
Source: U.S. Bureau of the Census (Sept., 1999): www.census.gov> ipc> www> worldpop.html.
The solution to this initial value problem gives the population function P = 4454e 0.017t . In year 1999 (so t = 19), the solution predicts the world population in midyear to be about 6152 million, or 6.15 billion (Figure 15.8), which is more than the actual population of 6001 million from the U.S. Bureau of the Census. In Section 15.5 we propose a more realistic model considering environmental factors affecting the growth rate.
dH
FIGURE 15.9 An orthogonal trajectory intersects the family of curves at right angles, or orthogonally.
ass a
Orthogonal trajectory
Mu
An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family at right angles, or orthogonally (Figure 15.9). For instance, each straight line through the origin is an orthogonal trajectory of the family of circles x 2 + y 2 = a 2 , centered at the origin (Figure 15.10). Such mutually orthogonal systems of curves are of particular importance in physical problems related to electrical potential, where the curves in one family correspond to flow of electric current and those in the other family correspond to curves of constant potential. They also occur in hydrodynamics and heatflow problems.
ma
ham
y
Orthogonal Trajectories
EXAMPLE 2 x
FIGURE 15.10 Every straight line through the origin is orthogonal to the family of circles centered at the origin.
Find the orthogonal trajectories of the family of curves xy = a, where a Z 0 is an arbitrary constant. Solution The curves xy = a form a family of hyperbolas with asymptotes y = ; x. First
we find the slopes of each curve in this family, or their dy> dx values. Differentiating xy = a implicitly gives x
dy + y = 0 dx
or
y dy = x. dx
Thus the slope of the tangent line at any point (x, y) on one of the hyperbolas xy = a is y¿ = y>x. On an orthogonal trajectory the slope of the tangent line at this same point
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
i
must be the negative reciprocal, or x> y. Therefore, the orthogonal trajectories must satisfy the differential equation
y
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1522
dy x = y. dx
This differential equation is separable and we solve it as in Section 6.5: y dy = x dx
Separate variables.
y dy =
Integrate both sides.
x
0
x y a, a0
L
L
x dx
y 2  x 2 = b,
iaz
1 2 1 y = x2 + C 2 2
(5)
where b = 2C is an arbitrary constant. The orthogonal trajectories are the family of hyperbolas given by Equation (5) and sketched in Figure 15.11.
ass a
nR
FIGURE 15.11 Each curve is orthogonal to every curve it meets in the other family (Example 2).
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x2 y2 b b0
EXERCISES 15.3
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1. Coasting bicycle A 66kg cyclist on a 7kg bicycle starts coasting on level ground at 9 m> sec. The k in Equation (1) is about 3.9 kg> sec. a. About how far will the cyclist coast before reaching a complete stop?
b. How long will it take the cyclist’s speed to drop to 1 m> sec?
ma
2. Coasting battleship Suppose that an Iowa class battleship has mass around 51,000 metric tons (51,000,000 kg) and a k value in Equation (1) of about 59,000 kg> sec. Assume that the ship loses power when it is moving at a speed of 9 m> sec.
ham
a. About how far will the ship coast before it is dead in the water?
b. About how long will it take the ship’s speed to drop to 1 m> sec?
Mu
3. The data in Table 15.2 were collected with a motion detector and a CBL™ by Valerie Sharritts, a mathematics teacher at St. Francis DeSales High School in Columbus, Ohio. The table shows the distance s (meters) coasted on inline skates in t sec by her daughter Ashley when she was 10 years old. Find a model for Ashley’s position given by the data in Table 15.2 in the form of Equation (2). Her initial velocity was y0 = 2.75 m>sec , her mass m = 39.92 kg (she weighed 88 lb), and her total coasting distance was 4.91 m.
4. Coasting to a stop Table 15.3 shows the distance s (meters) coasted on inline skates in terms of time t (seconds) by Kelly Schmitzer. Find a model for her position in the form of Equation (2).
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Her initial velocity was y0 = 0.80 m>sec , her mass m = 49.90 kg (110 lb), and her total coasting distance was 1.32 m.
TABLE 15.2 Ashley Sharritts skating data
t (sec)
s (m)
t (sec)
s (m)
t (sec)
s (m)
0 0.16 0.32 0.48 0.64 0.80 0.96 1.12 1.28 1.44 1.60 1.76 1.92 2.08
0 0.31 0.57 0.80 1.05 1.28 1.50 1.72 1.93 2.09 2.30 2.53 2.73 2.89
2.24 2.40 2.56 2.72 2.88 3.04 3.20 3.36 3.52 3.68 3.84 4.00 4.16 4.32
3.05 3.22 3.38 3.52 3.67 3.82 3.96 4.08 4.18 4.31 4.41 4.52 4.63 4.69
4.48 4.64 4.80 4.96 5.12 5.28 5.44 5.60 5.76 5.92 6.08 6.24 6.40 6.56
4.77 4.82 4.84 4.86 4.88 4.89 4.90 4.90 4.91 4.90 4.91 4.90 4.91 4.91
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.4
s (m)
t (sec)
s (m)
2
1.30 1.31 1.32 1.32 1.32 1.32 1.32 1.32
Euler’s Method
15.4
HISTORICAL BIOGRAPHY Leonhard Euler (1703–1783) y
3.1 3.3 3.5 3.7 3.9 4.1 4.3 4.5
y L(x) y 0 f (x 0, y 0)(x x 0 )
10. y = e k x
9. y = ce x 2
11. Show that the curves 2x + 3y 2 = 5 and y 2 = x 3 are orthogonal. 12. Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families. a. x dx + y dy = 0
b. x dy  2y dx = 0
13. Suppose a and b are positive numbers. Sketch the parabolas y 2 = 4a 2  4ax
and
y 2 = 4b 2 + 4bx
in the same diagram. Show that they intersect at A a  b, ;22ab B , and that each “aparabola” is orthogonal to every “bparabola.”
If we do not require or cannot immediately find an exact solution for an initial value problem y¿ = ƒsx, yd, ysx0 d = y0 , we can often use a computer to generate a table of approximate numerical values of y for values of x in an appropriate interval. Such a table is called a numerical solution of the problem, and the method by which we generate the table is called a numerical method. Numerical methods are generally fast and accurate, and they are often the methods of choice when exact formulas are unnecessary, unavailable, or overly complicated. In this section, we study one such method, called Euler’s method, upon which many other numerical methods are based.
dH
y y (x)
8. 2x 2 + y 2 = c 2
7. k x + y = 1
You
0.89 0.97 1.05 1.11 1.17 1.22 1.25 1.28
iaz
1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9
nR
0 0.07 0.22 0.36 0.49 0.60 0.71 0.81
2
ass a
0 0.1 0.3 0.5 0.7 0.9 1.1 1.3
6. y = cx 2
5. y = mx
i
t (sec)
suf
s (m)
1523
In Exercises 5–10, find the orthogonal trajectories of the family of curves. Sketch several members of each family.
TABLE 15.3 Kelly Schmitzer skating data
t (sec)
Euler’s Method
Euler’s Method
y0
(x 0, y 0)
x
x0
ma
0
Given a differential equation dy>dx = ƒsx, yd and an initial condition ysx0 d = y0 , we can approximate the solution y = ysxd by its linearization
y
ham
FIGURE 15.12 The linearization L(x) of y = ysxd at x = x0 .
(x1, L(x1))
Lsxd = ysx0 d + y¿sx0 dsx  x0 d
y y(x)
y1 = Lsx1 d = y0 + ƒsx0 , y0 d dx
Mu
(x 0, y 0)
dx
x0
x1 x 0 dx
FIGURE 15.13 The first Euler step approximates ysx1 d with y1 = Lsx1 d .
Lsxd = y0 + ƒsx0 , y0 dsx  x0 d.
The function L(x) gives a good approximation to the solution y(x) in a short interval about x0 (Figure 15.12). The basis of Euler’s method is to patch together a string of linearizations to approximate the curve over a longer stretch. Here is how the method works. We know the point sx0 , y0 d lies on the solution curve. Suppose that we specify a new value for the independent variable to be x1 = x0 + dx. (Recall that dx = ¢x in the definition of differentials.) If the increment dx is small, then
(x1, y(x1))
0
or
x
is a good approximation to the exact solution value y = ysx1 d. So from the point sx0 , y0 d, which lies exactly on the solution curve, we have obtained the point sx1, y1 d, which lies very close to the point sx1, ysx1 dd on the solution curve (Figure 15.13). Using the point sx1, y1 d and the slope ƒsx1, y1 d of the solution curve through sx1, y1 d, we take a second step. Setting x2 = x1 + dx, we use the linearization of the solution curve through sx1, y1 d to calculate
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y2 = y1 + ƒsx1, y1 d dx.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
(x1, y1)
dx x1
dx x2
x3
x
FIGURE 15.14 Three steps in the Euler approximation to the solution of the initial value problem y¿ = ƒsx, yd, y sx0 d = y0 . As we take more steps, the errors involved usually accumulate, but not in the exaggerated way shown here.
and so on. We are literally building an approximation to one of the solutions by following the direction of the slope field of the differential equation. The steps in Figure 15.14 are drawn large to illustrate the construction process, so the approximation looks crude. In practice, dx would be small enough to make the red curve hug the blue one and give a good approximation throughout.
You
dx x0
y3 = y2 + ƒsx2, y2 d dx,
True solution curve y y(x)
(x 0, y 0)
0
Error
EXAMPLE 1
Find the first three approximations y1, y2 , y3 using Euler’s method for the initial value problem
iaz
Euler approximation (x , y ) 2 2
This gives the next approximation sx2 , y2 d to values along the solution curve y = ysxd (Figure 15.14). Continuing in this fashion, we take a third step from the point sx2 , y2 d with slope ƒsx2 , y2 d to obtain the third approximation
i
(x 3, y 3)
suf
y
y¿ = 1 + y, starting at x0 = 0 with dx = 0.1. Solution We
ys0d = 1,
nR
1524
have x0 = 0, y0 = 1, x1 = x0 + dx = 0.1, x2 = x0 + 2dx = 0.2, and x3 = x0 + 3 dx = 0.3. First:
y1 = y0 + ƒsx0 , y0 d dx
ass a
= y0 + s1 + y0 d dx
= 1 + s1 + 1ds0.1d = 1.2
Second:
y2 = y1 + ƒsx1, y1 d dx
dH
= y1 + s1 + y1 d dx = 1.2 + s1 + 1.2ds0.1d = 1.42
ma
Third:
y3 = y2 + ƒsx2 , y2 d dx = y2 + s1 + y2 d dx = 1.42 + s1 + 1.42ds0.1d = 1.662
Mu
ham
The stepbystep process used in Example 1 can be continued easily. Using equally spaced values for the independent variable in the table and generating n of them, set x1 = x0 + dx x2 = x1 + dx o xn = xn  1 + dx.
Then calculate the approximations to the solution, y1 = y0 + ƒsx0 , y0 d dx y2 = y1 + ƒsx1, y1 d dx o yn = yn  1 + ƒsxn  1, yn  1 d dx. The number of steps n can be as large as we like, but errors can accumulate if n is too large.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.4
Euler’s Method
1525
EXAMPLE 2
Use Euler’s method to solve y¿ = 1 + y,
You
suf
i
Euler’s method is easy to implement on a computer or calculator. A computer program generates a table of numerical solutions to an initial value problem, allowing us to input x0 and y0 , the number of steps n, and the step size dx. It then calculates the approximate solution values y1, y2 , Á , yn in iterative fashion, as just described. Solving the separable equation in Example 1, we find that the exact solution to the initial value problem is y = 2e x  1. We use this information in Example 2.
ys0d = 1,
iaz
on the interval 0 … x … 1, starting at x0 = 0 and taking (a) dx = 0.1, (b) dx = 0.05. Compare the approximations with the values of the exact solution y = 2e x  1. Solution
nR
(a) We used a computer to generate the approximate values in Table 15.4. The “error” column is obtained by subtracting the unrounded Euler values from the unrounded values found using the exact solution. All entries are then rounded to four decimal places.
ass a
TABLE 15.4 Euler solution of y¿ = 1 + y, ys0d = 1,
step size dx = 0.1
dH
x
ma
y 4
2 1
y (exact)
Error
1 1.2 1.42 1.662 1.9282 2.2210 2.5431 2.8974 3.2872 3.7159 4.1875
1 1.2103 1.4428 1.6997 1.9836 2.2974 2.6442 3.0275 3.4511 3.9192 4.4366
0 0.0103 0.0228 0.0377 0.0554 0.0764 0.1011 0.1301 0.1639 0.2033 0.2491
y 2e x 1
Mu
0
ham
3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
y (Euler)
1
x
FIGURE 15.15 The graph of y = 2e x  1 superimposed on a scatterplot of the Euler approximations shown in Table 15.4 (Example 2).
By the time we reach x = 1 (after 10 steps), the error is about 5.6% of the exact solution. A plot of the exact solution curve with the scatterplot of Euler solution points from Table 15.4 is shown in Figure 15.15. (b) One way to try to reduce the error is to decrease the step size. Table 15.5 shows the results and their comparisons with the exact solutions when we decrease the step size to 0.05, doubling the number of steps to 20. As in Table 15.4, all computations are performed before rounding. This time when we reach x = 1, the relative error is only about 2.9%.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
step size dx = 0.05
Error
1 1.1 1.205 1.3153 1.4310 1.5526 1.6802 1.8142 1.9549 2.1027 2.2578 2.4207 2.5917 2.7713 2.9599 3.1579 3.3657 3.5840 3.8132 4.0539 4.3066
1 1.1025 1.2103 1.3237 1.4428 1.5681 1.6997 1.8381 1.9836 2.1366 2.2974 2.4665 2.6442 2.8311 3.0275 3.2340 3.4511 3.6793 3.9192 4.1714 4.4366
0 0.0025 0.0053 0.0084 0.0118 0.0155 0.0195 0.0239 0.0287 0.0340 0.0397 0.0458 0.0525 0.0598 0.0676 0.0761 0.0853 0.0953 0.1060 0.1175 0.1300
iaz
You
y (exact)
dH
ass a
0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
y (Euler)
nR
x
suf
TABLE 15.5 Euler solution of y¿ = 1 + y, ys0d = 1,
i
1526
ham
ma
It might be tempting to reduce the step size even further in Example 2 to obtain greater accuracy. Each additional calculation, however, not only requires additional computer time but more importantly adds to the buildup of roundoff errors due to the approximate representations of numbers inside the computer. The analysis of error and the investigation of methods to reduce it when making numerical calculations are important but are appropriate for a more advanced course. There are numerical methods more accurate than Euler’s method, as you can see in a further study of differential equations. We study one improvement here.
Mu
HISTORICAL BIOGRAPHY Carl Runge (1856–1927)
Improved Euler’s Method We can improve on Euler’s method by taking an average of two slopes. We first estimate yn as in the original Euler method, but denote it by zn . We then take the average of ƒsxn  1, yn  1 d and ƒsxn , zn d in place of ƒsxn  1, yn  1 d in the next step. Thus, we calculate the next approximation yn using
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zn = yn  1 + ƒsxn  1, yn  1 d dx yn = yn  1 + c
ƒsxn  1, yn  1 d + ƒsxn , zn d d dx. 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.4
1527
y¿ = 1 + y,
ys0d = 1,
i
Use the improved Euler’s method to solve
suf
EXAMPLE 3
Euler’s Method
on the interval 0 … x … 1, starting at x0 = 0 and taking dx = 0.1. Compare the approximations with the values of the exact solution y = 2e x  1.
You
Solution We used a computer to generate the approximate values in Table 15.6. The “error” column is obtained by subtracting the unrounded improved Euler values from the unrounded values found using the exact solution. All entries are then rounded to four decimal places.
TABLE 15.6 Improved Euler solution of y¿ = 1 + y,
iaz
ys0d = 1, step size dx = 0.1 y (improved Euler)
y (exact)
Error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
1 1.21 1.4421 1.6985 1.9818 2.2949 2.6409 3.0231 3.4456 3.9124 4.4282
1 1.2103 1.4428 1.6997 1.9836 2.2974 2.6442 3.0275 3.4511 3.9192 4.4366
0 0.0003 0.0008 0.0013 0.0018 0.0025 0.0034 0.0044 0.0055 0.0068 0.0084
dH
ass a
nR
x
ma
By the time we reach x = 1 (after 10 steps), the relative error is about 0.19%.
ham
By comparing Tables 15.4 and 15.6, we see that the improved Euler’s method is considerably more accurate than the regular Euler’s method, at least for the initial value problem y¿ = 1 + y, ys0d = 1.
EXERCISES 15.4
Mu
In Exercises 1–6, use Euler’s method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. y 1. y¿ = 1  x , ys2d = 1, dx = 0.5 2. y¿ = xs1  yd,
ys1d = 0,
dx = 0.2
3. y¿ = 2xy + 2y,
ys0d = 3,
dx = 0.2
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4. y¿ = y 2s1 + 2xd, 2
T
5. y¿ = 2xe x ,
T
6. y¿ = y + e x  2,
ys 1d = 1,
ys0d = 2,
dx = 0.5
dx = 0.1
ys0d = 2,
dx = 0.5
7. Use the Euler method with dx = 0.2 to estimate y(1) if y¿ = y and ys0d = 1 . What is the exact value of y(1)? 8. Use the Euler method with dx = 0.2 to estimate y(2) if y¿ = y>x and ys1d = 2 . What is the exact value of y(2)?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations 19. y¿ = 2y 2sx  1d, ys2d = 1>2, x0 = 2, (See Exercise 17 for the exact solution.)
T 10. Use the Euler method with dx = 1>3 to estimate y(2) if y¿ = y  e 2x and ys0d = 1 . What is the exact value of y(2)?
20. y¿ = y  1, ys0d = 3, x0 = 0, x * = 1 (See Exercise 18 for the exact solution.)
Use a CAS to explore graphically each of the differential equations in Exercises 21–24. Perform the following steps to help with your explorations.
You
In Exercises 11 and 12, use the improved Euler’s method to calculate the first three approximations to the given initial value problem. Compare the approximations with the values of the exact solution.
x* = 3
suf
9. Use the Euler method with dx = 0.5 to estimate y(5) if y¿ = y 2> 2x and ys1d = 1 . What is the exact value of y(5)?
a. Plot a slope field for the differential equation in the given xywindow.
12. y¿ = xs1  yd, ys1d = 0, dx = 0.2 (See Exercise 2 for the exact solution.)
b. Find the general solution of the differential equation using your CAS DE solver.
COMPUTER EXPLORATIONS
c. Graph the solutions for the values of the arbitrary constant C = 2, 1, 0, 1, 2 superimposed on your slope field plot.
In Exercises 13–16, use Euler’s method with the specified step size to estimate the value of the solution at the given point x *. Find the value of the exact solution at x * .
d. Find and graph the solution that satisfies the specified initial condition over the interval [0, b].
ys0d = 2,
14. y¿ = y + e x  2, 15. y¿ = 2x>y, 2
16. y¿ = 1 + y ,
x* = 1
dx = 0.1,
ys0d = 2,
y 7 0,
dx = 0.5,
ys0d = 1,
ys0d = 0,
x* = 2 x* = 1
dx = 0.1,
dx = 0.1,
*
f. Repeat part (e) for 8, 16, and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e).
x = 1
ys0d = 3,
x0 = 0,
x0 = 2, x* = 1
x* = 3
dH
18. y¿ = y  1,
ys2d = 1>2,
ass a
In Exercises 17 and 18, (a) find the exact solution of the initial value problem. Then compare the accuracy of the approximation with ysx * d using Euler’s method starting at x0 with step size (b) 0.2, (c) 0.1, and (d) 0.05. 17. y¿ = 2y 2sx  1d,
In Exercises 19 and 20, compare the accuracy of the approximation with ysx * d using the improved Euler’s method starting at x0 with step size a. 0.2
b. 0.1
c. 0.05
e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the xinterval and plot the Euler approximation superimposed on the graph produced in part (d).
nR
2
13. y¿ = 2xe x ,
iaz
11. y¿ = 2ysx + 1d, ys0d = 3, dx = 0.2 (See Exercise 3 for the exact solution.)
g. Find the error s y sexactd  y sEulerdd at the specified point x = b for each of your four Euler approximations. Discuss the improvement in the percentage error.
21. y¿ = x + y, b = 1
ys0d = 7>10;
22. y¿ = x>y,
ys0d = 2;
4 … x … 4,
3 … x … 3,
23. A logistic equation y¿ = ys2  yd, 0 … x … 4, 0 … y … 3; b = 3 24. y¿ = ssin xdssin yd, b = 3p>2
ys0d = 2;
4 … y … 4;
3 … y … 3; b = 2
ys0d = 1>2;
6 … x … 6,
6 … y … 6;
ham
ma
d. Describe what happens to the error as the step size decreases.
15.5
Graphical Solutions of Autonomous Equations
Mu
T
i
1528
In Chapter 4 we learned that the sign of the first derivative tells where the graph of a function is increasing and where it is decreasing. The sign of the second derivative tells the concavity of the graph. We can build on our knowledge of how derivatives determine the shape of a graph to solve differential equations graphically. The starting ideas for doing so are the notions of phase line and equilibrium value. We arrive at these notions by investigating what happens when the derivative of a differentiable function is zero from a point of view different from that studied in Chapter 4.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.5
Graphical Solutions of Autonomous Equations
1529
i
Equilibrium Values and Phase Lines
suf
When we differentiate implicitly the equation 1 ln s5y  15d = x + 1 5
You
we obtain
dy 5 1 a = 1. b 5 5y  15 dx
nR
iaz
Solving for y¿ = dy>dx we find y¿ = 5y  15 = 5s y  3d. In this case the derivative y¿ is a function of y only (the dependent variable) and is zero when y = 3. A differential equation for which dy> dx is a function of y only is called an autonomous differential equation. Let’s investigate what happens when the derivative in an autonomous equation equals zero. We assume any derivatives are continuous.
ass a
DEFINITION If dy>dx = gs yd is an autonomous differential equation, then the values of y for which dy>dx = 0 are called equilibrium values or rest points.
Thus, equilibrium values are those at which no change occurs in the dependent variable, so y is at rest. The emphasis is on the value of y where dy>dx = 0, not the value of x, as we studied in Chapter 4. For example, the equilibrium values for the autonomous differential equation
dH
dy = s y + 1ds y  2d dx
Mu
ham
ma
are y = 1 and y = 2. To construct a graphical solution to an autonomous differential equation, we first make a phase line for the equation, a plot on the yaxis that shows the equation’s equilibrium values along with the intervals where dy> dx and d 2y>dx 2 are positive and negative. Then we know where the solutions are increasing and decreasing, and the concavity of the solution curves. These are the essential features we found in Section 4.4, so we can determine the shapes of the solution curves without having to find formulas for them.
EXAMPLE 1
Draw a phase line for the equation dy = s y + 1ds y  2d dx
and use it to sketch solutions to the equation. Solution
1.
Draw a number line for y and mark the equilibrium values y = 1 and y = 2, where dy>dx = 0.
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y –1
2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1530
Chapter 15: FirstOrder Differential Equations
Identify and label the intervals where y¿ 7 0 and y¿ 6 0. This step resembles what we did in Section 4.3, only now we are marking the yaxis instead of the xaxis.
y' 0
y' 0
y' 0
–1
y
You
2
suf
i
2.
–1
iaz
We can encapsulate the information about the sign of y¿ on the phase line itself. Since y¿ 7 0 on the interval to the left of y = 1, a solution of the differential equation with a yvalue less than 1 will increase from there toward y = 1. We display this information by drawing an arrow on the interval pointing to 1. y
2
3.
ass a
nR
Similarly, y¿ 6 0 between y = 1 and y = 2, so any solution with a value in this interval will decrease toward y = 1. For y 7 2, we have y¿ 7 0, so a solution with a yvalue greater than 2 will increase from there without bound. In short, solution curves below the horizontal line y = 1 in the xyplane rise toward y = 1. Solution curves between the lines y = 1 and y = 2 fall away from y = 2 toward y = 1. Solution curves above y = 2 rise away from y = 2 and keep going up. Calculate y– and mark the intervals where y– 7 0 and y– 6 0. To find y–, we differentiate y¿ with respect to x, using implicit differentiation.
dH
y¿ = s y + 1ds y  2d = y 2  y  2
y– =
y
y' 0 y'' 0 y' 0 y'' 0
differentiated implicitly with respect to x.
= s2y  1dy¿ = s2y  1ds y + 1ds y  2d.
From this formula, we see that y– changes sign at y = 1, y = 1>2, and y = 2. We add the sign information to the phase line.
x
ham
1 2 0 –1
ma
2
d d s y¿d = s y 2  y  2d dx dx
= 2yy¿  y¿
y' 0 y'' 0
y' 0 y'' 0
y' 0 y'' 0
Mu
FIGURE 15.16 Graphical solutions from Example 1 include the horizontal lines y = 1 and y = 2 through the equilibrium values. From Theorem 1, no two solution curves will ever cross or touch each other.
Formula for y¿ . . .
y' 0 y'' 0
y' 0 y'' 0
y' 0 y'' 0 y
–1
4.
1 2
2
Sketch an assortment of solution curves in the xyplane. The horizontal lines y = 1, y = 1>2, and y = 2 partition the plane into horizontal bands in which we know the signs of y¿ and y– . In each band, this information tells us whether the solution curves rise or fall and how they bend as x increases (Figure 15.16). The “equilibrium lines” y = 1 and y = 2 are also solution curves. (The constant functions y = 1 and y = 2 satisfy the differential equation.) Solution curves
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.5
Graphical Solutions of Autonomous Equations
1531
Stable and Unstable Equilibria
You
suf
i
that cross the line y = 1>2 have an inflection point there. The concavity changes from concave down (above the line) to concave up (below the line). As predicted in Step 2, solutions in the middle and lower bands approach the equilibrium value y = 1 as x increases. Solutions in the upper band rise steadily away from the value y = 2.
ass a
nR
iaz
Look at Figure 15.16 once more, in particular at the behavior of the solution curves near the equilibrium values. Once a solution curve has a value near y = 1, it tends steadily toward that value; y = 1 is a stable equilibrium. The behavior near y = 2 is just the opposite: all solutions except the equilibrium solution y = 2 itself move away from it as x increases. We call y = 2 an unstable equilibrium. If the solution is at that value, it stays, but if it is off by any amount, no matter how small, it moves away. (Sometimes an equilibrium value is unstable because a solution moves away from it only on one side of the point.) Now that we know what to look for, we can already see this behavior on the initial phase line. The arrows lead away from y = 2 and, once to the left of y = 2, toward y = 1. We now present several applied examples for which we can sketch a family of solution curves to the differential equation models using the method in Example 1. In Section 6.5 we solved analytically the differential equation dH = ksH  HS d, dt
k 7 0
dH
modeling Newton’s law of cooling. Here H is the temperature (amount of heat) of an object at time t and HS is the constant temperature of the surrounding medium. Our first example uses a phase line analysis to understand the graphical behavior of this temperature model over time.
EXAMPLE 2
ma
What happens to the temperature of the soup when a cup of hot soup is placed on a table in a room? We know the soup cools down, but what does a typical temperature curve look like as a function of time?
ham
Solution Suppose that the surrounding medium has a constant Celsius temperature of
15°C. We can then express the difference in temperature as Hstd  15. Assuming H is a differentiable function of time t, by Newton’s law of cooling, there is a constant of proportionality k 7 0 such that dH = ksH  15d dt
dH 0 dt
dH 0 dt
Mu
15
H
FIGURE 15.17 First step in constructing the phase line for Newton’s law of cooling in Example 2. The temperature tends towards the equilibrium (surroundingmedium) value in the long run.
(1)
(minus k to give a negative derivative when H 7 15). Since dH>dt = 0 at H = 15, the temperature 15°C is an equilibrium value. If H 7 15, Equation (1) tells us that sH  15d 7 0 and dH>dt 6 0. If the object is hotter than the room, it will get cooler. Similarly, if H 6 15, then sH  15d 6 0 and dH>dt 7 0. An object cooler than the room will warm up. Thus, the behavior described by Equation (1) agrees with our intuition of how temperature should behave. These observations are captured in the initial phase line diagram in Figure 15.17. The value H = 15 is a stable equilibrium.
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Chapter 15: FirstOrder Differential Equations
We determine the concavity of the solution curves by differentiating both sides of Equation (1) with respect to t:
dH 0 dt
d 2H 0 dt 2
d 2H 0 dt 2 H
Temperature of surrounding medium
15
Initial temperature t
FIGURE 15.19 Temperature versus time. Regardless of initial temperature, the object’s temperature H(t) tends toward 15°C, the temperature of the surrounding medium.
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H
Since k is negative, we see that d 2H>dt 2 is positive when dH>dt 6 0 and negative when dH>dt 7 0. Figure 15.18 adds this information to the phase line. The completed phase line shows that if the temperature of the object is above the equilibrium value of 15°C, the graph of H(t) will be decreasing and concave upward. If the temperature is below 15°C (the temperature of the surrounding medium), the graph of H(t) will be increasing and concave downward. We use this information to sketch typical solution curves (Figure 15.19). From the upper solution curve in Figure 15.19, we see that as the object cools down, the rate at which it cools slows down because dH> dt approaches zero. This observation is implicit in Newton’s law of cooling and contained in the differential equation, but the flattening of the graph as time advances gives an immediate visual representation of the phenomenon. The ability to discern physical behavior from graphs is a powerful tool in understanding realworld systems.
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FIGURE 15.18 The complete phase line for Newton’s law of cooling (Example 2).
dH d 2H . = k 2 dt dt
ass a
15
Initial temperature
d dH d a b = s ksH  15dd dt dt dt
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dH 0 dt
EXAMPLE 3 Galileo and Newton both observed that the rate of change in momentum encountered by a moving object is equal to the net force applied to it. In mathematical terms,
dH
F =
d smyd dt
(2)
ma
where F is the force and m and y the object’s mass and velocity. If m varies with time, as it will if the object is a rocket burning fuel, the righthand side of Equation (2) expands to m
dy dm + y dt dt
using the Product Rule. In many situations, however, m is constant, dm>dt = 0, and Equation (2) takes the simpler form
ham
Fr ky
m
y0
Mu
FIGURE 15.20 An object falling under the influence of gravity with a resistive force assumed to be proportional to the velocity.
dy dt
or
F = ma,
(3)
known as Newton’s second law of motion (see Section 15.3). In free fall, the constant acceleration due to gravity is denoted by g and the one force acting downward on the falling body is
y positive
Fp mg
F = m
Fp = mg, the propulsion due to gravity. If, however, we think of a real body falling through the air— say, a penny from a great height or a parachutist from an even greater height—we know that at some point air resistance is a factor in the speed of the fall. A more realistic model of free fall would include air resistance, shown as a force Fr in the schematic diagram in Figure 15.20.
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For low speeds well below the speed of sound, physical experiments have shown that Fr is approximately proportional to the body’s velocity. The net force on the falling body is therefore
i
y mg k
F = Fp  Fr ,
m dy 0 dt
d 2y 0 dt 2
d 2y 0 dt 2
y
mg k
dy = mg  ky dt k dy = g  m y. dt
We can use a phase line to analyze the velocity functions that solve this differential equation. The equilibrium point, obtained by setting the righthand side of Equation (4) equal to zero, is
mg k
Initial velocity
y
mg k
d 2y d k k dy = ag  m yb =  m . dt dt dt 2
t
We see that d 2y>dt 2 6 0 when y 6 mg>k and d 2y>dt 2 7 0 when y 7 mg>k. Figure 15.22 adds this information to the phase line. Notice the similarity to the phase line for Newton’s law of cooling (Figure 15.18). The solution curves are similar as well (Figure 15.23). Figure 15.23 shows two typical solution curves. Regardless of the initial velocity, we see the body’s velocity tending toward the limiting value y = mg>k. This value, a stable equilibrium point, is called the body’s terminal velocity. Skydivers can vary their terminal velocity from 95 mph to 180 mph by changing the amount of body area opposing the fall.
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Initial velocity
If the body is initially moving faster than this, dy> dt is negative and the body slows down. If the body is moving at a velocity below mg>k, then dy>dt 7 0 and the body speeds up. These observations are captured in the initial phase line diagram in Figure 15.21. We determine the concavity of the solution curves by differentiating both sides of Equation (4) with respect to t:
ass a
y
mg . k
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y =
FIGURE 15.22 The completed phase line for Example 3.
(4)
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dy 0 dt
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giving
FIGURE 15.21 Initial phase line for Example 3.
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dy 0 dt
dy 0 dt
Graphical Solutions of Autonomous Equations
ma
FIGURE 15.23 Typical velocity curves in Example 3. The value y = mg>k is the terminal velocity.
Mu
ham
EXAMPLE 4
In Section 15.3 we examined population growth using the model of exponential change. That is, if P represents the number of individuals and we neglect departures and arrivals, then dP = kP, dt
(5)
where k 7 0 is the birthrate minus the death rate per individual per unit time. Because the natural environment has only a limited number of resources to sustain life, it is reasonable to assume that only a maximum population M can be accommodated. As the population approaches this limiting population or carrying capacity, resources become less abundant and the growth rate k decreases. A simple relationship exhibiting this behavior is
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k = rsM  Pd,
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Chapter 15: FirstOrder Differential Equations
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where r 7 0 is a constant. Notice that k decreases as P increases toward M and that k is negative if P is greater than M. Substituting rsM  Pd for k in Equation (5) gives the differential equation dP = rsM  PdP = rMP  rP 2 . dt P M
0
The model given by Equation (6) is referred to as logistic growth. We can forecast the behavior of the population over time by analyzing the phase line for Equation (6). The equilibrium values are P = M and P = 0, and we can see that dP>dt 7 0 if 0 6 P 6 M and dP>dt 6 0 if P 7 M. These observations are recorded on the phase line in Figure 15.24. We determine the concavity of the population curves by differentiating both sides of Equation (6) with respect to t:
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FIGURE 15.24 The initial phase line for Equation 6.
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dP 0 dt
dP 0 dt
(6)
d 2P 0 dt 2
= rsM  2Pd
M 2
M
(7)
If P = M>2, then d 2P>dt 2 = 0. If P 6 M>2, then sM  2Pd and dP> dt are positive and d 2P>dt 2 7 0. If M>2 6 P 6 M, then sM  2Pd 6 0, dP>dt 7 0, and d 2P>dt 2 6 0. If P 7 M, then sM  2Pd and dP> dt are both negative and d 2P>dt 2 7 0. We add this information to the phase line (Figure 15.25). The lines P = M>2 and P = M divide the first quadrant of the tPplane into horizontal bands in which we know the signs of both dP> dt and d 2P>dt 2. In each band, we know how the solution curves rise and fall, and how they bend as time passes. The equilibrium lines P = 0 and P = M are both population curves. Population curves crossing the line P = M>2 have an inflection point there, giving them a sigmoid shape (curved in two directions like a letter S). Figure 15.26 displays typical population curves.
Mu
P
Population
ham
ma
dH
FIGURE 15.25 The completed phase line for logistic growth (Equation 6).
dP . dt
ass a
d 2P 0 dt 2
P 0
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dP 0 dt
dP 0 dt d 2P 0 dt 2
d 2P d = srMP  rP 2 d 2 dt dt dP dP = rM  2rP dt dt
Limiting population
M
M 2
Time
t
FIGURE 15.26 Population curves in Example 4.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.5
Graphical Solutions of Autonomous Equations
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EXERCISES 15.5 In Exercises 1–8,
b. Explain how this model differs from the logistic model dP>dt = rPsM  Pd. Is it better or worse than the logistic model?
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a. Identify the equilibrium values. Which are stable and which are unstable?
c. Show that if P 7 M for all t, then limt: q Pstd = M.
b. Construct a phase line. Identify the signs of y¿ and y–.
d. What happens if P 6 m for all t ?
c. Sketch several solution curves. dy dy 1. 2. = s y + 2ds y  3d = y2  4 dx dx
5. y¿ = 2y,
4.
dy = y 2  2y dx
6. y¿ = y  2y,
y 7 0
15. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of velocity, then the body’s velocity t seconds into the fall satisfies the equation.
y 7 0
8. y¿ = y 3  y 2
7. y¿ = s y  1ds y  2ds y  3d
The autonomous differential equations in Exercises 9–12 represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for P(t), selecting different starting values P(0) (as in Example 4). Which equilibria are stable, and which are unstable?
11.
dP = 1  2P dt
10.
dP = Ps1  2Pd dt
dP = 2PsP  3d dt
12.
dP 1 = 3Ps1  Pd aP  b 2 dt
m
dy = mg  ky 2, dt
k 7 0
where k is a constant that depends on the body’s aerodynamic properties and the density of the air. (We assume that the fall is too short to be affected by changes in the air’s density.) a. Draw a phase line for the equation.
ass a
9.
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dy = y3  y dx
e. Discuss the solutions to the differential equation. What are the equilibrium points of the model? Explain the dependence of the steadystate value of P on the initial values of P. About how many permits should be issued?
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3.
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13. Catastrophic continuation of Example 4 Suppose that a healthy population of some species is growing in a limited environment and that the current population P0 is fairly close to the carrying capacity M0 . You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenly a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destroys a significant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity M1 considerably less than M0 and, in fact, less than the current population P0 . Starting at some time before the catastrophe, sketch a “beforeandafter” curve that shows how the fish population responds to the change in environment.
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14. Controlling a population The fish and game department in a certain state is planning to issue hunting permits to control the deer population (one deer per permit). It is known that if the deer population falls below a certain level m, the deer will become extinct. It is also known that if the deer population rises above the carrying capacity M, the population will decrease back to M through disease and malnutrition.
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a. Discuss the reasonableness of the following model for the growth rate of the deer population as a function of time: dP = rPsM  PdsP  md , dt
where P is the population of the deer and r is a positive constant of proportionality. Include a phase line.
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b. Sketch a typical velocity curve. c. For a 160lb skydiver smg = 160d and with time in seconds and distance in feet, a typical value of k is 0.005. What is the diver’s terminal velocity?
16. Resistance proportional to 2Y A body of mass m is projected vertically downward with initial velocity y0 . Assume that the resisting force is proportional to the square root of the velocity and find the terminal velocity from a graphical analysis.
17. Sailing A sailboat is running along a straight course with the wind providing a constant forward force of 50 lb. The only other force acting on the boat is resistance as the boat moves through the water. The resisting force is numerically equal to five times the boat’s speed, and the initial velocity is 1 ft> sec. What is the maximum velocity in feet per second of the boat under this wind? 18. The spread of information Sociologists recognize a phenomenon called social diffusion, which is the spreading of a piece of information, technological innovation, or cultural fad among a population. The members of the population can be divided into two classes: those who have the information and those who do not. In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it. If X denotes the number of individuals who have the information in a population of N people, then a mathematical model for social diffusion is given by dX = k X sN  X d , dt where t represents time in days and k is a positive constant.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 15: FirstOrder Differential Equations
c. Sketch representative solution curves. d. Predict the value of X for which the information is spreading most rapidly. How many people eventually receive the information? 19. Current in an RLcircuit The accompanying diagram represents an electrical circuit whose total resistance is a constant R ohms and whose selfinductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. From Section 15.2, we have di L + Ri = V , dt where i is the intensity of the current in amperes and t is the time in seconds.
i
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b. Construct a phase line identifying the signs of X ¿ and X –.
Use a phase line analysis to sketch the solution curve assuming that the switch in the RLcircuit is closed at time t = 0 . What happens to the current as t : q ? This value is called the steadystate solution. 20. A pearl in shampoo Suppose that a pearl is sinking in a thick fluid, like shampoo, subject to a frictional force opposing its fall and proportional to its velocity. Suppose that there is also a resistive buoyant force exerted by the shampoo. According to Archimedes’ principle, the buoyant force equals the weight of the fluid displaced by the pearl. Using m for the mass of the pearl and P for the mass of the shampoo displaced by the pearl as it descends, complete the following steps.
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a. Discuss the reasonableness of the model.
a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Figure 15.20.
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b. Using y(t) for the pearl’s velocity as a function of time t, write a differential equation modeling the velocity of the pearl as a falling body.
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c. Construct a phase line displaying the signs of y¿ and y–.
V
d. Sketch typical solution curves.
e. What is the terminal velocity of the pearl?
Switch b
15.6
L
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a i
Systems of Equations and Phase Planes
Mu
ham
ma
In some situations we are led to consider not one, but several firstorder differential equations. Such a collection is called a system of differential equations. In this section we present an approach to understanding systems through a graphical procedure known as a phaseplane analysis. We present this analysis in the context of modeling the populations of trout and bass living in a common pond.
Phase Planes A general system of two firstorder differential equations may take the form dx = F(x, y), dt dy = G(x, y). dt Such a system of equations is called autonomous because dx>dt and dy>dt do not depend on the independent variable time t, but only on the dependent variables x and y. A solution
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.6
Systems of Equations and Phase Planes
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A CompetitiveHunter Model
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of such a system consists of a pair of functions x(t) and y(t) that satisfies both of the differential equations simultaneously for every t over some time interval (finite or infinite). We cannot look at just one of these equations in isolation to find solutions x(t) or y(t) since each derivative depends on both x and y. To gain insight into the solutions, we look at both dependent variables together by plotting the points (x(t), y(t)) in the xyplane starting at some specified point. Therefore the solution functions are considered as parametric equations (with parameter t ), and a corresponding solution curve through the specified point is called a trajectory of the system. The xyplane itself, in which these trajectories reside, is referred to as the phase plane. Thus we consider both solutions together and study the behavior of all the solution trajectories in the phase plane. It can be proved that two trajectories can never cross or touch each other.
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ass a
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Imagine two species of fish, say trout and bass, competing for the same limited resources in a certain pond. We let x(t) represent the number of trout and y(t) the number of bass living in the pond at time t. In reality x(t) and y(t) are always integer valued, but we will approximate them with realvalued differentiable functions. This allows us to apply the methods of differential equations. Several factors affect the rates of change of these populations. As time passes, each species breeds, so we assume its population increases proportionally to its size. Taken by itself, this would lead to exponential growth in each of the two populations. However, there is a countervailing effect from the fact that the two species are in competition. A large number of bass tends to cause a decrease in the number of trout, and viceversa. Our model takes the size of this effect to be proportional to the frequency with which the two species interact, which in turn is proportional to xy, the product of the two populations. These considerations lead to the following model for the growth of the trout and bass in the pond: dx = (a  by)x, dt dy = (m  nx)y. dt
(1a) (1b)
Mu
ham
ma
Here x(t) represents the trout population, y(t) the bass population, and a, b, m, n are positive constants. A solution of this system then consists of a pair of functions x(t) and y(t) that gives the population of each fish species at time t. Each equation in (1) contains both of the unknown functions x and y, so we are unable to solve them individually. Instead, we will use a graphical analysis to study the solution trajectories of this competitivehunter model. We now examine the nature of the phase plane in the troutbass population model. We will be interested in the 1st quadrant of the xyplane, where x Ăš 0 and y Ăš 0, since populations cannot be negative. First, we determine where the bass and trout populations are both constant. Noting that the (x(t), y(t)) values remain unchanged when dx>dt = 0 and dy>dt = 0, Equations (1a and 1b) then become (a  by)x = 0, (m  nx)y = 0. This pair of simultaneous equations has two solutions: (x, y) = (0, 0) and (x, y) = (m>n, a>b). At these (x, y) values, called equilibrium or rest points, the two populations remain at constant values over all time. The point (0, 0) represents a pond containing no members of either fish species; the point (m>n, a>b) corresponds to a pond with an unchanging number of each fish species.
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Chapter 15: FirstOrder Differential Equations
y Bass
FIGURE 15.28 To the left of the line x = m>n the trajectories move upward, and to the right they move downward.
a b
dx = 0 dt dy =0 dt Trout (a)
x
m n (b)
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Trout
Trout
a b
x
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m n
y Bass
y Bass
y Bass x
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Next, we note that if y = a>b, then Equation (1a) implies dx>dt = 0, so the trout population x(t) is constant. Similarly, if x = m>n, then Equation (1b) implies dy>dt = 0, and the bass population y(t) is constant. This information is recorded in Figure 15.27.
( mn , ab )
m n (c)
Trout
FIGURE 15.27 Rest points in the competitivehunter model given by Equations (1a and 1b).
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y Bass
Trout
x
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FIGURE 15.29 Above the line y = a>b the trajectories move to the left, and below it they move to the right.
y Bass B
a b C
D Trout
x
ham
m n
ma
A
In setting up our competitivehunter model, precise values of the constants a, b, m, n will not generally be known. Nonetheless, we can analyze the system of Equations (1) to learn the nature of its solution trajectories. We begin by determining the signs of dx>dt and dy>dt throughout the phase plane. Although x(t) represents the number of trout and y(t) the number of bass at time t, we are thinking of the pair of values (x(t), y(t)) as a point tracing out a trajectory curve in the phase plane. When dx>dt is positive, x(t) is increasing and the point is moving to the right in the phase plane. If dx>dt is negative, the point is moving to the left. Likewise, the point is moving upward where dy>dt is positive and downward where dy>dt is negative. We saw that dy>dt = 0 along the vertical line x = m>n. To the left of this line, dy>dt is positive since dy>dt = (m  nx)y and x 6 m>n. So the trajectories on this side of the line are directed upward. To the right of this line, dy>dt is negative and the trajectories point downward. The directions of the associated trajectories are indicated in Figure 15.28. Similarly, above the horizontal line y = a>b, we have dx>dt 6 0 and the trajectories head leftward; below this line they head rightward, as shown in Figure 15.29. Combining this information gives four distinct regions in the plane A, B, C, D, with their respective trajectory directions shown in Figure 15.30. Next, we examine what happens near the two equilibrium points. The trajectories near (0, 0) point away from it, upward and to the right. The behavior near the equilibrium point (m>n, a>b) depends on the region in which a trajectory begins. If it starts in region B, for instance, then it will move downward and leftward towards the equilibrium point. Depending on where the trajectory begins, it may move downward into region D, leftward into region A, or perhaps straight into the equilibrium point. If it enters into regions A or D, then it will continue to move away from the rest point. We say that both rest points are unstable, meaning (in this setting) there are trajectories near each point that head away from them. These features are indicated in Figure 15.31. It turns out that in each of the halfplanes above and below the line y = a>b, there is exactly one trajectory approaching the equilibrium point (m>n, a>b) (see Exercise 7). Above these two trajectories the bass population increases and below them it decreases. The two trajectories approaching the equilibrium point are suggested in Figure 15.32.
ass a
a b
(0, 0)
x
Mu
FIGURE 15.30 Composite graphical analysis of the trajectory directions in the four regions determined by x = m>n and y = a>b.
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D
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C m n
(0, 0)
Trout
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a b
Limitations of the PhasePlane Analysis Method
x
Unlike the situation for the competitivehunter model, it is not always possible to determine the behavior of trajectories near a rest point. For example, suppose we know that the trajectories near a rest point, chosen here to be the origin (0, 0), behave as in Figure 15.33. The information provided by Figure 15.33 is not sufficient to distinguish between the three possible trajectories shown in Figure 15.34. Even if we could determine that a trajectory near an equilibrium point resembles that of Figure 15.34c, we would still not know how the other trajectories behave. It could happen that a trajectory closer to the origin behaves like the motions displayed in Figure 15.34a or 15.34b. The spiraling trajectory in Figure 15.34b can never actually reach the rest point in a finite time period.
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FIGURE 15.31 Motion along the trajectories near the rest points (0, 0) and (m>n, a>b). y Bass
Bass win
(mn , ab) Trout
(0, 0)
y
x
y
ass a
Trout win
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B
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Our graphical analysis leads us to conclude that, under the assumptions of the competitivehunter model, it is unlikely that both species will reach equilibrium levels. This is because it would be almost impossible for the fish populations to move exactly along one of the two approaching trajectories for all time. Furthermore, the initial populations point (x0, y0) determines which of the two species is likely to survive over time, and mutual coexistence of the species is highly improbable.
y Bass A
Systems of Equations and Phase Planes
(x0, y0)
FIGURE 15.32 Qualitative results of analyzing the competitivehunter model. There are exactly two trajectories approaching the point (m>n, a>b).
(x0, y0)
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x
(a)
y
(x0, y0) x
(b)
x
(c)
FIGURE 15.34 Three possible trajectory motions: (a) periodic motion, (b) motion toward an asymptotically stable rest point, and (c) motion near an unstable rest point.
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FIGURE 15.33 Trajectory direction near the rest point (0, 0).
Another Type of Behavior
y
The system
ham
x2 + y2 = 1
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(x1, y1)
x
(x0, y0)
dx = y + x  x(x 2 + y 2), dt dy = x + y  y(x 2 + y 2) dt
(2a) (2b)
can be shown to have only one equilibrium point at (0, 0). Yet any trajectory starting on the unit circle traverses it clockwise because, when x 2 + y 2 = 1, we have dy>dx = x>y (see Exercise 2). If a trajectory starts inside the unit circle, it spirals outward, asymptotically approaching the circle as t : q . If a trajectory starts outside the unit circle, it spirals inward, again asymptotically approaching the circle as t : q . The circle x 2 + y 2 = 1 is called a limit cycle of the system (Figure 15.35). In this system, the values of x and y eventually become periodic.
FIGURE 15.35 The solution x 2 + y 2 = 1 is a limit cycle.
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Chapter 15: FirstOrder Differential Equations
4. How might the competitivehunter model be validated? Include a discussion of how the various constants a, b, m, and n might be estimated. How could state conservation authorities use the model to ensure the survival of both species?
7. Show that the two trajectories leading to (m>n, a>b) shown in Figure 15.32 are unique by carrying out the following steps. a. From system (1a and 1b) derive the following equation: dy (m  nx)y = . dx (a  by)x b. Separate variables, integrate, and exponentiate to obtain where K is a constant of integration.
c. Let ƒ( y) = y a>e by and g(x) = x m>e nx. Show that ƒ( y) has a unique maximum of My = (a>eb) a when y = a>b as shown in Figure 15.36. Similarly, show that g(x) has a unique maximum Mx = (m>en) m when x = m>n, also shown in Figure 15.36.
ass a
dx x = a a1  b x  bxy, dt k1
where x and y represent trout and bass populations, respectively.
a. What assumptions are implicitly being made about the growth of trout and bass in the absence of competition?
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b. Interpret the constants a, b, m, n, k1, and k2 in terms of the physical problem. c. Perform a graphical analysis:
c. Give an economic interpretation of the curves that determine the equilibrium points.
y ae by = Kx me nx
5. Consider another competitivehunter model defined by
dy y = m a1  b y  nxy, dt k2
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3. Develop a model for the growth of trout and bass assuming that in isolation trout demonstrate exponential decay [so that a 6 0 in Equations (1a and 1b)] and that the bass population grows logistically with a population limit M. Analyze graphically the motion in the vicinity of the rest points in your model. Is coexistence possible?
b. Perform a graphical stability analysis to determine what will happen to the levels of P and Q as time increases.
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2. For the system (2a and 2b), show that any trajectory starting on the unit circle x 2 + y 2 = 1 will traverse the unit circle in a periodic solution. First introduce polar coordinates and rewrite the system as dr>dt = r(1  r 2) and du>dt = 1.
a. If a = 1, b = 20,000, c = 1, and ƒ = 30, find the equilibrium points of this system. If possible, classify each equilibrium point with respect to its stability. If a point cannot be readily classified, give some explanation.
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1. List three important considerations that are ignored in the competitivehunter model as presented in the text.
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EXERCISES 15.6
f (y)
My
i. Find the possible equilibrium levels.
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iv. Interpret the outcomes predicted by your graphical analysis in terms of the constants a, b, m, n, k1, and k2.
y
a b
ii. Determine whether coexistence is possible.
iii. Pick several typical starting points and sketch typical trajectories in the phase plane.
y ae–by
g(x)
Mx x me–nx
ham
Note: When you get to part (iii), you should realize that five cases exist. You will need to analyze all five cases.
Mu
6. Consider the following economic model. Let P be the price of a single item on the market. Let Q be the quantity of the item available on the market. Both P and Q are functions of time. If one considers price and quantity as two interacting species, the following model might be proposed: b dP = aP a  Pb , Q dt dQ = cQ(ƒP  Q), dt
m n
x
FIGURE 15.36 Graphs of the functions ƒ( y) = y a>e by and g(x) = x m>e nx. d. Consider what happens as ( x, y) approaches (m>n, a>b). Take limits in part (b) as x : m>n and y : a>b to show that either
where a, b, c, and ƒ are positive constants. Justify and discuss the adequacy of the model.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 15.6 ya e
by
ba
e nx bd = K xm
y Bass
i
x: m>n y : a>b
ca
or My>Mx = K. Thus any solution trajectory that approaches (m>n, a>b) must satisfy
e by
= a
a b y0
My xm b a b. Mx e nx
You
ya
1541
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lim
Systems of Equations and Phase Planes
Unique x 0
e. Show that only one trajectory can approach (m>n, a>b) from below the line y = a>b. Pick y0 6 a>b. From Figure 15.36 you can see that ƒ(y0) 6 My, which implies that
m n
Trout
x
FIGURE 15.37 For any y 6 a>b only one solution trajectory leads to the rest point (m>n, a>b).
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My x m a b = y0 a>e by0 6 My. Mx e nx
xm 6 Mx. e nx
dy = z, dx dz = F(x, y, z). dx
Can something similar be done to the nthorder differential equation y (n) = F A x, y, y¿, y–, Á , y (n  1) B ?
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Figure 15.36 tells you that for g(x) there is a unique value x0 6 m>n satisfying this last inequality. That is, for each y 6 a>b there is a unique value of x satisfying the equation in part (d). Thus there can exist only one trajectory solution approaching (m>n, a>b) from below, as shown in Figure 15.37.
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8. Show that the secondorder differential equation y– = F(x, y, y¿) can be reduced to a system of two firstorder differential equations
This in turn implies that
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ma
dH
f. Use a similar argument to show that the solution trajectory leading to (m>n, a>b) is unique if y0 7 a>b.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Chapter
You
16
SECONDORDER DIFFERENTIAL EQUATIONS
16.1
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OVERVIEW In this chapter we extend our study of differential equations to those of second order. Secondorder differential equations arise in many applications in the sciences and engineering. For instance, they can be applied to the study of vibrating springs and electric circuits. You will learn how to solve such differential equations by several methods in this chapter.
SecondOrder Linear Equations An equation of the form
P(x)y–(x) + Q(x)y¿(x) + R(x)y(x) = G(x),
(1)
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which is linear in y and its derivatives, is called a secondorder linear differential equation. We assume that the functions P, Q, R, and G are continuous throughout some open interval I. If G(x) is identically zero on I, the equation is said to be homogeneous; otherwise it is called nonhomogeneous. Therefore, the form of a secondorder linear homogeneous differential equation is P(x)y– + Q(x)y¿ + R(x)y = 0.
(2)
We also assume that P(x) is never zero for any x H I . Two fundamental results are important to solving Equation (2). The first of these says that if we know two solutions y1 and y2 of the linear homogeneous equation, then any linear combination y = c1 y1 + c2 y2 is also a solution for any constants c1 and c2.
THEOREM 1—The Superposition Principle If y1(x) and y2(x) are two solutions to the linear homogeneous equation (2), then for any constants c1 and c2, the function y(x) = c1 y1(x) + c2 y2(x) is also a solution to Equation (2).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 162
Chapter 16: SecondOrder Differential Equations
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Q(x)y¿ + R(x)y P(x)(c1 y1 + c2 y2)– + Q(x)(c1 y1 + c2 y2)¿ + R(x)(c1 y1 + c2 y2) P(x)(c1 y1 – + c2 y2 –) + Q(x)(c1 y1 ¿ + c2 y2 ¿) + R(x)(c1 y1 + c2 y2) c1(P(x)y1 – + Q(x)y1 ¿ + R(x)y1) + c2(P(x)y2 – + Q(x)y2 ¿ + R(x)y2) 144442444443 144442444443 = 0, y1 is a solution
= c1(0) + c2(0) = 0.
You
P(x)y– + = = =
i
Proof Substituting y into Equation (2), we have
0, y2 is a solution
Therefore, y = c1 y1 + c2 y2 is a solution of Equation (2).
2. 3.
A sum of two solutions y1 + y2 to Equation (2) is also a solution. (Choose c1 = c2 = 1.) A constant multiple ky1 of any solution y1 to Equation (2) is also a solution. (Choose c1 = k and c2 = 0.) The trivial solution y(x) K 0 is always a solution to the linear homogeneous equation. (Choose c1 = c2 = 0.)
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1.
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Theorem 1 immediately establishes the following facts concerning solutions to the linear homogeneous equation.
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The second fundamental result about solutions to the linear homogeneous equation concerns its general solution or solution containing all solutions. This result says that there are two solutions y1 and y2 such that any solution is some linear combination of them for suitable values of the constants c1 and c2. However, not just any pair of solutions will do. The solutions must be linearly independent, which means that neither y1 nor y2 is a constant multiple of the other. For example, the functions ƒ(x) = e x and g(x) = xe x are linearly independent, whereas ƒ(x) = x 2 and g(x) = 7x 2 are not (so they are linearly dependent). These results on linear independence and the following theorem are proved in more advanced courses.
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THEOREM 2 If P, Q, and R are continuous over the open interval I and P(x) is never zero on I, then the linear homogeneous equation (2) has two linearly independent solutions y1 and y2 on I. Moreover, if y1 and y2 are any two linearly independent solutions of Equation (2), then the general solution is given by y(x) = c1 y1(x) + c2 y2(x),
where c1 and c2 are arbitrary constants.
We now turn our attention to finding two linearly independent solutions to the special case of Equation (2), where P, Q, and R are constant functions.
ConstantCoefficient Homogeneous Equations Suppose we wish to solve the secondorder homogeneous differential equation
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ay– + by¿ + cy = 0,
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(3)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.1
SecondOrder Linear Equations
163
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ar 2e rx + bre rx + ce rx = 0.
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where a, b, and c are constants. To solve Equation (3), we seek a function which when multiplied by a constant and added to a constant times its first derivative plus a constant times its second derivative sums identically to zero. One function that behaves this way is the exponential function y = e rx, when r is a constant. Two differentiations of this exponential function give y¿ = re rx and y– = r 2e rx, which are just constant multiples of the original exponential. If we substitute y = e rx into Equation (3), we obtain
Since the exponential function is never zero, we can divide this last equation through by e rx. Thus, y = e rx is a solution to Equation (3) if and only if r is a solution to the algebraic equation
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ar 2 + br + c = 0.
(4)
r1 =
nR
Equation (4) is called the auxiliary equation (or characteristic equation) of the differential equation ay– + by¿ + cy = 0. The auxiliary equation is a quadratic equation with roots b + 2b 2  4ac 2a
and
r2 =
b  2b 2  4ac . 2a
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There are three cases to consider which depend on the value of the discriminant b 2  4ac. Case 1: b2 4ac>0. In this case the auxiliary equation has two real and unequal roots
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r1 and r2. Then y1 = e r1 x and y2 = e r2 x are two linearly independent solutions to Equation (3) because e r2 x is not a constant multiple of e r1 x (see Exercise 61). From Theorem 2 we conclude the following result.
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THEOREM 3 If r1 and r2 are two real and unequal roots to the auxiliary equation ar 2 + br + c = 0, then y = c1e r1 x + c2e r2 x
is the general solution to ay– + by¿ + cy = 0.
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EXAMPLE 1
Find the general solution of the differential equation y–  y¿  6y = 0.
Solution Substitution of y = e rx into the differential equation yields the auxiliary
equation r 2  r  6 = 0, which factors as (r  3)(r + 2) = 0. The roots are r1 = 3 and r2 = 2. Thus, the general solution is
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y = c1e 3x + c2e 2x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 164
Chapter 16: SecondOrder Differential Equations Case 2: b2 4ac 0. In this case r1 = r2 = b>2a. To simplify the notation, let
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r = b>2a. Then we have one solution y1 = e rx with 2ar + b = 0. Since multiplication of e rx by a constant fails to produce a second linearly independent solution, suppose we try multiplying by a function instead. The simplest such function would be u(x) = x, so let’s see if y2 = xe rx is also a solution. Substituting y2 into the differential equation gives
You
ay2 – + by2 ¿ + cy2 = a(2re rx + r 2xe rx ) + b(e rx + rxe rx ) + cxe rx = (2ar + b)e rx + (ar 2 + br + c)xe rx = 0(e rx ) + (0)xe rx = 0.
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The first term is zero because r = b>2a; the second term is zero because r solves the auxiliary equation. The functions y1 = e rx and y2 = xe rx are linearly independent (see Exercise 62). From Theorem 2 we conclude the following result.
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THEOREM 4 If r is the only (repeated) real root to the auxiliary equation ar 2 + br + c = 0, then y = c1e rx + c2 xe rx
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is the general solution to ay– + by¿ + cy = 0.
EXAMPLE 2
Find the general solution to y– + 4y¿ + 4y = 0.
Solution The auxiliary equation is
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r 2 + 4r + 4 = 0,
which factors into
(r + 2) 2 = 0.
ma
Thus, r = 2 is a double root. Therefore, the general solution is y = c1e 2x + c2 xe 2x.
Case 3: b2 4ac<0. In this case the auxiliary equation has two complex roots
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r1 = a + ib and r2 = a  ib , where a and b are real numbers and i 2 = 1. (These real
numbers are a = b>2a and b = 24ac  b 2>2a.) These two complex roots then give rise to two linearly independent solutions y1 = e (a + ib)x = e ax(cos bx + i sin bx) and
y2 = e (a  ib)x = e ax(cos bx  i sin bx).
(The expressions involving the sine and cosine terms follow from Euler’s identity in Section 8.9.) However, the solutions y1 and y2 are complex valued rather than real valued. Nevertheless, because of the superposition principle (Theorem 1), we can obtain from them the two realvalued solutions y3 =
1 1 y + y2 = e ax cos bx 2 1 2
and
y4 =
1 1 y y = e ax sin bx. 2i 1 2i 2
The functions y3 and y4 are linearly independent (see Exercise 63). From Theorem 2 we conclude the following result.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.1
SecondOrder Linear Equations
165
y = e ax(c1 cos bx + c2 sin bx)
EXAMPLE 3
You
is the general solution to ay– + by¿ + cy = 0.
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THEOREM 5 If r1 = a + ib and r2 = a  ib are two complex roots to the auxiliary equation ar 2 + br + c = 0, then
Find the general solution to the differential equation
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y–  4y¿ + 5y = 0. Solution The auxiliary equation is
r 2  4r + 5 = 0.
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The roots are the complex pair r = (4 ; 216  20)>2 or r1 = 2 + i and r2 = 2  i. Thus, a = 2 and b = 1 give the general solution
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y = e 2x(c1 cos x + c2 sin x).
Initial Value and Boundary Value Problems
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To determine a unique solution to a firstorder linear differential equation, it was sufficient to specify the value of the solution at a single point. Since the general solution to a secondorder equation contains two arbitrary constants, it is necessary to specify two conditions. One way of doing this is to specify the value of the solution function and the value of its derivative at a single point: y(x0) = y0 and y¿(x0) = y1. These conditions are called initial conditions. The following result is proved in more advanced texts and guarantees the existence of a unique solution for both homogeneous and nonhomogeneous secondorder linear initial value problems.
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THEOREM 6 If P, Q, R, and G are continuous throughout an open interval I, then there exists one and only one function y(x) satisfying both the differential equation P(x)y–(x) + Q(x)y¿(x) + R(x)y(x) = G(x) on the interval I, and the initial conditions y(x0) = y0
and
y¿(x0) = y1
at the specified point x0 H I .
It is important to realize that any real values can be assigned to y0 and y1 and Theorem 6 applies. Here is an example of an initial value problem for a homogeneous equation.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
Find the particular solution to the initial value problem y–  2y¿ + y = 0,
y(0) = 1,
y¿(0) = 1.
Solution The auxiliary equation is
r 2  2r + 1 = (r  1) 2 = 0.
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y
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EXAMPLE 4
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166
The repeated real root is r = 1, giving the general solution –4 –3
–2
–1
0
1
y = c1e x + c2 xe x.
x
Then,
–2 y = e x – 2xe x
y¿ = c1e x + c2(x + 1)e x.
–4
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From the initial conditions we have –6
1 = c1 + c2 # 0
–8
and
1 = c1 + c2 # 1.
FIGURE 16.1 Particular solution curve for Example 4.
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Thus, c1 = 1 and c2 = 2. The unique solution satisfying the initial conditions is y = e x  2xe x.
The solution curve is shown in Figure 16.1.
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Another approach to determine the values of the two arbitrary constants in the general solution to a secondorder differential equation is to specify the values of the solution function at two different points in the interval I. That is, we solve the differential equation subject to the boundary values y(x1) = y1
and
y(x2) = y2,
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where x1 and x2 both belong to I. Here again the values for y1 and y2 can be any real numbers. The differential equation together with specified boundary values is called a boundary value problem. Unlike the result stated in Theorem 6, boundary value problems do not always possess a solution or more than one solution may exist (see Exercise 65). These problems are studied in more advanced texts, but here is an example for which there is a unique solution.
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EXAMPLE 5
Solve the boundary value problem y– + 4y = 0,
y(0) = 0,
ya
p b = 1. 12
Solution The auxiliary equation is r 2 + 4 = 0, which has the complex roots r = ;2i.
The general solution to the differential equation is y = c1 cos 2x + c2 sin 2x. The boundary conditions are satisfied if y(0) = c1 # 1 + c2 # 0 = 0 ya
p p p b = c1 cos a b + c2 sin a b = 1. 12 6 6
It follows that c1 = 0 and c2 = 2. The solution to the boundary value problem is
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y = 2 sin 2x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.1
SecondOrder Linear Equations
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EXERCISES 16.1 In Exercises 41–55, find the general solution.
In Exercises 1–30, find the general solution of the given equation. 2. 3y–  y¿ = 0
41. y–  2y¿  3y = 0
42. 6y–  y¿  y = 0
3. y– + 3y¿  4y = 0
4. y–  9y = 0
43. 4y– + 4y¿ + y = 0
44. 9y– + 12y¿ + 4y = 0
5. y–  4y = 0
6. y–  64y = 0
45. 4y– + 20y = 0
46. y– + 2y¿ + 2y = 0
8. 9y–  y = 0
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1. y–  y¿  12y = 0
48. 6y– + 13y¿  5y = 0
49. 4y– + 4y¿ + 5y = 0
50. y– + 4y¿ + 6y = 0
11. y– + 9y = 0
12. y– + 4y¿ + 5y = 0
51. 16y–  24y¿ + 9y = 0
52. 6y–  5y¿  6y = 0
13. y– + 25y = 0
14. y– + y = 0
53. 9y– + 24y¿ + 16y = 0
54. 4y– + 16y¿ + 52y = 0
15. y–  2y¿ + 5y = 0
16. y– + 16y = 0
55. 6y–  5y¿  4y = 0
17. y– + 2y¿ + 4y = 0
18. y–  2y¿ + 3y = 0
In Exercises 56–60, solve the initial value problem.
19. y– + 4y¿ + 9y = 0
20. 4y–  4y¿ + 13y = 0
56. y–  2y¿ + 2y = 0,
21. y– = 0 dy d 2y 23. + 4 + 4y = 0 dx dx 2 2 dy d y + 6 25. + 9y = 0 2 dx dx
22. y– + 8y¿ + 16y = 0 d 2y dy 24.  6 + 9y = 0 dx dx 2 2 d y dy 26. 4 2  12 + 9y = 0 dx dx
57. y– + 2y¿ + y = 0,
27. 4
2
dx d 2y
+ 4
dy + y = 0 dx
dy + y = 0 29. 9 2 + 6 dx dx
28. 4
d 2y 2
dx d 2y
 4
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d 2y
58. 4y–  4y¿ + y = 0,
dy + y = 0 dx
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y(0) = 0, y¿(0) = 3
y(0) = 2, y¿(0) = 2
33. y– + 12y = 0,
y(0) = 0, y¿(0) = 1
35. y– + 8y = 0,
y(0) = 1, y¿(0) = 1
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34. 12y– + 5y¿  2y = 0,
y(0) = 1, y¿(0) = 2
36. y– + 4y¿ + 4y = 0, 37. y–  4y¿ + 4y = 0,
y(0) = 2, y¿(0) = 1
60. 4y– + 4y¿ + 5y = 0,
y(p) = 1, y¿(p) = 0
62. Prove that the two solution functions in Theorem 4 are linearly independent.
dy + 4y = 0 30. 9 2  12 dx dx
32. y– + 16y = 0,
y(0) = 1, y¿(0) = 2
59. 3y– + y¿  14y = 0,
61. Prove that the two solution functions in Theorem 3 are linearly independent.
In Exercises 31– 40, find the unique solution of the secondorder initial value problem. 31. y– + 6y¿ + 5y = 0,
y(0) = 0, y¿(0) = 2
y(0) = 1, y¿(0) = 1
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9. 8y–  10y¿  3y = 0
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47. 25y– + 10y¿ + y = 0
10. 3y–  20y¿ + 12y = 0
7. 2y–  y¿  3y = 0
167
y(0) = 0, y¿(0) = 1 y(0) = 1, y¿(0) = 0
64. Prove that if y1 and y2 are linearly independent solutions to the homogeneous equation (2), then the functions y3 = y1 + y2 and y4 = y1  y2 are also linearly independent solutions. 65. a. Show that there is no solution to the boundary value problem y– + 4y = 0,
y(0) = 0, y(p) = 1.
b. Show that there are infinitely many solutions to the boundary value problem y– + 4y = 0,
y(0) = 0, y(p) = 0.
66. Show that if a, b, and c are positive constants, then all solutions of the homogeneous differential equation ay– + by¿ + cy = 0 approach zero as x : q .
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38. 4y–  4y¿ + y = 0, y(0) = 4, y¿(0) = 4 d 2y dy dy + 9y = 0, y(0) = 2, (0) = 1 39. 4 2 + 12 dx dx dx dy dy d 2y + 4y = 0, y(0) = 1, (0) = 1 40. 9 2  12 dx dx dx
63. Prove that the two solution functions in Theorem 5 are linearly independent.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
Nonhomogeneous Linear Equations
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16.2
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In this section we study two methods for solving secondorder linear nonhomogeneous differential equations with constant coefficients. These are the methods of undetermined coefficients and variation of parameters. We begin by considering the form of the general solution.
Form of the General Solution
Suppose we wish to solve the nonhomogeneous equation
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ay– + by¿ + cy = G(x),
(1)
where a, b, and c are constants and G is continuous over some open interval I. Let yc = c1 y1 + c2 y2 be the general solution to the associated complementary equation
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ay– + by¿ + cy = 0.
(2)
(We learned how to find yc in Section 16.1.) Now suppose we could somehow come up with a particular function yp that solves the nonhomogeneous equation (1). Then the sum y = yc + yp
(3)
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also solves the nonhomogeneous equation (1) because a(yc + yp)– + b(yc + yp)¿ + c(yc + yp) = (ayc – + byc ¿ + cyc) + (ayp – + byp ¿ + cyp)
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= 0 + G(x)
yc solves Eq. (2) and yp solves Eq. (1)
= G(x).
Moreover, if y = y(x) is the general solution to the nonhomogeneous equation (1), it must have the form of Equation (3). The reason for this last statement follows from the observation that for any function yp satisfying Equation (1), we have
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a(y  yp)– + b(y  yp)¿ + c(y  yp) = (ay– + by¿ + cy)  (ayp – + byp ¿ + cyp) = G(x)  G(x) = 0.
Thus, yc = y  yp is the general solution to the homogeneous equation (2). We have established the following result.
THEOREM 7 The general solution y = y(x) to the nonhomogeneous differential equation (1) has the form y = yc + yp, where the complementary solution yc is the general solution to the associated homogeneous equation (2) and yp is any particular solution to the nonhomogeneous equation (1).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2
Nonhomogeneous Linear Equations
169
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The Method of Undetermined Coefficients
p2(x)e ax cos bx,
p3(x)e ax sin bx.
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p1(x)e rx,
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This method for finding a particular solution yp to nonhomogeneous equation (1) applies to special cases for which G(x) is a sum of terms of various polynomials p(x) multiplying an exponential with possibly sine or cosine factors. That is, G(x) is a sum of terms of the following forms:
EXAMPLE 1
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For instance, 1  x, e 2x, xe x, cos x, and 5e x  sin 2x represent functions in this category. (Essentially these are functions solving homogeneous linear differential equations with constant coefficients, but the equations may be of order higher than two.) We now present several examples illustrating the method. Solve the nonhomogeneous equation y–  2y¿  3y = 1  x 2.
Solution The auxiliary equation for the complementary equation y–  2y¿  3y = 0 is
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r 2  2r  3 = (r + 1)(r  3) = 0.
It has the roots r = 1 and r = 3 giving the complementary solution yc = c1e  x + c2e 3x.
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Now G(x) = 1  x 2 is a polynomial of degree 2. It would be reasonable to assume that a particular solution to the given nonhomogeneous equation is also a polynomial of degree 2 because if y is a polynomial of degree 2, then y–  2y¿  3y is also a polynomial of degree 2. So we seek a particular solution of the form
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yp = Ax 2 + Bx + C.
We need to determine the unknown coefficients A, B, and C. When we substitute the polynomial yp and its derivatives into the given nonhomogeneous equation, we obtain 2A  2(2Ax + B)  3(Ax 2 + Bx + C) = 1  x 2
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or, collecting terms with like powers of x, 3Ax 2 + (4A  3B)x + (2A  2B  3C) = 1  x 2.
This last equation holds for all values of x if its two sides are identical polynomials of degree 2. Thus, we equate corresponding powers of x to get 3A = 1,
4A  3B = 0,
and
2A  2B  3C = 1.
These equations imply in turn that A 1>3, B 4>9, and C 5>27. Substituting these values into the quadratic expression for our particular solution gives yp =
5 1 2 4 x  x + . 3 9 27
By Theorem 7, the general solution to the nonhomogeneous equation is
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y = yc + yp = c1e  x + c2e 3x +
5 1 2 4 x  x + . 3 9 27
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
Find a particular solution of y–  y¿ = 2 sin x.
Solution If we try to find a particular solution of the form
yp = A sin x
i
EXAMPLE 2
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1610
You
and substitute the derivatives of yp in the given equation, we find that A must satisfy the equation A sin x + A cos x = 2 sin x
for all values of x. Since this requires A to equal both  2 and 0 at the same time, we conclude that the nonhomogeneous differential equation has no solution of the form A sin x. It turns out that the required form is the sum
iaz
yp = A sin x + B cos x.
The result of substituting the derivatives of this new trial solution into the differential equation is
nR
A sin x  B cos x  (A cos x  B sin x) = 2 sin x or
(B  A) sin x  (A + B) cos x = 2 sin x.
ass a
This last equation must be an identity. Equating the coefficients for like terms on each side then gives B  A = 2
and
A + B = 0.
Simultaneous solution of these two equations gives A = 1 and B = 1. Our particular solution is
dH
yp = cos x  sin x.
EXAMPLE 3
Find a particular solution of y–  3y¿ + 2y = 5e x.
ma
Solution If we substitute
yp = Ae x
and its derivatives in the differential equation, we find that Ae x  3Ae x + 2Ae x = 5e x
Mu
ham
or
0 = 5e x.
However, the exponential function is never zero. The trouble can be traced to the fact that y = e x is already a solution of the related homogeneous equation y–  3y¿ + 2y = 0. The auxiliary equation is r 2  3r + 2 = (r  1)(r  2) = 0, which has r = 1 as a root. So we would expect Ae x to become zero when substituted into the lefthand side of the differential equation. The appropriate way to modify the trial solution in this case is to multiply Ae x by x. Thus, our new trial solution is yp = Axe x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2
Nonhomogeneous Linear Equations
1611
or
suf
(Axe x + 2Ae x)  3(Axe x + Ae x) + 2Axe x = 5e x
i
The result of substituting the derivatives of this new candidate into the differential equation is
Ae x = 5e x.
You
Thus, A = 5 gives our soughtafter particular solution yp = 5xe x.
EXAMPLE 4
Find a particular solution of y–  6y¿ + 9y = e 3x.
Solution The auxiliary equation for the complementary equation
iaz
r 2  6r + 9 = (r  3) 2 = 0 has r = 3 as a repeated root. The appropriate choice for yp in this case is neither Ae 3x nor Axe 3x because the complementary solution contains both of those terms already. Thus, we choose a term containing the next higher power of x as a factor. When we substitute
nR
yp = Ax 2e 3x
and its derivatives in the given differential equation, we get (9Ax 2e 3x + 12Axe 3x + 2Ae 3x)  6(3Ax 2e 3x + 2Axe 3x) + 9Ax 2e 3x = e 3x
ass a
or
2Ae 3x = e 3x.
Thus, A = 1>2, and the particular solution is yp =
1 2 3x x e . 2
dH
When we wish to find a particular solution of Equation (1) and the function G(x) is the sum of two or more terms, we choose a trial function for each term in G(x) and add them.
EXAMPLE 5
Find the general solution to y–  y¿ = 5e x  sin 2x.
ma
Solution We first check the auxiliary equation
r 2  r = 0.
Mu
ham
Its roots are r = 1 and r = 0. Therefore, the complementary solution to the associated homogeneous equation is yc = c1e x + c2.
We now seek a particular solution yp. That is, we seek a function that will produce 5e x  sin 2x when substituted into the lefthand side of the given differential equation. One part of yp is to produce 5e x, the other sin 2x. Since any function of the form c1e x is a solution of the associated homogeneous equation, we choose our trial solution yp to be the sum yp = Axe x + B cos 2x + C sin 2x, including xe x where we might otherwise have included only e x. When the derivatives of yp are substituted into the differential equation, the resulting equation is (Axe x + 2Ae x  4B cos 2x  4C sin 2x)  (Axe x + Ae x  2B sin 2x + 2C cos 2x) = 5e x  sin 2x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1612
Chapter 16: SecondOrder Differential Equations
i
or
This equation will hold if A = 5,
4B + 2C = 0,
2B  4C = 1,
You
or A = 5, B = 1>10, and C = 1>5. Our particular solution is yp = 5xe x 
suf
Ae x  (4B + 2C) cos 2x + (2B  4C) sin 2x = 5e x  sin 2x.
1 1 cos 2x + sin 2x. 5 10
The general solution to the differential equation is
1 1 cos 2x + sin 2x. 5 10
iaz
y = yc + yp = c1e x + c2 + 5xe x 
TABLE 16.1
nR
You may find the following table helpful in solving the problems at the end of this section.
The method of undetermined coefficients for selected equations of the form
ass a
ayâ€“ + byÂż + cy = G(x).
If G(x) has a term that is a constant multiple of . . .
ma
dH
e
rx
sin kx, cos kx
Mu
ham
px 2 + qx + m
And if
r is not a root of the auxiliary equation r is a single root of the auxiliary equation r is a double root of the auxiliary equation k is not a root of the auxiliary equation 0 is not a root of the auxiliary equation 0 is a single root of the auxiliary equation 0 is a double root of the auxiliary equation
Then include this expression in the trial function for yp. Ae rx Axe rx Ax 2e rx B cos kx + C sin kx Dx 2 + Ex + F Dx 3 + Ex 2 + Fx Dx 4 + Ex 3 + Fx 2
The Method of Variation of Parameters This is a general method for finding a particular solution of the nonhomogeneous equation (1) once the general solution of the associated homogeneous equation is known. The method consists of replacing the constants c1 and c2 in the complementary solution by functions y1 = y1(x) and y2 = y2(x) and requiring (in a way to be explained) that the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2
Nonhomogeneous Linear Equations
1613
y1 ¿y1 + y2 ¿y2 = 0. y = y1 y1 + y2 y2,
(4)
You
Then we have
suf
i
resulting expression satisfy the nonhomogeneous equation (1). There are two functions to be determined, and requiring that Equation (1) be satisfied is only one condition. As a second condition, we also require that
y¿ = y1 y1 ¿ + y2 y2 ¿, y– = y1 y1 – + y2 y2 – + y1 ¿y1 ¿ + y2 ¿y2 ¿.
If we substitute these expressions into the lefthand side of Equation (1), we obtain y1(ay1 – + by1 ¿ + cy1) + y2(ay2 – + by2 ¿ + cy2) + a(y1 ¿y1 ¿ + y2 ¿y2 ¿) = G(x).
iaz
The first two parenthetical terms are zero since y1 and y2 are solutions of the associated homogeneous equation (2). So the nonhomogeneous equation (1) is satisfied if, in addition to Equation (4), we require that
nR
a(y1 ¿y1 ¿ + y2 ¿y2 ¿) = G(x).
(5)
Equations (4) and (5) can be solved together as a pair y1 ¿y1 + y2 ¿y2 = 0, G(x) a
ass a
y1 ¿y1 ¿ + y2 ¿y2 ¿ =
dH
for the unknown functions y1 ¿ and y2 ¿ . The usual procedure for solving this simple system is to use the method of determinants (also known as Cramer’s Rule), which will be demonstrated in the examples to follow. Once the derivative functions y1 ¿ and y2 ¿ are known, the two functions y1 = y1(x) and y2 = y2(x) can be found by integration. Here is a summary of the method.
Mu
ham
ma
Variation of Parameters Procedure To use the method of variation of parameters to find a particular solution to the nonhomogeneous equation ay– + by¿ + cy = G(x), we can work directly with the Equations (4) and (5). It is not necessary to rederive them. The steps are as follows. 1.
2.
Solve the associated homogeneous equation ay– + by¿ + cy = 0 to find the functions y1 and y2. Solve the equations y1 ¿y1 + y2 ¿y2 = 0, G(x) y1 ¿y1 ¿ + y2 ¿y2 ¿ = a
3. 4.
simultaneously for the derivative functions y1 ¿ and y2 ¿ . Integrate y1 ¿ and y2 ¿ to find the functions y1 = y1(x) and y2 = y2(x). Write down the particular solution to nonhomogeneous equation (1) as yp = y1y1 + y2 y2.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
EXAMPLE 6
Find the general solution to the equation
i
1614
suf
y– + y = tan x. Solution The solution of the homogeneous equation
You
y– + y = 0 is given by
yc = c1 cos x + c2 sin x.
Since y1(x) = cos x and y2(x) = sin x, the conditions to be satisfied in Equations (4) and (5) are
iaz
y1 ¿ cos x + y2 ¿ sin x = 0,
y1 ¿ sin x + y2 ¿ cos x = tan x. Solution of this system gives `
ass a
Likewise,
sin x ` cos x tan x sin x sin2 x = = cos x . 2 2 sin x cos x + sin x ` cos x
nR
0 tan x y1 ¿ = cos x ` sin x
`
cos x sin x y2 ¿ = cos x ` sin x
dH
a = 1
0 ` tan x = sin x. sin x ` cos x
ma
After integrating y1 ¿ and y2 ¿, we have y1(x) =
L
sin2 x cos x dx
(sec x  cos x) dx L = ln ƒ sec x + tan x ƒ + sin x, = 
Mu
ham
and
y2(x) =
L
sin x dx = cos x.
Note that we have omitted the constants of integration in determining y1 and y2. They would merely be absorbed into the arbitrary constants in the complementary solution. Substituting y1 and y2 into the expression for yp in Step 4 gives yp = [ln ƒ sec x + tan x ƒ + sin x] cos x + (cos x) sin x = (cos x) ln ƒ sec x + tan x ƒ. The general solution is y = c1 cos x + c2 sin x  (cos x) ln ƒ sec x + tan x ƒ.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.2
1615
Solve the nonhomogeneous equation
i
EXAMPLE 7
Nonhomogeneous Linear Equations
suf
y– + y¿  2y = xe x. Solution The auxiliary equation is
r 2 + r  2 = (r + 2)(r  1) = 0
You
giving the complementary solution
yc = c1e  2x + c2e x.
The conditions to be satisfied in Equations (4) and (5) are
a = 1
iaz
y1 ¿e  2x + y2 ¿e x = 0, 2y1 ¿e  2x + y2 ¿e x = xe x. Solving the above system for y1 ¿ and y2 ¿ gives
0 ex ` x xe ex xe 2x 1 = =  xe 3x. y1 ¿ =  2x x 3 3e  x e e `  2x x` 2e e `
e  2x 0 `  2x 2e xe x xe x x = = . x 3 3e 3e x
ass a
Likewise,
nR
`
y2 ¿ =
y1(x) =
L

1 3x xe dx 3
e 3x 1 xe 3x =  a dxb 3 3 L 3 =
ma
dH
Integrating to obtain the parameter functions, we have
1 (1  3x)e 3x, 27
Mu
ham
and
y2(x) =
x x2 dx = . 6 L3
Therefore, yp = c =
(1  3x)e 3x 2x x2 de + a be x 27 6
1 x 1 1 e  xe x + x 2e x. 27 9 6
The general solution to the differential equation is y = c1e 2x + c2e x 
1 x 1 xe + x 2e x, 9 6
where the term (1>27)e x in yp has been absorbed into the term c2e x in the complementary solution.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1616
Chapter 16: SecondOrder Differential Equations
suf
i
EXERCISES 16.2
2. y–  3y¿  10y = 2x  3
3. y–  y¿ = sin x
4. y– + 2y¿ + y = x 2
37. y– + y = cot x,
0 6 x 6 p
38. y– + y = csc x,
0 6 x 6 p
5. y– + y = cos 3x
6. y– + y = e
9. y–  y = e + x
39. y–  8y¿ = e
8. y– + y = 2x + 3e x
7. y–  y¿  2y = 20 cos x x
2x
2
10. y– + 2y¿ + y = 6 sin 2x
8x
40. y– + 4y = sin x
41. y–  y¿ = x 3
42. y– + 4y¿ + 5y = x + 2
2
x
11. y–  y¿  6y = e x  7 cos x
43. y– + 2y¿ = x  e
12. y– + 3y¿ + 2y = e + e  x d 2y dy dy d 2y 13. 14. + 5 = 15x 2 = 8x + 3 2 2 dx dx dx dx d 2y dy d 2y dy 3x 15. 16. 3 12x + 7 = e = 42x 2 + 5x + 1 2 2 dx dx dx dx
45. y– + y = sec x tan x,
Solve the equations in Exercises 17–28 by variation of parameters.
47. y¿  3y = e x
17. y– + y¿ = x p p 6 x 6 2 2 20. y– + 2y¿ + y = e x
19. y– + y = sin x 21. y– + 2y¿ + y = e 23. y–  y = e
22. y–  y = x
x
x
24. y–  y = sin x x
30. y–  y¿ = cos x + sin x,
ma
dH
25. y– + 4y¿ + 5y = 10 26. y–  y¿ = 2 2 d y p p + y = sec x,  6 x 6 27. 2 2 dx 2 d 2y dy = e x cos x, x 7 0 28. dx dx 2
31. y– + y = 2 cos x + sin x,
yp = Ax cos x + Bx sin x
In each of Exercises 29–32, the given differential equation has a particular solution yp of the form given. Determine the coefficients in yp. Then solve the differential equation. yp = Ax 2e 5x + Bxe 5x
yp = A cos x + B sin x
ham
29. y–  5y¿ = xe 5x,
x
32. y– + y¿  2y = xe ,
yp = Ax 2e x + Bxe x
Mu
35.
d 2y dx
2
 4
dy  5y = e x + 4 dx
36.
d 2y dx
2
 9
3x
dy = 9e 9x dx
52.
d 2y dx 2
+ y = e 2x;
50. y¿ + y = sin x
y(0) = 0, y¿(0) =
2 5
In Exercises 53–58, verify that the given function is a particular solution to the specified nonhomogeneous equation. Find the general solution and evaluate its arbitrary constants to find the unique solution satisfying the equation and the given initial conditions. 53. y– + y¿ = x, 54. y– + y = x, 55.
yp =
x2  x, 2
y(0) = 0, y¿(0) = 0
yp = 2 sin x + x,
y(0) = 0, y¿(0) = 0
1 y– + y¿ + y = 4e x(cos x  sin x), 2 yp = 2e x cos x, y(0) = 0, y¿(0) = 1
56. y–  y¿  2y = 1  2x,
yp = x  1,
yp = x 2e x,
y(0) = 0, y¿(0) = 1
y(0) = 1, y¿(0) = 0
58. y–  2y¿ + y = x 1e x, x 7 0, yp = xe x ln x,
y(1) = e, y¿(1) = 0
In Exercises 59 and 60, two linearly independent solutions y1 and y2 are given to the associated homogeneous equation of the variablecoefficient nonhomogeneous equation. Use the method of variation of parameters to find a particular solution to the nonhomogeneous equation. Assume x 7 0 in each exercise. 59. x 2y– + 2xy¿  2y = x 2, 60. x 2y– + xy¿  y = x,
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48. y¿ + 4y = x
Solve the differential equations in Exercises 51 and 52 subject to the given initial conditions. d 2y p p + y = sec2 x,  6 x 6 ; y (0) = y¿(0) = 1 51. 2 2 dx 2
57. y–  2y¿ + y = 2e x,
In Exercises 33–36, solve the given differential equations (a) by variation of parameters, and (b) by the method of undetermined coefficients. d 2y d 2y dy dy  4 = e x + e x + 4y = 2e 2x 33. 34. 2 2 dx dx dx dx
p p 6 x 6 2 2
The method of undetermined coefficients can sometimes be used to solve firstorder ordinary differential equations. Use the method to solve the equations in Exercises 47–50.
ass a


46. y–  3y¿ + 2y = e x  e 2x
49. y¿  3y = 5e
18. y– + y = tan x,
44. y– + 9y = 9x  cos x
iaz
2x
nR
x
You
1. y–  3y¿  10y = 3
Solve the differential equations in Exercises 37–46. Some of the equations can be solved by the method of undetermined coefficients, but others cannot.
Solve the equations in Exercises 1–16 by the method of undetermined coefficients.
y1 = x  2, y2 = x
y1 = x  1, y2 = x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.3
Applications
suf
i
Applications
16.3
1617
In this section we apply secondorder differential equations to the study of vibrating springs and electric circuits.
You
Vibrations
A spring has its upper end fastened to a rigid support, as shown in Figure 16.2. An object of mass m is suspended from the spring and stretches it a length s when the spring comes to rest in an equilibrium position. According to Hooke’s Law (Section 6.6), the tension force in the spring is ks, where k is the spring constant. The force due to gravity pulling down on the spring is mg, and equilibrium requires that
s y⫽0
iaz
mass m at equilibrium
ks = mg.
nR
FIGURE 16.2 Mass m stretches a spring by length s to the equilibrium position at y = 0.
Suppose that the object is pulled down an additional amount y0 beyond the equilibrium position and then released. We want to study the object’s motion, that is, the vertical position of its center of mass at any future time. Let y, with positive direction downward, denote the displacement position of the object away from the equilibrium position y = 0 at any time t after the motion has started. Then the forces acting on the object are (see Figure 16.3) Fp = mg,
the propulsion force due to gravity,
ass a
y
(1)
Fs = k(s + y), Fr = d
the restoring force of the spring’s tension,
dy , dt
a frictional force assumed proportional to velocity.
dH
The frictional force tends to retard the motion of the object. The resultant of these forces is F = Fp  Fs  Fr, and by Newton’s second law F = ma, we must then have
s
m
d 2y dt
2
= mg  ks  ky  d
dy . dt
By Equation (1), mg  ks = 0, so this last equation becomes
ma
y⫽0 Fs
Fr
y0
ham
a position after release
y
start position
y
m
Fp
Mu
FIGURE 16.3 The propulsion force (weight) Fp pulls the mass downward, but the spring restoring force Fs and frictional force Fr pull the mass upward. The motion starts at y = y0 with the mass vibrating up and down.
d 2y dt
2
+ d
dy + ky = 0, dt
(2)
subject to the initial conditions y(0) = y0 and y¿(0) = 0. (Here we use the prime notation to denote differentiation with respect to time t.) You might expect that the motion predicted by Equation (2) will be oscillatory about the equilibrium position y = 0 and eventually damp to zero because of the retarding frictional force. This is indeed the case, and we will show how the constants m, d, and k determine the nature of the damping. You will also see that if there is no friction (so d = 0), then the object will simply oscillate indefinitely.
Simple Harmonic Motion Suppose first that there is no retarding frictional force. Then d = 0 and there is no damping. If we substitute v = 2k>m to simplify our calculations, then the secondorder equation (2) becomes
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y– + v2y = 0,
with
y(0) = y0
and
y¿(0) = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1618
Chapter 16: SecondOrder Differential Equations
r 2 + v2 = 0,
suf
i
The auxiliary equation is
having the imaginary roots r = ;vi. The general solution to the differential equation in (2) is
To fit the initial conditions, we compute
(3)
You
y = c1 cos vt + c2 sin vt.
y¿ = c1v sin vt + c2v cos vt
and then substitute the conditions. This yields c1 = y0 and c2 = 0. The particular solution (4)
iaz
y = y0 cos vt
nR
describes the motion of the object. Equation (4) represents simple harmonic motion of amplitude y0 and period T = 2p>v. The general solution given by Equation (3) can be combined into a single term by using the trigonometric identity sin (vt + f) = cos vt sin f + sin vt cos f. C=
To apply the identity, we take (see Figure 16.4)
兹 c12 + c22
c1 = C sin f where
c2
and
ass a
c1
C = 2c12 + c2 2
c1 f = tan1 c2 .
Then the general solution in Equation (3) can be written in the alternative form y = C sin (vt + f).
dH
FIGURE 16.4 c1 = C sin f and c2 = C cos f.
and
c2 = C cos f,
(5)
Mu
ham
ma
Here C and f may be taken as two new arbitrary constants, replacing the two constants c1 and c2. Equation (5) represents simple harmonic motion of amplitude C and period T = 2p>v. The angle vt + f is called the phase angle, and f may be interpreted as its initial value. A graph of the simple harmonic motion represented by Equation (5) is given in Figure 16.5.
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y
C C sin
Period T = 2
t
0 –C y = C sin (t + )
FIGURE 16.5 Simple harmonic motion of amplitude C and period T with initial phase angle f (Equation 5).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.3
1619
Applications
i
Damped Motion
suf
Assume now that there is friction in the spring system, so d Z 0. If we substitute v = 2k>m and 2b = d>m, then the differential equation (2) is y– + 2by¿ + v2y = 0.
(6)
You
The auxiliary equation is r 2 + 2br + v2 = 0,
with roots r = b ; 2b 2  v2. Three cases now present themselves, depending upon the relative sizes of b and v.
iaz
Case 1: b V . The double root of the auxiliary equation is real and equals r = v. The general solution to Equation (6) is
y = (c1 + c2t)e  vt.
nR
This situation of motion is called critical damping and is not oscillatory. Figure 16.6a shows an example of this kind of damped motion. Case 2: b>V . The roots of the auxiliary equation are real and unequal, given by
r1 = b + 2b 2  v2 and r2 = b  2b 2  v2. The general solution to Equation (6) is given by y = c1e A  b + 2b
ass a
2
 v2 B t
+ c2e A  b  2b
2
 v2Bt
.
Here again the motion is not oscillatory and both r1 and r2 are negative. Thus y approaches zero as time goes on. This motion is referred to as overdamping (see Figure 16.6b). Case 3: b<V . The roots to the auxiliary equation are complex and given by
dH
r = b ; i2v2  b 2. The general solution to Equation (6) is given by y = e  bt A c1 cos2v2  b 2 t + c2 sin2v2  b 2 t B .
Mu
ham
ma
This situation, called underdamping, represents damped oscillatory motion. It is analogous to simple harmonic motion of period T = 2p> 2v2  b 2 except that the amplitude is not constant but damped by the factor e  bt. Therefore, the motion tends to zero as t increases, so the vibrations tend to die out as time goes on. Notice that the period T = 2p> 2v2  b 2 is larger than the period T0 = 2p>v in the frictionfree system. Moreover, the larger the value of b = d>2m in the exponential damping factor, the more quickly the vibrations tend to become unnoticeable. A curve illustrating underdamped motion is shown in Figure 16.6c. y
y
t
0 y = (1 +
t)e–t
(a) Critical damping
y
t
0 y=
2e –2t – e –t
(b) Overdamping
t
0 y=
e –t sin (5t
+ /4)
(c) Underdamping
FIGURE 16.6 Three examples of damped vibratory motion for a spring system with friction, so d Z 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1620
Chapter 16: SecondOrder Differential Equations
m
d 2y dt
2
+ d
You
suf
i
An external force F(t) can also be added to the spring system modeled by Equation (2). The forcing function may represent an external disturbance on the system. For instance, if the equation models an automobile suspension system, the forcing function might represent periodic bumps or potholes in the road affecting the performance of the suspension system; or it might represent the effects of winds when modeling the vertical motion of a suspension bridge. Inclusion of a forcing function results in the secondorder nonhomogeneous equation dy + ky = F(t). dt
(7)
We leave the study of such spring systems to a more advanced course.
iaz
Electric Circuits
dH
ass a
nR
The basic quantity in electricity is the charge q (analogous to the idea of mass). In an electric field we use the flow of charge, or current I = dq>dt, as we might use velocity in a gravitational field. There are many similarities between motion in a gravitational field and the flow of electrons (the carriers of charge) in an electric field. Consider the electric circuit shown in Figure 16.7. It consists of four components: voltage source, resistor, inductor, and capacitor. Think of electrical flow as being like a fluid flow, where the voltage source is the pump and the resistor, inductor, and capacitor tend to block the flow. A battery or generator is an example of a source, producing a voltage that causes the current to flow through the circuit when the switch is closed. An electric light bulb or appliance would provide resistance. The inductance is due to a magnetic field that opposes any change in the current as it flows through a coil. The capacitance is normally created by two metal plates that alternate charges and thus reverse the current flow. The following symbols specify the quantities relevant to the circuit: q: charge at a cross section of a conductor measured in coulombs (abbreviated c);
Mu
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I: current or rate of change of charge dq/dt (flow of electrons) at a cross section of a conductor measured in amperes (abbreviated A); E: electric (potential) source measured in volts (abbreviated V); V: difference in potential between two points along the conductor measured in volts (V). R, Resistor
Voltage E source
L, Inductor
C, Capacitor
FIGURE 16.7 An electric circuit.
Ohm observed that the current I flowing through a resistor, caused by a potential difference across it, is (approximately) proportional to the potential difference (voltage drop). He named his constant of proportionality 1>R and called R the resistance. So Ohmâ€™s law is
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I =
1 V. R
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.3
Applications
1621
L
dI dt
and
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Similarly, it is known from physics that the voltage drops across an inductor and a capacitor are q , C
You
where L is the inductance and C is the capacitance (with q the charge on the capacitor). The German physicist Gustav R. Kirchhoff (1824–1887) formulated the law that the sum of the voltage drops in a closed circuit is equal to the supplied voltage E(t). Symbolically, this says that q dI + RI + L = E(t). C dt
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Since I = dq>dt, Kirchhoff’s law becomes d 2q dq 1 L 2 + R + q = E(t). C dt dt
(8)
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The secondorder differential equation (8), which models an electric circuit, has exactly the same form as Equation (7) modeling vibratory motion. Both models can be solved using the methods developed in Section 16.2.
Summary
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The following chart summarizes our analogies for the physics of motion of an object in a spring system versus the flow of charged particles in an electrical circuit.
Linear SecondOrder ConstantCoefficient Models
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Mechanical System
Electrical System
Lq– + Rq¿ +
y: y¿ : y– : m: d: k: F(t):
q: q¿ : q– : L: R: 1> C: E(t):
displacement velocity acceleration mass damping constant spring constant forcing function
ham
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my– + dy¿ + ky = F(t)
1 q = E(t) C
charge current change in current inductance resistance where C is the capacitance voltage source
EXERCISES 16.3
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1. A 16lb weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 1 lb> ft. The resistance in the springmass system is numerically equal to the instantaneous velocity. At t = 0 the weight is set in motion from a position 2 ft below its equilibrium position by giving it a downward velocity of 2 ft> sec. Write an initial value problem that models the given situation.
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2. An 8lb weight stretches a spring 4 ft. The spring–mass system resides in a medium offering a resistance to the motion that is numerically equal to 1.5 times the instantaneous velocity. If the weight is released at a position 2 ft above its equilibrium position with a downward velocity of 3 ft> sec, write an initial value problem modeling the given situation.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
5. An (open) electrical circuit consists of an inductor, a resistor, and a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is present and a voltage of Estd = 20 cos t is applied. In this circuit the voltage drop across the resistor is 4 times the instantaneous change in the charge, the voltage drop across the capacitor is 10 times the charge, and the voltage drop across the inductor is 2 times the instantaneous change in the current. Write an initial value problem to model the circuit.
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11. A 10lb weight is suspended by a spring that is stretched 2 in. by the weight. Assume a resistance whose magnitude is 40> 1g lb times the instantaneous velocity in feet per second. If the weight is pulled down 3 in. below its equilibrium position and released, find the time required to reach the equilibrium position for the first time. 12. A weight stretches a spring 6 in. It is set in motion at a point 2 in. below its equilibrium position with a downward velocity of 2 in.> sec. a. When does the weight return to its starting position? b. When does it reach its highest point?
c. Show that the maximum velocity is 212g + 1 in.> sec.
13. A weight of 10 lb stretches a spring 10 in. The weight is drawn down 2 in. below its equilibrium position and given an initial velocity of 4 in.> sec. An identical spring has a different weight attached to it. This second weight is drawn down from its equilibrium position a distance equal to the amplitude of the first motion and then given an initial velocity of 2 ft> sec. If the amplitude of the second motion is twice that of the first, what weight is attached to the second spring?
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6. An inductor of 2 henrys is connected in series with a resistor of 12 ohms, a capacitor of 1> 16 farad, and a 300 volt battery. Initially, the charge on the capacitor is zero and the current is zero. Formulate an initial value problem modeling this electrical circuit.
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4. A 10lb weight is suspended by a spring that is stretched 2 in. by the weight. Assume a resistance whose magnitude is 20> 1g lb times the instantaneous velocity y in feet per second. If the weight is pulled down 3 in. below its equilibrium position and released, formulate an initial value problem modeling the behavior of the spring–mass system.
10. A mass of 1 slug is attached to a spring whose constant is 25> 4 lb> ft. Initially the mass is released 1 ft above the equilibrium position with a downward velocity of 3 ft> sec, and the subsequent motion takes place in a medium that offers a damping force numerically equal to 3 times the instantaneous velocity. An external force ƒ(t) is driving the system, but assume that initially ƒ(t) K 0. Formulate and solve an initial value problem that models the given system. Interpret your results.
iaz
3. A 20lb weight is hung on an 18in. spring and stretches it 6 in. The weight is pulled down 5 in. and 5 lb are added to the weight. If the weight is now released with a downward velocity of y0 in.> sec, write an initial value problem modeling the vertical displacement.
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1622
Mechanical units in the British and metric systems may be helpful in doing the following problems. British System
MKS System
Distance Mass Time Force g(earth)
Feet (ft) Slugs Seconds (sec) Pounds (lb) 32 ft> sec2
Meters (m) Kilograms (kg) Seconds (sec) Newtons (N) 9.81 m> sec2
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Unit
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7. A 16lb weight is attached to the lower end of a coil spring suspended from the ceiling and having a spring constant of 1 lb> ft. The resistance in the spring–mass system is numerically equal to the instantaneous velocity. At t = 0 the weight is set in motion from a position 2 ft below its equilibrium position by giving it a downward velocity of 2 ft> sec. At the end of p sec, determine whether the mass is above or below the equilibrium position and by what distance.
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8. An 8lb weight stretches a spring 4 ft. The spring–mass system resides in a medium offering a resistance to the motion equal to 1.5 times the instantaneous velocity. If the weight is released at a position 2 ft above its equilibrium position with a downward velocity of 3 ft> sec, find its position relative to the equilibrium position 2 sec later. 9. A 20lb weight is hung on an 18in. spring stretching it 6 in. The weight is pulled down 5 in. and 5 lb are added to the weight. If the weight is now released with a downward velocity of y0 in.> sec, find the position of mass relative to the equilibrium in terms of y0 and valid for any time t Ú 0.
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14. A weight stretches one spring 3 in. and a second weight stretches another spring 9 in. If both weights are simultaneously pulled down 1 in. below their respective equilibrium positions and then released, find the first time after t = 0 when their velocities are equal. 15. A weight of 16 lb stretches a spring 4 ft. The weight is pulled down 5 ft below the equilibrium position and then released. What initial velocity y0 given to the weight would have the effect of doubling the amplitude of the vibration? 16. A mass weighing 8 lb stretches a spring 3 in. The spring–mass system resides in a medium with a damping constant of 2 lbsec> ft. If the mass is released from its equilibrium position with a velocity of 4 in.> sec in the downward direction, find the time required for the mass to return to its equilibrium position for the first time. 17. A weight suspended from a spring executes damped vibrations with a period of 2 sec. If the damping factor decreases by 90% in 10 sec, find the acceleration of the weight when it is 3 in. below its equilibrium position and is moving upward with a speed of 2 ft> sec. 18. A 10lb weight stretches a spring 2 ft. If the weight is pulled down 6 in. below its equilibrium position and released, find the highest point reached by the weight. Assume the spring–mass system resides in a medium offering a resistance of 10> 1g lb times the instantaneous velocity in feet per second.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4 Euler Equations 19. An LRC circuit is set up with an inductance of 1> 5 henry, a resistance of 1 ohm, and a capacitance of 5> 6 farad. Assuming the initial charge is 2 coulombs and the initial current is 4 amperes, find the solution function describing the charge on the capacitor at any time. What is the charge on the capacitor after a long period of time?
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You
25. Suppose L = 10 henrys, R = 10 ohms, C = 1>500 farads, E = 100 volts, q(0) = 10 coulombs, and q¿(0) = i(0) = 0. Formulate and solve an initial value problem that models the given LRC circuit. Interpret your results. 26. A series circuit consisting of an inductor, a resistor, and a capacitor is open. There is an initial charge of 2 coulombs on the capacitor, and 3 amperes of current is present in the circuit at the instant the circuit is closed. A voltage given by E(t) = 20 cos t is applied. In this circuit the voltage drops are numerically equal to the following: across the resistor to 4 times the instantaneous change in the charge, across the capacitor to 10 times the charge, and across the inductor to 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. Determine the charge on the capacitor and the current at time t = 10.
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22. A 10kg mass is attached to a spring having a spring constant of 140 N> m. The mass is started in motion from the equilibrium position with an initial velocity of 1 m> sec in the upward direction and with an applied external force given by ƒ(t) = 5 sin t (in newtons). The mass is in a viscous medium with a coefficient of resistance equal to 90 Nsec> m. Formulate an initial value problem that models the given system; solve the model and interpret the results.
24. An 8lb weight stretches a spring 4 ft. The spring–mass system resides in a medium offering a resistance to the motion equal to 1.5 times the instantaneous velocity, and an external force given by ƒ(t) = 6 + e  t (in pounds) is being applied. If the weight is released at a position 2 ft above its equilibrium position with downward velocity of 3 ft> sec, find its position relative to the equilibrium after 2 sec have elapsed.
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21. A 16lb weight stretches a spring 4 ft. This spring–mass system is in a medium with a damping constant of 4.5 lbsec> ft, and an external force given by ƒ(t) = 4 + e  2t (in pounds) is being applied. What is the solution function describing the position of the mass at any time if the mass is released from 2 ft below the equilibrium position with an initial velocity of 4 ft> sec downward?
position thereby stretching the spring 1.96 m. The mass is in a viscous medium that offers a resistance in newtons numerically equal to 4 times the instantaneous velocity measured in meters per second. The mass is then pulled down 2 m below its equilibrium position and released with a downward velocity of 3 m> sec. At this same instant an external force given by ƒ(t) = 20 cos t (in newtons) is applied to the system. At the end of p sec determine if the mass is above or below its equilibrium position and by how much.
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20. An (open) electrical circuit consists of an inductor, a resistor, and a capacitor. There is an initial charge of 2 coulombs on the capacitor. At the instant the circuit is closed, a current of 3 amperes is present but no external voltage is being applied. In this circuit the voltage drops at three points are numerically related as follows: across the capacitor, 10 times the charge; across the resistor, 4 times the instantaneous change in the charge; and across the inductor, 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time.
1623
Euler Equations
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16.4
dH
23. A 2kg mass is attached to the lower end of a coil spring suspended from the ceiling. The mass comes to rest in its equilibrium
Mu
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In Section 16.1 we introduced the secondorder linear homogeneous differential equation P(x)y–(x) + Q(x)y¿(x) + R(x)y(x) = 0
and showed how to solve this equation when the coefficients P, Q, and R are constants. If the coefficients are not constant, we cannot generally solve this differential equation in terms of elementary functions we have studied in calculus. In this section you will learn how to solve the equation when the coefficients have the special forms P(x) = ax 2,
Q(x) = bx,
and
R(x) = c,
where a, b, and c are constants. These special types of equations are called Euler equations, in honor of Leonhard Euler who studied them and showed how to solve them. Such equations arise in the study of mechanical vibrations.
The General Solution of Euler Equations Consider the Euler equation ax 2y– + bxy¿ + cy = 0,
To Read it Online & Download:
x 7 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1624
Chapter 16: SecondOrder Differential Equations
and
y(x) = Y(z).
We next use the chain rule to find the derivatives y¿(x) and y–(x): d dz d 1 Y(z) = Y(z) = Y ¿(z) x dx dz dx
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y¿(x) =
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z = ln x
i
To solve Equation (1), we first make the change of variables
and
d d dz 1 1 1 1 1 y¿(x) = Y¿(z) x =  2 Y ¿(z) + x Y –(z) =  2 Y ¿(z) + 2 Y –(z). dx dx dx x x x Substituting these two derivatives into the lefthand side of Equation (1), we find y–(x) =
1 1 1 Y ¿(z) + 2 Y –(z)b + bx a x Y ¿(z)b + cY(z) 2 x x
iaz
ax 2y– + bxy¿ + cy = ax 2 a
= aY –(z) + (b  a)Y ¿(z) + cY(z).
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Therefore, the substitutions give us the secondorder linear differential equation with constant coefficients aY –(z) + (b  a)Y ¿(z) + cY(z) = 0.
(2)
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We can solve Equation (2) using the method of Section 16.1. That is, we find the roots to the associated auxiliary equation ar 2 + (b  a)r + c = 0
(3)
to find the general solution for Y(z). After finding Y(z), we can determine y(x) from the substitution z = ln x. Find the general solution of the equation x 2 y– + 2xy¿  2y = 0.
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EXAMPLE 1
Solution This is an Euler equation with a = 1, b = 2, and c = 2. The auxiliary equa
tion (3) for Y(z) is
r 2 + (2  1)r  2 = (r  1)(r + 2) = 0,
ma
with roots r = 2 and r = 1. The solution for Y(z) is given by Y(z) = c1e  2z + c2e z.
Mu
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Substituting z = ln x gives the general solution for y(x):
EXAMPLE 2
y(x) = c1e 2 ln x + c2e ln x = c1 x 2 + c2 x. Solve the Euler equation x 2y–  5xy¿ + 9y = 0.
Solution Since a = 1, b = 5, and c = 9, the auxiliary equation (3) for Y(z) is
r 2 + (5  1)r + 9 = (r  3) 2 = 0. The auxiliary equation has the double root r = 3 giving Y(z) = c1e 3z + c2 ze 3z. Substituting z = ln x into this expression gives the general solution y(x) = c1e 3 ln x + c2 ln x e 3 ln x = c1 x 3 + c2 x 3 ln x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.4 Euler Equations
1625
Find the particular solution to x 2y–  3xy¿ + 68y = 0 that satisfies the initial conditions y(1) = 0 and y¿(1) = 1.
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EXAMPLE 3
Solution Here a = 1, b = 3, and c = 68 substituted into the auxiliary equation (3)
gives
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r 2  4r + 68 = 0.
The roots are r = 2 + 8i and r = 2  8i giving the solution
Y(z) = e 2z(c1 cos 8z + c2 sin 8z). Substituting z = ln x into this expression gives
iaz
y(x) = e 2 ln x A c1 cos (8 ln x) + c2 sin (8 ln x) B.
From the initial condition y(1) = 0, we see that c1 = 0 and y(x) = c2 x 2 sin (8 ln x).
y
To fit the second initial condition, we need the derivative
10
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y¿(x) = c2 A 8x cos (8 ln x) + 2x sin (8 ln x) B .
5
2
4
6
8
10
x
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0
Since y¿(1) = 1, we immediately obtain c2 = 1>8. Therefore, the particular solution satisfying both initial conditions is y(x) =
y = 18 x 2 sin (8 ln x)
–5
1 2 x sin (8 ln x). 8
Since 1 … sin (8 ln x) … 1, the solution satisfies –10

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Graph of the solution to
A graph of the solution is shown in Figure 16.8.
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FIGURE 16.8 Example 3.
x2 x2 … y(x) … . 8 8
16.4 EXERCISES
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In Exercises 1–24, find the general solution to the given Euler equation. Assume x 7 0 throughout. 1. x 2y– + 2xy¿  2y = 0 2
2. x 2y– + xy¿  4y = 0 2
3. x y–  6y = 0
4. x y– + xy¿  y = 0
5. x 2y–  5xy¿ + 8y = 0
6. 2x 2y– + 7xy¿ + 2y = 0
7. 3x 2y– + 4xy¿ = 0
8. x 2y– + 6xy¿ + 4y = 0
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9. x 2y–  xy¿ + y = 0
10. x 2y–  xy¿ + 2y = 0
11. x 2y–  xy¿ + 5y = 0
12. x 2y– + 7xy¿ + 13y = 0
13. x 2y– + 3xy¿ + 10y = 0
14. x 2y–  5xy¿ + 10y = 0
15. 4x 2y– + 8xy¿ + 5y = 0
16. 4x 2y–  4xy¿ + 5y = 0
17. x 2y– + 3xy¿ + y = 0
18. x 2y–  3xy¿ + 9y = 0
19. x 2y– + xy¿ = 0
20. 4x 2y– + y = 0
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21. 9x 2y– + 15xy¿ + y = 0 22. 16x 2y–  8xy¿ + 9y = 0 23. 16x 2y– + 56xy¿ + 25y = 0 24. 4x 2y–  16xy¿ + 25y = 0 In Exercises 25–30, solve the given initial value problem. 25. x 2y– + 3xy¿  3y = 0,
y(1) = 1, y¿(1) = 1
26. 6x 2y– + 7xy¿  2y = 0,
y(1) = 0, y¿(1) = 1
2
27. x y–  xy¿ + y = 0,
y(1) = 1, y¿(1) = 1
28. x 2y– + 7xy¿ + 9y = 0, 2
y(1) = 1, y¿(1) = 0
29. x y–  xy¿ + 2y = 0,
y(1) = 1, y¿(1) = 1
30. x 2y– + 3xy¿ + 5y = 0,
y(1) = 1, y¿(1) = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
PowerSeries Solutions
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16.5
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1626
You
In this section we extend our study of secondorder linear homogeneous equations with variable coefficients. With the Euler equations in Section 16.4, the power of the variable x in the nonconstant coefficient had to match the order of the derivative with which it was paired: x 2 with y– , x 1 with y¿ , and x 0 (=1) with y. Here we drop that requirement so we can solve more general equations.
Method of Solution
The powerseries method for solving a secondorder homogeneous differential equation consists of finding the coefficients of a power series
iaz
q
y(x) = a cn x n = c0 + c1x + c2 x 2 + Á n=0
(1)
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which solves the equation. To apply the method we substitute the series and its derivatives into the differential equation to determine the coefficients c0, c1, c2, Á . The technique for finding the coefficients is similar to that used in the method of undetermined coefficients presented in Section 16.2. In our first example we demonstrate the method in the setting of a simple equation whose general solution we already know. This is to help you become more comfortable with solutions expressed in series form.
EXAMPLE 1
Solve the equation y– + y = 0 by the powerseries method.
Solution We assume the series solution takes the form of q
y = a cn x n n=0
dH
and calculate the derivatives q
q
y¿ = a ncn x n  1
and
n=1
y– = a n(n  1)cn x n  2. n=2
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Substitution of these forms into the secondorder equation gives us q
q
n2 + a cn x n = 0. a n(n  1)cn x
n=2
n=0
Mu
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Next, we equate the coefficients of each power of x to zero as summarized in the following table.
Power of x
Coefficient Equation
x0
2(1)c2 + c0 = 0
or
x1
3(2)c3 + c1 = 0
or
x2
4(3)c4 + c2 = 0
or
x3
5(4)c5 + c3 = 0
or
x4
6(5)c6 + c4 = 0
or
o n(n  1)cn + cn  2 = 0
or
o xn2
To Read it Online & Download:
1 c2 =  c0 2 1 c3 =  # c1 3 2 1 c4 =  # c2 4 3 1 c5 =  # c3 5 4 1 c6 =  # c4 6 5 o 1 c cn = n(n  1) n  2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5
PowerSeries Solutions
1627
2k(2k  1)c2k + c2k  2 = 0
c2k = 
You
or
1 c . 2k(2k  1) 2k  2
From this recursive relation we find
1 1 1 1 d cd Á cd c dc0 2 2k(2k  1) (2k  2)(2k  3) 4(3)
iaz
c2k = c
(1) k c. (2k)! 0
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=
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i
From the table we notice that the coefficients with even indices (n = 2k, k = 1, 2, 3, Á ) are related to each other and the coefficients with odd indices (n = 2k + 1) are also interrelated. We treat each group in turn. Even indices: Here n = 2k , so the power is x 2k  2. From the last line of the table, we have
Odd indices: Here n = 2k + 1, so the power is x 2k  1. Substituting this into the last line of the table yields
or
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(2k + 1)(2k)c2k + 1 + c2k  1 = 0
c2k + 1 = 
dH
Thus,
c2k + 1 = c
ma
=
1 . c (2k + 1)(2k) 2k  1
1 1 1 1 d cd Á cd cdc (2k + 1)(2k) (2k  1)(2k  2) 5(4) 3(2) 1
(1) k c. (2k + 1)! 1
Mu
ham
Writing the power series by grouping its even and odd powers together and substituting for the coefficients yields q
y = a cn x n n=0 q
q
= a c2k x 2k + a c2k + 1 x 2k + 1 k=0
k=0
q q (1) k 2k (1) k = c0 a x + c1 a x 2k + 1. (2k)! (2k + 1)! k=0 k=0
From Table 8.1 in Section 8.10, we see that the first series on the righthand side of the last equation represents the cosine function and the second series represents the sine. Thus, the general solution to y– + y = 0 is y = c0 cos x + c1 sin x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 16: SecondOrder Differential Equations
EXAMPLE 2
Find the general solution to y– + xy¿ + y = 0.
i
1628
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Solution We assume the series solution form q
y = a cn x n n=0
You
and calculate the derivatives q
q
y¿ = a ncn x n  1
y– = a n(n  1)cn x n  2.
and
n=2
n=1
Substitution of these forms into the secondorder equation yields q
q
q
n2 + a ncn x n + a cn x n = 0. a n(n  1)cn x n=1
n=0
iaz
n=2
We equate the coefficients of each power of x to zero as summarized in the following table.
xn
2(1)c2 3(2)c3 4(3)c4 5(4)c5 6(5)c6
ass a
x0 x1 x2 x3 x4 o
Coefficient Equation
nR
Power of x
+ + + +
c1 2c2 3c3 4c4
+ + + + +
c0 c1 c2 c3 c4
= = = = = o
0 0 0 0 0
(n + 2)(n + 1)cn + 2 + (n + 1)cn = 0
or or or or or
c2 c3 c4 c5 c6
= = = = = o
 12 c0  13 c1  14 c2  15 c3  16 c4
or cn + 2 = 
1 c n + 2 n
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From the table notice that the coefficients with even indices are interrelated and the coefficients with odd indices are also interrelated. Even indices: Here n = 2k  2, so the power is x 2k  2. From the last line in the table, we have c2k = 
1 c . 2k 2k  2
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From this recurrence relation we obtain c2k = a=
1 1 1 1 1 b ab Á a b a b a bc0 6 4 2 2k 2k  2
(1) k c. (2)(4)(6) Á (2k) 0
Odd indices: Here n = 2k  1, so the power is x 2k  1. From the last line in the table, we have c2k + 1 = 
1 . c 2k + 1 2k  1
From this recurrence relation we obtain c2k + 1 = a=
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1 1 1 1 b ab Á a b a bc1 5 3 2k + 1 2k  1
(1) k c. (3)(5) Á (2k + 1) 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5
PowerSeries Solutions
1629
q
suf
i
Writing the power series by grouping its even and odd powers and substituting for the coefficients yields q
y = a c2k x 2k + a c2k + 1 x 2k + 1 k=0
k=0
q (1) k (1) k 2k 2k + 1 x = c0 a + c . 1 a Á (2k) Á (2k + 1) x k = 0 (2)(4) k = 0 (3)(5)
EXAMPLE 3
Find the general solution to
You
q
(1  x 2)y–  6xy¿  4y = 0,
iaz
x 6 1.
Solution Notice that the leading coefficient is zero when x = ;1. Thus, we assume the
solution interval I: 1 6 x 6 1. Substitution of the series form q
y = a cn x n
nR
n=0
and its derivatives gives us
q
q
q
(1  x 2) a n(n  1)cn x n  2  6 a ncn x n  4 a cn x n = 0,
ass a
n=2
q
n=1
n=0
q
q
q
n2  a n(n  1)cn x n  6 a ncn x n  4 a cn x n = 0. a n(n  1)cn x
n=2
n=2
n=1
n=0
dH
Next, we equate the coefficients of each power of x to zero as summarized in the following table. Power of x
2(1)c2 3(2)c3  6(1)c1 4(3)c4  2(1)c2  6(2)c2 5(4)c5  3(2)c3  6(3)c3 
(n + 2)(n + 1)cn + 2
4c0 4c1 4c2 4c3
= = = = o  [n(n  1) + 6n + 4]cn =
0 0 0 0
or or or or
c2 c3 c4 c5
= = = = o
4 2 c0 5 3 c1 6 4 c2 7 5 c3
0
(n + 2)(n + 1)cn + 2  (n + 4)(n + 1)cn = 0
or cn + 2 =
n + 4 c n + 2 n
Again we notice that the coefficients with even indices are interrelated and those with odd indices are interrelated. Even indices: Here n = 2k  2, so the power is x 2k. From the righthand column and last line of the table, we get
Mu
ham
ma
x0 x1 x2 x3 o xn
Coefficient Equation
c2k =
2k + 2 c2k  2 2k
= a
2k + 2 2k 2k  2 Á 6 4 ba ba b a bc 4 2 0 2k 2k  2 2k  4
= (k + 1)c0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 1630
Chapter 16: SecondOrder Differential Equations
2k + 3 c 2k + 1 2k  1
= a =
2k + 3 2k + 1 2k  1 Á 7 5 a bc ba ba b 5 3 1 2k + 1 2k  1 2k  3
You
c2k + 1 =
suf
i
Odd indices: Here n = 2k  1, so the power is x 2k + 1. The righthand column and last line of the table gives us
2k + 3 c1. 3
The general solution is
n=0 q
iaz
q
y = a cn x n
q
= a c2k x 2k + a c2k + 1 x 2k + 1 k=0
nR
k=0
q
q
= c0 a (k + 1)x 2k + c1 a k=0
Find the general solution to y–  2xy¿ + y = 0.
ass a
EXAMPLE 4
k=0
2k + 3 2k + 1 . x 3
Solution Assuming that
q
y = a cn x n, n=0
dH
substitution into the differential equation gives us q
q
q
n2  2 a ncn x n + a cn x n = 0. a n(n  1)cn x
n=2
n=1
n=0
ma
We next determine the coefficients, listing them in the following table.
Mu
ham
Power of x
Coefficient Equation
x0
2(1)c2
+ c0 = 0
or
x1
3(2)c3  2c1 + c1 = 0
or
x2
4(3)c4  4c2 + c2 = 0
or
x3
5(4)c5  6c3 + c3 = 0
or
x4
6(5)c6  8c4 + c4 = 0
or
o xn
o (n + 2)(n + 1)cn + 2  (2n  1)cn = 0
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1 c2 =  c0 2 1 c3 = # c1 3 2 3 c4 = # c2 4 3 5 c5 = # c3 5 4 7 c6 = # c4 6 5 o
or
cn + 2 =
2n  1 c (n + 2)(n + 1) n
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 16.5
PowerSeries Solutions
1631
2n  1 c, (n + 2)(n + 1) n
suf
cn + 2 =
i
From the recursive relation
we write out the first few terms of each series for the general solution: 3 4 1 2 21 6 x x x  Áb 2 4! 6!
In Exercises 1–18, use power series to find the general solution of the differential equation. 1. y– + 2y¿ = 0
9. (x 2  1)y– + 2xy¿  2y = 0
nR
EXERCISES 16.5
5 5 45 7 1 3 x + x + x + Á b. 3! 5! 7!
iaz
+ c1 ax +
You
y = c0 a1 
10. y– + y¿  x 2y = 0 11. (x 2  1)y–  6y = 0
2. y– + 2y¿ + y = 0
12. xy–  (x + 2)y¿ + 2y = 0
3. y– + 4y = 0 5. x 2y–  2xy¿ + 2y = 0 6. y–  xy¿ + y = 0
ass a
13. (x 2  1)y– + 4xy¿ + 2y = 0
4. y–  3y¿ + 2y = 0
14. y–  2xy¿ + 4y = 0 15. y–  2xy¿ + 3y = 0 16. (1  x 2)y–  xy¿ + 4y = 0
7. (1 + x)y–  y = 0
17. y–  xy¿ + 3y = 0 18. x 2y–  4xy¿ + 6y = 0
Mu
ham
ma
dH
8. (1  x 2)y–  4xy¿ + 6y = 0
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suf
i
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Mathematical Induction Many formulas, like
nR
A.1
iaz
APPENDICES
nsn + 1d 1 + 2 +Á+ n = , 2
Check that the formula holds for n = 1. Prove that if the formula holds for any positive integer n = k, then it also holds for the next integer, n = k + 1.
dH
1. 2.
ass a
can be shown to hold for every positive integer n by applying an axiom called the mathematical induction principle. A proof that uses this axiom is called a proof by mathematical induction or a proof by induction. The steps in proving a formula by induction are the following:
Mu
ham
ma
The induction axiom says that once these steps are completed, the formula holds for all positive integers n. By Step 1 it holds for n = 1. By Step 2 it holds for n = 2, and therefore by Step 2 also for n = 3, and by Step 2 again for n = 4, and so on. If the first domino falls, and the kth domino always knocks over the sk + 1dst when it falls, all the dominoes fall. From another point of view, suppose we have a sequence of statements S1, S2 , Á , Sn , Á , one for each positive integer. Suppose we can show that assuming any one of the statements to be true implies that the next statement in line is true. Suppose that we can also show that S1 is true. Then we may conclude that the statements are true from S1 on.
EXAMPLE 1
Use mathematical induction to prove that for every positive integer n, nsn + 1d 1 + 2 +Á+ n = . 2
Solution
1.
We accomplish the proof by carrying out the two steps above.
The formula holds for n = 1 because 1 =
1s1 + 1d . 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP2
Appendices
If the formula holds for n = k, does it also hold for n = k + 1? The answer is yes, as we now show. If ksk + 1d 1 + 2 +Á+ k = , 2
You
then
suf
i
2.
ksk + 1d k 2 + k + 2k + 2 1 + 2 + Á + k + sk + 1d = + sk + 1d = 2 2 =
sk + 1dssk + 1d + 1d sk + 1dsk + 2d = . 2 2
iaz
The last expression in this string of equalities is the expression nsn + 1d>2 for n = sk + 1d.
nR
The mathematical induction principle now guarantees the original formula for all positive integers n. In Example 4 of Section 5.2 we gave another proof for the formula giving the sum of the first n integers. However, proof by mathematical induction is more general. It can be used to find the sums of the squares and cubes of the first n integers (Exercises 9 and 10). Here is another example. Show by mathematical induction that for all positive integers n,
ass a
EXAMPLE 2
1 1 1 1 + 2 + Á + n = 1  n. 1 2 2 2 2
Solution
We accomplish the proof by carrying out the two steps of mathematical
dH
induction. 1.
1 1 = 1  1. 1 2 2
If
ma
2.
The formula holds for n = 1 because
1 1 1 1 + 2 + Á + k = 1  k, 1 2 2 2 2
Mu
ham
then
1 1 1 1 1 1 1#2 1 + 2 + Á + k + k+1 = 1  k + k+1 = 1  k + k+1 1 # 2 2 2 2 2 2 2 2 2 2 1 1 = 1  k+1 + k+1 = 1  k+1 . 2 2 2
Thus, the original formula holds for n = sk + 1d whenever it holds for n = k. With these steps verified, the mathematical induction principle now guarantees the formula for every positive integer n.
Other Starting Integers Instead of starting at n = 1 some induction arguments start at another integer. The steps for such an argument are as follows.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.1
AP3
i
Check that the formula holds for n = n1 (the first appropriate integer). Prove that if the formula holds for any integer n = k Ú n1, then it also holds for n = sk + 1d.
suf
1. 2.
Mathematical Induction
EXAMPLE 3
Show that n! 7 3n if n is large enough.
How large is large enough? We experiment: n n! 3n
1 1 3
2 2 9
3 6 27
4 24 81
5 120 243
iaz
Solution
You
Once these steps are completed, the mathematical induction principle guarantees the formula for all n Ú n1.
6 720 729
7 5040 2187
nR
It looks as if n! 7 3n for n Ú 7. To be sure, we apply mathematical induction. We take n1 = 7 in Step 1 and complete Step 2. Suppose k! 7 3k for some k Ú 7. Then sk + 1d! = sk + 1dsk!d 7 sk + 1d3k 7 7 # 3k 7 3k + 1. Thus, for k Ú 7,
ass a
k! 7 3k
implies
sk + 1d! 7 3k + 1.
EXERCISES A.1
dH
The mathematical induction principle now guarantees n! Ú 3n for all n Ú 7.
1. Assuming that the triangle inequality ƒ a + b ƒ … ƒ a ƒ + ƒ b ƒ holds for any two numbers a and b, show that
5. Show that 2 2 2 1 + 2 +Á+ n = 1  n 3 3 31 3
for any n numbers.
ma
ƒ x1 + x2 + Á + xn ƒ … ƒ x1 ƒ + ƒ x2 ƒ + Á + ƒ xn ƒ
for all positive integers n. 6. Show that n! 7 n 3 if n is large enough.
2. Show that if r Z 1, then
7. Show that 2n 7 n 2 if n is large enough.
1  rn+1 1 + r + r2 + Á + rn = 1  r
ham
8. Show that 2n Ú 1>8 for n Ú 3 . 9. Sums of squares Show that the sum of the squares of the first n positive integers is
for every positive integer n.
Mu
dy du d suyd = u + y , and the fact that 3. Use the Product Rule, dx dx dx d d n sxd = 1 to show that sx d = nx n  1 for every positive dx dx integer n. 4. Suppose that a function ƒ(x) has the property that ƒsx1 x2 d = ƒsx1 d + ƒsx2 d for any two positive numbers x1 and x2 . Show that ƒsx1 x2 Á xn d = ƒsx1 d + ƒsx2 d + Á + ƒsxn d
for the product of any n positive numbers x1, x2 Á , xn .
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n an +
1 bsn + 1d 2 . 3
10. Sums of cubes Show that the sum of the cubes of the first n positive integers is snsn + 1d>2d2 . 11. Rules for finite sums Show that the following finite sum rules hold for every positive integer n. n
n
n
a. a sak + bk d = a ak + a bk k=1
k=1
k=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP4
Appendices n
n
n
n
k=1
(if ak has the constant value c)
k=1
n
c. a cak = c # a ak
12. Show that ƒ x n ƒ = ƒ x ƒ n for every positive integer n and every real number x.
(Any number c)
k=1
Mu
ham
ma
dH
ass a
nR
iaz
You
k=1
k=1
suf
k=1
d. a ak = n # c
i
n
b. a sak  bk d = a ak  a bk
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Appendices
Proofs of Limit Theorems
fi
A.2
usu
AP4
THEOREM 1 Limit Laws If L, M, c, and k are real numbers and lim ƒsxd = L
x:c
and
2.
Difference Rule:
3.
Product Rule:
4.
Constant Multiple Rule:
x:c
lim A ƒsxd  gsxd B = L  M
x:c
nR
lim A ƒsxd # gsxd B = L # M
x:c
lim A kƒsxd B = kL
x:c
ƒsxd L , = M gsxd
Ha ssa
6.
Power Rule:
then
iaz
Sum Rule:
Quotient Rule:
lim gsxd = M,
x:c
lim A ƒsxd + gsxd B = L + M
1.
5.
Yo
This appendix proves Theorem 1, Parts 2–5, and Theorem 4 from Section 2.2.
lim
x:c
sany number kd if M Z 0
If r and s are integers with no common factor and s Z 0, then lim A ƒ(x) B r>s = L r>s
x:c
ad
provided that L r>s is a real number. (If s is even, we assume that L 7 0.)
Mu
ha
mm
We proved the Sum Rule in Section 2.3 and the Power Rule is proved in more advanced texts. We obtain the Difference Rule by replacing gsxd by gsxd and M by M in the Sum Rule. The Constant Multiple Rule is the special case gsxd = k of the Product Rule. This leaves only the Product and Quotient Rules.
Proof of the Limit Product Rule We show that for any P 7 0 there exists a d 7 0 such that for all x in the intersection D of the domains of ƒ and g, 0 6 ƒx  cƒ 6 d
Q
ƒ ƒsxdgsxd  LM ƒ 6 P.
Suppose then that P is a positive number, and write ƒ(x) and g(x) as ƒsxd = L + sƒsxd  Ld,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.2
Proofs of Limit Theorems
AP5
suf
ƒsxd # gsxd  LM = sL + sƒsxd  LddsM + sgsxd  Mdd  LM
i
Multiply these expressions together and subtract LM:
= LM + Lsgsxd  Md + Msƒsxd  Ld + sƒsxd  Ldsgsxd  Md  LM
You
= Lsgsxd  Md + Msƒsxd  Ld + sƒsxd  Ldsgsxd  Md. (1) Since ƒ and g have limits L and M as x : c, there exist positive numbers d1, d2, d3, and d4 such that for all x in D Q
ƒ ƒsxd  L ƒ 6 2P>3
0 6 ƒ x  c ƒ 6 d2
Q
ƒ gsxd  M ƒ 6 2P>3
iaz
0 6 ƒ x  c ƒ 6 d1
0 6 ƒ x  c ƒ 6 d3
Q
ƒ ƒsxd  L ƒ 6 P>s3s1 + ƒ M ƒ dd
0 6 ƒ x  c ƒ 6 d4
Q
ƒ gsxd  M ƒ 6 P>s3s1 + ƒ L ƒ dd
(2)
ƒ ƒsxd # gsxd  LM ƒ
nR
If we take d to be the smallest numbers d1 through d4, the inequalities on the righthand side of the Implications (2) will hold simultaneously for 0 6 ƒ x  c ƒ 6 d. Therefore, for all x in D, 0 6 ƒ x  c ƒ 6 d implies Triangle inequality applied to Equation (1)
ass a
… ƒ L ƒ ƒ gsxd  M ƒ + ƒ M ƒ ƒ ƒsxd  L ƒ + ƒ ƒsxd  L ƒ ƒ gsxd  M ƒ … s1 + ƒ L ƒ d ƒ gsxd  M ƒ + s1 + ƒ M ƒ d ƒ ƒsxd  L ƒ + ƒ ƒsxd  L ƒ ƒ gsxd  M ƒ 6
P P P P + + = P. 3 3 A 3A 3
Values from (2)
dH
This completes the proof of the Limit Product Rule.
ma
Proof of the Limit Quotient Rule We show that limx:c s1>gsxdd = 1>M. We can then conclude that lim
x:c
ƒsxd 1 1 1 L = L# = = lim aƒsxd # b = lim ƒsxd # lim M M gsxd gsxd x:c x:c x:c g(x)
Mu
ham
by the Limit Product Rule. Let P 7 0 be given. To show that limx:c s1>gsxdd = 1>M, we need to show that there exists a d 7 0 such that for all x. 0 6 ƒx  cƒ 6 d
Q
`
1 1 ` 6 P. M gsxd
Since ƒ M ƒ 7 0, there exists a positive number d1 such that for all x 0 6 ƒ x  c ƒ 6 d1
Q
M ƒ gsxd  M ƒ 6 2 .
(3)
For any numbers A and B it can be shown that ƒ A ƒ  ƒ B ƒ … ƒ A  B ƒ and ƒ B ƒ  ƒ A ƒ … ƒ A  B ƒ , from which it follows that ƒ ƒ A ƒ  ƒ B ƒ ƒ … ƒ A  B ƒ . With A = gsxd and B = M, this becomes
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP6
Appendices
i
which can be combined with the inequality on the right in Implication (3) to get, in turn,

suf
ƒMƒ ƒ ƒ gsxd ƒ  ƒ M ƒ ƒ 6 2 ƒMƒ ƒMƒ 6 ƒ gsxd ƒ  ƒ M ƒ 6 2 2
You
3ƒ M ƒ ƒMƒ 6 ƒ gsxd ƒ 6 2 2 ƒ M ƒ 6 2 ƒ gsxd ƒ 6 3 ƒ M ƒ
3 2 1 6 6 ƒMƒ ƒ gsxd ƒ ƒ gsxd ƒ
(4)
M  gsxd 1 1 1 # 1 # ` = ` ` … ƒ M  gsxd ƒ M M gsxd Mgsxd ƒ ƒ ƒ gsxd ƒ 1 ƒMƒ
2 ƒMƒ
nR
`
iaz
Therefore, 0 6 ƒ x  c ƒ 6 d1 implies that
6
#
#
ƒ M  gsxd ƒ .
Inequality (4)
(5)
Since s1>2d ƒ M ƒ 2P 7 0, there exists a number d2 7 0 such that for all x Q
ass a
0 6 ƒ x  c ƒ 6 d2
P ƒ M  gsxd ƒ 6 2 ƒ M ƒ 2.
(6)
If we take d to be the smaller of d1 and d2, the conclusions in (5) and (6) both hold for all x such that 0 6 ƒ x  c ƒ 6 d. Combining these conclusions gives Q
`
1 1 ` 6 P. M gsxd
dH
0 6 ƒx  cƒ 6 d
This concludes the proof of the Limit Quotient Rule.
Mu
ham
ma
THEOREM 4 The Sandwich Theorem Suppose that gsxd … ƒsxd … hsxd for all x in some open interval I containing c, except possibly at x = c itself. Suppose also that limx:c gsxd = limx:c hsxd = L. Then limx:c ƒsxd = L.
Proof for RightHand Limits Suppose limx:c+ gsxd = limx:c+ hsxd = L. Then for any P 7 0 there exists a d 7 0 such that for all x the interval c 6 x 6 c + d is contained in I and the inequality implies L  P 6 gsxd 6 L + P
and
L  P 6 hsxd 6 L + P.
These inequalities combine with the inequality gsxd … ƒsxd … hsxd to give L  P 6 gsxd … ƒsxd … hsxd 6 L + P, L  P 6 ƒsxd 6 L + P,  P 6 ƒsxd  L 6 P. Therefore, for all x, the inequality c 6 x 6 c + d implies ƒ ƒsxd  L ƒ 6 P.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.2 Proofs of Limit Theorems
AP7
L  P 6 gsxd 6 L + P
suf
i
Proof for LeftHand Limits Suppose limx:c gsxd = limx:c hsxd = L. Then for any P 7 0 there exists a d 7 0 such that for all x the interval c  d 6 x 6 c is contained in I and the inequality implies and
L  P 6 hsxd 6 L + P.
You
We conclude as before that for all x, c  d 6 x 6 c implies ƒ ƒsxd  L ƒ 6 P.
Proof for TwoSided Limits If limx:c gsxd = limx:c hsxd = L, then g(x) and h(x) both approach L as x : c + and as x : c  ; so limx:c+ ƒsxd = L and limx:c ƒsxd = L. Hence limx:c ƒsxd exists and equals L.
lim
x:c
ƒsxd ƒscd = . gsxd gscd
6. Composites of continuous functions Figure A.1 gives the diagram for a proof that the composite of two continuous functions is continuous. Reconstruct the proof from the diagram. The statement to be proved is this: If ƒ is continuous at x = c and g is continuous at ƒ(c), then g ƒ is continuous at c. Assume that c is an interior point of the domain of ƒ and that ƒ(c) is an interior point of the domain of g. This will make the limits involved twosided. (The arguments for the cases that involve onesided limits are similar.)
ass a
2. Use mathematical induction and the Limit Product Rule in Theorem 1 to show that if functions ƒ1sxd, ƒ2sxd, Á , ƒnsxd have limits L1, L2 , Á , Ln as x : c , then
5. Limits of rational functions Use Theorem 1 and the result of Exercise 4 to show that if ƒ(x) and g(x) are polynomial functions and gscd Z 0 , then
nR
1. Suppose that functions ƒ1sxd, ƒ2sxd, and ƒ3sxd have limits L1, L2 , and L3 , respectively, as x : c . Show that their sum has limit L1 + L2 + L3 . Use mathematical induction (Appendix 1) to generalize this result to the sum of any finite number of functions.
iaz
EXERCISES A.2
lim ƒ1sxdƒ2sxd # Á # ƒnsxd = L1 # L2 # Á # Ln .
x: c
3. Use the fact that limx:c x = c and the result of Exercise 2 to show that limx:c x n = c n for any integer n 7 1.
dH
4. Limits of polynomials Use the fact that limx:cskd = k for any number k together with the results of Exercises 1 and 3 to show that limx:c ƒsxd = ƒscd for any polynomial function
ma
ƒsxd = an x n + an1 x n  1 + Á + a1 x + a0 .
ham
f
g f
f
f
c
g g
g f(c)
g( f(c))
Mu
FIGURE A.1 The diagram for a proof that the composite of two continuous functions is continuous.
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i f u s You
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
A.3 Commonly Occurring Limits
A.3
Commonly Occurring Limits
z a i an R
s s a H d a m ham
AP7
This appendix verifies limits (4)–(6) in Theorem 5 of Section 11.1. Limit 4: If ƒ x ƒ 6 1, lim x n = 0 We need to show that to each P 7 0 there corresn: ˆ ponds an integer N so large that ƒ x n ƒ 6 P for all n greater than N. Since P1>n : 1, while
Mu
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP8
Appendices
This is the integer we seek because, if ƒ x ƒ 6 1, then ƒ x n ƒ 6 ƒ x N ƒ for all n 7 N.
(1)
suf
ƒ x N ƒ = ƒ x ƒ N 6 P.
i
ƒ x ƒ 6 1, there exists an integer N for which P1>N 7 ƒ x ƒ . In other words,
(2)
x n Limit 5: For any number x, lim A 1 + n B = e x n: ˆ
You
Combining (1) and (2) produces ƒ x n ƒ 6 P for all n 7 N, concluding the proof.
Let n
iaz
x an = a1 + n b . Then
n
nR
x x ln an = ln a1 + n b = n ln a1 + n b : x, as we can see by the following application of l’Hôpital’s Rule, in which we differentiate with respect to n:
ass a
lns1 + x>nd x lim n ln a1 + n b = lim 1>n n: q n: q a
= lim
n: q
x 1 b # a 2 b 1 + x>n n x = lim = x. n: q 1 + x>n 1>n2
ma
dH
Apply Theorem 4, Section 11.1, with ƒsxd = e x to conclude that n
x a1 + n b = an = e ln an : e x .
n: ˆ
xn = 0 Since n! 
ƒ x ƒn ƒ x ƒn xn … … , n! n! n!
all we need to show is that ƒ x ƒ n>n! : 0. We can then apply the Sandwich Theorem for Sequences (Section 11.1, Theorem 2) to conclude that x n>n! : 0. The first step in showing that ƒ x ƒ n>n! : 0 is to choose an integer M 7 ƒ x ƒ , so that s ƒ x ƒ >Md 6 1. By Limit 4, just proved, we then have s ƒ x ƒ >Mdn : 0. We then restrict our attention to values of n 7 M. For these values of n, we can write
Mu
ham
Limit 6: For any number x, lim
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ƒ x ƒn ƒ x ƒn = # # Á # # n! 1 2 M sM + 1dsM + 2d # Á # n ('''''')''''''* sn  Md factors
…
n ƒ x ƒn ƒ x ƒ nM M MM ƒ x ƒ = = a b . M! M M!M n M!M n  M
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.3 Commonly Occurring Limits
AP9
n ƒ x ƒn MM ƒ x ƒ a b . … n! M! M
suf
0 …
i
Thus,
Mu
ham
ma
dH
ass a
nR
iaz
You
Now, the constant M M>M! does not change as n increases. Thus the Sandwich Theorem tells us that ƒ x ƒ n>n! : 0 because s ƒ x ƒ >Mdn : 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
A.4
AP9
i f u s u
A.4 Theory of the Real Numbers
Theory of the Real Numbers
A rigorous development of calculus is based on properties of the real numbers. Many results about functions, derivatives, and integrals would be false if stated for functions defined only on the rational numbers. In this appendix we briefly examine some basic concepts of the theory of the reals that hint at what might be learned in a deeper, more theoretical study of calculus. Three types of properties make the real numbers what they are. These are the algebraic, order, and completeness properties. The algebraic properties involve addition and multiplication, subtraction and division. They apply to rational or complex numbers as well as to the reals. The structure of numbers is built around a set with addition and multiplication operations. The following properties are required of addition and multiplication. A1 A2 A3 A4 M1 M2 M3 M4 D
m a
d a m
h u M
z ia
o Y
R n
a + sb + cd = sa + bd + c for all a, b, c. a + b = b + a for all a, b, c. There is a number called “0” such that a + 0 = a for all a. For each number a, there is a b such that a + b = 0. a(bc) = (ab)c for all a, b, c. ab = ba for all a, b. There is a number called “1” such that a # 1 = a for all a.
a ss
a H
For each nonzero a, there is a b such that ab = 1. asb + cd = ab + bc for all a, b, c.
A1 and M1 are associative laws, A2 and M2 are commutativity laws, A3 and M3 are identity laws, and D is the distributive law. Sets that have these algebraic properties are examples of fields, and are studied in depth in the area of theoretical mathematics called abstract algebra. The order properties allow us to compare the size of any two numbers. The order properties are O1 O2 O3 O4 O5
For any a and b, either a … b or b … a or both. If a … b and b … a then a = b. If a … b and b … c then a … c. If a … b then a + c … b + c. If a … b and 0 … c then ac … bc.
O3 is the transitivity law, and O4 and O5 relate ordering to addition and multiplication.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Appendices
We can order the reals, the integers, and the rational numbers, but we cannot order the complex numbers (see Appendix A.5). There is no reasonable way to decide whether a number like i = 21 is bigger or smaller than zero. A field in which the size of any two elements can be compared as above is called an ordered field. Both the rational numbers and the real numbers are ordered fields, and there are many others. We can think of real numbers geometrically, lining them up as points on a line. The completeness property says that the real numbers correspond to all points on the line, with no “holes” or “gaps.” The rationals, in contrast, omit points such as 22 and p, and the integers even leave out fractions like 1> 2. The reals, having the completeness property, omit no points. What exactly do we mean by this vague idea of missing holes? To answer this we must give a more precise description of completeness. A number M is an upper bound for a set of numbers if all numbers in the set are smaller than or equal to M. M is a least upper bound if it is the smallest upper bound. For example, M = 2 is an upper bound for the negative numbers. So is M = 1, showing that 2 is not a least upper bound. The least upper bound for the set of negative numbers is M = 0. We define a complete ordered field to be one in which every nonempty set bounded above has a least upper bound. If we work with just the rational numbers, the set of numbers less than 22 is bounded, but it does not have a rational least upper bound, since any rational upper bound M can be replaced by a slightly smaller rational number that is still larger than 22. So the rationals are not complete. In the real numbers, a set that is bounded above always has a least upper bound. The reals are a complete ordered field. The completeness property is at the heart of many results in calculus. One example occurs when searching for a maximum value for a function on a closed interval [a, b], as in Section 4.1. The function y = x  x 3 has a maximum value on [0, 1] at the point x satisfying 1  3x 2 = 0, or x = 11>3. If we limited our consideration to functions defined only on rational numbers, we would have to conclude that the function has no maximum, since 11>3 is irrational (Figure A.2). The Extreme Value Theorem (Section 4.1), which implies that continuous functions on closed intervals [a, b] have a maximum value, is not true for functions defined only on the rationals. The Intermediate Value Theorem implies that a continuous function ƒ on an interval [a, b] with ƒsad 6 0 and ƒsbd 7 0 must be zero somewhere in [a, b]. The function values cannot jump from negative to positive without there being some point x in [a, b] where ƒsxd = 0. The Intermediate Value Theorem also relies on the completeness of the real numbers and is false for continuous functions defined only on the rationals. The function ƒsxd = 3x 2  1 has ƒs0d = 1 and ƒs1d = 2, but if we consider ƒ only on the rational numbers, it never equals zero. The only value of x for which ƒsxd = 0 is x = 21>3 , an irrational number. We have captured the desired properties of the reals by saying that the real numbers are a complete ordered field. But we’re not quite finished. Greek mathematicians in the school of Pythagoras tried to impose another property on the numbers of the real line, the condition that all numbers are ratios of integers. They learned that their effort was doomed when they discovered irrational numbers such as 12. How do we know that our efforts to specify the real numbers are not also flawed, for some unseen reason? The artist Escher drew optical illusions of spiral staircases that went up and up until they rejoined themselves at the bottom. An engineer trying to build such a staircase would find that no structure realized the plans the architect had drawn. Could it be that our design for the reals contains some subtle contradiction, and that no construction of such a number system can be made? We resolve this issue by giving a specific description of the real numbers and verifying that the algebraic, order, and completeness properties are satisfied in this model. This
y 0.5 y x x3 0.3
0.1 0.1
0.3
0.5
0.7
0.9 1
x
兹1/3
Mu
ham
ma
FIGURE A.2 The maximum value of y = x  x 3 on [0, 1] occurs at the irrational number x = 21>3 .
dH
ass a
nR
iaz
You
suf
i
AP10
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.4
Theory of the Real Numbers
AP11
a.d1d2d3d4 Á
suf
i
is called a construction of the reals, and just as stairs can be built with wood, stone, or steel, there are several approaches to constructing the reals. One construction treats the reals as all the infinite decimals,
iaz
You
In this approach a real number is an integer a followed by a sequence of decimal digits d1, d2 , d3 , Á , each between 0 and 9. This sequence may stop, or repeat in a periodic pattern, or keep going forever with no pattern. In this form, 2.00, 0.3333333 Á and 3.1415926535898 Á represent three familiar real numbers. The real meaning of the dots “ Á ” following these digits requires development of the theory of sequences and series, as in Chapter 11. Each real number is constructed as the limit of a sequence of rational numbers given by its finite decimal approximations. An infinite decimal is then the same as a series a +
d2 d1 + + Á. 10 100
Mu
ham
ma
dH
ass a
nR
This decimal construction of the real numbers is not entirely straightforward. It’s easy enough to check that it gives numbers that satisfy the completeness and order properties, but verifying the algebraic properties is rather involved. Even adding or multiplying two numbers requires an infinite number of operations. Making sense of division requires a careful argument involving limits of rational approximations to infinite decimals. A different approach was taken by Richard Dedekind (1831–1916), a German mathematician, who gave the first rigorous construction of the real numbers in 1872. Given any real number x, we can divide the rational numbers into two sets: those less than or equal to x and those greater. Dedekind cleverly reversed this reasoning and defined a real number to be a division of the rational numbers into two such sets. This seems like a strange approach, but such indirect methods of constructing new structures from old are common in theoretical mathematics. These and other approaches (see Appendix A.5) can be used to construct a system of numbers having the desired algebraic, order, and completeness properties. A final issue that arises is whether all the constructions give the same thing. Is it possible that different constructions result in different number systems satisfying all the required properties? If yes, which of these is the real numbers? Fortunately, the answer turns out to be no. The reals are the only number system satisfying the algebraic, order, and completeness properties. Confusion about the nature of real numbers and about limits caused considerable controversy in the early development of calculus. Calculus pioneers such as Newton, Leibniz, and their successors, when looking at what happens to the difference quotient ƒsx + ¢xd  ƒsxd ¢y = ¢x ¢x as each of ¢y and ¢x approach zero, talked about the resulting derivative being a quotient of two infinitely small quantities. These “infinitesimals,” written dx and dy, were thought to be some new kind of number, smaller than any fixed number but not zero. Similarly, a definite integral was thought of as a sum of an infinite number of infinitesimals ƒsxd # dx as x varied over a closed interval. While the approximating difference quotients ¢y> ¢x were understood much as today, it was the quotient of infinitesimal quantities, rather than
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP12
Appendices
Mu
ham
ma
dH
ass a
nR
iaz
You
suf
i
a limit, that was thought to encapsulate the meaning of the derivative. This way of thinking led to logical difficulties, as attempted definitions and manipulations of infinitesimals ran into contradictions and inconsistencies. The more concrete and computable difference quotients did not cause such trouble, but they were thought of merely as useful calculation tools. Difference quotients were used to work out the numerical value of the derivative and to derive general formulas for calculation, but were not considered to be at the heart of the question of what the derivative actually was. Today we realize that the logical problems associated with infinitesimals can be avoided by defining the derivative to be the limit of its approximating difference quotients. The ambiguities of the old approach are no longer present, and in the standard theory of calculus, infinitesimals are neither needed nor used.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP12
Appendices
A.5
Complex Numbers
i f u s u o
Complex numbers are expressions of the form a + ib, where a and b are real numbers and i is a symbol for 11 . Unfortunately, the words “real” and “imaginary” have connotations that somehow place 11 in a less favorable position in our minds than 12. As a matter of fact, a good deal of imagination, in the sense of inventiveness, has been required to construct the real number system, which forms the basis of the calculus (see Appendix A.4). In this appendix we review the various stages of this invention. The further invention of a complex number system is then presented.
The Development of the Real Numbers
n a sa s H d
Y z
iR a
The earliest stage of number development was the recognition of the counting numbers 1, 2, 3, Á , which we now call the natural numbers or the positive integers. Certain simple arithmetical operations can be performed with these numbers without getting outside the system. That is, the system of positive integers is closed under the operations of addition and multiplication. By this we mean that if m and n are any positive integers, then m + n = p
mn = q
(1)
are also positive integers. Given the two positive integers on the left side of either equation in (1), we can find the corresponding positive integer on the right side. More than this, we can sometimes specify the positive integers m and p and find a positive integer n such that m + n = p. For instance, 3 + n = 7 can be solved when the only numbers we know are the positive integers. But the equation 7 + n = 3 cannot be solved unless the number system is enlarged. The number zero and the negative integers were invented to solve equations like 7 + n = 3. In a civilization that recognizes all the integers
a m m
a h u M
and
Á , 3, 2, 1, 0, 1, 2, 3, Á ,
(2)
an educated person can always find the missing integer that solves the equation m + n = p when given the other two integers in the equation. Suppose our educated people also know how to multiply any two of the integers in the list (2). If, in Equations (1), they are given m and q, they discover that sometimes they can find n and sometimes they cannot. Using their imagination, they may be
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.5
Complex Numbers
AP13
1
1
1. (a) addition (b) subtraction
2. (a) multiplication (b) division
on any two numbers in the system, except that they cannot divide by zero because it is meaningless. The geometry of the unit square (Figure A.3) and the Pythagorean theorem showed that they could construct a geometric line segment that, in terms of some basic unit of length, has length equal to 22. Thus they could solve the equation
iaz
FIGURE A.3 With a straightedge and compass, it is possible to construct a segment of irrational length.
You
兹2
suf
i
inspired to invent still more numbers and introduce fractions, which are just ordered pairs m> n of integers m and n. The number zero has special properties that may bother them for a while, but they ultimately discover that it is handy to have all ratios of integers m> n, excluding only those having zero in the denominator. This system, called the set of rational numbers, is now rich enough for them to perform the rational operations of arithmetic:
x2 = 2
ass a
nR
by a geometric construction. But then they discovered that the line segment representing 22 is an incommensurable quantity. This means that 22 cannot be expressed as the ratio of two integer multiples of some unit of length. That is, our educated people could not find a rational number solution of the equation x 2 = 2. There is no rational number whose square is 2. To see why, suppose that there were such a rational number. Then we could find integers p and q with no common factor other than 1, and such that p 2 = 2q 2 .
(3)
ma
dH
Since p and q are integers, p must be even; otherwise its product with itself would be odd. In symbols, p = 2p1 , where p1 is an integer. This leads to 2p 12 = q 2 which says q must be even, say q = 2q1 , where q1 is an integer. This makes 2 a factor of both p and q, contrary to our choice of p and q as integers with no common factor other than 1. Hence there is no rational number whose square is 2. Although our educated people could not find a rational solution of the equation x 2 = 2, they could get a sequence of rational numbers 1 , 1
7 , 5
41 , 29
239 , 169
Á,
(4)
Mu
ham
whose squares form a sequence 1 , 1
49 , 25
1681 , 841
57,121 , 28,561
Á,
(5)
that converges to 2 as its limit. This time their imagination suggested that they needed the concept of a limit of a sequence of rational numbers. If we accept the fact that an increasing sequence that is bounded from above always approaches a limit (Theorem 6, Section 11.1) and observe that the sequence in (4) has these properties, then we want it to have a limit L. This would also mean, from (5), that L 2 = 2, and hence L is not one of our rational numbers. If to the rational numbers we further add the limits of all bounded increasing sequences of rational numbers, we arrive at the system of all “real” numbers. The word real is placed in quotes because there is nothing that is either “more real” or “less real” about this system than there is about any other mathematical system.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP14
Appendices
i
The Complex Numbers
2. 3.
The first invented system: the set of all integers as constructed from the counting numbers. The second invented system: the set of rational numbers m> n as constructed from the integers. The third invented system: the set of all real numbers x as constructed from the rational numbers.
You
1.
suf
Imagination was called upon at many stages during the development of the real number system. In fact, the art of invention was needed at least three times in constructing the systems we have discussed so far:
In the system of all integers, we can solve all equations of the form
nR
1.
iaz
These invented systems form a hierarchy in which each system contains the previous system. Each system is also richer than its predecessor in that it permits additional operations to be performed without going outside the system:
x + a = 0,
2.
(6)
where a can be any integer. In the system of all rational numbers, we can solve all equations of the form
3.
ass a
ax + b = 0,
(7)
provided a and b are rational numbers and a Z 0. In the system of all real numbers, we can solve all of Equations (6) and (7) and, in addition, all quadratic equations
dH
ax 2 + bx + c = 0
having
a Z 0
and
b 2  4ac Ăš 0.
(8)
ma
You are probably familiar with the formula that gives the solutions of Equation (8), namely, x =
b ; 2b2  4ac , 2a
(9)
Mu
ham
and are familiar with the further fact that when the discriminant, b 2  4ac, is negative, the solutions in Equation (9) do not belong to any of the systems discussed above. In fact, the very simple quadratic equation x2 + 1 = 0
is impossible to solve if the only number systems that can be used are the three invented systems mentioned so far. Thus we come to the fourth invented system, the set of all complex numbers a + ib. We could dispense entirely with the symbol i and use the ordered pair notation (a, b). Since, under algebraic operations, the numbers a and b are treated somewhat differently, it is essential to keep the order straight. We therefore might say that the complex number system consists of the set of all ordered pairs of real numbers (a, b), together with the rules by which they are to be equated, added, multiplied, and so on, listed below. We will use both the (a, b) notation and the notation a + ib in the discussion that follows. We call a the real part and b the imaginary part of the complex number (a, b).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.5
Complex Numbers
AP15
i
We make the following definitions.
suf
Equality a + ib = c + id if and only if a = c and b = d.
Two complex numbers (a, b) and (c, d) are equal if and only if a = c and b = d .
The sum of the two complex numbers (a, b) and (c, d) is the complex number sa + c, b + dd.
The product of two complex numbers (a, b) and (c, d) is the complex number sac  bd, ad + bcd.
iaz
Multiplication sa + ibdsc + idd = sac  bdd + isad + bcd
The product of a real number c and the complex number (a, b) is the complex number (ac, bc).
nR
csa + ibd = ac + isbcd
You
Addition sa + ibd + sc + idd = sa + cd + isb + dd
The set of all complex numbers (a, b) in which the second number b is zero has all the properties of the set of real numbers a. For example, addition and multiplication of (a, 0) and (c, 0) give
ass a
sa, 0d + sc, 0d = sa + c, 0d, sa, 0d # sc, 0d = sac, 0d,
which are numbers of the same type with imaginary part equal to zero. Also, if we multiply a â€œreal numberâ€? (a, 0) and the complex number (c, d), we get
dH
sa, 0d # sc, dd = sac, add = asc, dd.
ma
In particular, the complex number (0, 0) plays the role of zero in the complex number system, and the complex number (1, 0) plays the role of unity or one. The number pair (0, 1), which has real part equal to zero and imaginary part equal to one, has the property that its square, s0, 1ds0, 1d = s 1, 0d,
Mu
ham
has real part equal to minus one and imaginary part equal to zero. Therefore, in the system of complex numbers (a, b) there is a number x = s0, 1d whose square can be added to unity = s1, 0d to produce zero = s0, 0d, that is, s0, 1d2 + s1, 0d = s0, 0d. The equation x2 + 1 = 0 therefore has a solution x = s0, 1d in this new number system. You are probably more familiar with the a + ib notation than you are with the notation (a, b). And since the laws of algebra for the ordered pairs enable us to write sa, bd = sa, 0d + s0, bd = as1, 0d + bs0, 1d, while (1, 0) behaves like unity and (0, 1) behaves like a square root of minus one, we need not hesitate to write a + ib in place of (a, b). The i associated with b is like a tracer element
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP16
Appendices
You
suf
i
that tags the imaginary part of a + ib. We can pass at will from the realm of ordered pairs (a, b) to the realm of expressions a + ib, and conversely. But there is nothing less “real” about the symbol s0, 1d = i than there is about the symbol s1, 0d = 1, once we have learned the laws of algebra in the complex number system of ordered pairs (a, b). To reduce any rational combination of complex numbers to a single complex number, we apply the laws of elementary algebra, replacing i 2 wherever it appears by 1. Of course, we cannot divide by the complex number s0, 0d = 0 + i0. But if a + ib Z 0, then we may carry out a division as follows: sc + iddsa  ibd sac + bdd + isad  bcd c + id = = . a + ib sa + ibdsa  ibd a2 + b2 The result is a complex number x + iy with ac + bd , a2 + b2
ad  bc , a2 + b2
iaz
x =
y =
nR
and a 2 + b 2 Z 0, since a + ib = sa, bd Z s0, 0d. The number a  ib that is used as multiplier to clear the i from the denominator is called the complex conjugate of a + ib. It is customary to use z (read “z bar”) to denote the complex conjugate of z; thus z = a + ib,
z = a  ib.
ass a
Multiplying the numerator and denominator of the fraction sc + idd>sa + ibd by the complex conjugate of the denominator will always replace the denominator by a real number.
EXAMPLE 1
Arithmetic Operations with Complex Numbers
dH
(a) s2 + 3id + s6  2id = s2 + 6d + s3  2di = 8 + i (b) s2 + 3id  s6  2id = s2  6d + s3  s 2ddi = 4 + 5i (c) s2 + 3ids6  2id = s2ds6d + s2ds 2id + s3ids6d + s3ids 2id = 12  4i + 18i  6i 2 = 12 + 14i + 6 = 18 + 14i
2 + 3i 2 + 3i 6 + 2i = 6  2i 6  2i 6 + 2i
ma
(d)
12 + 4i + 18i + 6i2 36 + 12i  12i  4i2 6 + 22i 3 11 = = + i 40 20 20 =
ham
y
P(x, y)
r
y
Mu
O
x
x
FIGURE A.4 This Argand diagram represents z = x + iy both as a point § P(x, y) and as a vector OP .
Argand Diagrams There are two geometric representations of the complex number z = x + iy: 1. 2.
as the point P(x, y) in the xyplane § as the vector OP from the origin to P.
In each representation, the xaxis is called the real axis and the yaxis is the imaginary axis. Both representations are Argand diagrams for x + iy (Figure A.4). In terms of the polar coordinates of x and y, we have
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.5
Complex Numbers
AP17
i
and z = x + iy = rscos u + i sin ud.
suf
(10)
We define the absolute value of a complex number x + iy to be the length r of a vector § OP from the origin to P(x, y). We denote the absolute value by vertical bars; thus,
You
ƒ x + iy ƒ = 2x 2 + y 2 .
If we always choose the polar coordinates r and u so that r is nonnegative, then r = ƒ x + iy ƒ .
iaz
The polar angle u is called the argument of z and is written u = arg z. Of course, any integer multiple of 2p may be added to u to produce another appropriate angle. The following equation gives a useful formula connecting a complex number z, its conjugate z, and its absolute value ƒ z ƒ , namely,
Euler’s Formula The identity
nR
z # z = ƒ z ƒ2.
e iu = cos u + i sin u,
ass a
called Euler’s formula, enables us to rewrite Equation (10) as z = re iu .
Mu
ham
ma
dH
This formula, in turn, leads to the following rules for calculating products, quotients, powers, and roots of complex numbers. It also leads to Argand diagrams for e iu . Since cos u + i sin u is what we get from Equation (10) by taking r = 1, we can say that e iu is represented by a unit vector that makes an angle u with the positive xaxis, as shown in Figure A.5. y
y
ei cos i sin
r1 arg z O
ei cos i sin (cos , sin )
x
(a)
O
x
(b)
FIGURE A.5 Argand diagrams for e iu = cos u + i sin u (a) as a vector and (b) as a point.
Products To multiply two complex numbers, we multiply their absolute values and add their angles. Let
To Read it Online & Download:
z1 = r1 eiu1,
z2 = r2 eiu2 ,
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(11)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP18
Appendices
arg z1 = u1;
ƒ z2 ƒ = r2,
arg z2 = u2 .
Then z1 z2 = r1 e iu1 # r2 e iu2 = r1 r2 e isu1 + u2d
z2 r2
z1
r1
2
and hence
You
r1r 2
ƒ z1 z2 ƒ = r1 r2 = ƒ z1 ƒ # ƒ z2 ƒ
1
arg sz1 z2 d = u1 + u2 = arg z1 + arg z2 .
x
O
z1 1 i z 1z 2
z1 = 22e ip>4,
4
兹3 1
12 – 6
1 兹3
Then
z1 z2 = 222 exp a
x
1
ip ip ip b = 222 exp a b 4 6 12 p p + i sin b L 2.73 + 0.73i. 12 12
The notation exp (A) stands for e A .
Quotients
ma
FIGURE A.7 To multiply two complex numbers, multiply their absolute values and add their arguments.
= 222 acos
dH
z 2 兹3 i
z2 = 2e ip>6 .
ass a
2兹2
2 –1
iaz
Let z1 = 1 + i, z2 = 23  i. We plot these complex numbers in an Argand diagram (Figure A.7) from which we read off the polar representations
y
0
Finding a Product of Complex Numbers
nR
EXAMPLE 2
兹2
(12)
Thus, the product of two complex numbers is represented by a vector whose length is the product of the lengths of the two factors and whose argument is the sum of their arguments (Figure A.6). In particular, from Equation (12) a vector may be rotated counterclockwise through an angle u by multiplying it by e iu . Multiplication by i rotates 90°, by 1 rotates 180°, by i rotates 270°, and so on.
FIGURE A.6 When z1 and z2 are multiplied, ƒ z1 z2 ƒ = r1 # r2 and arg sz1 z2 d = u1 + u2 .
1
suf
ƒ z1 ƒ = r1,
1
i
so that
y z 1z 2
Mu
ham
Suppose r2 Z 0 in Equation (11). Then z1 r1 e iu1 r1 isu1  u2d . z2 = r e iu2 = r2 e 2
Hence
z1 r1 ƒ z1 ƒ ` z2 ` = r2 = ƒ z2 ƒ
and
z1 arg a z2 b = u1  u2 = arg z1  arg z2.
That is, we divide lengths and subtract angles for the quotient of complex numbers.
EXAMPLE 3
Let z1 = 1 + i and z2 = 23  i, as in Example 2. Then
1 + i 23  i
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=
22 5pi>12 22e ip>4 5p 5p = e L 0.707 acos + i sin b 2 12 12 2e ip>6
L 0.183 + 0.683i.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.5
Complex Numbers
AP19
i
Powers z n = z # z # Á # z.
n factors
iu
With z = re , we obtain Á + ud
n summands
You
zn = sreiu dn = rneisu + u +
suf
If n is a positive integer, we may apply the product formulas in Equation (12) to find
n inu
= re .
(13)
De Moivre’s Theorem
iaz
The length r = ƒ z ƒ is raised to the nth power and the angle u = arg z is multiplied by n. If we take r = 1 in Equation (13), we obtain De Moivre’s Theorem.
(14)
nR
scos u + i sin udn = cos nu + i sin nu.
ass a
If we expand the left side of De Moivre’s equation above by the Binomial Theorem and reduce it to the form a + ib, we obtain formulas for cos nu and sin nu as polynomials of degree n in cos u and sin u.
EXAMPLE 4
If n = 3 in Equation (14), we have scos u + i sin ud3 = cos 3u + i sin 3u.
The left side of this equation expands to
dH
cos3 u + 3i cos2 u sin u  3 cos u sin2 u  i sin3 u.
The real part of this must equal cos 3u and the imaginary part must equal sin 3u. Therefore, cos 3u = cos3 u  3 cos u sin2 u, sin 3u = 3 cos2 u sin u  sin3 u.
ma
Roots
Mu
ham
If z = re iu is a complex number different from zero and n is a positive integer, then there are precisely n different complex numbers w0 , w1 , Á , wn  1 , that are nth roots of z. To see why, let w = re ia be an nth root of z = re iu , so that wn = z or rne ina = re iu . Then n
r = 2r is the real, positive nth root of r. For the argument, although we cannot say that na and u must be equal, we can say that they may differ only by an integer multiple of 2p. That is,
To Read it Online & Download:
na = u + 2kp,
k = 0, ;1, ;2, Á .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP20
Appendices
z rei
Hence, all the nth roots of z = re iu are given by
r
x
3
O 2 3
u 2p n n 2re iu = 2r exp i a n + k n b,
2 3 w2
You
w0
r 1/3
suf
u 2p a = n + k n .
2 3
w1
i
Therefore,
y
k = 0, ;1, ;2, Á .
(15)
iaz
There might appear to be infinitely many different answers corresponding to the infinitely many possible values of k, but k = n + m gives the same answer as k = m in Equation (15). Thus, we need only take n consecutive values for k to obtain all the different nth roots of z. For convenience, we take
FIGURE A.8 The three cube roots of z = re iu .
nR
k = 0, 1, 2, Á , n  1.
ass a
All the nth roots of re iu lie on a circle centered at the origin and having radius equal to the real, positive nth root of r. One of them has argument a = u>n. The others are uniformly spaced around the circle, each being separated from its neighbors by an angle equal to 2p>n. Figure A.8 illustrates the placement of the three cube roots, w0, w1, w2 , of the complex number z = re iu .
EXAMPLE 5
y
Finding Fourth Roots
Find the four fourth roots of 16. 2
2
2
–16
As our first step, we plot the number 16 in an Argand diagram (Figure A.9) and determine its polar representation re iu . Here, z = 16, r = +16, and u = p. One of the fourth roots of 16e ip is 2e ip>4 . We obtain others by successive additions of 2p>4 = p>2 to the argument of this first one. Hence,
Solution w0 4
x 2
2
w3
p 3p 5p 7p 4 2 16 exp ip = 2 exp i a , , , b, 4 4 4 4
ma
w2
dH
w1
and the four roots are
Mu
ham
FIGURE A.9 The four fourth roots of 16 .
w0 = 2 ccos
p p + i sin d = 22s1 + id 4 4
w1 = 2 ccos
3p 3p + i sin d = 22s 1 + id 4 4
w2 = 2 ccos
5p 5p + i sin d = 22s 1  id 4 4
w3 = 2 ccos
7p 7p + i sin d = 22s1  id. 4 4
The Fundamental Theorem of Algebra One might say that the invention of 2 1 is all well and good and leads to a number system that is richer than the real number system alone; but where will this process end? Are
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.5
Complex Numbers
AP21
The Fundamental Theorem of Algebra Every polynomial equation of the form
You
suf
i
4 6 11, we also going to invent still more systems so as to obtain 11, and so on? But it turns out this is not necessary. These numbers are already expressible in terms of the complex number system a + ib. In fact, the Fundamental Theorem of Algebra says that with the introduction of the complex numbers we now have enough numbers to factor every polynomial into a product of linear factors and so enough numbers to solve every possible polynomial equation.
iaz
an z n + an  1 z n  1 + Á + a1 z + a0 = 0,
nR
in which the coefficients a0, a1, Á , an are any complex numbers, whose degree n is greater than or equal to one, and whose leading coefficient an is not zero, has exactly n roots in the complex number system, provided each multiple root of multiplicity m is counted as m roots.
ass a
A proof of this theorem can be found in almost any text on the theory of functions of a complex variable.
dH
EXERCISES A.5 Operations with Complex Numbers
1. How computers multiply complex numbers = sac  bd, ad + bcd . c.
b.
(This is how complex numbers are multiplied by computers.)
ham 2
1 + i 1 b + = 1 + i x + iy 1  i
c. s3  2idsx + iyd = 2sx  2iyd + 2i  1
Mu
4. Show that the distance between the two points z1 and z2 in an Argand diagram is ƒ z1  z2 ƒ .
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9. ƒ z + i ƒ = ƒ z  1 ƒ
1 + i23
14. s2 + 3ids1  2id
Powers and Roots Use De Moivre’s Theorem to express the trigonometric functions in Exercises 15 and 16 in terms of cos u and sin u .
b. s zd d. 1> z
7. ƒ z + 1 ƒ = 1
8. ƒ z + 1 ƒ = ƒ z  1 ƒ
1  i23
3. How may the following complex numbers be obtained from z = x + iy geometrically? Sketch. c. z
6. ƒ z  1 ƒ = 2
13.
Graphing and Geometry
a. z
c. ƒ z ƒ 7 2
Express the complex numbers in Exercises 11–14 in the form re iu , with r Ú 0 and p 6 u … p . Draw an Argand diagram for each calculation. 1 + i 11. A 1 + 23 B 2 12. 1  i
a. s3 + 4id  2sx  iyd = x + iy b. a
b. ƒ z ƒ 6 2
10. ƒ z + 1 ƒ Ú ƒ z ƒ
2. Solve the following equations for the real numbers, x and y. 2
In Exercises 5–10, graph the points z = x + iy that satisfy the given conditions. 5. a. ƒ z ƒ = 2
s2, 1d # s 2, 3d
ma
a.
s2, 3d # s4, 2d s 1, 2d # s2, 1d
Find sa, bd # sc, dd
15. cos 4u
16. sin 4u
17. Find the three cube roots of 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Appendices
18. Find the two square roots of i.
a. Extend the results of Exercise 26 to show that ƒszd = ƒszd if
i
19. Find the three cube roots of 8i .
23. Find all solutions of the equation x 4 + 4x 2 + 16 = 0 . 24. Solve the equation x 4 + 1 = 0 .
Theory and Examples 25. Complex numbers and vectors in the plane Show with an Argand diagram that the law for adding complex numbers is the same as the parallelogram law for adding vectors. 26. Complex arithmetic with conjugates Show that the conjugate of the sum (product, or quotient) of two complex numbers, z1 and z2, is the same as the sum (product, or quotient) of their conjugates.
a. z + z = 2Reszd
b. z  z = 2iImszd
c. ƒ Reszd ƒ … ƒ z ƒ d. ƒ z1 + z2 ƒ 2 = ƒ z1 ƒ 2 + ƒ z2 ƒ 2 + 2Resz1z2 d e. ƒ z1 + z2 ƒ … ƒ z1 ƒ + ƒ z2 ƒ
Mu
ham
ma
dH
ass a
27. Complex roots of polynomials with real coefficients come in complexconjugate pairs
You
22. Find the six solutions of the equation z 6 + 2z 3 + 2 = 0 .
is a polynomial with real coefficients a0, Á , an .
b. If z is a root of the equation ƒszd = 0 , where ƒ(z) is a polynomial with real coefficients as in part (a), show that the conjugate z is also a root of the equation. (Hint: Let ƒszd = u + iy = 0 ; then both u and y are zero. Use the fact that ƒszd = ƒszd = u  iy .) 28. Absolute value of a conjugate Show that ƒ z ƒ = ƒ z ƒ . 29. When z = z If z and z are equal, what can you say about the location of the point z in the complex plane? 30. Real and imaginary parts Let Re(z) denote the real part of z and Im(z) the imaginary part. Show that the following relations hold for any complex numbers z, z1 , and z2 .
iaz
21. Find the four solutions of the equation z 4  2z 2 + 4 = 0 .
suf
ƒszd = an z n + an  1 z n  1 + Á + a1 z + a0
20. Find the six sixth roots of 64.
nR
AP22
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP22
i f su
Appendices
A.6
The Distributive Law for Vector Cross Products In this appendix, we prove the Distributive Law
z a i R
u * sv + wd = u * v + u * w which is Property 2 in Section 12.4.
u o Y
Proof To derive the Distributive Law, we construct u * v a new way. We draw u and v from the common point O and construct a plane M perpendicular to u at O (Figure A.10). We then project v orthogonally onto M, yielding a vector v¿ with length ƒ v ƒ sin u. We rotate v¿ 90° about u in the positive sense to produce a vector v– . Finally, we multiply v– by the
m m
d a
a h u M
n a ss
a H
M'
v u O
M v'
v''
uv
90˚
FIGURE A.10 As explained in the text, u * v = ƒ u ƒ v–.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.7 The Mixed Derivative Theorem and the Increment Theorem
AP23
suf
i
length of u. The resulting vector ƒ u ƒ v– is equal to u * v since v– has the same direction as u * v by its construction (Figure A.10) and ƒ u ƒ ƒ v– ƒ = ƒ u ƒ ƒ v¿ ƒ = ƒ u ƒ ƒ v ƒ sin u = ƒ u * v ƒ. Now each of these three operations, namely, 2. 3.
projection onto M rotation about u through 90° multiplication by the scalar ƒ u ƒ
You
1.
when applied to a triangle whose plane is not parallel to u, will produce another triangle. If we start with the triangle whose sides are v, w, and v + w (Figure A.11) and apply these three steps, we successively obtain the following: A triangle whose sides are v¿, w¿, and sv + wd¿ satisfying the vector equation
iaz
1.
v¿ + w¿ = sv + wd¿
A triangle whose sides are v–, w– , and sv + wd– satisfying the vector equation
nR
2.
v– + w– = sv + wd–
ass a
(the double prime on each vector has the same meaning as in Figure A.10)
w
u
vw v' (v w)'
dH
M
w'
v
FIGURE A.11 The vectors, v, w, v + w , and their projections onto a plane perpendicular to u.
A triangle whose sides are ƒ u ƒ v–, ƒ u ƒ w– , and ƒ u ƒ sv + wd– satisfying the vector equation ƒ u ƒ v– + ƒ u ƒ w– = ƒ u ƒ sv + wd–.
Substituting ƒ u ƒ v– = u * v, ƒ u ƒ w– = u * w, and ƒ u ƒ sv + wd– = u * sv + wd from our discussion above into this last equation gives u * v + u * w = u * sv + wd, which is the law we wanted to establish.
Mu
ham
ma
3.
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i f u s You
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
A.7 The Mixed Derivative Theorem and the Increment Theorem
A.7
z a i an R
AP23
The Mixed Derivative Theorem and the Increment Theorem
s s a H d a m ham
This appendix derives the Mixed Derivative Theorem (Theorem 2, Section 14.3) and the Increment Theorem for Functions of Two Variables (Theorem 3, Section 14.3). Euler first published the Mixed Derivative Theorem in 1734, in a series of papers he wrote on hydrodynamics.
Mu
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP24
Appendices
You
suf
i
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 ƒxysa, bd = ƒyxsa, bd.
iaz
Proof The equality of ƒxysa, bd and ƒyxsa, bd can be established by four applications of the Mean Value Theorem (Theorem 4, Section 4.2). By hypothesis, the point (a, b) lies in the interior of a rectangle R in the xyplane on which ƒ, ƒx , ƒy , ƒxy , and ƒyx are all defined. We let h and k be the numbers such that the point sa + h, b + kd also lies in R, and we consider the difference ¢ = Fsa + hd  Fsad,
(1)
Fsxd = ƒsx, b + kd  ƒsx, bd.
(2)
nR
where
We apply the Mean Value Theorem to F, which is continuous because it is differentiable. Then Equation (1) becomes ¢ = hF¿sc1 d,
(3)
ass a
where c1 lies between a and a + h. From Equation (2). F¿sxd = ƒxsx, b + kd  ƒxsx, bd,
y
so Equation (3) becomes
¢ = h[ƒxsc1, b + kd  ƒxsc1, bd].
R
(a, b)
Now we apply the Mean Value Theorem to the function gs yd = fxsc1, yd and have
R' h
dH
k
(4)
gsb + kd  gsbd = kg¿sd1 d,
or
x
ƒxsc1, b + kd  ƒxsc1, bd = kƒxysc1, d1 d
ma
0
for some d1 between b and b + k. By substituting this into Equation (4), we get
Mu
ham
FIGURE A.12 The key to proving ƒxysa, bd = ƒyxsa, bd is that no matter how small R¿ is, ƒxy and ƒyx take on equal values somewhere inside R¿ (although not necessarily at the same point).
¢ = hkƒxysc1, d1 d
(5)
for some point sc1, d1 d in the rectangle R¿ whose vertices are the four points (a, b), sa + h, bd, sa + h, b + kd, and sa, b + kd. (See Figure A.12.) By substituting from Equation (2) into Equation (1), we may also write ¢ = ƒsa + h, b + kd  ƒsa + h, bd  ƒsa, b + kd + ƒsa, bd = [ƒsa + h, b + kd  ƒsa, b + kd]  [ƒsa + h, bd  ƒsa, bd] (6)
= fsb + kd  fsbd, where fs yd = ƒsa + h, yd  ƒsa, yd.
(7)
The Mean Value Theorem applied to Equation (6) now gives
To Read it Online & Download:
¢ = kf¿sd2 d
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(8)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.7
The Mixed Derivative Theorem and the Increment Theorem
AP25
Substituting from Equation (9) into Equation (8) gives
(9)
suf
f¿s yd = ƒysa + h, yd  ƒysa, yd.
i
for some d2 between b and b + k. By Equation (7),
¢ = k[ƒysa + h, d2 d  ƒysa, d2 d].
You
Finally, we apply the Mean Value Theorem to the expression in brackets and get ¢ = khƒyxsc2, d2 d
(10)
for some c2 between a and a + h. Together, Equations (5) and (10) show that
(11)
iaz
ƒxysc1, d1 d = ƒyxsc2, d2 d,
nR
where sc1, d1 d and sc2, d2 d both lie in the rectangle R¿ (Figure A.12). Equation (11) is not quite the result we want, since it says only that ƒxy has the same value at sc1, d1 d that ƒyx has at sc2, d2 d. The numbers h and k in our discussion, however, may be made as small as we wish. The hypothesis that ƒxy and ƒyx are both continuous at (a, b) means that ƒxysc1, d1 d = ƒxysa, bd + P1 and ƒyxsc2, d2 d = ƒyxsa, bd + P2 , where each of P1, P2 : 0 as both h, k : 0. Hence, if we let h and k : 0, we have ƒxysa, bd = ƒyxsa, bd.
ass a
The equality of ƒxysa, bd and ƒyxsa, bd can be proved with hypotheses weaker than the ones we assumed. For example, it is enough for ƒ, ƒx , and ƒy to exist in R and for ƒxy to be continuous at (a, b). Then ƒyx will exist at (a, b) and equal ƒxy at that point.
dH ma
C(x0 , x, y0 ,y)
THEOREM 3 The Increment Theorem for Functions of Two Variables Suppose that the first partial derivatives of z = ƒsx, yd 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 sx0 + ¢x, y0 + ¢yd in R satisfies an equation of the form ¢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.
ham
A(x0, y0 )
B(x0 , x, y0 )
Mu
T
FIGURE A.13 The rectangular region T in the proof of the Increment Theorem. The figure is drawn for ¢x and ¢y positive, but either increment might be zero or negative.
Proof We work within a rectangle T centered at Asx0 , y0 d and lying within R, and we assume that ¢x and ¢y are already so small that the line segment joining A to Bsx0 + ¢x, y0 d and the line segment joining B to Csx0 + ¢x, y0 + ¢yd lie in the interior of T (Figure A.13). We may think of ¢z as the sum ¢z = ¢z1 + ¢z2 of two increments, where ¢z1 = ƒsx0 + ¢x, y0 d  ƒsx0 , y0 d is the change in the value of ƒ from A to B and ¢z2 = ƒsx0 + ¢x, y0 + ¢yd  ƒsx0 + ¢x, y0 d is the change in the value of ƒ from B to C (Figure A.14).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP26
Appendices
i
z
suf
S Q z f (x, y)
P0
Q'
P''
P'
B
nR
y 0 ,y (x0 , x, y 0 )
, z1
iaz
A(x0 , y 0 )
x
You
,z
0
y0
, z2
y
C(x0 , x, y 0 ,y)
ass a
FIGURE A.14 Part of the surface z = ƒsx, yd near P0sx0, y0, ƒsx0, y0 dd . The points P0 , P¿, and P– have the same height z0 = ƒsx0, y0 d above the xyplane. The change in z is ¢z = P¿S . The change ¢z1 = ƒsx0 + ¢x, y0 d  ƒsx0 , y0 d,
dH
shown as P–Q = P¿Q¿ , is caused by changing x from x0 to x0 + ¢x while holding y equal to y0 . Then, with x held equal to x0 + ¢x , ¢z2 = ƒsx0 + ¢x, y0 + ¢yd  ƒsx0 + ¢x, y0 d
is the change in z caused by changing y0 from y0 + ¢y , which is represented by Q¿S? The total change in z is the sum of ¢z1 and ¢z2 .
Mu
ham
ma
On the closed interval of xvalues joining x0 to x0 + ¢x, the function Fsxd = ƒsx, y0 d is a differentiable (and hence continuous) function of x, with derivative F¿sxd = ƒxsx, y0 d.
By the Mean Value Theorem (Theorem 4, Section 4.2), there is an xvalue c between x0 and x0 + ¢x at which Fsx0 + ¢xd  Fsx0 d = F¿scd¢x or ƒsx0 + ¢x, y0 d  ƒsx0, y0 d = ƒxsc, y0 d¢x or ¢z1 = ƒxsc, y0 d¢x.
(12)
Similarly, Gs yd = ƒsx0 + ¢x, yd is a differentiable (and hence continuous) function of y on the closed yinterval joining y0 and y0 + ¢y, with derivative
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.7
The Mixed Derivative Theorem and the Increment Theorem
AP27
suf
Gs y0 + ¢yd  Gs y0 d = G¿sdd¢y
i
Hence, there is a yvalue d between y0 and y0 + ¢y at which
or
ƒsx0 + ¢x, y0 + ¢yd  ƒsx0 + ¢x, yd = ƒysx0 + ¢x, dd¢y
You
or
¢z2 = ƒysx0 + ¢x, dd¢y.
(13)
Now, as both ¢x and ¢y : 0, we know that c : x0 and d : y0 . Therefore, since ƒx and ƒy are continuous at sx0 , y0 d, the quantities P1 = ƒxsc, y0 d  ƒxsx0, y0 d,
iaz
(14)
P2 = ƒysx0 + ¢x, dd  ƒysx0, y0 d
nR
both approach zero as both ¢x and ¢y : 0. Finally, ¢z = ¢z1 + ¢z2
= ƒxsc, y0 d¢x + ƒysx0 + ¢x, dd¢y
From Equations (12) and (13)
= [ƒxsx0 , y0 d + P1]¢x + [ƒysx0 , y0 d + P2]¢y
From Equation (14)
ass a
= ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y + P1 ¢x + P2 ¢y,
where both P1 and P2 : 0 as both ¢x and ¢y : 0, which is what we set out to prove.
dH
Analogous results hold for functions of any finite number of independent variables. Suppose that the first partial derivatives of w = ƒsx, y, zd are defined throughout an open region containing the point sx0 , y0 , z0 d and that ƒx , ƒy , and ƒz are continuous at sx0 , y0 , z0 d. Then ¢w = ƒsx0 + ¢x, y0 + ¢y, z0 + ¢zd  ƒsx0 , y0 , z0 d = ƒx ¢x + ƒy ¢y + ƒz ¢z + P1 ¢x + P2 ¢y + P3 ¢z,
(15)
Mu
ham
ma
where P1, P2, P3 : 0 as ¢x, ¢y, and ¢z : 0. The partial derivatives ƒx , ƒy , ƒz in Equation (15) are to be evaluated at the point sx0 , y0 , z0 d. Equation (15) can be proved by treating ¢w as the sum of three increments, ¢w1 = ƒsx0 + ¢x, y0 , z0 d  ƒsx0 , y0 , z0 d
(16)
¢w2 = ƒsx0 + ¢x, y0 + ¢y, z0 d  ƒsx0 + ¢x, y0 , z0 d
(17)
¢w3 = ƒsx0 + ¢x, y0 + ¢y, z0 + ¢zd  ƒsx0 + ¢x, y0 + ¢y, z0 d,
(18)
and applying the Mean Value Theorem to each of these separately. Two coordinates remain constant and only one varies in each of these partial increments ¢w1, ¢w2, ¢w3 . In Equation (17), for example, only y varies, since x is held equal to x0 + ¢x and z is held equal to z0 . Since ƒsx0 + ¢x, y, z0 d is a continuous function of y with a derivative ƒy , it is subject to the Mean Value Theorem, and we have ¢w2 = ƒysx0 + ¢x, y1, z0 d¢y for some y1 between y0 and y0 + ¢y.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP28
Appendices
suf
i
The Area of a Parallelogram’s Projection on a Plane
A.8
This appendix proves the result needed in Section 16.5 that ƒ su * vd # p ƒ is the area of the projection of the parallelogram with sides determined by u and v onto any plane whose normal is p. (See Figure A.15.)
You
P u
v
Q
THEOREM The area of the orthogonal projection of the parallelogram determined by two vectors u and v in space onto a plane with unit normal vector p is
S R
S' R'
FIGURE A.15 The parallelogram determined by two vectors u and v in space and the orthogonal projection of the parallelogram onto a plane. The projection lines, orthogonal to the plane, lie parallel to the unit normal vector p.
Proof In the notation of Figure A.15, which shows a typical parallelogram determined by vectors u and v and its orthogonal projection onto a plane with unit normal vector p, § § u = PP ¿ + u¿ + Q¿Q § § § § sQ¿Q = QQ ¿d = u¿ + PP ¿  QQ ¿
nR
Q'
iaz
p
v'
= u¿ + s p.
Similarly,
ass a
u' P'
Area = ƒ su * vd # p ƒ .
(For some scalar s because § § (PP ¿  QQ ¿ ) is parallel to p)
v = v¿ + tp
for some scalar t. Hence,
u * v = su¿ + spd * sv¿ + tpd
dH
= su¿ * v¿d + ssp * v¿d + tsu¿ * pd + stsp * pd.
(1)
(')'* 0
z
Q(2, –1, 2)
k
S(1, 3, 2)
y
Mu
su * vd # p = su¿ * v¿d # p.
In particular,
ham
P(0, 0, 3)
ma
The vectors p * v¿ and u¿ * p are both orthogonal to p. Hence, when we dot both sides of Equation (1) with p, the only nonzero term on the right is su¿ * v¿d # p. That is,
R(3, 2, 1)
ƒ su * vd # p ƒ = ƒ su¿ * v¿d # p ƒ .
(2)
The absolute value on the right is the volume of the box determined by u¿, v¿ , and p. The height of this particular box is ƒ p ƒ = 1, so the box’s volume is numerically the same as its base area, the area of parallelogram P¿Q¿R¿S¿ . Combining this observation with Equation (2) gives Area of P¿Q¿R¿S¿ = ƒ su¿ * v¿d # p ƒ = ƒ su * vd # p ƒ , which says that the area of the orthogonal projection of the parallelogram determined by u and v onto a plane with unit normal vector p is ƒ su * vd # p ƒ . This is what we set out to prove.
x
FIGURE A.16 Example 1 calculates the area of the orthogonal projection of parallelogram PQRS on the xyplane.
EXAMPLE 1
Finding the Area of a Projection
Find the area of the orthogonal projection onto the xyplane of the parallelogram determined by the points P(0, 0, 3), Qs2, 1, 2d, R(3, 2, 1), and S(1, 3, 2) (Figure A.16).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.9 Basic Algebra, Geometry, and Trigonometry Formulas
§ u = PQ = 2i  j  k,
§ v = PS = i + 3j  k,
we have 2 31 0
1 2 1 3 = ` 1 1
1 ` = 7, 3
Mu
ham
ma
dH
ass a
nR
iaz
so the area is ƒ su * vd # p ƒ = ƒ 7 ƒ = 7.
1 3 0
You
su * vd # p =
and
i
With
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p = k,
suf
Solution
AP29
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
A.9 Basic Algebra, Geometry, and Trigonometry Formulas
A.9
Basic Algebra, Geometry, and Trigonometry Formulas Algebra
Y z
Arithmetic Operations asb + cd = ab + ac,
iR a
a c ad + bc + = , b d bd
n a sa s H d
Laws of Signs
s ad = a,
AP29
a#c ac = b d bd
a>b a d = #c b c>d
a a a =  = b b b
Zero Division by zero is not defined. 0 If a Z 0: a = 0,
a 0 = 1,
0a = 0
For any number a: a # 0 = 0 # a = 0
Laws of Exponents
a m m
a ma n = a m + n,
sabdm = a mb m,
sa m dn = a mn,
n
am>n = 2am =
If a Z 0,
a h u M
am = a m  n, an
a 0 = 1,
a m =
1 . am
The Binomial Theorem For any positive integer n, sa + bdn = an + nan  1b +
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+
nsn  1d n  2 2 a b 1#2
nsn  1dsn  2d n  3 3 a b + Ă + nabn  1 + bn . 1#2#3
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n aBm A2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP30
Appendices
sa  bd2 = a2  2ab + b2
sa + bd3 = a3 + 3a2b + 3ab2 + b3,
suf
sa + bd2 = a2 + 2ab + b2,
i
For instance,
sa  bd3 = a2  3a2b + 3ab2  b3 .
You
Factoring the Difference of Like Integer Powers, n>1
a n  b n = sa  bdsa n  1 + a n  2b + a n  3b 2 + Ă + ab n  2 + b n  1 d For instance, a2  b2 = sa  bdsa + bd,
a3  b3 = sa  bdsa2 + ab + b2 d,
iaz
a4  b4 = sa  bdsa3 + a2b + ab2 + b3 d. Completing the Square If a Z 0,
nR
b ax 2 + bx + c = a ax 2 + a xb + c
b b2 b2 = a ax 2 + a x + b + c 4a 2 4a 2
ass a
b b2 b2 = a ax 2 + a x + b + a ab + c 4a 2 4a 2 b b2 b2 = a ax 2 + a x + b + c 4a 4a 2
dH
('''')''''* This is ax +
2
b b . 2a
= au 2 + C
(')'* Call this part C.
su = x + sb>2add
ma
The Quadratic Formula If a Z 0 and ax 2 + bx + c = 0, then b ; 2b 2  4ac . 2a
x =
Mu
ham
Geometry
Formulas for area, circumference, and volume: sA = area, B = area of base, C = circumference, S = lateral area or surface area, V = volumed Triangle
Similar Triangles c h
b
A 1 bh 2
To Read it Online & Download:
c'
a'
Pythagorean Theorem a
c
b
b' b a' b' c' abc
a a2 b2 c2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.9
Parallelogram
AP31
Basic Algebra, Geometry, and Trigonometry Formulas
Trapezoid
Circle
suf
i
a h h b
b A bh
You
A 1 (a b)h 2
Any Cylinder or Prism with Parallel Bases
A r 2, C 2r
r
Right Circular Cylinder
iaz
r
h
nR
h
h
V Bh
B
ass a
B
Right Circular Cone
dH
Any Cone or Pyramid
s r V 1 r 2h 3 S rs Area of side
ma V 1 Bh 3
Mu
ham
B
Sphere
h
h
h
V r 2h S 2rh Area of side
B
V 4 r 3, S 4r 2 3
Trigonometry Formulas
y
Definitions and Fundamental Identities Sine: Cosine: Tangent:
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P(x, y)
y 1 sin u = r = csc u x 1 cos u = r = sec u y 1 tan u = x = cot u
O
r x
y
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x
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH AP32
Appendices
sin2 u + cos2 u = 1,
sec2 u = 1 + tan2 u,
cos 2u = cos2 u  sin2 u
sin 2u = 2 sin u cos u, cos2 u =
csc2 u = 1 + cot2 u
1 + cos 2u , 2
1  cos 2u 2
sin2 u =
sin sA + Bd = sin A cos B + cos A sin B sin sA  Bd = sin A cos B  cos A sin B cos sA + Bd = cos A cos B  sin A sin B
suf
cos s ud = cos u
You
sin s ud = sin u,
i
Identities
tan A + tan B , 1  tan A tan B
sin aA 
p b = cos A, 2
sin aA +
p b = cos A, 2
tan sA  Bd =
cos aA 
tan A  tan B 1 + tan A tan B
p b = sin A 2
nR
tan sA + Bd =
iaz
cos sA  Bd = cos A cos B + sin A sin B
cos aA +
p b = sin A 2
1 1 cos sA  Bd  cos sA + Bd 2 2
ass a
sin A sin B =
1 1 cos sA  Bd + cos sA + Bd 2 2
sin A cos B =
1 1 sin sA  Bd + sin sA + Bd 2 2
dH
cos A cos B =
1 1 sA + Bd cos sA  Bd 2 2
sin A  sin B = 2 cos
1 1 sA + Bd sin sA  Bd 2 2
ma
sin A + sin B = 2 sin
1 1 sA + Bd cos sA  Bd 2 2
cos A  cos B = 2 sin
1 1 sA + Bd sin sA  Bd 2 2
Mu
ham
cos A + cos B = 2 cos
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH A.9
Basic Algebra, Geometry, and Trigonometry Formulas
AP33
Trigonometric Functions Radians
suf
i
Degrees
Radian Measure
45 兹2
1
θ
45
C ir
1
e
it cir cl
cle of ra diu
1
4
90
r
Un
1
sr
1
6
iaz
30
u s or r = 1 = u 180° = p radians.
s u = r,
2
You
s
4
兹2
兹3
2
60
3
90
1
nR
兹3
2 2
1
The angles of two common triangles, in degrees and radians.
y
y
ass a
y sin x
– – 2
2
0
3 2 2
x
y
2
0
3 2 2
Domain: (–, ) Range: [–1, 1]
y
y tan x
0 3 2 2
x
Domain: All real numbers except odd integer multiples of /2 Range: (–, )
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– – 2
y sec x
1
– 3 – – 2 2
Mu
ham
ma
dH
y sinx Domain: (–, ) Range: [–1, 1]
y cos x
y
3 2 2
x
Domain: x , 3 , . . . 2 2 Range: (–, –1] h [1, ) y
y csc x
1 – – 0 2
– 3 – – 0 2 2
y cot x
1 2
3 2 2
Domain: x 0, , 2, . . . Range: (–, –1] h [1, )
x
– – 0 2
2
3 2 2
Domain: x 0, , 2, . . . Range: (–, )
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x
x
Published on Nov 23, 2012
Calculus (11 ed.Text Book) by Thomas (Ch16+Appendixes) for BSSE