Bismillah hir Rehman nir Raheem ------------------------------------Assalat o Wasalam o Allika Ya RasoolALLAH

Calculus

th

(7 Ed. Text Book)

Howard Anton, Bivens, Davis

Ch Ch

Published By: Muhammad Hassan Riaz Yousufi

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

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A NALYTIC G EOMETRY IN C ALCULUS

n this chapter we will study aspects of analytic geometry that are important in applications of calculus. We will begin by introducing polar coordinate systems, which are used, for example, in tracking the motion of planets and satellites, in identifying the locations of objects from information on radar screens, and in the design of antennas. We will then discuss relationships between curves in polar coordinates and parametric curves in rectangular coordinates, and we will discuss methods for finding areas in polar coordinates and tangent lines to curves given in polar coordinates or parametrically in rectangular coordinates. We will then review the basic properties of parabolas, ellipses, and hyperbolas and discuss these curves in the context of polar coordinates. Finally, we will give some basic applications of our work in astronomy.

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Analytic Geometry in Calculus

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726

11.1 POLAR COORDINATES

••••••••••••••••••••••••••••••••••••••

P(r, u)

Ri

r u

O

A polar coordinate system in a plane consists of a fixed point O, called the pole (or origin), and a ray emanating from the pole, called the polar axis. In such a coordinate system we can associate with each point P in the plane a pair of polar coordinates (r, θ ), where r is the distance from P to the pole and θ is an angle from the polar axis to the ray OP (Figure 11.1.1). The number r is called the radial coordinate of P and the number θ the angular coordinate (or polar angle) of P . In Figure 11.1.2, the points (6, 45 ◦ ), (5, 120 ◦ ), (3, 225 ◦ ), and (4, 330 ◦ ) are plotted in polar coordinate systems. If P is the pole, then r = 0, but there is no clearly defined polar angle. We will agree that an arbitrary angle can be used in this case; that is, (0, θ) are polar coordinates of the pole for all choices of θ .

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POLAR COORDINATE SYSTEMS

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Up to now we have specified the location of a point in the plane by means of coordinates relative to two perpendicular coordinate axes. However, sometimes a moving point has a special affinity for some fixed point, such as a planet moving in an orbit under the central attraction of the Sun. In such cases, the path of the particle is best described by its angular direction and its distance from the fixed point. In this section we will discuss a new kind of coordinate system that is based on this idea.

Pole

Polar axis

Figure 11.1.1

(5, 120°)

225°

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n

(6, 45°)

120°

45°

Ha

Figure 11.1.2

330° (4, 330°)

(3, 225°)

The polar coordinates of a point are not unique. For example, the polar coordinates (1, 315 ◦ ),

(1, −45 ◦ ),

and

(1, 675 ◦ )

all represent the same point (Figure 11.1.3). In general, if a point P has polar coordinates (r, θ ), then

ad

(r, θ + n · 360 ◦ )

and

(r, θ − n · 360 ◦ )

m

are also polar coordinates of P for any nonnegative integer n. Thus, every point has infinitely many pairs of polar coordinates.

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315°

225°

Polar axis

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id

ls

a in

rm

Te

P(3, 225°)

de

l na

si

i

rm

Te

P(–3, 45°) Figure 11.1.4

45°

Polar axis

675° –45° 1

1 (1, 315°)

1 (1, –45°)

(1, 675°)

Figure 11.1.3

As defined above, the radial coordinate r of a point P is nonnegative, since it represents the distance from P to the pole. However, it will be convenient to allow for negative values of r as well. To motivate an appropriate definition, consider the point P with polar coordinates (3, 225 ◦ ). As shown in Figure 11.1.4, we can reach this point by rotating the polar axis through an angle of 225 ◦ and then moving 3 units from the pole along the terminal side of the angle, or we can reach the point P by rotating the polar axis through an angle of 45 ◦ and then moving 3 units from the pole along the extension of the terminal side. This suggests that the point (3, 225 ◦ ) might also be denoted by (−3, 45 ◦ ), with the minus sign serving to indicate that the point is on the extension of the angle’s terminal side rather than on the terminal side itself.

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Polar Coordinates

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11.1

In general, the terminal side of the angle θ + 180 ◦ is the extension of the terminal side of θ , so we define negative radial coordinates by agreeing that (−r, θ )

and

(r, θ + 180 ◦ )

are polar coordinates of the same point. • FOR THE READER.

RELATIONSHIP BETWEEN POLAR AND RECTANGULAR COORDINATES

p/ 2

r

x = r cos θ,

(x, y) (r, u)

u

x

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(1)

These equations are well suited for finding x and y when r and θ are known. However, to find r and θ when x and y are known, it is preferable to use the identities sin2 θ + cos2 θ = 1 and tan θ = sin θ / cos θ to rewrite (1) as

y = r sin u

0

x = r cos u

y = r sin θ

r 2 = x2 + y2,

n

P

tan θ =

y x

(2)

sa

y

Frequently, it will be useful to superimpose a rectangular xy-coordinate system on top of a polar coordinate system, making the positive x-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates (x, y) and polar coordinates (r, θ ). As suggested by Figure 11.1.5, these coordinates are related by the equations

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••••••••••••••••••••••••••••••••••••••

For many purposes it does not matter whether polar angles are measured in degrees or radians. However, in problems that involve derivatives or integrals they must be measured in radians, since the derivatives of the trigonometric functions were derived under this assumption. Henceforth, we will use radian measure for polar angles, except in certain applications where it is not required and degree measure is more convenient.

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• • • • • • • • • • • • • • • • • • • • • • •

Figure 11.1.5

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Example 1 Find the rectangular coordinates of the point P whose polar coordinates are (6, 2π/3).

Solution. Substituting the polar coordinates r = 6 and θ = 2π/3 in (1) yields 2π 1 x = 6 cos =6 − = −3 3 2 √ √ 2π 3 y = 6 sin =6 =3 3 3 2

p/ 2

√ Thus, the rectangular coordinates of P are (−3, 3 3) (Figure 11.1.6).

2p/ 3

m

r=6

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y

P

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x

Figure 11.1.6

0

Example √ 2 (−2, 2 3 ).

Find polar coordinates of the point P whose rectangular coordinates are

Solution. We will find the polar coordinates (r, θ ) of P that satisfy the conditions r > 0 and 0 ≤ θ < 2π. From the first equation in (2), √ r 2 = x 2 + y 2 = (−2)2 + (2 3 )2 = 4 + 12 = 16 so r = 4. From the second equation in (2), √ √ y 2 3 tan θ = = =− 3 x −2 √ From this and the fact that (−2, 2 3 ) lies in the second quadrant, it follows that the angle satisfying the requirement 0 ≤ θ < 2π is θ = 2π/3. Thus, (4, 2π/3) are polar coordinates of P . All other polar coordinates of P are expressible in the form 2π 5π + 2nπ 4, + 2nπ or −4, 3 3 where n is an integer.

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Analytic Geometry in Calculus

••••••••••••••••••••••••••••••••••••••

We will now consider the problem of graphing equations of the form r = f(θ) in polar coordinates, where θ is assumed to be measured in radians. Some examples of such equations are 4 r = 2 cos θ, r = , r=θ 1 − 3 sin θ In a rectangular coordinate system the graph of an equation y = f(x) consists of all points whose coordinates (x, y) satisfy the equation. However, in a polar coordinate system, points have infinitely many different pairs of polar coordinates, so that a given point may have some polar coordinates that satisfy the equation r = f(θ) and others that do not. Taking this into account, we define the graph of r = f (θ) in polar coordinates to consist of all points with at least one pair of coordinates (r, θ ) that satisfy the equation. The most elementary way to graph an equation r = f(θ) in polar coordinates is to plot points. The idea is to choose some typical values of θ, calculate the corresponding values of r, and then plot the resulting pairs (r, θ ) in a polar coordinate system. Here are some examples. Example 3 points.

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GRAPHS IN POLAR COORDINATES

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Sketch the graph of the equation r = sin θ in polar coordinates by plotting

Solution. Table 11.1.1 shows the coordinates of points on the graph at increments of

n

π/6 (= 30 ◦ ).

0

2

4

r = sin u

0

1 2

√3

(r, u)

(0, 0)

6

1

2

8

a

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u (radians)

sa

Table 11.1.1

√3 2

1 2

(12 , a)

(12 , 2 )

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k

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0

–1 2

– √3

–1

– √3 2

–1 2

0

(0, c)

2

(– 12 , 76π ) (– √23 , 43π ) (–1, 32π ) (– √23 , 53π ) (– 12 , 116π )

(0, 2π )

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(√23 , 4 )

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These points are plotted in Figure 11.1.7. Note, however, that there are 13 points listed in the table but only 6 distinct plotted points. This is because the pairs from θ = π on yield duplicates of the preceding points. For example, (−1/2, 7π/6) and (1/2, π/6) represent the same point.

(1, 6 ) (√23 , 8)

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( 12 , π6 ) ( √23 , π3 ) (1, π2 ) ( √23 , 23π ) ( 12 , 56π )

c

(0, 0)

r = sin u

Figure 11.1.7

r

1

r = sin u

u

6

-1

Figure 11.1.8

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i

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Observe that the points in Figure 11.1.7 appear to lie on a circle. We can confirm that this is so by expressing the polar equation r = sin θ in terms of x and y. To do this, we multiply the equation through by r to obtain r 2 = r sin θ which now allows us to apply Formulas (1) and (2) to rewrite the equation as x2 + y2 = y

Rewriting this equation as x 2 + y 2 − y = 0 and then completing the square yields 2 x 2 + y − 12 = 14 which is a circle of radius 12 centered at the point 0, 12 in the xy-plane. Just because an equation r = f(θ) involves the variables r and θ does not mean that it has to be graphed in a polar coordinate system. When useful, this equation can also be graphed in a rectangular coordinate system. For example, Figure 11.1.8 shows the graph of r = sin θ in a rectangular θr-coordinate system. This graph can actually help to visualize how the polar graph in Figure 11.1.7 is generated:

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Polar Coordinates

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11.1

At θ = 0 we have r = 0, which corresponds to the pole (0, 0) on the polar graph. As θ varies from 0 to π/2, the value of r increases from 0 to 1, so the point (r, θ ) moves along the circle from the pole to the high point at (1, π/2). • As θ varies from π/2 to π, the value of r decreases from 1 back to 0, so the point (r, θ ) moves along the circle from the high point back to the pole. • As θ varies from π to 3π/2, the values of r are negative, varying from 0 to −1. Thus, the point (r, θ ) moves along the circle from the pole to the high point at (1, π/2), which is the same as the point (−1, 3π/2). This duplicates the motion that occurred for 0 ≤ θ ≤ π/2. • As θ varies from 3π/2 to 2π, the value of r varies from −1 to 0. Thus, the point (r, θ ) moves along the circle from the high point back to the pole, duplicating the motion that occurred for π/2 ≤ θ ≤ π. •

r 1

u c

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•

Example 4 Sketch the graph of r = cos 2θ in polar coordinates.

Solution. Instead of plotting points, we will use the graph of r = cos 2θ in rectangular

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-1

coordinates (Figure 11.1.9) to visualize how the polar graph of this equation is generated. The analysis and the resulting polar graph are shown in Figure 11.1.10. This curve is called a four-petal rose.

r = cos 2u

r varies from 0 to –1 as u

varies from 0 to p/ 4.

p/ 4 to p/ 2.

varies from

••••••••••••••••••••••••••••••••••••••

r varies from 0 to 1 as u

r varies from 1 to 0 as u

r varies from 0 to –1 as u

varies from

varies from 3p/ 4 to p.

varies from p to 5p/ 4.

varies from

varies from

varies from

5p/ 4 to 3p/ 2.

3p/ 2 to 7p/ 4.

7p/ 4 to 2p.

p/ 2 to 3p/ 4.

r varies from –1 to 0 as u

r varies from 0 to 1 as u

Observe that the polar graph of r = cos 2θ in Figure 11.1.10 is symmetric about the xaxis and the y-axis. This symmetry could have been predicted from the following theorem, which is suggested by Figure 11.1.11 (we omit the proof).

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SYMMETRY TESTS

r varies from –1 to 0 as u

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Figure 11.1.10

r varies from 1 to 0 as u

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n

Figure 11.1.9

Figure 11.1.11

11.1.1 THEOREM (Symmetry Tests). (a) A curve in polar coordinates is symmetric about the x-axis if replacing θ by −θ in its equation produces an equivalent equation (Figure 11.1.11a). (b) A curve in polar coordinates is symmetric about the y-axis if replacing θ by π − θ in its equation produces an equivalent equation (Figure 11.1.11b). (c) A curve in polar coordinates is symmetric about the origin if replacing r by −r in its equation produces an equivalent equation (Figure 11.1.11c). p/ 2

p/ 2 (r, p – u)

(r, u)

(r, u)

0

(r, u) 0

(r, –u)

(a)

p/ 2

0

(–r, u)

(b)

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(c)

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Analytic Geometry in Calculus

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Example 5 Use Theorem 11.1.1 to confirm that the graph of r = cos 2θ in Figure 11.1.10 is symmetric about the x-axis and y-axis.

Solution. To test for symmetry about the x-axis, we replace θ by −θ. This yields r = cos(−2θ ) = cos 2θ

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Thus, replacing θ by −θ does not alter the equation. To test for symmetry about the y-axis, we replace θ by π − θ. This yields r = cos 2(π − θ) = cos(2π − 2θ) = cos(−2θ) = cos 2θ Thus, replacing θ by π − θ does not alter the equation.

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Example 6 Sketch the graph of r = a(1 − cos θ) in polar coordinates, assuming a to be a positive constant.

Solution. Observe first that replacing θ by −θ does not alter the equation, so we know in

•

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advance that the graph is symmetric about the polar axis. Thus, if we graph the upper half of the curve, then we can obtain the lower half by reflection about the polar axis. As in our previous examples, we will first graph the equation in rectangular coordinates. This graph, which is shown in Figure 11.1.12a, can be obtained by rewriting the given equation as r = a − a cos θ, from which we see that the graph in rectangular coordinates can be obtained by first reflecting the graph of r = a cos θ about the x-axis to obtain the graph of r = −a cos θ, and then translating that graph up a units to obtain the graph of r = a − a cos θ. Now we can see that:

•

As θ varies from 2π/3 to π, r increases from 3a /2 to 2a.

•

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•

As θ varies from 0 to π/3, r increases from 0 to a /2. As θ varies from π/3 to π/2, r increases from a /2 to a. As θ varies from π/2 to 2π/3, r increases from a to 3a /2.

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This produces the polar curve shown in Figure 11.1.12b. The rest of the curve can be obtained by continuing the preceding analysis from π to 2π or, as noted above, by reflecting the portion already graphed about the x-axis (Figure 11.1.12c). This heart-shaped curve is called a cardioid (from the Greek word “kardia” for heart).

m

r

M uh am

2a 3a 2 a a 2

( 3a2 , 2p3 )

a

(

c

o

p 2

)

2a

( 2a , p3 )

u 4 68

a,

a

(2a, p) r = a(1 – cos u)

(a)

(b)

Figure 11.1.12

Example 7 Sketch the curves π (a) r = 1 (b) θ = 4 in polar coordinates.

(c) r = θ

(θ ≥ 0)

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(c)

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Polar Coordinates

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11.1

Solution (a). For all values of θ, the point (1, θ) is 1 unit away from the pole. Thus, the graph is the circle of radius 1 centered at the pole (Figure 11.1.13a).

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Solution (b). For all values of r, the point (r, π/4) lies on a line that makes an angle of π/4 with the polar axis (Figure 11.1.13b). Positive values of r correspond to points on the line in the first quadrant and negative values of r to points on the line in the third quadrant. Thus, in absence of any restriction on r, the graph is the entire line. Observe, however, that had we imposed the restriction r ≥ 0, the graph would have been just the ray in the first quadrant.

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Solution (c). Observe that as θ increases, so does r; thus, the graph is a curve that spirals out from the pole as θ increases. A reasonably accurate sketch of the spiral can be obtained by plotting the intersections with the x- and y-axes for values of θ that are multiples of π/2, keeping in mind that the value of r is always equal to the value of θ (Figure 11.1.13c). p/ 2

Ri

p/ 2

p/4

0

r=u u≥0

r=1

Ha

(a)

sa

n

1

0

9p / 2 5p / 2 p/2 3p

2p

p

4p

3p / 2 7p / 2

u = p/4

r=u

(b)

(c)

Figure 11.1.13 • REMARK.

r=u u≤0

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Example 8 Sketch the graph of r 2 = 4 cos 2θ in polar coordinates.

M uh am r=u –∞ < u < +∞

Figure 11.1.14

The spiral in Figure 11.1.13c, which belongs to the family of Archimedean spirals r = aθ, coils counterclockwise around the pole because of the restriction θ ≥ 0. Had we made the restriction θ ≤ 0, the spiral would have coiled clockwise, and had we allowed both positive and negative values of θ, the clockwise and counterclockwise spirals would have been superimposed to form a double Archimedean spiral (Figure 11.1.14).

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• • • • • • • • • • • • • • • • • • • • • • •

Solution. This equation does not express r as a function of θ, since solving for r in terms of θ yields two functions: √ √ r = 2 cos 2θ and r = −2 cos 2θ Thus, to graph the equation r 2 = 4 cos 2θ we will have to graph the two functions separately and then combine those graphs. √ We will start with the graph of r = 2 cos 2θ. Observe first that this equation is not changed if we replace θ by −θ or if we replace θ by π − θ . Thus, the graph is symmetric about the x-axis and the y-axis. This means that the entire graph can be obtained by graphing the portion in the first quadrant, reflecting that portion about the y-axis to obtain the portion in the second quadrant and then reflecting those two portions about the x-axis to obtain the portions in the third and fourth quadrants. √ To begin the analysis, we will graph the equation r = 2 cos 2θ in rectangular coordinates (see Figure 11.1.15a). Note that there are gaps in that graph over the intervals π/4 < θ < 3π/4 and 5π/4 < θ < 7π/4 because cos 2θ is negative for those values of θ. From this graph we can see that:

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Analytic Geometry in Calculus • •

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732

As θ varies from 0 to π/4, r decreases from 2 to 0. As θ varies from π/4 to π/2, no points are generated on the polar graph.

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This produces the portion of the graph shown in Figure 11.1.15b. As noted above, we can complete the graph by a reflection about the y-axis followed by a reflection about the x-axis (11.1.15c). The resulting propeller-shaped graph is called a lemniscate (from the Greek word “lemniscos” for a looped √ribbon resembling the number 8). We√leave it for you to verify that the equation r = 2 cos 2θ has the same graph as r = −2 cos 2θ, but traced in a diagonally opposite manner. Thus, the graph of the equation r 2 = 4 cos 2θ consists of two identical superimposed lemniscates. p/ 2

p/ 2

az

u = p/4

r

2

2

3

9

f

l

0

-2

Ri

u

r = 2√cos 2u

o

(a)

(b)

2

0

(c)

••••••••••••••••••••••••••••••••••••••

If θ0 is a fixed angle, then for all values of r the point (r, θ0 ) lies on the line that makes an angle of θ = θ0 with the polar axis; and, conversely, every point on this line has a pair of polar coordinates of the form (r, θ0 ). Thus, the equation θ = θ0 represents the line that passes through the pole and makes an angle of θ0 with the polar axis (Figure 11.1.16a). If r is restricted to be nonnegative, then the graph of the equation θ = θ0 is the ray that emanates from the pole and makes an angle of θ0 with the polar axis (Figure 11.1.16b). Thus, as θ0 varies, the equation θ = θ0 produces either a family of lines through the pole or a family of rays through the pole, depending on the restrictions on r.

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sa

FAMILIES OF LINES AND RAYS THROUGH THE POLE

n

Figure 11.1.15

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m

ad

p/ 2

••••••••••••••••••••••••••••••••••••••

FAMILIES OF CIRCLES

p/ 2

u0

u0

0

u = u0

u = u0 (r ≥ 0)

(a)

(b)

0

Figure 11.1.16

We will consider three families of circles in which a is assumed to be a positive constant: r=a

r = 2a cos θ

r = 2a sin θ

(3–5)

The equation r = a represents a circle of radius a centered at the pole (Figure 11.1.17a). Thus, as a varies, this equation produces a family of circles centered at the pole. For families (4) and (5), recall from plane geometry that a triangle that is inscribed in a circle with a diameter of the circle for a side must be a right triangle. Thus, as indicated in Figures 11.1.17b and 11.1.17c, the equation r = 2a cos θ represents a circle of radius a, centered on the x-axis and tangent to the y-axis at the origin; similarly, the equation r = 2a sin θ represents a circle

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p/ 2

p/ 2

p/ 2

p/ 2

a

0

P(r, u)

u 2a

0

733

p/ 2

P(r, u) r u 2a

P(a, u)

Polar Coordinates

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11.1

0

r u

Yo

0

0

r=a

r = 2a cos u

r = 2a sin u

r = 2a cos u

r = 2a sin u

(a)

(b)

(c)

(d)

(e)

az

Figure 11.1.17

of radius a, centered on the y-axis and tangent to the x-axis at the origin. Thus, as a varies, Equations (4) and (5) produce the families illustrated in Figures 11.1.17d and 11.1.17e.

••••••••••••••••••••••••••••••••••••••

Observe that replacing θ by −θ does not change the equation r = 2a cos θ, and replacing θ by π − θ does not change the equation r = 2a sin θ. This explains why the circles in Figure 11.1.17d are symmetric about the x-axis and those in Figure 11.1.17e are symmetric about the y-axis.

n

• • • • • • • • • • • • • • • • • •

Ri

• REMARK.

In polar coordinates, equations of the form

FAMILIES OF ROSE CURVES

r = a cos nθ

sa

r = a sin nθ

(6–7)

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in which a > 0 and n is a positive integer represent families of flower-shaped curves called roses (Figure 11.1.18). The rose consists of n equally spaced petals of radius a if n is odd and 2n equally spaced petals of radius a if n is even. It can be shown that a rose with an even number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < 2π and a rose with an odd number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < π (Exercise 73). A four-petal rose of radius 1 was graphed in Example 4.

n=2

n=3

n=4

n=5

n=6

m

ad

rose curves

M uh am

r = a sin nu

r = a cos nu

Figure 11.1.18

••••••••••••••••••••••••••••••••••••••

FAMILIES OF CARDIOIDS AND LIMAC¸ONS

•

FOR THE READER.

What do the graphs of the one-petal roses look like?

Equations with any of the four forms r = a ± b sin θ

r = a ± b cos θ

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(8–9)

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Analytic Geometry in Calculus

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734

Yo

in which a > 0 and b > 0 represent polar curves called lima¸cons (from the Latin word “limax” for a snail-like creature that is commonly called a slug). There are four possible shapes for a limac¸ on that are determined by the ratio a /b (Figure 11.1.19). If a = b (the case a /b = 1), then the limac¸ on is called a cardioid because of its heart-shaped appearance, as noted in Example 6.

a/b = 1

1 < a/b < 2

a/b ≥ 2

Limaçon with inner loop

Cardioid

Dimpled limaçon

Convex limaçon

az

a/b < 1

Ri

Figure 11.1.19

a = 0.25

Ha

sa

n

Example 9 Figure 11.1.20 shows the family of limac¸ ons r = a + cos θ with the constant a varying from 0.25 to 2.50 in steps of 0.25. In keeping with Figure 11.1.19, the limac¸ ons evolve from the loop type to the convex type. As a increases from the starting value of 0.25, the loops get smaller and smaller until the cardioid is reached at a = 1. As a increases further, the limac¸ ons evolve through the dimpled type into the convex type.

a = 0.5

a = 0.75

a=1

a = 1.25

a = 1.50

a = 1.75

a = 2.00

a = 2.25

a = 2.50

ad

r = a + cos u Figure 11.1.20

••••••••••••••••••••••••••••••••••••••

A spiral is a curve that coils around a central point. As illustrated in Figure 11.1.14, spirals generally have “left-hand” and “right-hand” versions that coil in opposite directions, depending on the restrictions on the polar angle and the signs of constants that appear in their equations. Some of the more common types of spirals are shown in Figure 11.1.21 for nonnegative values of θ, a, and b.

M uh am

m

FAMILIES OF SPIRALS

p/ 2

p/ 2

p/ 2

p/ 2

p/ 2

0

0

0

0 0

Archimedean spiral

Parabolic spiral

Logarithmic spiral

Lituus spiral

Hyperbolic spiral

r = au

r = a √u

r = ae bu

r = a /√u

r = a /u

Figure 11.1.21

••••••••••••••••••••••••••••••••••••••

SPIRALS IN NATURE

Spirals of many kinds occur in nature. For example, the shell of the chambered nautilus (below) forms a logarithmic spiral, and a coiled sailor’s rope forms an Archimedean spiral. Spirals also occur in flowers, the tusks of certain animals, and in the shapes of galaxies.

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Polar Coordinates

735

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11.1

GENERATING POLAR CURVES WITH GRAPHING UTILITIES

az

For polar curves that are too complicated for hand computation, graphing utilities must be used. Although many graphing utilities are capable of graphing polar curves directly, some are not. However, if a graphing utility is capable of graphing parametric equations, then it can be used to graph a polar curve r = f(θ) by converting this equation to parametric form. This can be done by substituting f(θ) for r in (1). This yields y = f(θ) sin θ

(10)

sa

x = f(θ) cos θ,

n

••••••••••••••••••••••••••••••••••••••

A sailor’s coiled rope forms an Archimedean spiral.

Ri

The shell of the chambered nautilus reveals a logarithmic spiral. The animal lives in the outermost chamber.

which is a pair of parametric equations for the polar curve in terms of the parameter θ.

Ha

Example 10 Express the polar equation 5θ 2 parametrically, and generate the polar graph from the parametric equations using a graphing utility. r = 2 + cos

M uh am

m

ad

Solution. Substituting the given expression for r in x = r cos θ and y = r sin θ yields

p/ 2

0

r = 2 + cos

Figure 11.1.22

5u 2

the parametric equations 5θ x = 2 + cos cos θ, 2

5θ y = 2 + cos sin θ 2

Next, we need to find an interval over which to vary θ to produce the entire graph. To find such an interval, we will look for the smallest number of complete revolutions that must occur until the value of r begins to repeat. Algebraically, this amounts to finding the smallest positive integer n such that 5(θ + 2nπ) 5θ 2 + cos = 2 + cos 2 2 or 5θ 5θ cos + 5nπ = cos 2 2 For this equality to hold, the quantity 5nπ must be an even multiple of π; the smallest n for which this occurs is n = 2. Thus, the entire graph will be traced in two revolutions, which means it can be generated from the parametric equations 5θ 5θ x = 2 + cos cos θ, y = 2 + cos sin θ (0 ≤ θ ≤ 4π) 2 2 This yields the graph in Figure 11.1.22.

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Analytic Geometry in Calculus

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• FOR THE READER.

EXERCISE SET 11.1

Some graphing utilities require that t be used for the parameter. If this is true of your graphing utility, then you will have to replace θ by t in (10) to generate graphs in polar coordinates. Use a graphing utility to duplicate the curve in Figure 11.1.22.

Yo

• • • • • • • • • • • •

Graphing Utility

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

In Exercises 1 and 2, plot the points in polar coordinates.

In Exercises 11 and 12, express the given equations in polar coordinates.

(c) (1, π/2) (f ) (−1, 9π/4)

2. (a) (2, −π/3) (d) (−5, −π/6)

(b) (3/2, −7π/4) (c) (−3, 3π/2) (e) (2, 4π/3) (f ) (0, π)

az

(b) (5, 2π/3) (e) (−6, −π)

(c) (−6, −5π/6) (f ) (−5, 0)

4. (a) (−8, π/4) (d) (5, 0)

(b) (7, −π/4) (e) (−2, −3π/2)

(c) (8, 9π/4) (f ) (0, π)

12. (a) y = −3 (c) x 2 + y 2 + 4x = 0

(b) x 2 + y 2 = 5 (d) x 2 (x 2 + y 2 ) = y 2

In Exercises 13–16, a graph is given in a rectangular θrcoordinate system. Sketch the corresponding graph in polar coordinates.

sa

(b) (7, 2π/3) (e) (7, 17π/6)

(b) x 2 + y 2 = 9 (d) 4xy = 9

n

In Exercises 3 and 4, find the rectangular coordinates of the points whose polar coordinates are given. 3. (a) (6, π/6) (d) (0, −π)

11. (a) x = 7 (c) x 2 + y 2 − 6y = 0

Ri

1. (a) (3, π/4) (d) (4, 7π/6)

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m

ad

6. In each part find polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are √ (− 3, 1). (a) r ≥ 0 and 0 ≤ θ < 2π (b) r ≤ 0 and 0 ≤ θ < 2π (c) r ≥ 0 and −2π < θ ≤ 0 (d) r ≤ 0 and −π < θ ≤ π In Exercises 7 and 8, use a calculating utility, where needed, to approximate the polar coordinates of the points whose rectangular coordinates are given.

r

2 u c

(b) (2, −5)

8. (a) (−3, 4)

(b) (−3, 1.7)

10. (a) r = 5 sec θ (c) r = 4 cos θ + 4 sin θ

2

r

15.

6

c

a

-2

-3

r

16.

7

4 2

3 u -1

2

4

6

u 6

8

c

i

o

In Exercises 17–20, find an equation for the given polar graph. 17. (a)

(b)

(c)

−1

(c) (1, tan 1) (c) 2, sin−1 12

5

(b) r sin θ = 4 6 (d) r = 3 cos θ + 2 sin θ (b) r = 2 sin θ (d) r = sec θ tan θ

6

Circle

In Exercises 9 and 10, identify the curve by transforming the given polar equation to rectangular coordinates.

(c) r = 3 cos θ

u

o

1

7. (a) (4, 3)

9. (a) r = 2

r

14.

3

Ha

5. In each part, a point is given in rectangular coordinates. Find two pairs of polar coordinates for the point, one pair satisfying r ≥ 0 and 0 ≤ θ < 2π, and the second pair satisfying r ≥ 0 and −π <√θ ≤ π. (a) (−5, 0) (b) (2 3, −2) (c) (0, −2) √ (f ) (1, 1) (d) (−8, −8) (e) (−3, 3 3 )

13.

18. (a)

2

Circle

(b)

Cardioid

(c) 1

3

1 3

Limaçon

Circle

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Three-petal rose

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19. (a)

3

(b)

(c)

5

p/ 2 3

3 1

20. (a)

Limaçon

Lemniscate

(b)

(c)

0

4

3 1 6

Figure Ex-56 Five-petal rose

Circle

57. The accompanying figure shows the Archimedean spiral r = θ /2 produced with a graphing calculator. (a) What interval of values for θ do you think was used to generate the graph? (b) Duplicate the graph with your own graphing utility.

az

Cardioid

In Exercises 21–50, sketch the curve in polar coordinates. 3π 4 24. r = 4 sin θ

π 6 23. r = 3

22. θ = −

25. r = 6 cos θ

26. r = 1 + sin θ

27. 2r = cos θ

28. r − 2 = 2 cos θ

29. r = 3(1 − sin θ)

30. r = −5 + 5 sin θ

31. r = 4 − 4 cos θ

32. r = 1 + 2 sin θ

33. r = −1 − cos θ

34. r = 4 + 3 cos θ

35. r = 2 + sin θ

36. r = 3 − cos θ

37. r = 3 + 4 cos θ

38. r − 5 = 3 sin θ

39. r = 5 − 2 cos θ

40. r = −3 − 4 sin θ

41. r = 9 cos 2θ

42. r 2 = sin 2θ

43. r 2 = 16 sin 2θ

44. r = 4θ

n sa

Ha

(θ ≤ 0)

Figure Ex-57

58. The accompanying figure shows graphs of the√Archimedean spiral r = θ and the parabolic spiral r = θ. Which is which? Explain your reasoning. p/ 2

0

m

48. r = 3 sin 2θ

49. r = 9 sin 4θ

p/ 2

(θ ≥ 0)

46. r = 4θ

47. r = cos 2θ

0

50. r = 2 cos 3θ

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In Exercises 51–55, use a graphing utility to generate the polar graph. Be sure to choose the parameter interval so that a complete graph is generated.

θ 51. r = cos 2

53. r = 1 + 2 cos

θ 4

I

II

Figure Ex-58

59. (a) Show that if a varies, then the polar equation θ 52. r = sin 2 54. r = 0.5 + cos

r = a sec θ θ 3

θ 5 56. The accompanying figure shows the graph of the “butterfly curve” θ r = ecos θ − 2 cos 4θ + sin3 4 Generate the complete butterfly with a graphing utility, and state the parameter interval you used. 55. r = cos

[–9, 9] × [–6, 6] xScl = 1, yScl = 1

ad

45. r = 4θ

Ri

21. θ =

2

737

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Four-petal rose

Polar Coordinates

us uf i

11.1

(−π/2 < θ < π/2)

describes a family of lines perpendicular to the polar axis. (b) Show that if b varies, then the polar equation r = b csc θ

(0 < θ < π)

describes a family of lines parallel to the polar axis. 60. Show that if the polar graph of r = f(θ ) is rotated counterclockwise around the origin through an angle α, then r = f(θ − α) is an equation for the rotated curve. [Hint: If (r0 , θ0 ) is any point on the original graph, then (r0 , θ0 + α) is a point on the rotated graph.]

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Analytic Geometry in Calculus

(a) Show that if a = b, then the polar equation of the Cassini oval is r 2 = 2a 2 cos 2θ , which is a lemniscate. (b) Use the formula in Exercise 67(a) to show that the lemniscate in part (a) is the curve traced by a point that moves in such a way that the product of its distances from the polar points (a, 0) and (a, π) is a 2 .

61. Use the result in Exercise 60 to find an equation for the cardioid r = 1 + cos θ after it has been rotated through the given angle, and check your answer with a graphing utility. π 5π π (a) (c) π (d) (b) 2 4 4 62. Use the result in Exercise 60 to find an equation for the lemniscate that results when the lemniscate in Example 8 is rotated counterclockwise through an angle of π/2.

y

63. Sketch the polar graph of the equation (r − 1)(θ − 1) = 0.

Yo

738

x

64. (a) Show that if A and B are not both zero, then the graph of the polar equation

a=b

r = A sin θ + B cos θ

a>b

az

a<b Figure Ex-68

is a circle. Find its radius. (b) Derive Formulas (4) and (5) from the formula given in part (a).

Ri

Vertical and horizontal asymptotes of polar curves can often be detected by investigating the behavior of x = r cos θ and y = r sin θ as θ varies. This idea is used in Exercises 69–72.

65. Find the highest point on the cardioid r = 1 + cos θ.

sa

67. (a) Show that in a polar coordinate system the distance d between the points (r1 , θ1 ) and (r2 , θ2 ) is d = r12 + r22 − 2r1 r2 cos(θ1 − θ2 )

69. Show that the hyperbolic spiral r = 1/θ (θ > 0) has a horizontal asymptote at y = 1 by showing that y → 1 and x → +⬁ as θ → 0+ . Confirm this result by generating the spiral with a graphing utility. 70. Show that the spiral r = 1/θ 2 does not have any horizontal asymptotes.

n

66. Find the leftmost point on the upper half of the cardioid r = 1 + cos θ.

Ha

(b) Show that if 0 ≤ θ1 < θ2 ≤ π and if r1 and r2 are positive, then the area A of the triangle with vertices (0, 0), (r1 , θ1 ), and (r2 , θ2 ) is A = 12 r1 r2 sin(θ2 − θ1 )

ad

(c) Find the distance between the points whose polar coordinates are (3, π/6) and (2, π/3). (d) Find the area of the triangle whose vertices in polar coordinates are (0, 0), (1, 5π/6), and (2, π/3).

68. In the late seventeenth century the Italian astronomer Giovanni Domenico Cassini (1625–1712) introduced the family of curves

m

(x 2 + y 2 + a 2 )2 − b4 − 4a 2 x 2 = 0

(a > 0, b > 0)

M uh am

in his studies of the relative motions of the Earth and the Sun. These curves, which are called Cassini ovals, have one of the three basic shapes shown in the accompanying figure.

••••••••••••••••••••••••••••••••••••••

TANGENT LINES TO PARAMETRIC CURVES

71. (a) Show that the kappa curve r = 4 tan θ (0 ≤ θ ≤ 2π) has a vertical asymptote at x = 4 by showing that x → 4 and y → +⬁ as θ → π/2− and that x → 4 and y → −⬁ as θ → π/2+ . (b) Use the method in part (a) to show that the kappa curve also has a vertical asymptote at x = −4. (c) Confirm the results in parts (a) and (b) by generating the kappa curve with a graphing utility. 72. Use a graphing utility to make a conjecture about the existence of asymptotes for the cissoid r = 2 sin θ tan θ, and then confirm your conjecture by calculating appropriate limits.

73. Prove that a rose with an even number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < 2π and a rose with an odd number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < π.

11.2 TANGENT LINES AND ARC LENGTH FOR PARAMETRIC AND POLAR CURVES In this section we will derive the formulas required to find slopes, tangent lines, and arc lengths of parametric and polar curves. We will be concerned in this section with curves that are given by parametric equations x = f(t),

y = g(t)

in which f(t) and g(t) have continuous first derivatives with respect to t. It can be proved that if dx /dt = 0, then y is a differentiable function of x, in which case the chain rule

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Tangent Lines and Arc Length for Parametric and Polar Curves

739

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11.2

implies that dy/dt dy = dx dx/dt

(1)

Yo

This formula makes it possible to find dy/dx directly from the parametric equations without eliminating the parameter.

y

Example 1 Find the slope of the tangent line to the unit circle x = cos t,

1 p/6

x

y = sin t

(0 ≤ t ≤ 2π)

at the point where t = π/6 (Figure 11.2.1).

Ri

dy dy/dt cos t = = = − cot t dx dx/dt − sin t Thus, the slope at t = π/6 is

√ dy

π = − cot = − 3

dx t=π/6 6

az

Solution. From (1), the slope at a general point on the circle is

Figure 11.2.1

• REMARK.

P(cos t, sin t) x

It follows from Formula (1) that the tangent line to a parametric curve will be horizontal at those points where dy/dt = 0 and dx/dt = 0, since dy/dx = 0 at such points. Two different situations occur when dx/dt = 0. At points where dx/dt = 0 and dy/dt = 0, the right side of (1) has a nonzero numerator and a zero denominator; we will agree that the curve has infinite slope and a vertical tangent line at such points. At points where dx/dt and dy/dt are both zero, the right side of (1) becomes an indeterminate form; we call such points singular points. No general statement can be made about the behavior of parametric curves at singular points; they must be analyzed case by case.

Ha

1 t

Note that Formula (2) makes sense geometrically because the radius to the point P (cos t, sin t) has slope m = tan t; hence, the tangent line at P , being perpendicular to the radius, has slope −1/m = −1/ tan t = − cot t (Figure 11.2.2).

n

• • • • • • • • • • • •

sa

y

(2)

O

ad

Radius OP has slope m = tan t.

Figure 11.2.2

M uh am

m

Example 2 In a disastrous first flight, an experimental paper airplane follows the trajectory x = t − 3 sin t,

y = 4 − 3 cos t

(t ≥ 0)

but crashes into a wall at time t = 10 (Figure 11.2.3). (a) (b)

At what times was the airplane flying horizontally? At what times was it flying vertically?

Solution (a). The airplane was flying horizontally at those times when dy/dt = 0 and

dx/dt = 0. From the given trajectory we have

dy dx = 1 − 3 cos t (3) = 3 sin t and dt dt Setting dy/dt = 0 yields the equation 3 sin t = 0, or, more simply, sin t = 0. This equation has four solutions in the time interval 0 ≤ t ≤ 10: t = 0,

t = π,

t = 2π,

t = 3π

Since dx/dt = 1 − 3 cos t = 0 for these values of t (verify), the airplane was flying horizontally at times t = 0,

t = π ≈ 3.14,

t = 2π ≈ 6.28,

and

t = 3π ≈ 9.42

which is consistent with Figure 11.2.3.

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Analytic Geometry in Calculus

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740

y 8

t=3

t=9 t = 10

t=4

t=2

t=8

t=7

t=0

t=6

Yo

t=5 t=1

x

-2

12

y

1 3

o

1 3

dy/dt = 0. Setting dx/dt = 0 in (3) yields the equation

x 10

1 âˆ’ 3 cos t = 0

or

cos t =

1 3

This equation has three solutions in the time interval 0 â‰¤ t â‰¤ 10 (Figure 11.2.4): t = cosâˆ’1 13 , t = 2Ď€ âˆ’ cosâˆ’1 13 , t = 2Ď€ + cosâˆ’1 13 Since dy/dt = 3 sin t is not zero at these points (why?), it follows that the airplane was flying vertically at times

-1

Figure 11.2.4

t = cosâˆ’1

1 3

â‰ˆ 1.23,

t â‰ˆ 2Ď€ âˆ’ 1.23 â‰ˆ 5.05,

n

cosâ€“1

Solution (b). The airplane was flying vertically at those times when dx/dt = 0 and

y = cos t

Ri

1

az

Figure 11.2.3

sa

which again is consistent with Figure 11.2.3.

t â‰ˆ 2Ď€ + 1.23 â‰ˆ 7.51

Example 3 The curve represented by the parametric equations

y

x = t 2,

y = t3

(âˆ’âŹ < t < +âŹ )

Ha

6

is called a semicubical parabola. The parameter t can be eliminated by cubing x and squaring y, from which it follows that y 2 = x 3 . The graph of this equation, shown in Figure 11.2.5, consists of two branches: an upper branch obtained by graphing y = x 3/2 and a lower branch obtained by graphing y = âˆ’x 3/2 . The two branches meet at the origin, which corresponds to t = 0 in the parametric equations. This is a singular point because the derivatives dx/dt = 2t and dy/dt = 3t 2 are both zero there.

x

ad

5

m

Example 4 Without eliminating the parameter, find dy/dx and d 2 y/dx 2 at the points (1, 1) and (1, âˆ’1) on the semicubical parabola given by the parametric equations in Example 3.

Solution. From (1) we have

â€“6

M uh am

x = t 2, y = t 3 (â€“âˆž < t < +âˆž)

Figure 11.2.5

dy/dt 3t 2 3 dy = = = t (t = 0) (4) / dx dx dt 2t 2 and from (1) applied to y = dy/dx we have d 2y dy /dt 3/2 3 dy = = = (5) = 2 dx dx dx/dt 2t 4t Since the point (1, 1) on the curve corresponds to t = 1 in the parametric equations, it follows from (4) and (5) that

d 2 y

dy

3 3 and = = dx t=1 2 dx 2 t=1 4 Similarly, the point (1, âˆ’1) corresponds to t = âˆ’1 in the parametric equations, so applying (4) and (5) again yields

dy

3 3 d 2 y

=âˆ’ =âˆ’ and dx t=âˆ’1 2 dx 2 t=âˆ’1 4 Note that the values we obtained for the first and second derivatives are consistent with the graph in Figure 11.2.5, since at (1, 1) on the upper branch the tangent line has positive

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Tangent Lines and Arc Length for Parametric and Polar Curves

741

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11.2

TANGENT LINES TO POLAR CURVES

Our next objective is to find a method for obtaining slopes of tangent lines to polar curves of the form r = f(Î¸) in which r is a differentiable function of Î¸. We showed in the last section that a curve of this form can be expressed parametrically in terms of the parameter Î¸ by substituting f(Î¸) for r in the equations x = r cos Î¸ and y = r sin Î¸. This yields x = f(Î¸) cos Î¸,

y = f(Î¸) sin Î¸

az

â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

Yo

slope and the curve is concave up, and at (1, âˆ’1) on the lower branch the tangent line has negative slope and the curve is concave down. Finally, observe that we were able to apply Formulas (4) and (5) for both t = 1 and t = âˆ’1, even though the points (1, 1) and (1, âˆ’1) lie on different branches. In contrast, had we chosen to perform the same computations by eliminating the parameter, we would have had to obtain separate derivative formulas for y = x 3/2 and y = âˆ’x 3/2 .

n

Ri

from which we obtain dx dr cos Î¸ = âˆ’f (Î¸ ) sin Î¸ + f (Î¸ ) cos Î¸ = âˆ’r sin Î¸ + dÎ¸ dÎ¸ (6) dy dr = f (Î¸ ) cos Î¸ + f (Î¸ ) sin Î¸ = r cos Î¸ + sin Î¸ dÎ¸ dÎ¸ Thus, if dx/dÎ¸ and dy/dÎ¸ are continuous and if dx/dÎ¸ = 0, then y is a differentiable function of x, and Formula (1) with Î¸ in place of t yields

sa

dr r cos Î¸ + sin Î¸ dy/dÎ¸ dy dÎ¸ = = dr dx dx/dÎ¸ âˆ’r sin Î¸ + cos Î¸ dÎ¸

Ha

(7)

Example 5 Find the slope of the tangent line to the circle r = 4 cos Î¸ at the point where Î¸ = Ď€/4.

Solution. From (7) with r = 4 cos Î¸ we obtain (verify)

ad

p/ 2

m

Tangent

0

M uh am

p/ 4 4

r = 4 cos u

Figure 11.2.6

dy 4 cos2 Î¸ âˆ’ 4 sin2 Î¸ 4 cos 2Î¸ = = = âˆ’ cot 2Î¸ dx âˆ’8 sin Î¸ cos Î¸ âˆ’4 sin 2Î¸ Thus, at the point where Î¸ = Ď€/4 the slope of the tangent line is

dy

Ď€ m= = âˆ’ cot = 0 dx Î¸ =Ď€/4 2

which implies that the circle has a horizontal tangent line at the point where Î¸ = Ď€/4 (Figure 11.2.6). Example 6 Find the points on the cardioid r = 1 âˆ’ cos Î¸ at which there is a horizontal tangent line, a vertical tangent line, or a singular point.

Solution. A horizontal tangent line will occur where dy/dÎ¸ = 0 and dx/dÎ¸ = 0, a vertical

tangent line where dy/dÎ¸ = 0 and dx/dÎ¸ = 0, and a singular point where dy/dÎ¸ = 0 and dx/dÎ¸ = 0. We could find these derivatives from the formulas in (6). However, an alternative approach is to go back to basic principles and express the cardioid parametrically by substituting r = 1 âˆ’ cos Î¸ in the conversion formulas x = r cos Î¸ and y = r sin Î¸. This yields x = (1 âˆ’ cos Î¸) cos Î¸,

y = (1 âˆ’ cos Î¸) sin Î¸

(0 â‰¤ Î¸ â‰¤ 2Ď€)

Differentiating these equations with respect to Î¸ and then simplifying yields (verify) dx = sin Î¸(2 cos Î¸ âˆ’ 1), dÎ¸

dy = (1 âˆ’ cos Î¸)(1 + 2 cos Î¸) dÎ¸

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Thus, dx/dθ = 0 if sin θ = 0 or cos θ = 12 , and dy/dθ = 0 if cos θ = 1 or cos θ = − 12 . We leave it for you to solve these equations and show that the solutions of dx/dθ = 0 on the interval 0 ≤ θ ≤ 2π are dx π 5π = 0 : θ = 0, , π, , 2π dθ 3 3 and the solutions of dy/dθ = 0 on the interval 0 ≤ θ ≤ 2π are dy 2π 4π = 0 : θ = 0, , , 2π dθ 3 3 Thus, horizontal tangent lines occur at θ = 2π/3 and θ = 4π/3; vertical tangent lines occur at θ = π/3, π, and 5π/3; and singular points occur at θ = 0 and θ = 2π (Figure 11.2.7). Note, however, that r = 0 at both singular points, so there is really only one singular point on the cardioid—the pole.

p/ 2

( 32 , 2p3 ) ( 12 , p3 )

(2, p)

0

)

Yo

(

1 , 5p 2 3

( 32 , 4p3 ) r = 1 – cos u

Figure 11.2.7 ••••••••••••••••••••••••••••••••••••••

Formula (7) reveals some useful information about the behavior of a polar curve r = f(θ) that passes through the origin. If we assume that r = 0 and dr /dθ = 0 when θ = θ0 , then it follows from Formula (7) that the slope of the tangent line to the curve at θ = θ0 is dr 0 + sin θ0 sin θ0 dy dθ = = = tan θ0 dr dx cos θ0 0 + cos θ0 dθ (Figure 11.2.8). However, tan θ0 is also the slope of the line θ = θ0 , so we can conclude that this line is tangent to the curve at the origin. Thus, we have established the following result.

Ri

TANGENT LINES TO POLAR CURVES AT THE ORIGIN

p/ 2

n

r = f (u)

sa

u = u0

11.2.1 THEOREM. If the polar curve r = f(θ) passes through the origin at θ = θ0 , and if dr /dθ = 0 at θ = θ0 , then the line θ = θ0 is tangent to the curve at the origin.

Slope = tan u0

u0

Ha

0

This theorem tells us that equations of the tangent lines at the origin to the curve r = f(θ) can be obtained by solving the equation f(θ) = 0. It is important to keep in mind, however, that r = f(θ) may be zero for more than one value of θ, so there may be more than one tangent line at the origin. This is illustrated in the next example.

Figure 11.2.8

p/ 2

u = p/ 3

M uh am

m

0

Example 7 The three-petal rose r = sin 3θ in Figure 11.2.9 has three tangent lines at the origin, which can be found by solving the equation sin 3θ = 0 It was shown in Exercise 73 of Section 11.1 that the complete rose is traced once as θ varies over the interval 0 ≤ θ < π, so we need only look for solutions in this interval. We leave it for you to confirm that these solutions are 2π π θ = 0, θ = , and θ = 3 3 Since dr /dθ = 3 cos 3θ = 0 for these values of θ, these three lines are tangent to the rose at the origin, which is consistent with the figure.

ad

u = 2p/ 3

us uf i

Analytic Geometry in Calculus

az

742

r = sin 3u

Figure 11.2.9

••••••••••••••••••••••••••••••••••••••

ARC LENGTH OF A POLAR CURVE

A formula for the arc length of a polar curve r = f(θ) can be derived by expressing the curve in parametric form and applying Formula (6) of Section 6.4 for the arc length of a parametric curve. We leave it as an exercise to show the following. 11.2.2 ARC LENGTH FORMULA FOR POLAR CURVES. If no segment of the polar curve r = f(θ) is traced more than once as θ increases from α to β, and if dr /dθ is continuous for α ≤ θ ≤ β, then the arc length L from θ = α to θ = β is L=

β

r2 α

+

dr dθ

2 dθ

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p/ 2

743

Example 8 Find the arc length of the spiral r = eÎ¸ in Figure 11.2.10 between Î¸ = 0 and Î¸ = Ď€.

(1, 0)

0

Solution.

2

Ď€ dÎ¸ = (eÎ¸ )2 + (eÎ¸ )2 dÎ¸ Îą 0 Ď€ Ď€âˆš âˆš âˆš = 2 eÎ¸ dÎ¸ = 2 eÎ¸ = 2(eĎ€ âˆ’ 1) â‰ˆ 31.3 Î˛

L=

Figure 11.2.10

r2 +

dr dÎ¸

0

Yo

r = eu

(p, ep)

Tangent Lines and Arc Length for Parametric and Polar Curves

us uf i

11.2

0

Example 9 Find the total arc length of the cardioid r = 1 + cos Î¸ .

Solution. The cardioid is traced out once as Î¸ varies from Î¸ = 0 to Î¸ = 2Ď€. Thus, L=

Î˛

r2 + Îą

dr dÎ¸

2

2Ď€

dÎ¸ = 0

âˆš 2

2Ď€ âˆš

0

1 + cos Î¸ dÎ¸

2Ď€

1 cos2 Î¸ dÎ¸ 2 0

2Ď€

1

cos Î¸ dÎ¸ =2

2

0 =2

r = 1 + cos u

Identity (45) of Appendix E

n

p/ 2

(1 + cos Î¸)2 + (âˆ’ sin Î¸)2 dÎ¸

Ri

=

az

Since cos 12 Î¸ changes sign at Ď€, we must split the last integral into the sum of two integrals: the integral from 0 to Ď€ plus the integral from Ď€ to 2Ď€. However, the integral from Ď€ to 2Ď€ is equal to the integral from 0 to Ď€, since the cardioid is symmetric about the polar axis (Figure 11.2.11). Thus,

2Ď€

Ď€

1 Ď€ 1

cos 1 Î¸ dÎ¸ = 4 L=2 Î¸ dÎ¸ = 8 sin Î¸ cos =8

2

2 2 0 0 0

Ha

sa

0

EXERCISE SET 11.2

ad

Figure 11.2.11

Graphing Utility

m

â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

M uh am

1. (a) Find the slope of the tangent line to the parametric curve x = t 2 + 1, y = t /2 at t = âˆ’1 and at t = 1 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 2. (a) Find the slope of the tangent line to the parametric curve x = 3 cos t, y = 4 sin t at t = Ď€/4 and at t = 7Ď€/4 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

3. For the parametric curve in Exercise 1, make a conjecture about the sign of d 2 y/dx 2 at t = âˆ’1 and at t = 1, and confirm your conjecture without eliminating the parameter. 4. For the parametric curve in Exercise 2, make a conjecture about the sign of d 2 y/dx 2 at t = Ď€/4 and at t = 7Ď€/4, and confirm your conjecture without eliminating the parameter.

In Exercises 5â€“10, find dy/dx and d 2 y/dx 2 at the given point without eliminating the parameter. 5. x =

âˆš t, y = 2t + 4; t = 1

6. x = 12 t 2 , y = 13 t 3 ; t = 2 7. x = sec t, y = tan t; t = Ď€/3 8. x = sinh t, y = cosh t; t = 0 9. x = 2Î¸ + cos Î¸, y = 1 âˆ’ sin Î¸ ; Î¸ = Ď€/3 10. x = cos Ď†, y = 3 sin Ď†; Ď† = 5Ď€/6 11. (a) Find the equation of the tangent line to the curve x = et ,

y = eâˆ’t

at t = 1 without eliminating the parameter. (b) Check your answer in part (a) by eliminating the parameter.

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12. (a) Find the equation of the tangent line to the curve x = 2t + 4,

In Exercises 21–26, find the slope of the tangent line to the polar curve for the given value of θ.

y = 8t − 2t + 4 2

at t = 1 without eliminating the parameter. (b) Check your answer in part (a) by eliminating the parameter. In Exercises 13 and 14, find all values of t at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line. 13. x = 2 cos t, y = 4 sin t

21. r = 2 cos θ ; θ = π/3

22. r = 1 + sin θ; θ = π/4

23. r = 1/θ ; θ = 2

24. r = a sec 2θ ; θ = π/6

25. r = cos 3θ ; θ = 3π/4

26. r = 4 − 3 sin θ; θ = π

In Exercises 27 and 28, calculate the slopes of the tangent lines indicated in the accompanying figures. 27. r = 2 + 2 sin θ

(0 ≤ t ≤ 2π)

p/ 2

15. As shown in the accompanying figure, the Lissajous curve (0 ≤ t ≤ 2π)

Ri

y = sin 2t

crosses itself at the origin. Find equations for the two tangent lines at the origin. 16. As shown in the accompanying figure, the prolate cycloid y = 2t − π sin t

(−π ≤ t ≤ π)

y

x

x

Figure Ex-28

In Exercises 29 and 30, find polar coordinates of all points at which the polar curve has a horizontal or a vertical tangent line.

29. r = a(1 + cos θ )

30. r = a sin θ

In Exercises 31 and 32, use a graphing utility to make a conjecture about the number of points on the polar curve at which there is a horizontal tangent line, and confirm your conjecture by finding appropriate derivatives.

Figure Ex-16

ad

Figure Ex-15

Figure Ex-27

Ha

y

0

0

sa

crosses itself at a point on the x-axis. Find equations for the two tangent lines at that point.

p/ 2

n

x = 2 − π cos t,

28. r = 1 − 2 sin θ

az

14. x = 2t 3 − 15t 2 + 24t + 7, y = t 2 + t + 1 x = sin t,

us uf i

Analytic Geometry in Calculus

Yo

744

m

17. Show that the curve x = t 3 − 4t, y = t 2 intersects itself at the point (0, 4), and find equations for the two tangent lines to the curve at the point of intersection. 18. Show that the curve with parametric equations y = t 3 + t 2 − 10t + 9

M uh am

x = t 2 − 3t + 5,

intersects itself at the point (3, 1), and find equations for the two tangent lines to the curve at the point of intersection.

19. (a) Use a graphing utility to generate the graph of the parametric curve x = cos3 t,

y = sin3 t

(0 ≤ t ≤ 2π)

and make a conjecture about the values of t at which singular points occur. (b) Confirm your conjecture in part (a) by calculating appropriate derivatives.

20. (a) At what values of θ would you expect the cycloid in Figure 1.8.13 to have singular points? (b) Confirm your answer in part (a) by calculating appropriate derivatives.

31. r = sin θ cos2 θ

32. r = 1 − 2 sin θ

In Exercises 33–38, sketch the polar curve and find polar equations of the tangent lines to the curve at the pole. 33. r = 2 cos 3θ √ 35. r = 4 cos 2θ

36. r = sin 2θ

34. r = 4 cos θ

37. r = 1 + 2 cos θ

38. r = 2θ

In Exercises 39–44, use Formula (8) to calculate the arc length of the polar curve. 39. The entire circle r = a 40. The entire circle r = 2a cos θ 41. The entire cardioid r = a(1 − cos θ ) 42. r = sin2 (θ /2) from θ = 0 to θ = π 43. r = e3θ from θ = 0 to θ = 2 44. r = sin3 (θ /3) from θ = 0 to θ = π/2 45. (a) What is the slope of the tangent line at time t to the trajectory of the paper airplane in Example 2?

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Tangent Lines and Arc Length for Parametric and Polar Curves

(b) What was the airplaneâ€™s approximate angle of inclination when it crashed into the wall?

(t â‰Ľ 0)

Yo

51. Find the area of the surface generated by revolving the curve x = cos2 t, y = sin2 t (0 â‰¤ t â‰¤ Ď€/2) about the y-axis.

but lands on a wall at time t = 10. (a) At what times was the bee flying horizontally? (b) At what times was the bee flying vertically?

52. Find the area of the surface generated by revolving x = t, y = 2t 2 (0 â‰¤ t â‰¤ 1) about the y-axis.

47. (a) Show that the arc length of one petal of the rose r = cos nÎ¸ is given by Ď€/(2n) 2 1 + (n2 âˆ’ 1) sin2 nÎ¸ dÎ¸

53. By revolving the semicircle x = r cos t,

y = r sin t

(0 â‰¤ t â‰¤ Ď€)

about the x-axis, show that the surface area of a sphere of radius r is 4Ď€r 2 .

az

y = 2 âˆ’ 2 cos t

49. Find the area of the surface generated by revolving x = t 2 , y = 2t (0 â‰¤ t â‰¤ 4) about the x-axis. 50. Find the area of the surface generated by revolving the curve x = et cos t, y = et sin t (0 â‰¤ t â‰¤ Ď€/2) about the x-axis.

46. Suppose that a bee follows the trajectory x = t âˆ’ 2 sin t,

0

54. The equations

Ri

x = aĎ† âˆ’ a sin Ď†,

y = a âˆ’ a cos Ď†

(0 â‰¤ Ď† â‰¤ 2Ď€)

represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the x-axis is given by S = 64Ď€a 2 /3.

55. As illustrated in the accompanying figure, suppose that a rod with one end fixed at the pole of a polar coordinate system rotates counterclockwise at the constant rate of 1 rad/s. At time t = 0 a bug on the rod is 10 mm from the pole and is moving outward along the rod at the constant speed of 2 mm/s. (a) Find an equation of the form r = f(Î¸ ) for the path of motion of the bug, assuming that Î¸ = 0 when t = 0. (b) Find the distance the bug travels along the path in part (a) during the first 5 seconds. Round your answer to the nearest tenth of a millimeter.

n

(b) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the four-petal rose r = cos 2Î¸. (c) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the n-petal rose r = cos nÎ¸ for n = 2, 3, 4, . . . , 20; then make a conjecture about the limit of these arc lengths as n â†’ +âŹ .

Ha

sa

48. (a) Sketch the spiral r = eâˆ’Î¸ (0 â‰¤ Î¸ < +âŹ ). (b) Find an improper integral for the total arc length of the spiral. (c) Show that the integral converges and find the total arc length of the spiral. Exercises 49â€“54 require the formulas developed in the following discussion: If f (t) and g (t) are continuous functions and if no segment of the curve

m

ad

x = f(t), y = g(t) (a â‰¤ t â‰¤ b) is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the x-axis is

2 2 b dx dy S= 2Ď€y + dt dt dt a

Bug

and the area of the surface generated by revolving the curve about the y-axis is

2 2 b dx dy 2Ď€x + dt S= dt dt a

M uh am

745

us uf i

11.2

[The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.]

t=0s

t=5s Figure Ex-55

56. Use Formula (6) of Section 6.4 to derive Formula (8).

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Analytic Geometry in Calculus

us uf i

746

11.3 AREA IN POLAR COORDINATES

In this section we will show how to find areas of regions that are bounded by polar curves.

AREA IN POLAR COORDINATES

11.3.1 AREA PROBLEM IN POLAR COORDINATES. that satisfy the condition Îą < Î˛ â‰¤ Îą + 2Ď€

r = f (u)

az

u=a

In rectangular coordinates we solved Area Problem 5.1.1 by dividing the region into an increasing number of vertical strips, approximating the strips by rectangles, and taking a limit. In polar coordinates rectangles are clumsy to work with, and it is better to divide the region into wedges by using rays

Figure 11.3.1

Î¸ = Î¸1 , Î¸ = Î¸2 , . . . , Î¸ = Î¸nâˆ’1

u = u2 u = un â€“ 1 A2

such that Îą < Î¸1 < Î¸2 < Âˇ Âˇ Âˇ < Î¸nâˆ’1 < Î˛

u=a

(Figure 11.3.2). As shown in that figure, the rays divide the region R into n wedges with areas A1 , A2 , . . . , An and central angles Î¸1 , Î¸2 , . . . , Î¸n . The area of the entire region can be written as n A = A1 + A2 + Âˇ Âˇ Âˇ + An = Ak (1)

Figure 11.3.2

u = u *k

r = f (u)

k=1

If Î¸k is small, and if we assume for simplicity that f (Î¸ ) is nonnegative, then we can approximate the area Ak of the kth wedge by the area of a sector with central angle Î¸k and radius f(Î¸kâˆ— ), where Î¸ = Î¸kâˆ— is any ray that lies in the kth wedge (Figure 11.3.3). Thus, from (1) and Formula (5) of Appendix E for the area of a sector, we obtain n n 1 A= [f(Î¸kâˆ— )]2 Î¸k Ak â‰ˆ (2) 2 k=1 k=1

ad

âˆ†uk

Ha

sa

âˆ†u1

n

A1

âˆ†u2

... ..

u=b

u = u1

Ri

u=b

âˆ†un

Suppose that Îą and Î˛ are angles

and suppose that f(Î¸) is continuous for Îą â‰¤ Î¸ â‰¤ Î˛. Find the area of the region R enclosed by the polar curve r = f(Î¸) and the rays Î¸ = Îą and Î¸ = Î˛ (Figure 11.3.1).

R

An

Yo

â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

u *k

If we now increase n in such a way that max Î¸k â†’ 0, then the sectors will become better and better approximations of the wedges and it is reasonable to expect that (2) will approach the exact value of the area A (Figure 11.3.4); that is, Î˛ n 1 1 A = lim [f(Î¸kâˆ— )]2 Î¸k = [f(Î¸)]2 dÎ¸ max Î¸k â†’ 0 2 2 Îą k=1

M uh am

m

Figure 11.3.3

u=b

Figure 11.3.4

Thus, we have the following solution of Area Problem 11.3.1.

u=a

11.3.2

AREA IN POLAR COORDINATES.

If Îą and Î˛ are angles that satisfy the condition

Îą < Î˛ â‰¤ Îą + 2Ď€ and if f(Î¸) is continuous for Îą â‰¤ Î¸ â‰¤ Î˛, then the area A of the region R enclosed by the polar curve r = f(Î¸) and the rays Î¸ = Îą and Î¸ = Î˛ is A= Îą

Î˛

1 [f(Î¸)]2 dÎ¸ = 2

Îą

Î˛

1 2 r dÎ¸ 2

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Area in Polar Coordinates

747

us uf i

11.3

The hardest part of applying (3) is determining the limits of integration. This can be done as follows:

Step 2.

Draw an arbitrary “radial line” from the pole to the boundary curve r = f(θ).

Step 3.

Ask, “Over what interval of values must θ vary in order for the radial line to sweep out the region R?”

Step 4.

Your answer in Step 3 will determine the lower and upper limits of integration.

Yo

Sketch the region R whose area is to be determined.

az

r = 1 – cos u

Step 1.

p/ 2

Ri

Example 1 Find the area of the region in the first quadrant that is within the cardioid r = 1 − cos θ .

Solution. The region and a typical radial line are shown in Figure 11.3.5. For the radial

0

n

line to sweep out the region, θ must vary from 0 to π/2. Thus, from (3) with α = 0 and β = π/2, we obtain π/2 1 2 1 π/2 1 π/2 2 A= (1 − cos θ) dθ = (1 − 2 cos θ + cos2 θ ) dθ r dθ = 2 2 0 2 0 0

The shaded region is swept out by the radial line as u varies from 0 to p/ 2.

Figure 11.3.5

1 2

0

π/2

π/2 3 3 1 1 3 1 = π−1 − 2 cos θ + cos 2θ dθ = θ − 2 sin θ + sin 2θ 2 2 2 2 4 8 0

Ha

A=

sa

With the help of the identity cos2 θ = 12 (1 + cos 2θ), this can be rewritten as

Example 2 Find the entire area within the cardioid of Example 1.

Solution. For the radial line to sweep out the entire cardioid, θ must vary from 0 to 2π.

M uh am

m

ad

Thus, from (3) with α = 0 and β = 2π, 2π 1 2 1 2π A= r dθ = (1 − cos θ)2 dθ 2 2 0 0 If we proceed as in Example 1, this reduces to 1 2π 3 1 3π A= − 2 cos θ + cos 2θ dθ = 2 0 2 2 2

••••••••••••••••••••••••••••••••••••••

USING SYMMETRY

Alternative Solution. Since the cardioid is symmetric about the x-axis, we can calculate the portion of the area above the x-axis and double the result. In the portion of the cardioid above the x-axis, θ ranges from 0 to π, so that π π 1 2 3π A=2 (1 − cos θ)2 dθ = r dθ = 2 2 0 0 Although Formula (3) is applicable if r = f (θ ) is negative, area computations can sometimes be simplified by using symmetry to restrict the limits of integration to intervals where r ≥ 0. This is illustrated in the next example. Example 3 Find the area of the region enclosed by the rose curve r = cos 2θ.

Solution. Referring to Figure 11.1.10 and using symmetry, the area in the first quadrant that is swept out for 0 ≤ θ ≤ π/4 is one-eighth of the total area inside the rose. Thus, from

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Analytic Geometry in Calculus

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748

Yo

Formula (3) π/4 π/4 1 2 A=8 cos2 2θ dθ r dθ = 4 2 0 0 π/4 π/4 1 =4 (1 + cos 4θ) dθ = 2 (1 + cos 4θ) dθ 2 0 0 π/4 1 π = 2θ + sin 4θ = 2 2 0

Ha

sa

n

Ri

az

Sometimes the most natural way to satisfy the restriction α < β ≤ α + 2π required by Formula (3) is to use a negative value for α. For example, suppose that we are interested in finding the area of the shaded region in Figure 11.3.6a. The first step would be to determine the intersections of the cardioid r = 4 + 4 cos θ and the circle r = 6, since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 1 4 + 4 cos θ = 6 or cos θ = 2 which is satisfied by the positive angles π 5π θ= and θ = 3 3 However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region shown in Figure 11.3.6b as θ varies over the interval π/3 ≤ θ ≤ 5π/3. There are two ways to circumvent this problem—one is to take advantage of the symmetry by integrating over the interval 0 ≤ θ ≤ π/3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval −π/3 ≤ θ ≤ π/3 (Figure 11.3.6c). The two methods are illustrated in the next example. p/ 2

p/ 2

r=6

u=4

r = 4 + 4 cos u

(a)

(b)

0

u=k

u=4

0

u=$

(c)

p/ 2

u=4

0

u=$

(d)

u=$

(e)

m

Figure 11.3.6

p/ 2

u=4

0

ad

0

p/ 2

M uh am

Example 4 Find the area of the region that is inside of the cardioid r = 4 + 4 cos θ and outside of the circle r = 6.

Solution Using a Negative Angle. The area of the region can be obtained by subtracting the areas in Figures 11.3.6d and 11.3.6e: π/3 π/3 1 1 2 Area inside cardioid A= (4 + 4 cos θ)2 dθ − (6) dθ minus area inside circle. 2 2 / / −π 3 −π 3 π/3 π/3 1 (16 cos θ + 8 cos2 θ − 10) dθ [(4 + 4 cos θ)2 − 36] dθ = = −π/3 2 −π/3 √ π/3 = 16 sin θ + (4θ + 2 sin 2θ) − 10 θ −π/3 = 18 3 − 4π

Solution Using Symmetry. Using symmetry, we can calculate the area above the polar axis and double it. This yields (verify) π/3 √ √ 1 A=2 [(4 + 4 cos θ)2 − 36] dθ = 2(9 3 − 2π) = 18 3 − 4π 2 0 which agrees with the preceding result.

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••••••••••••••••••••••••••••••••••••••

INTERSECTIONS OF POLAR GRAPHS

p/ 2

In the last example we found the intersections of the cardioid and circle by equating their expressions for r and solving for θ. However, because a point can be represented in different ways in polar coordinates, this procedure will not always produce all of the intersections. For example, the cardioids r = 1 − cos θ

r = 1 + cos u

(1, 6 ) 0

r = 1 + cos θ

and

(4) / / intersect at three points: the pole, the point (1, π 2), and the point (1, 3π 2) (Figure 11.3.7). Equating the right-hand sides of the equations in (4) yields 1 − cos θ = 1 + cos θ or cos θ = 0, so π θ = + kπ, k = 0, ±1, ±2, . . . 2 Substituting any of these values in (4) yields r = 1, so that we have found only two distinct points of intersection, (1, π/2) and (1, 3π/2); the pole has been missed. This problem occurs because the two cardioids pass through the pole at different values of θ—the cardioid r = 1 − cos θ passes through the pole at θ = 0, and the cardioid r = 1 + cos θ passes through the pole at θ = π. The situation with the cardioids is analogous to two satellites circling the Earth in intersecting orbits (Figure 11.3.8). The satellites will not collide unless they reach the same point at the same time. In general, when looking for intersections of polar curves, it is a good idea to graph the curves to determine how many intersections there should be.

az

(1, i)

749

Yo

r = 1 – cos u

Area in Polar Coordinates

us uf i

11.3

sa

n

Ri

Figure 11.3.7

Figure 11.3.8

EXERCISE SET 11.3

Graphing Utility

Ha

The orbits intersect, but the satellites do not collide.

C

CAS

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(b)

M uh am

r = 1 – cos u

(d)

r=u

6. The region in the first quadrant within the cardioid r = 1 + sin θ. 7. The region enclosed by the rose r = 4 cos 3θ.

r = 2 cos u

(e)

In Exercises 5–10, find the area of the region described. 5. The region that is enclosed by the cardioid r = 2 + 2 cos θ.

(c)

m

(a)

ad

1. Write down, but do not evaluate, an integral for the area of each shaded region.

r = sin 2u

8. The region enclosed by the rose r = 2 sin 2θ. 9. The region enclosed by the inner loop of the limac¸ on r = 1 + 2 cos θ. [Hint: r ≤ 0 over the interval of integration.]

(f)

10. The region swept out by a radial line from the pole to the curve r = 2/θ as θ varies over the interval 1 ≤ θ ≤ 3. r = 1 – sin u

r = cos 2u

2. Evaluate the integrals you obtained in Exercise 1.

In Exercises 11–14, find the area of the shaded region. 11.

12.

3. In each part, find the area of the circle by integration. (a) r = a (b) r = 2a sin θ (c) r = 2a cos θ 4. (a) Show that r = sin θ + cos θ is a circle. (b) Find the area of the circle using a geometric formula and then by integration.

r = √cos 2u r = 2 cos u

r = 1 + cos u r = cos u

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Analytic Geometry in Calculus

13.

us uf i

750

14.

(b) State conditions under which these formulas hold.

In Exercises 27–30, sketch the surface, and use the formulas in Exercise 26 to find the surface area.

27. The surface generated by revolving the circle r = cos θ about the line θ = π/2.

Yo

r = 1 + cos t r = 3 cos t

In Exercises 15–22, find the area of the region described.

28. The surface generated by revolving the spiral r = eθ (0 ≤ θ ≤ π/2) about the line θ = π/2.

15. The region inside the circle r = 5 sin θ and outside the limac¸ on r = 2 + sin θ .

29. The “apple” generated by revolving the upper half of the cardioid r = 1 − cos θ (0 ≤ θ ≤ π) about the polar axis.

16. The region outside the cardioid r = 2 − 2 cos θ and inside the circle r = 4.

18. The region that is common to the circles r = 4 cos θ and r = 4 sin θ. 1 2

C

+ cos θ .

20. The region inside the cardioid r = 2 + 2 cos θ and to the right of the line r cos θ = 32 .

r=

3 sin θ cos θ

cos3 θ + sin3 θ

(b) Use a CAS to show that the area inside of the loop is (Figure 3.6.2).

C

3 2

32. (a) What is the area that is enclosed by one petal of the rose r = a cos nθ if n is an even integer? (b) What is the area that is enclosed by one petal of the rose r = a cos nθ if n is an odd integer? (c) Use a CAS to show that the total area enclosed by the rose r = a cos nθ is πa 2 /2 if the number of petals is even. [Hint: See Exercise 73 of Section 11.1.] (d) Use a CAS to show that the total area enclosed by the rose r = a cos nθ is πa 2 /4 if the number of petals is odd.

sa

21. The region inside the circle r = 10 and to the right of the line r = 6 sec θ .

31. (a) Show that the Folium of Descartes x 3 − 3xy + y 3 = 0 can be expressed in polar coordinates as

n

19. The region between the loops of the limac¸ on r =

30. The sphere of radius a generated by revolving the semicircle r = a in the upper half-plane about the polar axis.

Ri

17. The region inside the cardioid r = 2 + 2 cos θ and outside the circle r = 3.

az

r = 4 cos t r = 4√3 sin t

22. The region inside the rose r = 2a cos 2θ and outside the √ circle r = a 2.

ad

Ha

23. (a) Find the error: The area that is inside the lemniscate r 2 = a 2 cos 2θ is 2π 2π 1 2 1 2 A= r dθ = a cos 2θ dθ 2 2 0 0 2π 1 2 =0 = a sin 2θ 4 0 (b) Find the correct area. 2 (c) Find the area inside the √ lemniscate r = 4 cos 2θ and outside the circle r = 2.

m

24. Find the area inside the curve r 2 = sin 2θ.

M uh am

25. A radial line is drawn from the origin to the spiral r = aθ (a > 0 and θ ≥ 0). Find the area swept out during the second revolution of the radial line that was not swept out during the first revolution. 26. (a) In the discussion associated with Exercises 49–54 of Section 11.2, formulas were given for the area of the surface of revolution that is generated by revolving a parametric curve about the x-axis or y-axis. Use those formulas to derive the following formulas for the areas of the surfaces of revolution that are generated by revolving the portion of the polar curve r = f(θ ) from θ = α to θ = β about the polar axis and about the line θ = π/2:

dr 2 dθ dθ α

2 β dr 2πr cos θ r 2 + dθ S= dθ α

S=

β

2πr sin θ r 2 +

33. One of the most famous problems in Greek antiquity was “squaring the circle”; that is, using a straightedge and compass to construct a square whose area is equal to that of a given circle. It was proved in the nineteenth century that no such construction is possible. However, show that the shaded areas in the accompanying figure are equal, thereby “squaring the crescent.” p/ 2

0

Figure Ex-33

About θ = 0

34. Use a graphing utility to generate the polar graph of the equation r = cos 3θ + 2, and find the area that it encloses.

About θ = π/2

35. Use a graphing utility to generate the graph of the bifolium r = 2 cos θ sin2 θ , and find the area of the upper loop.

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11.4 CONIC SECTIONS IN CALCULUS

Conic Sections in Calculus

751

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11.4

Yo

In this section we will discuss some of the basic geometric properties of parabolas, ellipses, and hyperbolas. These curves play an important role in calculus and also arise naturally in a broad range of applications in such fields as planetary motion, design of telescopes and antennas, geodetic positioning, and medicine, to name a few.

••••••••••••••••••••••••••••••••••••••

Circles, ellipses, parabolas, and hyperbolas are called conic sections or conics because they can be obtained as intersections of a plane with a double-napped circular cone (Figure 11.4.1). If the plane passes through the vertex of the double-napped cone, then the intersection is a point, a pair of intersecting lines, or a single line. These are called degenerate conic sections.

ad

Ha

sa

n

Ri

CONIC SECTIONS

az

Some students may already be familiar with the material in this section, in which case it can be treated as a review. Instructors who want to spend some additional time on precalculus review may want to allocate more than one lecture on this material.

Ellipse

Parabola

Hyperbola

M uh am

m

Circle

••••••••••••••••••••••••••••••••••••••

DEFINITIONS OF THE CONIC SECTIONS

A point

A pair of intersecting lines

A single line

Figure 11.4.1

Although we could derive properties of parabolas, ellipses, and hyperbolas by defining them as intersections with a double-napped cone, it will be better suited to calculus if we begin with equivalent definitions that are based on their geometric properties.

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Analytic Geometry in Calculus

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752

11.4.1 DEFINITION. A parabola is the set of all points in the plane that are equidistant from a fixed line and a fixed point not on the line.

The line is called the directrix of the parabola, and the point is called the focus (Figure 11.4.2). A parabola is symmetric about the line that passes through the focus at right angles to the directrix. This line, called the axis or the axis of symmetry of the parabola, intersects the parabola at a point called the vertex.

Yo

All points on the parabola are equidistant from the focus and directrix.

11.4.2 DEFINITION. An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a given positive constant that is greater than the distance between the fixed points.

Axis

az

Focus

The two fixed points are called the foci (plural of “focus”) of the ellipse, and the midpoint of the line segment joining the foci is called the center (Figure 11.4.3a). To help visualize Definition 11.4.2, imagine that two ends of a string are tacked to the foci and a pencil traces a curve as it is held tight against the string (Figure 11.4.3b). The resulting curve will be an ellipse since the sum of the distances to the foci is a constant, namely the total length of the string. Note that if the foci coincide, the ellipse reduces to a circle. For ellipses other than circles, the line segment through the foci and across the ellipse is called the major axis (Figure 11.4.3c), and the line segment across the ellipse, through the center, and perpendicular to the major axis is called the minor axis. The endpoints of the major axis are called vertices.

Vertex Directrix

sa

n

Ri

Figure 11.4.2

The sum of the distances to the foci is constant.

Ha Center

ad

Focus

(a)

Minor axis

Major axis Vertex

Vertex

Focus

(b)

(c)

M uh am

m

Figure 11.4.3

11.4.3 DEFINITION. A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed distinct points is a given positive constant that is less than the distance between the fixed points.

The two fixed points are called the foci of the hyperbola, and the term “difference” that is used in the definition is understood to mean the distance to the farther focus minus the distance to the closer focus. As a result, the points on the hyperbola form two branches, each “wrapping around” the closer focus (Figure 11.4.4a). The midpoint of the line segment joining the foci is called the center of the hyperbola, the line through the foci is called the focal axis, and the line through the center that is perpendicular to the focal axis is called the conjugate axis. The hyperbola intersects the focal axis at two points called the vertices. Associated with every hyperbola is a pair of lines, called the asymptotes of the hyperbola. These lines intersect at the center of the hyperbola and have the property that as a point P moves along the hyperbola away from the center, the distance between P and one of the asymptotes approaches zero (Figure 11.4.4b).

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The distance from the farther focus minus the distance to the closer focus is constant.

Center

Focal axis

Focus

Focus Vertex

y

These distances approach zero as the point moves away from the center.

x These distances approach zero as the point moves away from the center.

az

Vertex

(a)

(b)

Figure 11.4.4 ••••••••••••••••••••••••••••••••••••••

Ri

It is traditional in the study of parabolas to denote the distance between the focus and the vertex by p. The vertex is equidistant from the focus and the directrix, so the distance between the vertex and the directrix is also p; consequently, the distance between the focus and the directrix is 2p (Figure 11.4.5). As illustrated in that figure, the parabola passes through two of the corners of a box that extends from the vertex to the focus along the axis of symmetry and extends 2p units above and 2p units below the axis of symmetry. The equation of a parabola is simplest if the vertex is the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations are shown in Figure 11.4.6. These are called the standard positions of a parabola, and the resulting equations are called the standard equations of a parabola.

n

EQUATIONS OF PARABOLAS IN STANDARD POSITION

2p p

sa

Axis

2p

Directrix

Ha

p

753

Yo

Conjugate axis

Conic Sections in Calculus

us uf i

11.4

parabolas in standard position

Figure 11.4.5

ad

y

x

( p, 0)

m M uh am

y

y

y=p x

(–p, 0)

x = –p

y 2 = 4px

y

(0, p) x

x=p

y 2 = –4px

x

(0, –p)

y = –p x 2 = 4py

x 2 = –4py

Figure 11.4.6

To illustrate how the equations in Figure 11.4.6 are obtained, we will derive the equation for the parabola with focus (p, 0) and directrix x = −p. Let P (x, y) be any point on the parabola. Since P is equidistant from the focus and directrix, the distances PF and PD in Figure 11.4.7 are equal; that is,

y

D(–p, y)

P(x, y)

F(p, 0)

x = –p

Figure 11.4.7

PF = PD x

(1)

where D(−p, y) is the foot of the perpendicular from P to the directrix. From the distance formula, the distances PF and PD are PF = (x − p)2 + y 2 and PD = (x + p)2 (2) Substituting in (1) and squaring yields (x − p)2 + y 2 = (x + p)2

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754

Analytic Geometry in Calculus

and after simplifying y 2 = 4px

(4)

The derivations of the other equations in Figure 11.4.6 are similar.

Determine whether the axis of symmetry is along the x-axis or the y-axis. Referring to Figure 11.4.6, the axis of symmetry is along the x-axis if the equation has a y 2 -term, and it is along the y-axis if it has an x 2 -term.

•

Determine which way the parabola opens. If the axis of symmetry is along the x-axis, then the parabola opens to the right if the coefficient of x is positive, and it opens to the left if the coefficient is negative. If the axis of symmetry is along the y-axis, then the parabola opens up if the coefficient of y is positive, and it opens down if the coefficient is negative.

•

Determine the value of p and draw a box extending p units from the origin along the axis of symmetry in the direction in which the parabola opens and extending 2p units on each side of the axis of symmetry.

•

Using the box as a guide, sketch the parabola so that its vertex is at the origin and it passes through the corners of the box (Figure 11.4.8).

y

az

•

Rough sketch

Figure 11.4.8

Yo

Parabolas can be sketched from their standard equations using four basic steps:

Ri

A TECHNIQUE FOR SKETCHING PARABOLAS

n

••••••••••••••••••••••••••••••••••••••

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Example 1 Sketch the graphs of the parabolas

-6

6

y = –3

x 2 = 12y

Solution (a). This equation involves x 2 , so the axis of symmetry is along the y-axis, and the coefficient of y is positive, so the parabola opens upward. From the coefficient of y, we obtain 4p = 12 or p = 3. Drawing a box extending p = 3 units up from the origin and 2p = 6 units to the left and 2p = 6 units to the right of the y-axis, then using corners of the box as a guide, yields the graph in Figure 11.4.9. The focus is p = 3 units from the vertex along the axis of symmetry in the direction in which the parabola opens, so its coordinates are (0, 3). The directrix is perpendicular to the axis of symmetry at a distance of p = 3 units from the vertex on the opposite side from the focus, so its equation is y = −3.

ad

Figure 11.4.9

and show the focus and directrix of each.

Ha

x

(b) y 2 + 8x = 0

sa

(a) x 2 = 12y

(0, 3)

y

m

Solution (b). We first rewrite the equation in the standard form

M uh am

4

x

(–2, 0)

-4

x=2

y 2 = –8x

Figure 11.4.10

y 2 = −8x

This equation involves y 2 , so the axis of symmetry is along the x-axis, and the coefficient of x is negative, so the parabola opens to the left. From the coefficient of x we obtain 4p = 8, so p = 2. Drawing a box extending p = 2 units left from the origin and 2p = 4 units above and 2p = 4 units below the x-axis, then using corners of the box as a guide, yields the graph in Figure 11.4.10. Example 2 Find an equation of the parabola that is symmetric about the y-axis, has its vertex at the origin, and passes through the point (5, 2).

Solution. Since the parabola is symmetric about the y-axis and has its vertex at the origin, the equation is of the form x 2 = 4py

or

x 2 = −4py

where the sign depends on whether the parabola opens up or down. But the parabola must open up, since it passes through the point (5, 2), which lies in the first quadrant. Thus, the equation is of the form x 2 = 4py

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Conic Sections in Calculus

Since the parabola passes through (5, 2), we must have 52 = 4p Âˇ 2 or 4p = (5) becomes x2 = â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

b

b a

from which it follows that a = b2 + c2

Figure 11.4.11

âˆšb 2 + c 2

or, equivalently, c = a 2 âˆ’ b2

âˆšb 2 + c 2

Q

sa

aâ€“c a

Ha

Figure 11.4.12

a

b

(7)

n

P

c

(6)

From (6), the distance from a focus to an end of the minor axis is a (Figure 11.4.13), which implies that for all points on the ellipse the sum of the distances to the foci is 2a. It also follows from (6) that a â‰Ľ b with the equality holding only when c = 0. Geometrically, this means that the major axis of an ellipse is at least as large as the minor axis and that the two axes have equal length only when the foci coincide, in which case the ellipse is a circle. The equation of an ellipse is simplest if the center of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure 11.4.14. These are called the standard positions of an ellipse, and the resulting equations are called the standard equations of an ellipse.

b c

It is traditional in the study of ellipses to denote the length of the major axis by 2a, the length of the minor axis by 2b, and the distance between the foci by 2c (Figure 11.4.11). The number a is called the semimajor axis and the number b the semiminor axis (standard but odd terminology, since a and b are numbers, not geometric axes). There is a basic relationship between the numbers a, b, and c that can be obtained by examining the sum of the distances to the foci from a point P at the end of the major axis and from a point Q at the end of the minor axis (Figure 11.4.12). From Definition 11.4.2, these sums must be equal, so we obtain 2 b2 + c2 = (a âˆ’ c) + (a + c)

az

c

a

Therefore,

25 y 2

Ri

c

25 . 2

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EQUATIONS OF ELLIPSES IN STANDARD POSITION

755

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11.4

ellipses in standard position

ad

c

M uh am

m

Figure 11.4.13

y

y

a (0, c) b x

â€“a (â€“c, 0)

(c, 0)

a

x

â€“b

b

â€“b

(0, â€“c) â€“a

x 2 y2 + =1 a 2 b2

x 2 y2 + =1 b2 a2

y

Figure 11.4.14

P(x, y)

Fâ€˛(â€“c, 0)

F(c, 0)

x

To illustrate how the equations in Figure 11.4.14 are obtained, we will derive the equation for the ellipse with foci on the x-axis. Let P (x, y) be any point on that ellipse. Since the sum of the distances from P to the foci is 2a, it follows (Figure 11.4.15) that PF + PF = 2a so

Figure 11.4.15

(x + c)2 + y 2 + (x âˆ’ c)2 + y 2 = 2a

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Analytic Geometry in Calculus

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756

Transposing the second radical to the right side of the equation and squaring yields (x + c)2 + y 2 = 4a 2 − 4a (x − c)2 + y 2 + (x − c)2 + y 2 and, on simplifying, c (x − c)2 + y 2 = a − x a Squaring again and simplifying yields

Yo

(8)

x2 y2 =1 + 2 2 a a − c2 which, by virtue of (6), can be written as

••••••••••••••••••••••••••••••••••••••

Ellipses can be sketched from their standard equations using three basic steps: •

Determine whether the major axis is on the x-axis or the y-axis. This can be ascertained from the sizes of the denominators in the equation. Referring to Figure 11.4.14, and keeping in mind that a 2 > b2 (since a > b), the major axis is along the x-axis if x 2 has the larger denominator, and it is along the y-axis if y 2 has the larger denominator. If the denominators are equal, the ellipse is a circle.

•

Determine the values of a and b and draw a box extending a units on each side of the center along the major axis and b units on each side of the center along the minor axis. Using the box as a guide, sketch the ellipse so that its center is at the origin and it touches the sides of the box where the sides intersect the coordinate axes (Figure 11.4.16).

Ha

•

sa

n

A TECHNIQUE FOR SKETCHING ELLIPSES

Ri

az

x2 y2 + =1 (9) a2 b2 Conversely, it can be shown that any point whose coordinates satisfy (9) has 2a as the sum of its distances from the foci, so that such a point is on the ellipse.

Example 3 Sketch the graphs of the ellipses y2 x2 + =1 (b) x 2 + 2y 2 = 4 9 16 showing the foci of each.

Rough sketch

ad

(a)

Figure 11.4.16

Solution (a). Since y 2 has the larger denominator, the major axis is along the y-axis.

m

Moreover, since a 2 > b2 , we must have a 2 = 16 and b2 = 9, so

y

(0, √7)

M uh am

4

x

-3

3

-4

and b = 3

Drawing a box extending 4 units on each side of the origin along the y-axis and 3 units on each side of the origin along the x-axis as a guide yields the graph in Figure 11.4.17. The foci lie c units on each side of the center along the major axis, where c is given by (7). From the values of a 2 and b2 above, we obtain √ √ c = a 2 − b2 = 16 − 9 = 7 ≈ 2.6 √ √ Thus, the coordinates of the foci are (0, 7 ) and (0, − 7 ), since they lie on the y-axis.

(0, – √7)

x2 y2 + =1 9 16

Figure 11.4.17

a=4

Solution (b). We first rewrite the equation in the standard form y2 x2 + =1 4 2 Since x 2 has the larger denominator, the major axis lies along the x-axis, and we have a 2 = 4 and b2 = 2. Drawing √ a box extending a = 2 on each side of the origin along the x-axis and extending b = 2 ≈ 1.4 units on each side of the origin along the y-axis as a guide yields the graph in Figure 11.4.18.

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y

(– √2, 0)

Conic Sections in Calculus

From (7), we obtain √ c = a 2 − b2 = 2 ≈ 1.4

√2

√ √ Thus, the coordinates of the foci are ( 2, 0) and (− 2, 0), since they lie on the x-axis.

(√2, 0)

x 2

Example 4 Find an equation for the ellipse with foci (0, ±2) and major axis with endpoints (0, ±4).

Yo

-2

757

us uf i

11.4

-√2

Solution. From Figure 11.4.14, the equation has the form

Figure 11.4.18

b2 = a 2 − c2 = 16 − 4 = 12

Ri

so the equation of the ellipse is x2 y2 + =1 12 16 EQUATIONS OF HYPERBOLAS IN STANDARD POSITION

sa

This relationship, which can also be expressed as c = a 2 + b2

c a

ad

Figure 11.4.19

[(c − a) + 2a] − (c − a) = 2a

M uh am

c

m

a

Figure 11.4.20

V

a

c–a Figure 11.4.21

(11)

is pictured geometrically in Figure 11.4.20. As illustrated in that figure, and as we will show later in this section, the asymptotes pass through the corners of a box extending b units on each side of the center along the conjugate axis and a units on each side of the center along the focal axis. The number a is called the semifocal axis of the hyperbola and the number b the semiconjugate axis. (As with the semimajor and semiminor axes of an ellipse, these are numbers, not geometric axes). If V is one vertex of a hyperbola, then, as illustrated in Figure 11.4.21, the distance from V to the farther focus minus the distance from V to the closer focus is

a

b

It is traditional in the study of hyperbolas to denote the distance between the vertices by 2a, the distance between the foci by 2c (Figure 11.4.19), and to define the quantity b as b = c2 − a 2 (10)

Ha

c

n

••••••••••••••••••••••••••••••••••••••

az

y2 x2 =1 + b2 a2 and from the given information, a = 4 and c = 2. It follows from (6) that

x 2 y2 + =1 4 2

a

c–a

Thus, for all points on a hyperbola, the distance to the farther focus minus the distance to the closer focus is 2a. The equation of a hyperbola is simplest if the center of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure 11.4.22. These are called the standard positions of a hyperbola, and the resulting equations are called the standard equations of a hyperbola. The derivations of these equations are similar to those already given for parabolas and ellipses, so we will leave them as exercises. However, to illustrate how the equations of the asymptotes are derived, we will derive those equations for the hyperbola x2 y2 − =1 a2 b2 We can rewrite this equation as b2 2 (x − a 2 ) a2 which is equivalent to the pair of equations b 2 b 2 y= x − a 2 and y = − x − a2 a a y2 =

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Analytic Geometry in Calculus

us uf i

758

hyperbolas in standard position y

y

b y = ax b x

a

x

–b

(c, 0)

b

–a

–b y = – ba x

–

y2 b2

y2

=1

a2

Figure 11.4.22

–

x2

b2

=1

Thus, in the first quadrant, the vertical distance between the line y = (b/a)x and the hyperbola can be written (Figure 11.4.23) as b b 2 x − a2 x− a a But this distance tends to zero as x → +⬁ since b 2 b b 2 = lim (x − x 2 − a 2 ) x− x −a lim x → +⬁ x → +⬁ a a a b (x − x 2 − a 2 )(x + x 2 − a 2 ) = lim x → +⬁ a x + x 2 − a2

b y = a √x 2 – a 2

n

y

y = –a x b

az

a2

(0, –c)

Ri

x2

a x b

Yo

a

–a

(–c, 0)

y=

(0, c)

sa

b y = ax

Ha

x

Figure 11.4.23

ab =0 x + x 2 − a2 The analysis in the remaining quadrants is similar.

ad

= lim

M uh am

A QUICK WAY TO FIND ASYMPTOTES

There is a trick that can be used to avoid memorizing the equations of the asymptotes of a hyperbola. They can be obtained, when needed, by substituting 0 for the 1 on the right side of the hyperbola equation, and then solving for y in terms of x. For example, for the hyperbola

m

••••••••••••••••••••••••••••••••••••••

x → +⬁

••••••••••••••••••••••••••••••••••••••

A TECHNIQUE FOR SKETCHING HYPERBOLAS

x2 y2 − 2 =1 2 a b we would write x2 y2 b2 2 b 2 − = 0 or y = x or y = ± x a2 b2 a2 a which are the equations for the asymptotes. Hyperbolas can be sketched from their standard equations using four basic steps: •

Determine whether the focal axis is on the x-axis or the y-axis. This can be ascertained from the location of the minus sign in the equation. Referring to Figure 11.4.22, the focal axis is along the x-axis when the minus sign precedes the y 2 -term, and it is along the y-axis when the minus sign precedes the x 2 -term.

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•

Draw the asymptotes along the diagonals of the box. Using the box and the asymptotes as a guide, sketch the graph of the hyperbola (Figure 11.4.24).

y2 x2 − =1 (b) y 2 − x 2 = 1 4 9 showing their vertices, foci, and asymptotes.

az

(a)

Yo

Determine the values of a and b and draw a box extending a units on either side of the center along the focal axis and b units on either side of the center along the conjugate axis. (The squares of a and b can be read directly from the equation.)

Example 5 Sketch the graphs of the hyperbolas Figure 11.4.24

759

•

•

Rough sketch

Conic Sections in Calculus

us uf i

11.4

Solution (a). The minus sign precedes the y 2 -term, so the focal axis is along the x-axis. y

y=

From the denominators in the equation we obtain

3 x 2

a2 = 4

5

n

sa

√13 5

x2 y2 3 − = 0 or y = ± x 2 4 9 The foci lie c units on each side of the center along the focal axis, where c is given by (11). From the values of a 2 and b2 above we obtain √ √ c = a 2 + b2 = 4 + 9 = 13 ≈ 3.6 √ √ Since the foci lie on the x-axis in this case, their coordinates are ( 13, 0) and (− 13, 0).

4

–

y2

9

Ha

-5 x2

b2 = 9

Since a and b are positive, we must have a = 2 and b = 3. Recalling that the vertices lie a units on each side of the center on the focal axis, it follows that their coordinates in this case are (2, 0) and (−2, 0). Drawing a box extending a = 2 units along the x-axis on each side of the origin and b = 3 units on each side of the origin along the y-axis, then drawing the asymptotes along the diagonals of the box as a guide, yields the graph in Figure 11.4.25. To obtain equations for the asymptotes, we substitute 0 for 1 in the given equation; this yields

x - √13

and

Ri

y = –3x 2

=1

ad

Figure 11.4.25

Solution (b). The minus sign precedes the x 2 -term, so the focal axis is along the y-axis. From the denominators in the equation we obtain a 2 = 1 and b2 = 1, from which it follows that

4

y = –x

y=x

M uh am

√2

m

y

-4

x 4

−√2

-4

Figure 11.4.26

a=1

and

b=1

Thus, the vertices are at (0, −1) and (0, 1). Drawing a box extending a = 1 unit on either side of the origin along the y-axis and b = 1 unit on either side of the origin along the x-axis, then drawing the asymptotes, yields the graph in Figure 11.4.26. Since the box is actually a square, the asymptotes are perpendicular and have equations y = ±x. This can also be seen by substituting 0 for 1 in the given equation, which yields y 2 − x 2 = 0 or y = ±x. Also, √ √ c = a 2 + b2 = 1 + 1 = 2 √ √ so the foci, which lie on the y-axis, are (0, − 2 ) and (0, 2 ).

• REMARK. • • • • • • •

A hyperbola in which a = b, as in part (b) of this example, is called an equilateral hyperbola. Such hyperbolas always have perpendicular asymptotes.

Example 6 y = ± 43 x.

Find the equation of the hyperbola with vertices (0, ±8) and asymptotes

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Analytic Geometry in Calculus

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760

Solution. Since the vertices are on the y-axis, the equation of the hyperbolas has the form

Yo

(y 2 /a 2 ) − (x 2 /b2 ) = 1 and the asymptotes are a y=± x b From the locations of the vertices we have a = 8, so the given equations of the asymptotes yield a 4 8 y=± x=± x=± x b 3 b from which it follows that b = 6. Thus, the hyperbola has the equation y2 x2 =1 − 64 36

Equations of conics that are translated from their standard positions can be obtained by replacing x by x − h and y by y − k in their standard equations. For a parabola, this translates the vertex from the origin to the point (h, k); and for ellipses and hyperbolas, this translates the center from the origin to the point (h, k).

Ri

TRANSLATED CONICS

az

••••••••••••••••••••••••••••••••••••••

Parabolas with vertex (h, k) and axis parallel to x-axis 4p(x − h)

(12)

[Opens left]

(13)

sa

(y − k) = −4p(x − h) 2

[Opens right]

n

(y − k)2 =

Parabolas with vertex (h, k) and axis parallel to y-axis 4p(y − k)

Ha

(x − h)2 =

(x − h)2 = −4p(y − k)

[Opens up]

(14)

[Opens down]

(15)

Ellipse with center (h, k) and major axis parallel to x-axis

ad

(x − h)2 (y − k)2 =1 + a2 b2

[b ≤ a]

(16)

M uh am

m

Ellipse with center (h, k) and major axis parallel to y-axis (x − h)2 (y − k)2 + =1 b2 a2

[b ≤ a]

(17)

Hyperbola with center (h, k) and focal axis parallel to x-axis (x − h)2 (y − k)2 − =1 2 a b2

(18)

Hyperbola with center (h, k) and focal axis parallel to y-axis (y − k)2 (x − h)2 − =1 a2 b2

(19)

Example 7 Find an equation for the parabola that has its vertex at (1, 2) and its focus at (4, 2).

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Conic Sections in Calculus

761

us uf i

11.4

Solution. Since the focus and vertex are on a horizontal line, and since the focus is to the right of the vertex, the parabola opens to the right and its equation has the form (y − k)2 = 4p(x − h)

Yo

Since the vertex and focus are 3 units apart, we have p = 3, and since the vertex is at (h, k) = (1, 2), we obtain (y − 2)2 = 12(x − 1)

Sometimes the equations of translated conics occur in expanded form, in which case we are faced with the problem of identifying the graph of a quadratic equation in x and y: Ax 2 + Cy 2 + Dx + Ey + F = 0

az

(20)

y

Ri

The basic procedure for determining the nature of such a graph is to complete the squares of the quadratic terms and then try to match up the resulting equation with one of the forms of a translated conic. Example 8 Describe the graph of the equation y 2 − 8x − 6y − 23 = 0

Solution. The equation involves quadratic terms in y but none in x, so we first take all of y 2 − 6y = 8x + 23

sa

(–4, 3)

n

the y-terms to one side: (–2, 3)

Next, we complete the square on the y-terms by adding 9 to both sides:

x

Ha

(y − 3)2 = 8x + 32

Finally, we factor out the coefficient of the x-term to obtain (y − 3)2 = 8(x + 4)

Directrix

x = –6 – 8x – 6y – 23 = 0

This equation is of form (12) with h = −4, k = 3, and p = 2, so the graph is a parabola with vertex (−4, 3) opening to the right. Since p = 2, the focus is 2 units to the right of the vertex, which places it at the point (−2, 3); and the directrix is 2 units to the left of the vertex, which means that its equation is x = −6. The parabola is shown in Figure 11.4.27.

ad

y2

Figure 11.4.27

M uh am

m

Example 9 Describe the graph of the equation 16x 2 + 9y 2 − 64x − 54y + 1 = 0

Solution. This equation involves quadratic terms in both x and y, so we will group the x-terms and the y-terms on one side and put the constant on the other: (16x 2 − 64x) + (9y 2 − 54y) = −1 Next, factor out the coefficients of x 2 and y 2 and complete the squares: 16(x 2 − 4x + 4) + 9(y 2 − 6y + 9) = −1 + 64 + 81 or 16(x − 2)2 + 9(y − 3)2 = 144 Finally, divide through by 144 to introduce a 1 on the right side: (x − 2)2 (y − 3)2 + =1 9 16 This is an equation of form (17), with h = 2, k = 3, a 2 = 16, and b2 = 9. Thus, the graph

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y

(2, 7)

of the equation is an ellipse with center (2, 3) and major axis parallel to the y-axis. Since a = 4, the major axis extends 4 units above and 4 units below the center, so its endpoints are (2, 7) and (2, −1) (Figure 11.4.28). Since b = 3, the minor axis extends 3 units to the left and 3 units to the right of the center, so its endpoints are (−1, 3) and (5, 3). Since √ √ c = a 2 − b2 = 16 − 9 = 7 √ √ the foci lie √7 units above and below the center, placing them at the points (2, 3 + 7 ) and (2, 3 − 7 ).

(2, 3 + √7 )

(2, 3)

(–1, 3)

us uf i

Analytic Geometry in Calculus

(5, 3)

x

Example 10 Describe the graph of the equation

(2, 3 – √7 )

x 2 − y 2 − 4x + 8y − 21 = 0

16x 2 + 9y 2 – 64x – 54y + 1 = 0

az

(2, –1)

Yo

762

Solution. This equation involves quadratic terms in both x and y, so we will group the

Figure 11.4.28

(x 2 − 4x) − (y 2 − 8y) = 21

Ri

x-terms and the y-terms on one side and put the constant on the other:

We leave it for you to verify by completing the squares that this equation can be written as (x − 2)2 (y − 4)2 =1 − 9 9

n

This is an equation of form (18) with h = 2, k = 4, a 2 = 9, and b2 = 9. Thus, the equation represents a hyperbola with center (2, 4) and focal axis parallel to the x-axis. Since a = 3, the vertices are located 3 units to the left √ and 3 units√to the right of √ the center, or at the points 2 + b2 = (−1, 4) and (5, 4). From (11), c = a 9 + 9 = 3 2, located √ so the foci are √ √ 3 2 units to the left and right of the center, or at the points (2 − 3 2, 4) and (2 + 3 2, 4). The equations of the asymptotes may be found using the trick of substituting 0 for 1 in (21) to obtain

y= x+2

sa

(2 – 3√2, 4)

y

(2 + 3√2, 4)

Ha

y = –x + 6

x

(x − 2)2 (y − 4)2 − =0 9 9 This can be written as y − 4 = ±(x − 2), which yields the asymptotes and

y = −x + 6

With the aid of a box extending a = 3 units left and right of the center and b = 3 units above and below the center, we obtain the sketch in Figure 11.4.29.

m

Figure 11.4.29

ad

y =x+2

x 2 – y 2 – 4x + 8y – 21 = 0

(21)

••••••••••••••••••••••••••••••••••••••

M uh am

REFLECTION PROPERTIES OF THE CONIC SECTIONS

Parabolas, ellipses, and hyperbolas have certain reflection properties that make them extremely valuable in various applications. In the exercises we will ask you to prove the following results. 11.4.4 THEOREM (Reflection Property of Parabolas). The tangent line at a point P on a parabola makes equal angles with the line through P parallel to the axis of symmetry and the line through P and the focus (Figure 11.4.30a).

11.4.5 THEOREM (Reflection Property of Ellipses). A line tangent to an ellipse at a point P makes equal angles with the lines joining P to the foci (Figure 11.4.30b).

11.4.6 THEOREM (Reflection Property of Hyperbolas). A line tangent to a hyperbola at a point P makes equal angles with the lines joining P to the foci (Figure 11.4.30c).

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Tangent line at P

Axis of symmetry

a

P a

a a

a a

Focus

P Tangent line at P

Tangent line at P

(a)

(b)

Figure 11.4.30

••••••••••••••••••••••••••••••••••••••

(c)

Fermat’s principle in optics states that light reflects off of a surface at an angle equal to its angle of incidence. (See Exercise 61 in Section 4.6.) In particular, if a reflecting surface is generated by revolving a parabola about its axis of symmetry, it follows from Theorem 11.4.4 that all light rays entering parallel to the axis will be reflected to the focus (Figure 11.4.31a); conversely, if a light source is located at the focus, then the reflected rays will all be parallel to the axis (Figure 11.4.31b). This principle is used in certain telescopes to reflect the approximately parallel rays of light from the stars and planets off of a parabolic mirror to an eyepiece at the focus; and the parabolic reflectors in flashlights and automobile headlights utilize this principle to form a parallel beam of light rays from a bulb placed at the focus. The same optical principles apply to radar signals and sound waves, which explains the parabolic shape of many antennas.

sa

n

Ri

az

APPLICATIONS OF THE CONIC SECTIONS

763

Yo

P

Conic Sections in Calculus

us uf i

11.4

Ha

Incoming signals are reflected by the parabolic antenna to the receiver at the focus.

(b)

ad

(a)

M uh am

m

Figure 11.4.31

Ship

Atlantic Ocean

Figure 11.4.32

Visitors to various rooms in the United States Capitol Building and in St. Paul’s Cathedral in Rome are often astonished by the “whispering gallery” effect in which two people at opposite ends of the room can hear one another’s whispers very clearly. Such rooms have ceilings with elliptical cross sections and common foci. Thus, when the two people stand at the foci, their whispers are reflected directly to one another off of the elliptical ceiling. Hyperbolic navigation systems, which were developed in World War II as navigational aids to ships, are based on the definition of a hyperbola. With these systems the ship receives synchronized radio signals from two widely spaced transmitters with known positions. The ship’s electronic receiver measures the difference in reception times between the signals and then uses that difference to compute the difference 2a in its distance between the two transmitters. This information places the ship somewhere on the hyperbola whose foci are at the transmitters and whose points have 2a as the difference in their distances from the foci. By repeating the process with a second set of transmitters, the position of the ship can be approximated as the intersection of two hyperbolas (Figure 11.4.32).

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Analytic Geometry in Calculus

EXERCISE SET 11.4

Graphing Utility

C

CAS

us uf i

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••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

11. (a) 9(x − 1)2 + 16(y − 3)2 = 144 (b) 3(x + 2)2 + 4(y + 1)2 = 12

1. In each part, find the equation of the conic. y

y

1 x 0

x

13. (a) x 2 + 9y 2 + 2x − 18y + 1 = 0 (b) 4x 2 + y 2 + 8x − 10y = −13

Yo

1

0 -1

14. (a) 9x 2 + 4y 2 + 18x − 24y + 9 = 0 (b) 5x 2 + 9y 2 − 20x + 54y = −56

-2

-1

-3 -3 -2 -1 3 y

(c) 3

2

3

In Exercises 15–20, sketch the hyperbola, and label the vertices, foci, and asymptotes.

2 x

1

0

0

-1

-1

-2

-2 1

0

2

-3 -3 -2 -1

3

y

(e) 3

0

1

2

3

2

1

x

0

1 -1

-2

-2 0

1

2

3

x

0

-1

(b) 9y 2 − 4x 2 = 36 (b) 16x 2 − 25y 2 = 400

(y − 4)2 (x − 2)2 − =1 9 4 2 2 (b) (y + 3) − 9(x + 2) = 36 (y + 4)2 (x − 2)2 18. (a) − =1 3 5 2 (b) 16(x + 1) − 8(y − 3)2 = 16

17. (a)

y

(f) 3

2

y2 x2 − =1 16 4 x2 y2 − =1 16. (a) 9 25 15. (a)

x

Ri

1

-3 -3 -2 -1

1

y

(d) 3

2

-3 -3 -2 -1

0

4

az

2

sa

1

Ha

-2

12. (a) (x + 3)2 + 4(y − 5)2 = 16 (b) 14 x 2 + 19 (y + 2)2 − 1 = 0

(b)

2

n

(a)

-3 -3 -2 -1

0

1

2

3

ad

2. (a) Find the focus and directrix for each parabola in Exercise 1. (b) Find the foci of the ellipses in Exercise 1. (c) Find the foci and the equations of the asymptotes of the hyperbolas in Exercise 1.

19. (a) x 2 − 4y 2 + 2x + 8y − 7 = 0 (b) 16x 2 − y 2 − 32x − 6y = 57

20. (a) 4x 2 − 9y 2 + 16x + 54y − 29 = 0 (b) 4y 2 − x 2 + 40y − 4x = −60 In Exercises 21–26, find an equation for the parabola that satisfies the given conditions. 21. (a) Vertex (0, 0); focus (3, 0). (b) Vertex (0, 0); directrix x = 7. 22. (a) Vertex (0, 0); focus (0, −4). (b) Vertex (0, 0); directrix y = 12 .

3. (a) y 2 = 6x

23. (a) Focus (0, −3); directrix y = 3. (b) Vertex (1, 1); directrix y = −2.

m

In Exercises 3–8, sketch the parabola, and label the focus, vertex, and directrix.

M uh am

(b) x 2 = −9y

4. (a) y 2 = −10x

(b) x 2 = 4y

5. (a) (y − 3)2 = 6(x − 2)

6. (a) (y + 1)2 = −7(x − 4)

(b) (x + 2)2 = −(y + 2) 2 (b) x − 12 = 2(y − 1)

7. (a) x 2 − 4x + 2y = 1

(b) x = y 2 − 4y + 2

8. (a) y 2 − 6y − 2x + 1 = 0

(b) y = 4x 2 + 8x + 5

In Exercises 9–14, sketch the ellipse, and label the foci, the vertices, and the ends of the minor axis.

x2 y2 9. (a) + =1 16 9

10. (a)

y2 x2 + =1 4 25

(b) 9x + y = 9 2

2

(b) 4x 2 + 9y 2 = 36

24. (a) Focus (6, 0); directrix x = −6. (b) Focus (−1, 4); directrix x = 5. 25. Axis y = 0; passes through (3, 2) and (2, −3). 26. Vertex (5, −3); axis parallel to the y-axis; passes through (9, 5). In Exercises 27–32, find an equation for the ellipse that satisfies the given conditions. 27. (a) Ends of major axis (±3, 0); ends of minor axis (0, ±2). (b) Length of major axis 26; foci (±5, 0). √ 28. (a) Ends of major axis (0, ± 5 ); ends of minor axis (±1, 0). (b) Length of minor axis 16; foci (0, ±6).

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√ 29. (a) Foci (±1, √ 0); b = 2. (b) c = 2 3; a = 4; center at the origin; foci on a coordinate axis (two answers).

31. (a) Ends of major axis (±6, 0); passes through (2, 3). (b) Foci (1, 2) and (1, 4); minor axis of length 2.

of 60 ◦ with the axis of the parabola when the comet is 40 million miles from the center of the Sun. Use the result in Exercise 41 to determine how close the comet will come to the center of the Sun. 43. For the parabolic reflector in the accompanying figure, how far from the vertex should the light source be placed to produce a beam of parallel rays?

32. (a) Center at (0, 0); major and minor axes along the coordinate axes; passes through (3, 2) and (1, 6). (b) Foci (2, 1) and (2, −3); major axis of length 6.

60°

1 ft

az

In Exercises 33–38, find an equation for a hyperbola that satisfies the given conditions. (In some cases there may be more than one hyperbola.)

1 ft

Ri

Figure Ex-42

33. (a) Vertices (±2, 0); foci (±3, 0). (b) Vertices (±1, 0); asymptotes y = ±2x.

y

(a)

n sa

36. (a) Asymptotes y = ± 34 x; c = 5. (b) Foci (±3, 0); asymptotes y = ±2x.

Ha

37. (a) Vertices (2, 4) and (10, 4); foci 10 units apart. (b) Asymptotes y = 2x + 1 and y = −2x + 3; passes through the origin. 38. (a) Foci (1, 8) and (1, −12); vertices 4 units apart. (b) Vertices (−3, −1) and (5, −1); b = 4.

(c)

y x

x

y 2 – 8x 2 = 5 y – 2x 2 = 0

3x 2 – 7y2 = 5 9y2 – 2x 2 = 1

x2 + y2 = 7 x2 – y2 = 1

45. (a) The accompanying figure shows an ellipse with semimajor axis a and semiminor axis b. Express the coordinates of the points P , Q, and R in terms of t. (b) How does the geometric interpretation of the parameter t differ between a circle x = a cos t,

y = a sin t

and an ellipse x = a cos t,

y = b sin t?

y

m

ad

39. (a) As illustrated in the accompanying figure, a parabolic arch spans a road 40 feet wide. How high is the arch if a center section of the road 20 feet wide has a minimum clearance of 12 feet? (b) How high would the center be if the arch were the upper half of an ellipse? 40. (a) Find an equation for the parabolic arch with base b and height h, shown in the accompanying figure. (b) Find the area under the arch.

M uh am

y

(b)

x

35. (a) Asymptotes y = ± 32 x; b = 4. (b) Foci (0, ±5); asymptotes y = ±2x.

Q

b P

R

x

t a

y

( 12 b, h) Figure Ex-45

12 ft

x

20 ft 40 ft

Figure Ex-39

Figure Ex-43

44. In each part, find the shaded area in the figure.

34. (a) Vertices (0, ±3); foci (0, ±5). (b) Vertices (0, ±3); asymptotes y = ±x.

12 ft

765

Yo

30. (a) Foci (±3, 0); a = 4. (b) b = 3; c = 4; center at the origin; foci on a coordinate axis (two answers).

Conic Sections in Calculus

us uf i

11.4

(b, 0) Figure Ex-40

41. Show that the vertex is the closest point on a parabola to the focus. [Suggestion: Introduce a convenient coordinate system and use Definition 11.4.1.]

42. As illustrated in the accompanying figure, suppose that a comet moves in a parabolic orbit with the Sun at its focus and that the line from the Sun to the comet makes an angle

46. (a) Show that the right and left branches of the hyperbola x2 y2 − 2 =1 2 b a can be represented parametrically as x=

a cosh t, y = b sinh t (−⬁ < t < +⬁) x = −a cosh t, y = b sinh t (−⬁ < t < +⬁) (b) Use a graphing utility to generate both branches of the hyperbola x 2 − y 2 = 1 on the same screen.

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47. (a) Show that the right and left branches of the hyperbola 2

57. Suppose that you want to draw an ellipse that has given values for the lengths of the major and minor axes by using the method shown in Figure 11.4.3b. Assuming that the axes are drawn, explain how a compass can be used to locate the positions for the tacks.

2

y x âˆ’ 2 =1 a2 b can be represented parametrically as x = a sec t, y = b tan t (âˆ’Ď€/2 < t < Ď€/2) x = âˆ’a sec t, y = b tan t (âˆ’Ď€/2 < t < Ď€/2) (b) Use a graphing utility to generate both branches of the hyperbola x 2 âˆ’ y 2 = 1 on the same screen.

Yo

58. The accompanying figure shows Keplerâ€™s method for constructing a parabola: a piece of string the length of the left edge of the drafting triangle is tacked to the vertex Q of the triangle and the other end to a fixed point F. A pencil holds the string taut against the base of the triangle as the edge opposite Q slides along a horizontal line L below F. Show that the pencil traces an arc of a parabola with focus F and directrix L.

az

48. Find an equation of the parabola traced by a point that moves so that its distance from (âˆ’1, 4) is the same as its distance to y = 1. 49. Find an equation of the ellipse traced by a point that moves so that the sum of its distances to (4, 1) and (4, 5) is 12. 50. Find the equation of the hyperbola traced by a point that moves so that the difference between its distances to (0, 0) and (1, 1) is 1.

Ri Figure Ex-58

59. The accompanying figure shows a method for constructing a hyperbola: a corner of a ruler is pinned to a fixed point F1 and the ruler is free to rotate about that point. A piece of string whose length is less than that of the ruler is tacked to a point F2 and to the free corner Q of the ruler on the same edge as F1 . A pencil holds the string taut against the top edge of the ruler as the ruler rotates about the point F1 . Show that the pencil traces an arc of a hyperbola with foci F1 and F2 .

Ha

sa

52. Suppose that the base of a solid is elliptical with a major axis of length 9 and a minor axis of length 4. Find the volume of the solid if the cross sections perpendicular to the minor axis are equilateral triangles (see the accompanying figure).

L

n

51. Suppose that the base of a solid is elliptical with a major axis of length 9 and a minor axis of length 4. Find the volume of the solid if the cross sections perpendicular to the major axis are squares (see the accompanying figure).

Q

F

Figure Ex-52

ad

Figure Ex-51

Q

53. Show that an ellipse with semimajor axis a and semiminor axis b has area A = Ď€ab.

M uh am

m

54. (a) Show that the ellipsoid that results when an ellipse with semimajor axis a and semiminor axis b is revolved about the major axis has volume V = 43 Ď€ab2 . (b) Show that the ellipsoid that results when an ellipse with semimajor axis a and semiminor axis b is revolved about the minor axis has volume V = 43 Ď€a 2 b.

F2

F1 Figure Ex-59

55. Show that the ellipsoid that results when an ellipse with semimajor axis a and semiminor axis b is revolved about the major axis has surface area b a âˆ’1 c S = 2Ď€ab + sin a c a âˆš 2 2 where c = a âˆ’ b .

60. Show that if a plane is not parallel to the axis of a right circular cylinder, then the intersection of the plane and cylinder is an ellipse (possibly a circle). [Hint: Let Î¸ be the angle shown in Figure Ex-60 (next page), introduce coordinate axes as shown, and express x and y in terms of x and y.]

56. Show that the ellipsoid that results when an ellipse with semimajor axis a and semiminor axis b is revolved about the minor axis has surface area a b a+c S = 2Ď€ab + ln b c b âˆš where c = a 2 âˆ’ b2 .

61. As illustrated in the accompanying figure, a carpenter needs to cut an elliptical hole in a sloped roof through which a circular vent pipe of diameter D is to be inserted vertically. The carpenter wants to draw the outline of the hole on the roof using a pencil, two tacks, and a piece of string (as in Figure 11.4.3b). The center point of the ellipse is known,

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occurred somewhere on the hyperbola y2 x2 − =1 v 2 t 2 /4 c2 − (v 2 t 2 /4)

and common sense suggests that its major axis must be perpendicular to the drip line of the roof. The carpenter needs to determine the length L of the string and the distance T between a tack and the center point. The architect’s plans show that the pitch of the roof is p (pitch = rise over run; see the accompanying figure). Find T and L in terms of D and p. [Note: This exercise is based on an article by William H. Enos, which appeared in the Mathematics Teacher, Feb. 1991, p. 148.]

Yo

y

x

F2 (–c, 0 )

y

F1(c, 0 )

x

Figure Ex-69 Vent pipe Drip line

70. As illustrated in the accompanying figure, suppose that two transmitting stations are positioned 100 km apart at points F1 (50, 0) and F2 (−50, 0) on a straight shoreline in an xycoordinate system. Suppose also that a ship is traveling parallel to the shoreline but 200 km at sea. Find the coordinates of the ship if the stations transmit a pulse simultaneously, but the pulse from station F1 is received by the ship 0.1 microsecond sooner than the pulse from station F2 . [Hint: Use the formula obtained in Exercise 69, assuming that the pulses travel at the speed of light (299,792,458 m/s).]

Rise Run

Ri

x′

u

az

y′

Figure Ex-61

Figure Ex-60

n

62. Prove: The line tangent to the parabola x 2 = 4py at the point (x0 , y0 ) is x0 x = 2p(y + y0 ).

Ha

y2 x2 + 2 =1 2 b a at the point (x0 , y0 ) has the equation yy0 xx0 + 2 =1 2 a b

sa

63. Prove: The line tangent to the ellipse

200 km x F2 (–50, 0 )

64. Prove: The line tangent to the hyperbola x2 y2 − =1 a2 b2 at the point (x0 , y0 ) has the equation yy0 xx0 − 2 =1 a2 b

F1(50, 0 )

m

ad

Figure Ex-70

65. Use the results in Exercises 63 and 64 to show that if an ellipse and a hyperbola have the same foci, then at each point of intersection their tangent lines are perpendicular.

M uh am

y

66. Find two values of k such that the line x + 2y = k is tangent to the ellipse x 2 + 4y 2 = 8. Find the points of tangency.

67. Find the coordinates of all points on the hyperbola 4x 2 − y 2 = 4

C

71. As illustrated in the accompanying figure, the tank of an oil truck is 18 feet long and has elliptical cross sections that are 6 feet wide and 4 feet high. (a) Show that the volume V of oil in the tank (in cubic feet) when it is filled to a depth of h feet is −1 h − 2 2 V = 27 4 sin + (h − 2) 4h − h + 2π 2 (b) Use the numerical root-finding capability of a CAS to determine how many inches from the bottom of a dipstick the calibration marks should be placed to indicate when the tank is 14 , 12 , and 34 full.

where the two lines that pass through the point and the foci are perpendicular.

Dipstick 18′

68. A line tangent to the hyperbola 4x − y = 36 intersects the y-axis at the point (0, 4). Find the point(s) of tangency. 2

2

69. As illustrated in the accompanying figure, suppose that two observers are stationed at the points F1 (c, 0) and F2 (−c, 0) in an xy-coordinate system. Suppose also that the sound of an explosion in the xy-plane is heard by the F1 observer t seconds before it is heard by the F2 observer. Assuming that the speed of sound is a constant v, show that the explosion

4′

h

Figure Ex-71

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Analytic Geometry in Calculus

tangent line at P (x0 , y0 ) intersects the y-axis at Q(0, −y0 ) and that the triangle whose three vertices are at P , Q, and the focus is isosceles.]

72. Consider the second-degree equation Ax + Cy + Dx + Ey + F = 0 where A and C are not both 0. Show by completing the square: (a) If AC > 0, then the equation represents an ellipse, a circle, a point, or has no graph. (b) If AC < 0, then the equation represents a hyperbola or a pair of intersecting lines. (c) If AC = 0, then the equation represents a parabola, a pair of parallel lines, or has no graph. 2

2

78. Given two intersecting lines, let L2 be the line with the larger angle of inclination φ2 , and let L1 be the line with the smaller angle of inclination φ1 . We define the angle θ between L1 and L2 by θ = φ2 − φ1 . (See the accompanying figure.) (a) Prove: If L1 and L2 are not perpendicular, then tan θ =

73. In each part, use the result in Exercise 72 to make a statement about the graph of the equation, and then check your conclusion by completing the square and identifying the graph. (a) x 2 − 5y 2 − 2x − 10y − 9 = 0 (b) x 2 − 3y 2 − 6y − 3 = 0 (c) 4x 2 + 8y 2 + 16x + 16y + 20 = 0 (d) 3x 2 + y 2 + 12x + 2y + 13 = 0 (e) x 2 + 8x + 2y + 14 = 0 (f ) 5x 2 + 40x + 2y + 94 = 0

az

Ri y

n

L1

Ha

77. Prove Theorem 11.4.4. [Hint: Choose coordinate axes so that the parabola has the equation x 2 = 4py. Show that the

m2 − m1 1 + m 1 m2

where L1 and L2 have slopes m1 and m2 . (b) Prove Theorem 11.4.5. [Hint: Introduce coordinate axes so that the ellipse has the equation x 2 /a 2 + y 2 /b2 = 1, and use part (a).] (c) Prove Theorem 11.4.6. [Hint: Introduce coordinate axes so that the hyperbola has the equation x 2 /a 2 − y 2 /b2 = 1, and use part (a).]

L2

sa

74. Derive the equation x 2 = 4py in Figure 11.4.6. 75. Derive the equation (x 2 /b2 ) + (y 2 /a 2 ) = 1 given in Figure 11.4.14. 76. Derive the equation (x 2 /a 2 ) − (y 2 /b2 ) = 1 given in Figure 11.4.22.

Yo

768

u

f1

f2

x

Figure Ex-78

ad

11.5 ROTATION OF AXES; SECOND-DEGREE EQUATIONS

m

In the preceding section we obtained equations of conic sections with axes parallel to the coordinate axes. In this section we will study the equations of conics that are “tilted” relative to the coordinate axes. This will lead us to investigate rotations of coordinate axes.

••••••••••••••••••••••••••••••••••••••

M uh am

QUADRATIC EQUATIONS IN x AND y

y

3 2

P(x, y)

(1, 2)

1

x

-3

-2

-1

1

-1

(–1, –2)

-2 -3

Figure 11.5.1

2

3

We saw in Examples 8–10 of the preceding section that equations of the form Ax 2 + Cy 2 + Dx + Ey + F = 0

(1)

can represent conic sections. Equation (1) is a special case of the more general equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0

(2)

which, if A, B, and C are not all zero, is called a second-degree equation or quadratic equation in x and y. We will show later in this section that the graph of any second-degree equation is a conic section (possibly a degenerate conic section). If B = 0, then (2) reduces to (1) and the conic section has its axis or axes parallel to the coordinate axes. However, if B = 0, then (2) contains a “cross-product” term Bxy, and the graph of the conic section represented by the equation has its axis or axes “tilted” relative to the coordinate axes. As an illustration, consider the ellipse with foci F1 (1, 2) and F2 (−1, −2) and such that the sum of the distances from each point P (x, y) on the ellipse to the foci is 6 units. Expressing this condition as an equation, we obtain (Figure 11.5.1) (x − 1)2 + (y − 2)2 + (x + 1)2 + (y + 2)2 = 6

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Rotation of Axes; Second-Degree Equations

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Squaring both sides, then isolating the remaining radical, then squaring again ultimately yields 8x 2 âˆ’ 4xy + 5y 2 = 36

â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

To study conics that are tilted relative to the coordinate axes it is frequently helpful to rotate the coordinate axes, so that the rotated coordinate axes are parallel to the axes of the conic. Before we can discuss the details, we need to develop some ideas about rotation of coordinate axes. In Figure 11.5.2a the axes of an xy-coordinate system have been rotated about the origin through an angle Î¸ to produce a new x y -coordinate system. As shown in the figure, each point P in the plane has coordinates (x , y ) as well as coordinates (x, y). To see how the two are related, let r be the distance from the common origin to the point P , and let Îą be the angle shown in Figure 11.5.2b. It follows that x = r cos(Î¸ + Îą),

y = r sin(Î¸ + Îą)

and x = r cos Îą,

Ri

az

ROTATION OF AXES

Yo

as the equation of the ellipse. This is of form (2) with A = 8, B = âˆ’4, C = 5, D = 0, E = 0, F = âˆ’36.

y = r sin Îą

(3) (4)

n

Using familiar trigonometric identities, the relationships in (3) can be written as

sa

x = r cos Î¸ cos Îą âˆ’ r sin Î¸ sin Îą y = r sin Î¸ cos Îą + r cos Î¸ sin Îą

Ha

and on substituting (4) in these equations we obtain the following relationships called the rotation equations:

ad

x = x cos Î¸ âˆ’ y sin Î¸ y = x sin Î¸ + y cos Î¸

y

y

(x, y) P (x', y' )

m

y'

M uh am

(5)

y'

P

x'

y' x'

r

y

a x

u

u x x'

(a)

x

(b)

Figure 11.5.2

Example 1 Suppose that the axes of an xy-coordinate system are rotated through an angle of Î¸ = 45 â—Ś to obtain an x y -coordinate system. Find the equation of the curve x 2 âˆ’ xy + y 2 âˆ’ 6 = 0 in x y -coordinates. âˆš

âˆš

Solution. Substituting sin Î¸ = sin 45 â—Ś = 1/ 2 and cos Î¸ = cos 45 â—Ś = 1/ 2 in (5) yields the rotation equations x y x=âˆš âˆ’âˆš 2 2

and

x y y=âˆš +âˆš 2 2

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Substituting these into the given equation yields y x y x y 2 x y 2 x + âˆš +âˆš âˆ’ âˆš âˆ’âˆš âˆ’6=0 âˆš +âˆš âˆš âˆ’âˆš 2 2 2 2 2 2 2 2 or

y y'

x'

45Â°

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Analytic Geometry in Calculus

x

x 2 âˆ’ 2x y + y 2 âˆ’ x 2 + y 2 + x 2 + 2x y + y 2 =6 2 or

x 2 â€“ xy + y 2 â€“ 6 = 0

az

x2 y2 + =1 12 4 which is the equation of an ellipse (Figure 11.5.3).

Yo

770

If the rotation equations (5) are solved for x and y in terms of x and y, one obtains (Exercise 14):

Figure 11.5.3

Ri

x = x cos Î¸ + y sin Î¸ y = âˆ’x sin Î¸ + y cos Î¸

(6)

n

Example 2 Find the new coordinates of the point (2, 4) if the coordinate axes are rotated through an angle of Î¸ = 30 â—Ś .

sa

Solution. Using the rotation equations in (6) with x = 2, y = 4, cos Î¸ = cos 30 â—Ś = âˆš

Ha

3/2, and sin Î¸ = sin 30 â—Ś = 1/2, we obtain âˆš âˆš x = 2( 3/2) + 4(1/2) = 3 + 2 âˆš âˆš y = âˆ’2(1/2) + 4( 3/2) = âˆ’1 + 2 3 âˆš âˆš Thus, the new coordinates are ( 3 + 2, âˆ’1 + 2 3 ).

In Example 1 we were able to identify the curve x 2 âˆ’ xy + y 2 âˆ’ 6 = 0 as an ellipse because the rotation of axes eliminated the xy-term, thereby reducing the equation to a familiar form. This occurred because the new x y -axes were aligned with the axes of the ellipse. The following theorem tells how to determine an appropriate rotation of axes to eliminate the cross-product term of a second-degree equation in x and y.

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ELIMINATING THE CROSS-PRODUCT TERM

11.5.1

THEOREM.

If the equation

Ax + Bxy + Cy + Dx + Ey + F = 0 2

2

(7)

is such that B = 0, and if an x y -coordinate system is obtained by rotating the xy-axes through an angle Î¸ satisfying Aâˆ’C cot 2Î¸ = (8) B then, in x y -coordinates, Equation (7 ) will have the form A x 2 + C y 2 + D x + E y + F = 0

Proof. Substituting (5) into (7) and simplifying yields A x 2 + B x y + C y 2 + D x + E y + F = 0

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Rotation of Axes; Second-Degree Equations

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11.5

where A = A cos2 Î¸ + B cos Î¸ sin Î¸ + C sin2 Î¸ B = B(cos2 Î¸ âˆ’ sin2 Î¸) + 2(C âˆ’ A) sin Î¸ cos Î¸ C = A sin2 Î¸ âˆ’ B sin Î¸ cos Î¸ + C cos2 Î¸

(9)

Yo

D = D cos Î¸ + E sin Î¸ E = âˆ’D sin Î¸ + E cos Î¸ F = F

Ri

az

(Verify.) To complete the proof we must show that B = 0 if Aâˆ’C cot 2Î¸ = B or equivalently, cos 2Î¸ Aâˆ’C = (10) sin 2Î¸ B However, by using the trigonometric double-angle formulas, we can rewrite B in the form B = B cos 2Î¸ âˆ’ (A âˆ’ C) sin 2Î¸

n

Thus, B = 0 if Î¸ satisfies (10). â€˘ REMARK.

It is always possible to satisfy (8) with an angle Î¸ in the range 0 < Î¸ < Ď€/2. We will always use such a value of Î¸ .

sa

â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘

Ha

Example 3 Identify and sketch the curve xy = 1.

Solution. As a first step, we will rotate the coordinate axes to eliminate the cross-product term. Comparing the given equation to (7), we have A = 0,

B = 1,

C=0

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m

ad

Thus, the desired angle of rotation must satisfy Aâˆ’C 0âˆ’0 cot 2Î¸ = =0 = 1 B â—Ś This condition âˆšcan be met by takingâ—Ś 2Î¸ =âˆšĎ€/2 or Î¸ = Ď€/4 = 45 . Substituting cos Î¸ = â—Ś cos 45 = 1/ 2 and sin Î¸ = sin 45 = 1/ 2 in (5) yields

y

y'

Figure 11.5.4

x y x y x=âˆš âˆ’âˆš and y = âˆš + âˆš 2 2 2 2 Substituting these in the equation xy = 1 yields y 2 x y x y x 2 âˆ’ =1 = 1 and âˆš âˆ’âˆš âˆš +âˆš 2 2 2 2 2 2

which isâˆš the equation in âˆšthe x y -coordinate system of an equilateral hyperbola with vertices at ( 2, 0) and (âˆ’ 2, 0) in that coordinate system (Figure 11.5.4).

xy = 1

x'

x

In problems where it is inconvenient to solve Aâˆ’C cot 2Î¸ = B for Î¸ , the values of sin Î¸ and cos Î¸ needed for the rotation equations can be obtained by first calculating cos 2Î¸ and then computing sin Î¸ and cos Î¸ from the identities 1 âˆ’ cos 2Î¸ 1 + cos 2Î¸ sin Î¸ = and cos Î¸ = 2 2

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Analytic Geometry in Calculus y

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772

Example 4 Identify and sketch the curve 153x 2 âˆ’ 192xy + 97y 2 âˆ’ 30x âˆ’ 40y âˆ’ 200 = 0

Solution. We have A = 153, B = âˆ’192, and C = 97, so

25

Aâˆ’C 56 7 =âˆ’ =âˆ’ B 192 24 Since Î¸ is to be chosen in the range 0 < Î¸ < Ď€/2, this relationship is represented by the 7 triangle in Figure 11.5.5. From that triangle we obtain cos 2Î¸ = âˆ’ 25 , which implies that

7 1 âˆ’ 25 1 + cos 2Î¸ 3 cos Î¸ = = = 2 2 5

7 1 + 25 1 âˆ’ cos 2Î¸ 4 sin Î¸ = = = 2 2 5 Substituting these values in (5) yields the rotation equations

24

cot 2Î¸ =

Figure 11.5.5

x = 35 x âˆ’ 45 y

y

and

y = 45 x + 35 y

Ri

x

â€“7

az

Yo

2u

and substituting these in turn in the given equation yields

y'

153 (3x 25

x'

âˆ’ 4y )2 âˆ’

192 (3x 25

âˆ’ 4y )(4x + 3y ) +

97 (4x 25

+ 3y )2

x

40 (4x 5

+ 3y ) âˆ’ 200 = 0

n

âˆ’ 30 (3x âˆ’ 4y ) âˆ’ 5 which simplifies to

sa

25x 2 + 225y 2 âˆ’ 50x âˆ’ 200 = 0 or

x 2 + 9y 2 âˆ’ 2x âˆ’ 8 = 0

(x' â€“ 1) 2 + y' 2 = 1 9 Figure 11.5.6

THE DISCRIMINANT

ad

Ha

Completing the square yields (x âˆ’ 1)2 + y2 = 1 9 which is the equation in the x y -coordinate system of an ellipse with center (1, 0) in that coordinate system and semiaxes a = 3 and b = 1 (Figure 11.5.6).

M uh am

m

It is possible to describe the graph of a second-degree equation without rotating coordinate axes. 11.5.2

THEOREM.

Consider a second-degree equation

Ax + Bxy + Cy + Dx + Ey + F = 0 2

2

(11)

If B âˆ’ 4AC < 0, the equation represents an ellipse, a circle, a point, or else has no graph. (b) If B 2 âˆ’ 4AC > 0, the equation represents a hyperbola or a pair of intersecting lines. (c) If B 2 âˆ’ 4AC = 0, the equation represents a parabola, a line, a pair of parallel lines, or else has no graph.

(a)

2

The quantity B 2 âˆ’ 4AC in this theorem is called the discriminant of the quadratic equation. To see why Theorem 11.5.2 is true, we need a fact about the discriminant. It can be shown (Exercise 19) that if the coordinate axes are rotated through any angle Î¸ , and if A x 2 + B x y + C y 2 + D x + E y + F = 0

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Rotation of Axes; Second-Degree Equations

is the equation resulting from (11) after rotation, then B 2 âˆ’ 4AC = B 2 âˆ’ 4A C

773

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11.5

(13)

A x 2 + C y 2 + D x + E y + F = 0

and since B = 0, (13) tells us that B 2 âˆ’ 4AC = âˆ’4A C

Yo

In other words, the discriminant of a quadratic equation is not altered by rotating the coordinate axes. For this reason the discriminant is said to be invariant under a rotation of coordinate axes. In particular, if we choose the angle of rotation to eliminate the crossproduct term, then (12) becomes (14) (15)

Proof of Theorem 11.5.2(a). If B 2 âˆ’ 4AC < 0, then from (15), A C > 0, so (14) can

Ri

az

be divided through by A C and written in the form F 1 E 1 D 2 2 = âˆ’ x + y + y + x A A C A C C Since A C > 0, the numbers A and C have the same sign. We assume that this sign is positive, since Equation (14) can be multiplied through by âˆ’1 to achieve this, if necessary. By completing the squares, we can rewrite the last equation in the form

Ha

sa

n

(x âˆ’ h)2 (y âˆ’ k)2 + =K âˆš âˆš ( C )2 ( A )2 There are three possibilities: K > 0, in which case the graph is either a circle or an ellipse, depending on whether or not the denominators are equal; K < 0, in which case there is no graph, since the left side is nonnegative for all x and y ; or K = 0, in which case the graph is the single point (h, k), since the equation is satisfied only by x = h and y = k. The proofs of parts (b) and (c) require a similar kind of analysis. Example 5 Use the discriminant to identify the graph of 8x 2 âˆ’ 3xy + 5y 2 âˆ’ 7x + 6 = 0

Solution. We have

ad

B 2 âˆ’ 4AC = (âˆ’3)2 âˆ’ 4(8)(5) = âˆ’151

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m

Since the discriminant is negative, the equation represents an ellipse, a point, or else has no graph. (Why canâ€™t the graph be a circle?)

EXERCISE SET 11.5

C

In cases where a quadratic equation represents a point, a line, a pair of parallel lines, a pair of intersecting lines, or has no graph, we say that equation represents a degenerate conic section. Thus, if we allow for possible degeneracy, it follows from Theorem 11.5.2 that every quadratic equation has a conic section as its graph.

CAS

â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

1. Let an x y -coordinate system be obtained by rotating an xy-coordinate system through an angle of Î¸ = 60 â—Ś . (a) Find the x y -coordinates of the point whose xy-coordinates are (âˆ’2, 6). âˆš (b) Find an equation of the curve 3xy + y 2 = 6 in x y coordinates. (c) Sketch the curve in part (b), showing both xy-axes and x y -axes.

2. Let an x y -coordinate system be obtained by rotating an xy-coordinate system through an angle of Î¸ = 30 â—Ś . (a) Find the x y -coordinates of the point whose xy-coorâˆš dinates are (1, âˆ’ 3). âˆš (b) Find an equation of the curve 2x 2 + 2 3xy = 3 in x y -coordinates. (c) Sketch the curve in part (b), showing both xy-axes and x y -axes.

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22. Prove: If B = 0, then the graph of x 2 + Bxy + F = 0 is a hyperbola if F = 0 and two intersecting lines if F = 0.

In Exercises 3â€“12, rotate the coordinate axes to remove the xy-term. Then name the conic and sketch its graph.

In Exercises 23â€“27, use the discriminant to identify the graph of the given equation.

4. x 2 âˆ’ xy + y 2 âˆ’ 2 = 0

3. xy = âˆ’9

5. x 2 + 4xy âˆ’ 2y 2 âˆ’ 6 = 0 âˆš 6. 31x 2 + 10 3xy + 21y 2 âˆ’ 144 = 0 âˆš âˆš 7. x 2 + 2 3xy + 3y 2 + 2 3x âˆ’ 2y = 0

23. x 2 âˆ’ xy + y 2 âˆ’ 2 = 0

24. x 2 + 4xy âˆ’ 2y 2 âˆ’ 6 = 0 âˆš âˆš 25. x 2 + 2 3xy + 3y 2 + 2 3x âˆ’ 2y = 0

8. 34x âˆ’ 24xy + 41y âˆ’ 25 = 0 2

2

9. 9x 2 âˆ’ 24xy + 16y 2 âˆ’ 80x âˆ’ 60y + 100 = 0 âˆš âˆš 10. 5x 2 âˆ’ 6xy + 5y 2 âˆ’ 8 2x + 8 2y = 8

26. 6x 2 + 24xy âˆ’ y 2 âˆ’ 12x + 26y + 11 = 0 27. 34x 2 âˆ’ 24xy + 41y 2 âˆ’ 25 = 0

az

11. 52x 2 âˆ’ 72xy + 73y 2 + 40x + 30y âˆ’ 75 = 0

28. Each of the following represents a degenerate conic section. Where possible, sketch the graph. (a) x 2 âˆ’ y 2 = 0 (b) x 2 + 3y 2 + 7 = 0 (c) 8x 2 + 7y 2 = 0 (d) x 2 âˆ’ 2xy + y 2 = 0 (e) 9x 2 + 12xy + 4y 2 âˆ’ 36 = 0 (f ) x 2 + y 2 âˆ’ 2x âˆ’ 4y = âˆ’5

12. 6x 2 + 24xy âˆ’ y 2 âˆ’ 12x + 26y + 11 = 0

Ri

13. Let an x y -coordinate system be obtained by rotating an xy-coordinate system through an angle Î¸. Prove: For every value of Î¸ , the equation x 2 + y 2 = r 2 becomes x 2 + y 2 = r 2 . Give a geometric explanation. 14. Derive (6) by solving the rotation equations in (5) for x and y in terms of x and y.

n

29. Prove parts (b) and (c) of Theorem 11.5.2.

C

30. Consider the conic whose equation is

sa

15. Let an x y -coordinate system be obtained by rotating an xycoordinate system through an angle of 45 â—Ś . Use (6) to find an equation of the curve 3x 2 + y 2 = 6 in xy-coordinates.

Ha

16. Let an x y -coordinate system be obtained by rotating an xy-coordinate system through an angle of 30 â—Ś . Use (5) to find an equation in x y -coordinates of the curve y = x 2 . 17. Show that the graph of the equation âˆš âˆš x+ y=1

is a portion of a parabola. [Hint: First rationalize the equation and then perform a rotation of axes.]

ad

2

19. Use (9) to prove that B âˆ’4AC = B âˆ’4A C for all values of Î¸. 20. Use (9) to prove that A + C = A + C for all values of Î¸ .

m

21. Prove: If A = C in (7), then the cross-product term can be eliminated by rotating through 45 â—Ś .

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C

x 2 + xy + 2y 2 âˆ’ x + 3y + 1 = 0

(a) Use the discriminant to identify the conic. (b) Graph the equation by solving for y in terms of x and graphing both solutions. (c) Your CAS may be able to graph the equation in the form given. If so, graph the equation in this way.

31. Consider the conic whose equation is 2x 2 + 9xy + y 2 âˆ’ 6x + y âˆ’ 4 = 0

18. Derive the expression for B in (9). 2

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Analytic Geometry in Calculus

Yo

774

(a) Use the discriminant to identify the conic. (b) Graph the equation by solving for y in terms of x and graphing both solutions. (c) Your CAS may be able to graph the equation in the form given. If so, graph the equation in this way.

11.6 CONIC SECTIONS IN POLAR COORDINATES It will be shown later in the text that if an object moves in a gravitational field that is directed toward a fixed point (such as the center of the Sun), then the path of that object must be a conic section with the fixed point at a focus. For example, planets in our solar system move along elliptical paths with the Sun at a focus, and the comets move along parabolic, elliptical, or hyperbolic paths with the Sun at a focus, depending on the conditions under which they were born. For applications of this type it is usually desirable to express the equations of the conic sections in polar coordinates with the pole at a focus. In this section we will show how to do this.

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••••••••••••••••••••••••••••••••••••••

Conic Sections in Polar Coordinates

775

us uf i

11.6

To obtain polar equations for the conic sections we will need the following theorem.

THE FOCUS–DIRECTRIX CHARACTERIZATION OF CONICS

Yo

11.6.1 THEOREM (Focus–Directrix Property of Conics). Suppose that a point P moves in the plane determined by a fixed point (called the focus) and a fixed line (called the directrix), where the focus does not lie on the directrix. If the point moves in such a way that its distance to the focus divided by its distance to the directrix is some constant e (called the eccentricity), then the curve traced by the point is a conic section. Moreover, the conic is a parabola if e = 1, an ellipse if 0 < e < 1, and a hyperbola if e > 1. • REMARK.

It is an unfortunate historical accident that the letter e is used for the base of the natural logarithms and the eccentricity of conic sections. However, the appropriate interpretation will usually be clear from the context in which the letter is used.

az

• • • • • • • • • • • •

sa

n

Ri

We will not give a formal proof of this theorem; rather, we will use the specific cases in Figure 11.6.1 to illustrate the basic ideas. For the parabola, we will take the directrix to be x = −p, as usual; and for the ellipse and the hyperbola we will take the directrix to be x = a 2 /c. We want to show in all three cases that if P is a point on the graph, F is the focus, and D is the directrix, then the ratio PF/PD is some constant e, where e = 1 for the parabola, 0 < e < 1 for the ellipse, and e > 1 for the hyperbola. We will give the arguments for the parabola and ellipse and leave the argument for the hyperbola as an exercise.

D

Ha

y

P(x, y)

y y

D P(x, y)

D x

x

ad

F(p, 0)

x = –p

P(x, y)

x

F(c, 0)

F(c, 0)

x = a2/c x = a2/c

M uh am

m

Figure 11.6.1

For the parabola, the distance PF to the focus is equal to the distance PD to the directrix, so that PF/PD = 1, which is what we wanted to show. For the ellipse, we rewrite Equation (8) of Section 11.4 as c c a2 −x (x − c)2 + y 2 = a − x = a c a But the expression on the left side is the distance PF, and the expression in the parentheses on the right side is the distance PD, so we have shown that c PF = PD a Thus, PF/PD is constant, and the eccentricity is e=

c a

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Analytic Geometry in Calculus

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776

ECCENTRICITY OF AN ELLIPSE AS A MEASURE OF FLATNESS

The eccentricity of an ellipse can be viewed as a measure of its flatness—as e approaches 0 the ellipses become more and more circular, and as e approaches 1 they become more and more flat (Figure 11.6.2). Table 11.6.1 shows the orbital eccentricities of various celestial objects. Note that most of the planets actually have fairly circular orbits.

az

••••••••••••••••••••••••••••••••••••••

Yo

If we rule out the degenerate case where a = 0 or c = 0, then it follows from Formula (7) of Section 11.4 that 0 < c < a, so 0 < e < 1, which is what we wanted to show. We will leave it as an exercise to show that the eccentricity of the hyperbola in Figure 11.6.1 is also given by Formula (1), but in this case it follows from Formula (11) of Section 11.4 that c > a, so e > 1.

Table 11.6.1

celestial body

e = 0.5 e = 0.8

e = 0.9

sa

n

F

Ellipses with a common focus and equal semimajor axes.

••••••••••••••••••••••••••••••••••••••

D

r F

u r cos u

M uh am

Pole

d

Directrix

Figure 11.6.3

Our next objective is to derive polar equations for the conic sections from their focus– directrix characterizations. We will assume that the focus is at the pole and the directrix is either parallel or perpendicular to the polar axis. If the directrix is parallel to the polar axis, then it can be above or below the pole; and if the directrix is perpendicular to the polar axis, then it can be to the left or right of the pole. Thus, there are four cases to consider. We will derive the formulas for the case in which the directrix is perpendicular to the polar axis and to the right of the pole. As illustrated in Figure 11.6.3, let us assume that the directrix is perpendicular to the polar axis and d units to the right of the pole, where the constant d is known. If P is a point on the conic and if the eccentricity of the conic is e, then it follows from Theorem 11.6.1 that PF/PD = e or, equivalently, that

m

P(r, u)

0.206 0.007 0.017 0.093 0.048 0.056 0.046 0.010 0.249 0.970

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POLAR EQUATIONS OF CONICS

eccentricity

Ha

Figure 11.6.2

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Halley's comet

Ri

e=0

PF = ePD

(2)

However, it is evident from Figure 11.6.3 that PF = r and PD = d − r cos θ. Thus, (2) can be written as r = e(d − r cos θ) which can be solved for r and expressed as r=

ed 1 + e cos θ

(verify). Observe that this single polar equation can represent a parabola, an ellipse, or a hyperbola, depending on the value of e. In contrast, the rectangular equations for these conics all have different forms. The derivations in the other three cases are similar.

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Conic Sections in Polar Coordinates

777

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11.6

r=

ed 1 + e cos θ

r=

Directrix right of pole

r=

Directrix left of pole

ed 1 + e sin θ

r=

ed 1 − e sin θ

(3–4)

(5–6)

Directrix below pole

az

Directrix above pole

Precise graphs of conic sections in polar coordinates can be generated with graphing utilities. However, it is often useful to be able to make quick sketches of these graphs that show their orientations and give some sense of their dimensions. The orientation of a conic relative to the polar axis can be deduced by matching its equation with one of the four forms in Theorem 11.6.2. The key dimensions of a parabola are determined by the constant p (Figure 11.4.5) and those of ellipses and hyperbolas by the constants a, b, and c (Figures 11.4.11 and 11.4.20). Thus, we need to show how these constants can be obtained from the polar equations.

Ri

••••••••••••••••••••••••••••••••••••••

ed 1 − e cos θ

Yo

11.6.2 THEOREM. If a conic section with eccentricity e is positioned in a polar coordinate system so that its focus is at the pole and the corresponding directrix is d units from the pole and is either parallel or perpendicular to the polar axis, then the equation of the conic has one of four possible forms, depending on its orientation:

sa

n

SKETCHING CONICS IN POLAR COORDINATES

2 in polar coordinates. 1 − cos θ

Ha

Example 1 Sketch the graph of r =

Solution. The equation is an exact match to (4) with d = 2 and e = 1. Thus, the graph is a parabola with the focus at the pole and the directrix 2 units to the left of the pole. This tells us that the parabola opens to the right along the polar axis and p = 1. Thus, the parabola looks roughly like that sketched in Figure 11.6.4.

Rough sketch

a

m

a a

b

M uh am

c

r1

Figure 11.6.5

All of the important geometric information about an ellipse can be obtained from the values of a, b, and c in Figure 11.6.5. One way to find these values from the polar equation of an ellipse is based on finding the distances from the focus to the vertices. As shown in the figure, let r0 be the distance from the focus to the closest vertex and r1 the distance to the farthest vertex. Thus,

ad

Figure 11.6.4

r0 = a − c

and

r1 = a + c

(7)

from which it follows that

r0

a = 12 (r1 + r0 )

(8)

c = 12 (r1 − r0 )

(9)

and

Moreover, it also follows from (7) that r0 r1 = a 2 − c2 = b2 Thus, b=

√ r 0 r1

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Analytic Geometry in Calculus

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• REMARK. • • • • • • • • • • • •

In words, Formula (8) states that a is the arithmetic average (also called the arithmetic mean) of r0 and r1 , and Formula (10) states that b is the geometric mean of r0 and r1 . 6 in polar coordinates. 2 + cos θ

Yo

Example 2 Sketch the graph of r =

Solution. This equation does not match any of the forms in Theorem 11.6.2 because they all require a constant term of 1 in the denominator. However, we can put the equation into one of these forms by dividing the numerator and denominator by 2 to obtain 3 r= 1 1 + 2 cos θ

Ri

az

This is an exact match to (3) with d = 6 and e = 12 , so the graph is an ellipse with the directrix 6 units to the right of the pole. The distance r0 from the focus to the closest vertex can be obtained by setting θ = 0 in this equation, and the distance r1 to the farthest vertex can be obtained by setting θ = π. This yields 3 3 3 3 = 1 =6 r0 = = 3 = 2, r1 = 1 1 1 + 2 cos 0 1 + 2 cos π 2 2 Rough sketch of

Thus, from Formulas (8), (10), and (9), respectively, we obtain √ √ a = 12 (r1 + r0 ) = 4, b = r0 r1 = 2 3, c = 12 (r1 − r0 ) = 2

3 1 + 12 cos u

n

r=

Thus, the ellipse looks roughly like that sketched in Figure 11.6.6.

sa

Figure 11.6.6

Ha

All of the important information about a hyperbola can be obtained from the values of a, b, and c in Figure 11.6.7. As with the ellipse, one way to find these values from the polar equation of a hyperbola is based on finding the distances from the focus to the vertices. As shown in the figure, let r0 be the distance from the focus to the closest vertex and r1 the distance to the farthest vertex. Thus, a b

r0 = c − a c

and

r1 = c + a

(11)

from which it follows that

(12)

c = 12 (r1 + r0 )

(13)

ad

a = 12 (r1 − r0 )

and

r1

Moreover, it also follows from (11) that

M uh am

Figure 11.6.7

m

r0

r0 r1 = c2 − a 2 = b2

from which it follows that √ b = r0 r1 Example 3 Sketch the graph of r =

(14) 2 in polar coordinates. 1 + 2 sin θ

Solution. This equation is an exact match to (5) with d = 1 and e = 2. Thus, the graph is a hyperbola with its directrix 1 unit above the pole. However, it is not so straightforward to compute the values of r0 and r1 , since hyperbolas in polar coordinates are generated in a strange way as θ varies from 0 to 2π. This can be seen from Figure 11.6.8a, which is the graph of the given equation in rectangular coordinates. It follows from this graph that the corresponding polar graph is generated in pieces (see Figure 11.6.8b): •

As θ varies over the interval 0 ≤ θ < 7π/6, the value of r is positive and varies down to 2/3 and then to +⬁, which generates part of the lower branch.

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Conic Sections in Polar Coordinates

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11.6

As θ varies over the interval 7π/6 < θ ≤ 3π/2, the value of r is negative and varies from −⬁ to −2, which generates the right part of the upper branch. • As θ varies over the interval 3π/2 ≤ θ < 11π/6, the value of r is negative and varies from −2 to −⬁, which generates the left part of the upper branch. • As θ varies over the interval 11π/6 < θ ≤ 2π, the value of r is positive and varies from +⬁ to 2, which fills in the missing piece of the lower right branch.

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•

It is now clear that we can obtain r0 by setting θ = π/2 and r1 by setting θ = 3π/2. Keeping in mind that r0 and r1 are positive, this yields

2

2 2 2

=

=2 r0 = = , r1 =

1 + 2 sin(3π/2) −1

1 + 2 sin(π/2) 3

Ri

az

Thus, from Formulas (12), (14), and (13), respectively, we obtain √ √ 1 1 2 2 3 4 a = (r1 − r0 ) = , b = r0 r1 = , c = (r1 + r0 ) = 3 3 2 3 2 Thus, the hyperbola looks roughly like that sketched in Figure 11.6.8c. r 2

u 6

e i m

sa

-1

o

n

1

-2

r=

2 1 + 2 sin u

(a) Figure 11.6.8 ••••••••••••••••••••••••••••••••••••••

Rough sketch

Rough sketch

(b)

(c)

∗

In 1609 Johannes Kepler published a book known as Astronomia Nova (or sometimes Commentaries on the Motions of Mars) in which he succeeded in distilling thousands of years of observational astronomy into three beautiful laws of planetary motion (Figure 11.6.9).

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m

ad

APPLICATIONS IN ASTRONOMY

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-3

∗

JOHANNES KEPLER (1571–1630). German astronomer and physicist, Kepler, whose work provided our contemporary view of planetary motion, led a fascinating but ill-starred life. His alcoholic father made him work in a family-owned tavern as a child, later withdrawing him from elementary school and hiring him out as a field laborer, where the boy contracted smallpox, permanently crippling his hands and impairing his eyesight. In later years, Kepler’s first wife and several children died, his mother was accused of witchcraft, and being a Protestant he was often subjected to persecution by Catholic authorities. He was often impoverished, eking out a living as an astrologer and prognosticator. Looking back on his unhappy childhood, Kepler described his father as “criminally inclined” and “quarrelsome” and his mother as “garrulous” and “bad-tempered.” However, it was his mother who left an indelible mark on the six-year-old Kepler by showing him the comet of 1577; and in later life he personally prepared her defense against the witchcraft charges. Kepler became acquainted with the work of Copernicus as a student at the University of T¨ubingen, where he received his master’s degree in 1591. He continued on as a theological student, but at the urging of the university officials he abandoned his clerical studies and accepted a position as a mathematician and teacher in Graz, Austria. However, he was expelled from the city when it came under Catholic control, and in 1600 he finally moved on to Prague, where he became an assistant at the observatory of the famous Danish astronomer Tycho Brahe. Brahe was a brilliant and meticulous astronomical observer who amassed the most accurate astronomical data known at that time; and when Brahe died in 1601 Kepler inherited the treasure-trove of data. After eight years of intense labor, Kepler deciphered the underlying principles buried in the data and in 1609 published his monumental work, Astronomia Nova, in which he stated his first two laws of planetary motion. Commenting on his discovery of elliptical orbits, Kepler wrote, “I was almost driven to madness in considering and calculating this matter. I could not find out why the planet would rather go on an elliptical orbit (rather than a circle). Oh ridiculous me!” It ultimately remained for Isaac Newton to discover the laws of gravitation that explained the reason for elliptical orbits.

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Analytic Geometry in Calculus

11.6.3

a

Sun

•

First law (Law of Orbits). Each planet moves in an elliptical orbit with the Sun at a focus.

•

Second law (Law of Areas). The radial line from the center of the Sun to the center of a planet sweeps out equal areas in equal times. Third law (Law of Periods). The square of a planet’s period (the time it takes the planet to complete one orbit about the Sun) is proportional to the cube of the semimajor axis of its orbit.

• Equal areas are swept out in equal times, and the square of the period T is proportional to a 3.

Kepler’s laws, although stated for planetary motion around the Sun, apply to all orbiting celestial bodies that are subjected to a single central gravitational force—artificial satellites subjected only to the central force of Earth’s gravity and moons subjected only to the central gravitational force of a planet, for example. Later in the text we will derive Kepler’s laws from basic principles, but for now we will show how they can be used in basic astronomical computations. In an elliptical orbit, the closest point to the focus is called the perigee and the farthest point the apogee (Figure 11.6.10). The distances from the focus to the perigee and apogee are called the perigee distance and apogee distance, respectively. For orbits around the Sun, it is more common to use the terms perihelion and aphelion, rather than perigee and apogee, and to measure time in Earth years and distances in astronomical units (AU), where 1 AU is the semimajor axis a of the Earth’s orbit (approximately 150×106 km or 92.9×106 mi). With this choice of units, the constant of proportionality in Kepler’s third law is 1, since a = 1 AU produces a period of T = 1 Earth year. In this case Kepler’s third law can be expressed as

Perigee

n

Apogee

Ri

az

Figure 11.6.9

KEPLER’S LAWS.

Yo

a

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780

sa

Figure 11.6.10

T = a 3/2

Shapes of elliptical orbits are often specified by giving the eccentricity e and the semimajor axis a, so it is useful to express the polar equations of an ellipse in terms of these constants. Figure 11.6.11, which can be obtained from the ellipse in Figure 11.6.1 and the relationship c = ea, implies that the distance d between the focus and the directrix is

Focus

ae

d=

a e

a a a(1 − e2 ) − c = − ea = e e e

(16)

from which it follows that ed = a(1−e2 ). Thus, depending on the orientation of the ellipse, the formulas in Theorem 11.6.2 can be expressed in terms of a and e as

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Figure 11.6.11

a

m

a

(15)

ad

Center

Ha

Directrix

r=

a(1 − e2 ) 1 ± e cos θ

+: Directrix right of pole −: Directrix left of pole

r=

a(1 − e2 ) 1 ± e sin θ

(17–18)

+: Directrix above pole −: Directrix below pole

Moreover, it is evident from Figure 11.6.11 that the distances from the focus to the closest and farthest vertices can be expressed in terms of a and e as r0 = a − ea = a(1 − e)

and

r1 = a + ea = a(1 + e)

(19–20)

Halley's comet p/ 2

Figure 11.6.12

Example 4 Halley’s comet (last seen in 1986) has an eccentricity of 0.97 and a semimajor axis of a = 18.1 AU. 0

(a) (b)

Find the equation of its orbit in the polar coordinate system shown in Figure 11.6.12. Find the period of its orbit.

(c)

Find its perihelion and aphelion distances.

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Conic Sections in Polar Coordinates

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11.6

Solution (a). From (17), the polar equation of the orbit has the form

a(1 − e2 ) 1 + e cos θ But a(1 − e2 ) = 18.1[1 − (0.97)2 ] ≈ 1.07. Thus, the equation of the orbit is 1.07 r= 1 + 0.97 cos θ

Yo

r=

Solution (b). From (15), with a = 18.1, the period of the orbit is

Halley’s comet photographed April 21, 1910 in Peru

T = (18.1)3/2 ≈ 77 years

Solution (c). Since the perihelion and aphelion distances are the distances to the closest

az

and farthest vertices, respectively, it follows from (19) and (20) that r0 = a − ea = a(1 − e) = 18.1(1 − 0.97) ≈ 0.543 AU r1 = a + ea = a(1 + e) = 18.1(1 + 0.97) ≈ 35.7 AU

Ri

or since 1 AU ≈ 150 × 106 km, the perihelion and aphelion distances in kilometers are r0 = 18.1(1 − 0.97)(150 × 106 ) ≈ 81,500,000 km

Minimum distance

r1 = 18.1(1 + 0.97)(150 × 106 ) ≈ 5,350,000,000 km • • • • • • •

Use the polar equation of the orbit of Halley’s comet to check the

values of r0 and r1 .

n

• FOR THE READER.

Ha

sa

Example 5 An Apollo lunar lander orbits the Moon in an elliptic orbit with eccentricity e = 0.12 and semimajor axis a = 2015 km. Assuming the Moon to be a sphere of radius 1740 km, find the minimum and maximum heights of the lander above the lunar surface (Figure 11.6.13).

Solution. If we let r0 and r1 denote the minimum and maximum distances from the center of the Moon, then the minimum and maximum distances from the surface of the Moon will be dmin = r0 − 1740 dmax = r1 − 1740

Maximum distance

ad

Figure 11.6.13

dmin = r0 − 1740 = a(1 − e) − 1740 = 2015(0.88) − 1740 = 33.2 km dmax = r1 − 1740 = a(1 + e) − 1740 = 2015(1.12) − 1740 = 516.8 km

Graphing Utility

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EXERCISE SET 11.6

m

or from Formulas (19) and (20)

For the conics in Exercises 1 and 2, find the eccentricity and the distance from the pole to the directrix, and sketch the graph in polar coordinates.

In Exercises 3 and 4, use Formulas (3)–(6) to name and describe the orientation of the conic, and then check your answer by generating the graph with a graphing utility.

3 2 − 2 cos θ 4 (c) r = 2 + 3 cos θ

(b) r =

3 2 + sin θ 5 (d) r = 3 + 3 sin θ

3. (a) r =

4 3 − 2 cos θ 1 (c) r = 3 + 3 sin θ

3 3 − 4 sin θ 1 (d) r = 2 + 6 sin θ

4. (a) r =

1. (a) r =

2. (a) r =

(b) r =

8 1 − sin θ 4 (c) r = 2 − 3 sin θ

16 4 + 3 sin θ 12 (d) r = 4 + cos θ

15 1 + cos θ 64 (c) r = 7 − 12 sin θ

(b) r =

(b) r =

2 3 + 3 cos θ 12 (d) r = 3 − 2 cos θ

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Analytic Geometry in Calculus

(b) Show that 1+e r1 = r0 1−e 16. (a) Show that the eccentricity of a hyperbola can be expressed in terms of r0 and r1 as r1 + r0 e= r1 − r 0

In Exercises 5–8, find a polar equation for the conic that has its focus at the pole and satisfies the stated conditions. Points are in polar coordinates and directrices in rectangular coordinates for simplicity. (In some cases there may be more than one conic that satisfies the conditions.) 5. (a) Ellipse; e = 23 ; directrix x = 1. (b) Parabola; directrix x = −1. (c) Hyperbola; e = 32 ; directrix y = 1.

(b) Show that e+1 r1 = r0 e−1

6. (a) Ellipse; e = 23 ; directrix y = −1. (b) Parabola; directrix y = 1. (c) Hyperbola; e = 43 ; directrix x = −1.

az

In Exercises 17–22, use the following values, where needed:

7. (a) Ellipse; vertices (6, 0) and (4, π). (b) Parabola; vertex (1, 3π/2). (c) Hyperbola; vertices (3, π/2) and (−7, 3π/2). 8. (a) Ellipse; ends of major axis (1, π/2) and (4, 3π/2). (b) Parabola; vertex (3, π). (c) Hyperbola; equilateral; vertex (5, 0).

1 2 − cos θ 8 (b) r = 4 − 3 sin θ

6 2 + sin θ 6 10. (a) r = 5 + 2 cos θ

(b) r =

10 6 − 9 cos θ 15 (b) r = 2 + 8 cos θ

ad

11. (a) r =

m

(b) r =

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In Exercises 13 and 14, find a polar equation for the ellipse that has its focus at the pole and satisfies the stated conditions. 13. (a) (b) (c) (d)

Directrix to the right of the pole; a = 8; e = 12 . Directrix below the pole; a = 4; e = 35 . Directrix to the left of the pole; b = 4; e = 35 . Directrix above the pole; c = 5; e = 15 .

14. (a) (b) (c) (d)

Directrix above the pole; a = 10; e = 12 . Directrix to the left of the pole; a = 6; e = 15 . Directrix below the pole; b = 4; e = 34 . Directrix to the right of the pole; c = 10; e = 45 .

Ri

17. The planet Pluto has eccentricity e = 0.249 and semimajor axis a = 39.5 AU. (a) Find the period T in years. (b) Find the perihelion and aphelion distances. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find a polar equation of Pluto’s orbit in that coordinate system. (d) Make a sketch of the orbit with reasonably accurate proportions.

n

Ha

In Exercises 11 and 12, find the distances from the pole to the vertices, and then apply Formulas (12)–(14) to find the equation of the hyperbola in rectangular coordinates. 2 1 + 3 sin θ 4 12. (a) r = 1 − 2 sin θ

radius of the Earth = 4000 mi = 6440 km 1 year (Earth year) = 365 days (Earth days) 1 AU = 92.9 × 106 mi = 150 × 106 km

sa

In Exercises 9 and 10, find the distances from the pole to the vertices, and then apply Formulas (8)–(10) to find the equation of the ellipse in rectangular coordinates. 9. (a) r =

Yo

782

15. (a) Show that the eccentricity of an ellipse can be expressed in terms of r0 and r1 as r1 − r0 e= r1 + r 0

18. (a) Let a be the semimajor axis of a planet’s orbit around the Sun, and let T be its period. Show that if T is measured in days and a in kilometers, then T = (365 × 10−9 )(a /150)3/2 . (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is a = 57.95 × 106 km. (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is e = 0.206. (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c). 19. The Hale–Bopp comet, discovered independently on July 23, 1995 by Alan Hale and Thomas Bopp, has an orbital eccentricity of e = 0.9951 and a period of 2380 years. (a) Find its semimajor axis in astronomical units (AU). (b) Find its perihelion and aphelion distances. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the Hale–Bopp orbit in that coordinate system. (d) Make a sketch of the Hale–Bopp orbit with reasonably accurate proportions. 20. Mars has a perihelion distance of 204,520,000 km and an aphelion distance of 246,280,000 km. (a) Use these data to calculate the eccentricity, and compare your answer to the value given in Table 11.6.1.

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(b) Find the period of Mars. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the orbit of Mars in that coordinate system. (d) Use a graphing utility to generate the orbit of Mars from the equation obtained in part (c).

Yo

21. Vanguard 1 was launched in March 1958 into an orbit around the Earth with eccentricity e = 0.21 and semimajor axis 8864.5 km. Find the minimum and maximum heights of Vanguard 1 above the surface of the Earth.

Not to scale

Figure Ex-22

az

23. What happens to the distance between the directrix and the center of an ellipse if the foci remain fixed and e → 0?

Ri

24. (a) Show that the coordinates of the point P on the hyperbola in Figure 11.6.1 satisfy the equation c (x − c)2 + y 2 = x − a a

(b) Use the result in part (a) to show that PF/PD = c/a.

n

22. The planet Jupiter is believed to have a rocky core of radius 10,000 km surrounded by two layers of hydrogen—a 40,000-km-thick layer of compressed metallic-like hydrogen and a 20,000-km-thick layer of ordinary molecular hydrogen. The visible features, such as the Great Red Spot, are at the outer surface of the molecular hydrogen layer. On November 6, 1997 the spacecraft Galileo was placed in a Jovian orbit to study the moon Europa. The orbit had eccentricity 0.814580 and semimajor axis 3,514,918.9 km. Find Galileo’s minimum and maximum heights above the molecular hydrogen layer (see the accompanying figure).

783

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Supplementary Exercises

C

CAS

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Graphing Utility

sa

SUPPLEMENTARY EXERCISES 1. Under what conditions does a parametric curve x = f(t), y = g(t) have a horizontal tangent line? A vertical tangent line? A singular point? 2. Express the point whose xy-coordinates are (−1, 1) in polar coordinates with (b) r < 0, 0 ≤ θ < 2π (d) r < 0, −π < θ ≤ π.

ad

(a) r > 0, 0 ≤ θ < 2π (c) r > 0, −π < θ ≤ π

5. The accompanying figure shows the polar graph of the equation r = f(θ ). Sketch the graph of π (a) r = f(−θ ) (b) r = f θ − 2 π (d) r = −f(θ ) (c) r = f θ + 2 (e) r = f(θ ) + 1. p/ 2

(1, p/ 4)

m

3. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a limac¸ on, a cardioid, a spiral, a lemniscate, or none of these. (b) r = cos 3θ

(g) r = (3 cos θ)2

(h) r = 1 + 3θ

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(a) r = 3 cos θ 3 (c) r = cos θ (e) r = 1 − 3 cos θ

0

(d) r = 3 − cos θ (f ) r = 3 cos θ

Figure Ex-5

2

4. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) find the distance from the pole to the directrix. 1 1 (a) r = (b) r = 3 + cos θ 1 − 3 cos θ 1 3 (c) r = (d) r = 3(1 + sin θ) 1 − sin θ

6. Find equations for the two families of circles in the accompanying figure. p/ 2

p/ 2 0

0

I Figure Ex-6

II

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Analytic Geometry in Calculus

7. In each part, identify the curve by converting the polar equation to rectangular coordinates. Assume that a > 0. θ (b) r 2 cos 2θ = a 2 (a) r = a sec2 2 π (c) r = 4 csc θ − (d) r = 4 cos θ + 8 sin θ 4 8. Use a graphing utility to investigate how the family of polar curves r = 1 + a cos nθ is affected by changing the values of a and n, where a is a positive real number and n is a positive integer. Write a brief paragraph to explain your conclusions.

is called a conchoid of Nicomedes (see the accompanying figure for the case 0 < a < b). (a) Describe how the conchoid x = cot t + 4 cos t,

y = 1 + 4 sin t

Yo

is generated as t varies over the interval 0 < t < 2π. (b) Find the horizontal asymptote of the conchoid given in part (a). (c) For what values of t does the conchoid in part (a) have a horizontal tangent line? A vertical tangent line? (d) Find a polar equation r = f(θ ) for the conchoid in part (a), and then find polar equations for the tangent lines to the conchoid at the pole.

az

In Exercises 9 and 10, find an equation in xy-coordinates for the conic section that satisfies the given conditions.

y

Ri

9. (a) Ellipse with eccentricity e = 27 and ends of the minor axis at the points (0, ±3). (b) Parabola with vertex at the origin, focus on the y-axis, and directrix passing through the point (7, 4). (c) Hyperbola that has the same foci as the ellipse 3x 2 + 16y 2 = 48 and asymptotes y = ±2x /3.

x

n

Figure Ex-14

15. Find the area of the region that is common to the circles r = 1, r = 2 cos θ, and r = 2 sin θ.

sa

10. (a) Ellipse with center (−3, 2), vertex (2, 2), and eccentricity e = 45 . (b) Parabola with focus (−2, −2) and vertex (−2, 0). (c) Hyperbola with vertex (−1, 7) and asymptotes y − 5 = ±8(x + 1).

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Ha

11. In each part, sketch the graph of the conic section with reasonably accurate proportions. (a) x 2 − 4x + 8y + 36 = 0 (b) 3x 2 + 4y 2 − 30x − 8y + 67 = 0 (c) 4x 2 − 5y 2 − 8x − 30y − 21 = 0 (d) x 2 + y 2 − 3xy − 3 = 0

ad

12. If you have a CAS that can graph implicit equations, use it to check your work in Exercise 11. 13. It can be shown that hanging cables form parabolic arcs rather than catenaries if they are subjected to uniformly distributed downward forces along their length. For example, if the weight of the roadway in a suspension bridge is assumed to be uniformly distributed along the supporting cables, then the cables can be modeled by parabolas. (a) Assuming a parabolic model, find an equation for the cable in the accompanying figure, taking the y-axis to be vertical and the origin at the low point of the cable. (b) Find the length of the cable between the supports.

M uh am

m

C

470 ft

4200 ft

Figure Ex-13

14. A parametric curve of the form x = a cot t + b cos t,

y = a + b sin t

(0 < t < 2π)

16. Find the area of the region that is inside the cardioid r = a(1 + sin θ ) and outside the circle r = a sin θ . 17. (a) Find the arc length of the polar curve r = 1/θ for π/4 ≤ θ ≤ π/2. (b) What can you say about the arc length of the portion of the curve that lies inside the circle r = 1? 18. (a) If a thread is unwound from a fixed circle while being held taut (i.e., tangent to the circle), then the end of the thread traces a curve called an involute of a circle. Show that if the circle is centered at the origin, has radius a, and the end of the thread is initially at the point (a, 0), then the involute can be expressed parametrically as x = a(cos θ + θ sin θ ),

y = a(sin θ − θ cos θ)

where θ is the angle shown in part (a) of Figure Ex-18 (next page). (b) Assuming that the dog in part (b) of Figure Ex-18 (next page) unwinds its leash while keeping it taut, for what values of θ in the interval 0 ≤ θ ≤ 2π will the dog be walking North? South? East? West? (c) Use a graphing utility to generate the curve traced by the dog, and show that it is consistent with your answer in part (b). 19. Let R be the region that is above the x-axis and enclosed 2 2 2 2 2 2 between √ the curve b x − a y = a b and the line 2 2 x = a +b . (a) Sketch the solid generated by revolving R about the x-axis, and find its volume.

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N

y

region R enclosed by the right branch of the hyperbola 1521x 2 −225y 2 = 342,225 and the lines x = 0, y = −h/2, and y = h/2 about the y-axis. (a) Find the volume of the tower. (b) Find the lateral surface area of the tower.

1 (a, 0)

u

W

E

S

(a)

27. The amusement park rides illustrated in the accompanying figure consist of two connected rotating arms of length 1— an inner arm that rotates counterclockwise at 1 radian per second and an outer arm that can be programmed to rotate either clockwise at 2 radians per second (the Scrambler ride) or counterclockwise at 2 radians per second (the Calypso ride). The center of the rider cage is at the end of the outer arm. (a) Show that in the Scrambler ride the center of the cage has parametric equations

Yo

x

u

(b)

Figure Ex-18

(b) Sketch the solid generated by revolving R about the y-axis, and find its volume.

az

a

785

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Supplementary Exercises

20. (a) Sketch the curves

x = cos t + cos 2t,

Ri

1 1 and r = 1 − cos θ 1 + cos θ (b) Find polar coordinates of the intersections of the curves in part (a). (c) Show that the curves are orthogonal, that is, their tangent lines are perpendicular at the points of intersection.

y = sin t − sin 2t

(b) Find parametric equations for the center of the cage in the Calypso ride, and use a graphing utility to confirm that the center traces the curve shown in the accompanying figure. (c) Do you think that a rider travels the same distance in one revolution of the Scrambler ride as in one revolution of the Calypso ride? Justify your conclusion.

n

r=

sa

21. How is the shape of a hyperbola affected as its eccentricity approaches 1? As it approaches +⬁? Draw some pictures to illustrate your conclusions.

Ha

22. Use the formula obtained in part (a) of Exercise 67 of Section 11.1 to find the distance between successive tips of the three-petal rose r = sin 3θ, and check your answer using trigonometry.

ad

23. (a) Find the minimum and maximum x-coordinates of points on the cardioid r = 1 + cos θ. (b) Find the minimum and maximum y-coordinates of points on the cardioid in part (a).

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m

24. (a) Show that the maximum √ value of the y-coordinate of points on the curve r = 1/ θ for θ in the interval (0, π] occurs when tan θ = 2θ. (b) Use Newton’s Method to solve the equation in part (a) for θ to at least four decimal-place accuracy. (c) Use the result of part (b) to approximate the maximum value of y for 0 < θ ≤ π. 25. Define the width of a petal of a rose curve to be the dimension shown in the accompanying figure. Show that the√ width w of a petal of the four-petal rose r = cos 2θ is w = 2 6/9. [Hint: Express y in terms of θ , and investigate the maximum value of y.]

Petal width

Figure Ex-25

26. A nuclear cooling tower is to have a height of h feet and the shape of the solid that is generated by revolving the

1

1

Scrambler ride

1

1

Calypso ride

Figure Ex-27

28. Use a graphing utility to explore the effect of changing the rotation rates and the arm lengths in Exercise 27. 29. Use the parametric equations x = a cos t, y = b sin t to show that the circumference C of an ellipse with semimajor axis a and eccentricity e is π/2 C = 4a 1 − e2 sin2 u du 0

30. Use Simpson’s rule or the numerical integration capability of a graphing utility to approximate the circumference of the ellipse 4x 2 + 9y 2 = 36 from the integral obtained in Exercise 29. 31. (a) Calculate the eccentricity of the Earth’s orbit, given that the ratio of the distance between the center of the Earth and the center of the Sun at perihelion to the distance between the centers at aphelion is 59 . 61 (b) Find the distance between the center of the Earth and the center of the Sun at perihelion, given that the average

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Analytic Geometry in Calculus

πu2 du 2 0 2 t πu sin du y = S(t) = 2 0

value of the perihelion and aphelion distances between the centers is 93 million miles.

cos

(−⬁ < t < +⬁)

Yo

(a) Use a CAS to graph the Cornu spiral. (b) Describe the behavior of the spiral as t → +⬁ and as t → −⬁. (c) Find the arc length of the spiral for −1 ≤ t ≤ 1.

32. It will be shown later in this text that if a projectile is launched with speed v0 at an angle α with the horizontal and at a height y0 above ground level, then the resulting trajectory relative to the coordinate system in the accompanying figure will have parametric equations

35. As illustrated in the accompanying figure, let P (r, θ ) be a point on the polar curve r = f(θ ), let ψ be the smallest counterclockwise angle from the extended radius OP to the tangent line at P , and let φ be the angle of inclination of the tangent line. Derive the formula r tan ψ = dr /dθ by substituting tan φ for dy/dx in Formula (7) of Section 11.2 and applying the trigonometric identity tan φ − tan θ tan(φ − θ ) = 1 + tan φ tan θ

az

y = y0 + (v0 sin α)t − 12 gt 2

Ri

where g is the acceleration due to gravity. (a) Show that the trajectory is a parabola. (b) Find the coordinates of the vertex. y

In Exercises 36 and 37, use the formula for ψ obtained in Exercise 35.

a

n

y0

t

x = C(t) =

(c) Use the result in Exercise 29 and Simpson’s rule or the numerical integration capability of a graphing utility to approximate the distance that the Earth travels in 1 year (one revolution around the Sun).

x = (v0 cos α)t,

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

sa

x

36. (a) Use the trigonometric identity

Ha

Figure Ex-32

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33. Mickey Mantle is recognized as baseball’s unofficial king of long home runs. On April 17, 1953 Mantle blasted a pitch by Chuck Stobbs of the hapless Washington Senators out of Griffith Stadium, just clearing the 50-ft wall at the 391-ft marker in left center. Assuming that the ball left the bat at a height of 3 ft above the ground and at an angle of 45 ◦ , use the parametric equations in Exercise 32 with g = 32 ft/s2 to find (a) the speed of the ball as it left the bat (b) the maximum height of the ball (c) the distance along the ground from home plate where the ball struck the ground. C

34. Recall from Section 7.5 that the Fresnel sine and cosine functions are defined as

S(x) =

x

sin

0

πt 2

2

dt and C(x) =

x

cos 0

πt 2

2

tan

to show that if (r, θ ) is a point on the cardioid r = 1 − cos θ

(0 ≤ θ < 2π)

then ψ = θ /2. (b) Sketch the cardioid and show the angle ψ at the points where the cardioid crosses the y-axis. (c) Find the angle ψ at the points where the cardioid crosses the y-axis. 37. Show that for a logarithmic spiral r = aebθ , the angle from the radial line to the tangent line is constant along the spiral (see the accompanying figure). [Note: For this reason, logarithmic spirals are sometimes called equiangular spirals.] c r = f (u)

P(r, u) Tangent line

dt

The following parametric curve, which is used to study amplitudes of light waves in optics, is called a clothoid or Cornu spiral in honor of the French scientist Marie Alfred Cornu (1841–1902):

1 − cos θ θ = 2 sin θ

u 0 Figure Ex-35

f

Figure Ex-37

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EXPANDING THE CALCULUS HORIZON

Comet Collision

Yo

The Earth lives in a cosmic shooting gallery of comets and asteroids. Although the probability that the Earth will be hit by a comet or asteroid in any given year is small, the consequences of such a collision are so catastrophic that the international community is now beginning to track near Earth objects (NEOs). Your job, as part of the international NEO tracking team, is to compute the orbits of incoming comets and asteroids, determine how close they will come to colliding with the Earth, and issue a notification if there is danger of a collision or near miss.

Ri

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At the time when the Earth is at its aphelion (its farthest point from the Sun), your NEO tracking team receives a notification from the NASA/Caltech Jet Propulsion Laboratory that a previously unknown comet (designation Rogue 2000) is traveling in the plane of Earth’s orbit and hurtling in the direction of the Earth. You immediately transmit a request to NASA for the orbital parameters and the current positions of the Earth and Rogue 2000 and receive the following report: orbital parameters

rogue 2000 Eccentricity: e2 = 0.98 Semimajor axis: a2 = 5 AU = 7.48 × 10 8 km Period: T2 = 5 √5 years

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Eccentricity: e1 = 0.017 Semimajor axis: a1 = 1 AU = 1.496 × 10 8 km Period: T1 = 1 year

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earth

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initial position information

ad

The major axes of Earth and Rogue 2000 lie on the same line. The aphelions of Earth and Rogue 2000 are on the same side of the Sun. Initial polar angle of Earth: u = 0 radians. Initial polar angle of Rogue 2000: u = 0.45 radian.

m

The Calculation Strategy Since the immediate concern is a possible collision at intersection A in Figure 1, your team works out the following plan: Find the polar equations for Earth and Rogue 2000. Find the polar coordinates of intersection A. Determine how long it will take the Earth to reach intersection A. Determine where Rogue 2000 will be when the Earth reaches intersection A. Determine how far Rogue 2000 will be from the Earth when the Earth is at intersection A.

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Step 1. Step 2. Step 3. Step 4. Step 5.

2

A

Comet

Earth

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Expanding the Calculus Horizon

0

Initial configuration of Earth and Rogue 2000

Figure 1

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Analytic Geometry in Calculus

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Polar Equations of the Orbits •••••••••••

Exercise 1

Write polar equations of the form

a(1 − e2 ) 1 − e cos θ for the orbits of Earth and Rogue 2000 using AU units for r.

Yo

r=

•••••••••••

Exercise 2

Use a graphing utility to generate the two orbits on the same screen.

•••••••••••

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Intersection of the Orbits The second step in your team’s calculation plan is to find the polar coordinates of intersection A in Figure 1. Exercise 3 For simplicity, let k1 = a1 (1 − e12 ) and k2 = a2 (1 − e22 ), and use the polar equations obtained in Exercise 1 to show that the angle θ at intersection A satisfies the equation k1 − k2 k 1 e2 − k 2 e 1

n

cos θ = •••••••••••

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Exercise 4 Use the result in Exercise 3 and the inverse cosine capability of a calculating utility to show that the angle θ at intersection A in Figure 1 is θ = 0.607 radian. •••••••••••

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Exercise 5 Use the result in Exercise 4 and either polar equation obtained in Exercise 1 to show that if r is in AU units, then the polar coordinates of intersection A are (r, θ ) = (1.014, 0.607).

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Time Required for Earth to Reach Intersection A According to Kepler’s second law (see 11.6.3), the radial line from the center of the Sun to the center of an object orbiting around it sweeps out equal areas in equal times. Thus, if t is the time that it takes for the radial line to sweep out an “elliptic sector” from some initial angle θI to some final angle θF (Figure 2), and if T is the period of the object (the time for one complete revolution), then t area of the “elliptic sector” = (1) T area of the entire ellipse

uF

uI

Figure 2

•••••••••••

Exercise 6

Use Formula (1) to show that

θF

T t=

r 2 dθ

2πa 2 1 − e2 θI

(2)

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Expanding the Calculus Horizon

•••••••••••

Exercise 7 Use a calculating utility with a numerical integration capability, Formula (2), and the polar equation for the orbit of the Earth obtained in Exercise 1 to find the time t (in years) required for the Earth to move from its initial position to intersection A.

Yo

Position of Rogue 2000 When the Earth Is at Intersection A The fourth step in your team’s calculation strategy is to determine the position of Rogue 2000 when the Earth reaches intersection A. •••••••••••

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Exercise 8 During the time that it takes for the Earth to move from its initial position to intersection A, the polar angle of Rogue 2000 will change from its initial value θI = 0.45 radian to some final value θF that remains to be determined. Apply Formula (2) using the orbital data for Rogue 2000 and the time t obtained in Exercise 7 to show that θF satisfies the equation 2 θF 2tπa22 1 − e22 a2 (1 − e22 ) dθ = (3) √ 5 5 0.45 1 − e2 cos θ

n

Your team is now faced with the problem of solving Equation (3) for the unknown upper limit θF . Some members of the team plan to use a CAS to perform the integration, some plan to use integration tables, and others plan to use hand calculation by making the substitution u = tan(θ /2) and applying the formulas in (5) of Section 8.6.

sa

•••••••••••

Exercise 9

(a) Evaluate the integral in (3) using a CAS or by hand calculation.

Ha

(b) Use the root-finding capability of a calculating utility to find the polar angle of Rogue 2000 when the Earth is at intersection A.

m

ad

Calculating the Critical Distance It is the policy of your NEO tracking team to issue a notification to various governmental agencies for any asteroid or comet that will be within 4 million kilometers of the Earth at an orbital intersection. (This distance is roughly 10 times that between the Earth and the Moon.) Accordingly, the final step in your team’s plan is to calculate the distance between the Earth and Rogue 2000 when the Earth is at intersection A, and then determine whether a notification should be issued. • • • • • • • • • • •

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Exercise 10 Use the polar equation of Rogue 2000 obtained in Exercise 1 and the result in Exercise 9(b) to find polar coordinates of Rogue 2000 with r in AU units when the Earth is at intersection A. • • • • • • • • • • •

Exercise 11 Use the distance formula in Exercise 67(a) of Section 11.1 to calculate the distance between the Earth and Rogue 2000 in AU units when the Earth is at intersection A, and then use the conversion factor 1 AU = 1.496 × 108 km to determine whether a government notification should be issued. Note: One of the closest near misses in recent history occurred on October 30, 1937 when the asteroid Hermes passed within 900,000 km of the Earth. More recently, on June 14, 1968 the asteroid Icarus passed within 23,000,000 km of the Earth.

.................................................................................................................................. Module by Mary Ann Connors, USMA, West Point, and Howard Anton, Drexel University

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T HREED IMENSIONAL S PACE; V ECTORS

n this chapter we will discuss rectangular coordinate systems in three dimensions, and we will study the analytic geometry of lines, planes, and other basic surfaces. The second theme of this chapter is the study of vectors. These are the mathematical objects that physicists and engineers use to study forces, displacements, and velocities of objects moving on curved paths. More generally, vectors are used to represent all physical entities that involve both a magnitude and a direction for their complete description. We will introduce various algebraic operations on vectors, and we will apply these operations to problems involving force, work, and rotational tendencies in two and three dimensions. Finally, we will discuss cylindrical and spherical coordinate systems, which are appropriate in problems that involve various kinds of symmetries and also have specific applications in navigation and celestial mechanics.

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Three-Dimensional Space; Vectors

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12.1 RECTANGULAR COORDINATES IN 3-SPACE; SPHERES; CYLINDRICAL SURFACES

••••••••••••••••••••••••••••••••••••••

In the remainder of this text we will call three-dimensional space 3-space, two-dimensional space (a plane) 2-space, and one-dimensional space (a line) 1-space. Just as points in 2space can be placed in one-to-one correspondence with pairs of real numbers using two perpendicular coordinate lines, so points in 3-space can be placed in one-to-one correspondence with triples of real numbers by using three mutually perpendicular coordinate lines, called the x-axis, the y-axis, and the z-axis, positioned so that their origins coincide (Figure 12.1.1). The three coordinate axes form a three-dimensional rectangular coordinate system (or Cartesian coordinate system). The point of intersection of the coordinate axes is called the origin of the coordinate system. Rectangular coordinate systems in 3-space fall into two categories: left-handed and right-handed. A right-handed system has the property that when the fingers of the right hand are cupped so that they curve from the positive x-axis toward the positive y-axis, the thumb points (roughly) in the direction of the positive z-axis (Figure 12.1.2a). Similarly for a left-handed coordinate system (Figure 12.1.2b). We will use only right-handed coordinate systems in this text.

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RECTANGULAR COORDINATE SYSTEMS

Yo

In this section we will discuss coordinate systems in three-dimensional space and some basic facts about surfaces in three dimensions.

Ha

z

z

z

x

y

y

O

Left-handed

Right-handed y

ad

x

(b)

(a)

x

Figure 12.1.1

Figure 12.1.2

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The coordinate axes, taken in pairs, determine three coordinate planes: the xy-plane, the xz-plane, and the yz-plane. To each point P in 3-space we can assign a triple of real numbers by passing three planes through P parallel to the coordinate planes and letting a, b, and c be the coordinates of the intersections of those planes with the x-axis, y-axis, and z-axis, respectively (Figure 12.1.3). We call a, b, and c the x-coordinate, y-coordinate, and z-coordinate of P , respectively, and we denote the point P by (a, b, c) or by P (a, b, c). Figure 12.1.4 shows the points (4, 5, 6) and (−3, 2, −4). z

z

z c

(4, 5, 6) P

(a, b, c) y

O a

y

y

b

(–3, 2, –4) x

x

x

Figure 12.1.3

Figure 12.1.4

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Rectangular Coordinates in 3-Space; Spheres; Cylindrical Surfaces

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12.1

description

xy-plane xz-plane yz-plane x-axis y-axis z-axis

Consists of all points of the form (x, y, 0) Consists of all points of the form (x, 0, z) Consists of all points of the form (0, y, z) Consists of all points of the form (x, 0, 0) Consists of all points of the form (0, y, 0) Consists of all points of the form (0, 0, z)

az

••••••••••••••••••••••••••••••••••••••

region

Yo

Just as the coordinate axes in a two-dimensional coordinate system divide 2-space into four quadrants, so the coordinate planes of a three-dimensional coordinate system divide 3-space into eight parts, called octants. The set of points with three positive coordinates forms the first octant; the remaining octants have no standard numbering. You should be able to visualize the following facts about three-dimensional rectangular coordinate systems:

Ri

To derive a formula for the distance between two points in 3-space, we start by considering a box whose sides have lengths a, b, and c (Figure 12.1.5). The length d of a diagonal of the box can be obtained by applying the Theorem of Pythagoras twice: first to show that a √ diagonal of the base has length a 2 + b2 , then again to show that a diagonal of the box has length d = ( a 2 + b 2 )2 + c 2 = a 2 + b 2 + c 2 (1)

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DISTANCE IN 3-SPACE

sa

c

We can now obtain a formula for the distance d between two points P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) in 3-space by finding the length of the diagonal of a box that has these points as diagonal corners (Figure 12.1.6). The sides of such a box have lengths

b

Ha

a

|x2 − x1 |,

√a 2 + b 2 Figure 12.1.5

|y2 − y1 |,

and

|z2 − z1 |

and hence from (1) the distance d between the points P1 and P2 is d=

z

P2 (x 2, y 2, z 2 )

(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2

(2)

ad

(where we have omitted the unnecessary absolute value signs).

|z 2 – z 1 | • REMARK.

|x 2 – x 1 |

x

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P1(x1, y1, z1) |y2 – y1 |

Figure 12.1.6

••••••••••••••••••••••••••••••••••••••

GRAPHS IN 3-SPACE

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

m

y

is

d=

Recall that in 2-space the distance d between points P1 (x1 , y1 ) and P2 (x2 , y2 )

(x2 − x1 )2 + (y2 − y1 )2

Thus, the distance formula in 3-space has the same form as the formula in 2-space, but it has a third term to account for the additional dimension. We will see that this is a common occurrence in extending formulas from 2-space to 3-space. Example 1 Find the distance d between the points (2, 3, −1) and (4, −1, 3).

Solution. From Formula (2) d=

√ (4 − 2)2 + (−1 − 3)2 + (3 + 1)2 = 36 = 6

Recall that in an xy-coordinate system, the set of points (x, y) whose coordinates satisfy an equation in x and y is called the graph of the equation. Analogously, in an xyz-coordinate system, the set of points (x, y, z) whose coordinates satisfy an equation in x, y, and z is called the graph of the equation. For example, consider the equation x 2 + y 2 + z2 = 25 The coordinates of a point (x, y, z) satisfy this equation if and only if the distance from

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Three-Dimensional Space; Vectors

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the origin to the point is 5 (why?). Thus, the graph of this equation is a sphere of radius 5 centered at the origin (Figure 12.1.7).

SPHERES z

The sphere with center (x0 , y0 , z0 ) and radius r consists of those points (x, y, z) whose coordinates satisfy (x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r

Yo

••••••••••••••••••••••••••••••••••••••

or, equivalently, (x, y, z)

(x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r 2

5 y

(3)

az

This is called the standard equation of the sphere with center (x0 , y0 , z0 ) and radius r. Some examples are given in the following table. x

Figure 12.1.7

graph

Sphere with center (3, 2, 1) and radius 3 Sphere with center (–1, 0, –4) and radius √5 Sphere with center (0, 0, 0) and radius 1

n

(x – 3)2 + (y – 2)2 + (z – 1)2 = 9 (x + 1)2 + y 2 + (z + 4)2 = 5 x2 + y2 + z2 = 1

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equation

sa

Recall that in 2-space the standard equation of the circle with center (x0 , y0 ) and radius r is (x − x0 )2 + (y − y0 )2 = r 2

Ha

Thus, the standard equation of a sphere in 3-space has the same form as the standard equation of a circle in 2-space, but with an additional term to account for the third coordinate.

ad

Example 2 A sphere S has center in the first octant and is tangent √to each of the three coordinate planes. The distance from the origin to the sphere is 3 − 3 units. What is the equation of the sphere?

Solution. Let P (x0 , y0 , z0 ) and r denote the center and radius of S, respectively. In order

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for S to be tangent to the xy-plane, the distance |z0 | from P (x0 , y0 , z0 ) to the xy-plane must equal r. Since P (x0 , y0 , z0 ) is in the first octant, we conclude that z0 = |z0 | = r. Simr and the center of√S is P (r, ilarly, x0 = y0 = √ √ r, r). The distance from the origin to the 2 + r2 + r2 = 2 = center of S is then r 3r 3r, from√which it follows √ √ √ that the distance √ the origin to S is given by 3r − r = ( 3 − 1)r. Solving the 3 − 3 = √ 3( 3 − 1)√from √ √ equation ( 3 − 1)r = 3( 3 − 1) yields the solution r = 3. Therefore, the equation of the sphere is √ √ √ (x − 3)2 + (y − 3)2 + (z − 3)2 = 3

If the terms in (3) are expanded and like terms are then collected, then the resulting equation has the form x 2 + y 2 + z2 + Gx + Hy + I z + J = 0

(4)

The following example shows how the center and radius of a sphere that is expressed in this form can be obtained by completing the squares. Example 3 Find the center and radius of the sphere x 2 + y 2 + z2 − 2x − 4y + 8z + 17 = 0

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12.1

Solution. We can put the equation in the form of (3) by completing the squares: (x 2 − 2x) + (y 2 − 4y) + (z2 + 8z) = −17 (x 2 − 2x + 1) + (y 2 − 4y + 4) + (z2 + 8z + 16) = −17 + 21 (x − 1)2 + (y − 2)2 + (z + 4)2 = 4

Yo

which is the equation of the sphere with center (1, 2, −4) and radius 2.

In general, completing the squares in (4) produces an equation of the form (x − x0 )2 + (y − y0 )2 + (z − z0 )2 = k

12.1.1

THEOREM.

az

√ If k > 0, then the graph of this equation is a sphere with center (x0 , y0 , z0 ) and radius k. If k = 0, then the sphere has radius zero, so the graph is the single point (x0 , y0 , z0 ). If k < 0, the equation is not satisfied by any values of x, y, and z (why?), so it has no graph. An equation of the form

x + y + z + Gx + Hy + I z + J = 0 represents a sphere, a point, or has no graph. 2

2

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2

••••••••••••••••••••••••••••••••••••••

Although it is natural to graph equations in two variables in 2-space and equations in three variables in 3-space, it is also possible to graph equations in two variables in 3-space. For example, the graph of the equation y = x 2 in an xy-coordinate system is a parabola; however, there is nothing to prevent us from writing this equation as y = x 2 + 0z and inquiring about its graph in an xyz-coordinate system. To obtain this graph we need only observe that the equation y = x 2 does not impose any restrictions on z. Thus, if we find values of x and y that satisfy this equation, then the coordinates of the point (x, y, z) will also satisfy the equation for arbitrary values of z. Geometrically, the point (x, y, z) lies on the vertical line through the point (x, y, 0) in the xy-plane, which means that we can obtain the graph of y = x 2 in an xyz-coordinate system by first graphing the equation in the xy-plane and then translating that graph parallel to the z-axis to generate the entire graph (Figure 12.1.8). The process of generating a surface by translating a plane curve parallel to some line is called extrusion, and surfaces that are generated by extrusion are called cylindrical surfaces. A familiar example is the surface of a right circular cylinder, which can be generated by translating a circle parallel to the axis of the cylinder. The following theorem provides basic information about graphing equations in two variables in 3-space:

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CYLINDRICAL SURFACES

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z

(x, y, z) y

ad

(x, y, 0 ) x

z

2-space

m

Figure 12.1.8

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x 2 + z2 = 1 x

Example 4 Sketch the graph of x 2 + z2 = 1 in 3-space.

Solution. Since y does not appear in this equation, the graph is a cylindrical surface

z

3-space

generated by extrusion parallel to the y-axis. In the xz-plane the graph of the equation x 2 + z2 = 1 is a circle (Figure 12.1.9). Thus, in 3-space the graph is a right circular cylinder along the y-axis.

x2 + z2 = 1

x

Figure 12.1.9

12.1.2 THEOREM. An equation that contains only two of the variables x, y, and z represents a cylindrical surface in an xyz-coordinate system. The surface can be obtained by graphing the equation in the coordinate plane of the two variables that appear in the equation and then translating that graph parallel to the axis of the missing variable.

y

Example 5 Sketch the graph of z = sin y in 3-space.

Solution. (See Figure 12.1.10.) •

FOR THE READER.

Describe the graph of x = 1 in an xyz-coordinate system.

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z z

z = sin y

z = sin y y

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y

x 2-space

3-space

EXERCISE SET 12.1

az

Figure 12.1.10

Graphing Utility

1. In each part, find the coordinates of the eight corners of the box. (b)

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z

(a)

9. In each part, find the standard equation of the sphere that satisfies the stated conditions. (a) Center (1, 0, −1); diameter = 8. (b) Center (−1, 3, 2) and passing through the origin. (c) A diameter has endpoints (−1, 2, 1) and (0, 2, 3).

z

11. In each part, find an equation of the sphere with center (2, −1, −3) and satisfying the given condition. (a) Tangent to the xy-plane (b) Tangent to the xz-plane (c) Tangent to the yz-plane

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x

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x

10. Find equations of two spheres that are centered at the origin and are tangent to the sphere of radius 1 centered at (3, −2, 4).

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y

2. A cube of side 4 has its geometric center at the origin and its faces parallel to the coordinate planes. Sketch the cube and give the coordinates of the corners.

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3. Suppose that a box has its faces parallel to the coordinate planes and the points (4, 2, −2) and (−6, 1, 1) are endpoints of a diagonal. Sketch the box and give the coordinates of the remaining six corners.

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4. Suppose that a box has its faces parallel to the coordinate planes and the points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are endpoints of a diagonal. (a) Find the coordinates of the remaining six corners. (b) Show that the midpoint of the line segment joining (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is 1 (x + x2 ), 12 (y1 + y2 ), 12 (z1 + z2 ) 2 1 [Suggestion: Apply Theorem D.2 in Appendix D to three appropriate edges of the box.]

5. Find the center and radius of the sphere that has (1, −2, 4) and (3, 4, −12) as endpoints of a diameter. [See Exercise 4.]

6. Show that (4, 5, 2), (1, 7, 3), and (2, 4, 5) are vertices of an equilateral triangle. 7. (a) Show that (2, 1, 6), (4, 7, 9), and (8, 5, −6) are the vertices of a right triangle. (b) Which vertex is at the 90 ◦ angle? (c) Find the area of the triangle.

8. Find the distance from the point (−5, 2, −3) to the (a) xy-plane (b) xz-plane (c) yz-plane (d) x-axis (e) y-axis (f ) z-axis.

12. (a) Find an equation of the sphere that is inscribed in the cube that is centered at the point (−2, 1, 3) and has sides of length 1 that are parallel to the coordinate planes. (b) Find an equation of the sphere that is circumscribed about the cube in part (a). In Exercises 13–18, describe the surface whose equation is given. 13. x 2 + y 2 + z2 + 10x + 4y + 2z − 19 = 0 14. x 2 + y 2 + z2 − y = 0 15. 2x 2 + 2y 2 + 2z2 − 2x − 3y + 5z − 2 = 0 16. x 2 + y 2 + z2 + 2x − 2y + 2z + 3 = 0 17. x 2 + y 2 + z2 − 3x + 4y − 8z + 25 = 0 18. x 2 + y 2 + z2 − 2x − 6y − 8z + 1 = 0 19. In each part, sketch the portion of the surface that lies in the first octant. (a) y = x (b) y = z (c) x = z 20. In each part, sketch the graph of the equation in 3-space. (a) x = 1 (b) y = 1 (c) z = 1 21. In each part, sketch the graph of the equation in 3-space. (a) x 2 + y 2 = 25 (b) y 2 + z2 = 25 (c) x 2 + z2 = 25 22. In each part, sketch the graph of the equation in 3-space. (a) x = y 2 (b) z = x 2 (c) y = z2

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Vectors

797

us uf i

12.2

23. In each part, write an equation for the surface. (a) The plane that contains the x-axis and the point (0, 1, 2). (b) The plane that contains the y-axis and the point (1, 0, 2). (c) The right circular cylinder that has radius 1 and is centered on the line parallel to the z-axis that passes through the point (1, 1, 0). (d) The right circular cylinder that has radius 1 and is centered on the line parallel to the y-axis that passes through the point (1, 0, 1).

38. Describe the set of all points in 3-space whose coordinates satisfy the inequality x 2 + y 2 + z2 − 2x + 8z ≤ 8.

24. Find equations for the following right circular cylinders. Each cylinder has radius a and is “tangent” to two coordinate planes.

41. As shown in the accompanying figure, a bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by packing a Styrofoam sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used. [Hint: Take the origin of a Cartesian coordinate system at a corner of the box with the coordinate axes along the edges.]

(b)

z

(c)

z

(0, a, a)

(a, 0, a)

y

y

y

(a, a, 0) x

x

Yo

az

z

40. The distance between a point P (x, y, z) and the point A(1, −2, 0) is twice the distance between P and the point B(0, 1, 1). Show that the set of all such points is a sphere, and find the center and radius of the sphere.

Ri

(a)

39. Describe the set of all points in 3-space whose coordinates satisfy the inequality y 2 + z2 + 6y − 4z > 3.

x

27. z = 1 − y

26. y = ex 28. z = cos x

2

29. 2x + z = 3

30. 2x + 3y = 6 √ 3−x

32. z =

33. y − 4z = 4

34. yz = 1

2

2

Ha

31. 4x 2 + 9z2 = 36

sa

25. y = sin x

n

In Exercises 25–34, sketch the surface in 3-space.

ad

35. Use a graphing utility to generate the curve y = x 3 /(1+x 2 ) in the xy-plane, and then use the graph to help sketch the surface z = y 3 /(1 + y 2 ) in 3-space. 36. Use a graphing utility to generate the curve y = x /(1 + x 4 ) in the xy-plane, and then use the graph to help sketch the surface z = y /(1 + y 4 ) in 3-space. 37. If a bug walks on the sphere

42. Consider the equation x 2 + y 2 + z2 + Gx + Hy + I z + J = 0 and let K = G2 + H 2 + I 2 − 4J . (a) Prove that the equation represents a sphere if K > 0, a point if K = 0, and has no graph if K < 0. (b) In the case where K > 0, find the center and radius of the sphere.

43. Show that for all values of θ and φ, the point (a sin φ cos θ, a sin φ sin θ, a cos φ) lies on the sphere x 2 + y 2 + z2 = a 2 .

M uh am

m

x 2 + y 2 + z2 + 2x − 2y − 4z − 3 = 0 how close and how far can it get from the origin?

Figure Ex-41

••••••••••••••••••••••••••••••••••••••

VECTORS IN PHYSICS AND ENGINEERING

12.2 VECTORS Many physical quantities such as area, length, mass, and temperature are completely described once the magnitude of the quantity is given. Such quantities are called “scalars.” Other physical quantities, called “vectors,” are not completely determined until both a magnitude and a direction are specified. For example, winds are usually described by giving their speed and direction, say 20 mi/h northeast. The wind speed and wind direction together form a vector quantity called the wind velocity. Other examples of vectors are force and displacement. In this section we will develop the basic mathematical properties of vectors. A particle that moves along a line can move in only two directions, so its direction of motion can be described by taking one direction to be positive and the other negative. Thus, the displacement or change in position of the point can be described by a signed real number. For

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Three-Dimensional Space; Vectors

us uf i

798

A

Ha

sa

B

n

Ri

az

Yo

example, a displacement of 3 (= +3) describes a position change of 3 units in the positive direction, and a displacement of −3 describes a position change of 3 units in the negative direction. However, for a particle that moves in two dimensions or three dimensions, a plus or minus sign is no longer sufficient to specify the direction of motion—other methods are required. One method is to use an arrow, called a vector, that points in the direction of motion and whose length represents the distance from the starting point to the ending point; this is called the displacement vector for the motion. For example, Figure 12.2.1a shows the displacement vector of a particle that moves from point A to point B along a circuitous path. Note that the length of the arrow describes the distance between the starting and ending points and not the actual distance traveled by the particle. Arrows are not limited to describing displacements—they can be used to describe any physical quantity that involves both a magnitude and direction. Two important examples are forces and velocities. For example, the arrow in Figure 12.2.1b shows a force vector of 10 lb acting in a specific direction on a block, and the arrows in Figure 12.2.1c show the velocity vector of a boat whose motor propels it parallel to the shore at 2 mi/h and the velocity vector of a 3 mi/h wind acting at an angle of 45 ◦ with the shoreline. Intuition suggests that the two velocity vectors will combine to produce some net velocity for the boat at an angle to the shoreline. Thus, our first objective in this section is to define mathematical operations on vectors that can be used to determine the combined effect of vectors.

Rope

3 mi/h

10 lb 45°

2 mi/h

A displacement vector

A force vector acting on a block

Two velocity vectors that affect the motion of the boat

(a)

(b)

(c)

ad

Figure 12.2.1

M uh am

VECTORS VIEWED GEOMETRICALLY

B

A

(a)

Figure 12.2.2

(b)

Vectors can be represented geometrically by arrows in 2-space or 3-space; the direction of the arrow specifies the direction of the vector and the length of the arrow describes its magnitude. The tail of the arrow is called the initial point of the vector, and the tip of the arrow the terminal point. We will denote vectors with lowercase boldface type such as a, k, v, w, and x. When discussing vectors, we will refer to real numbers as scalars. Scalars will be denoted by lowercase italic type such as a, k, v, w, and x. Two vectors, v and w, are considered to be equal (also called equivalent) if they have the same length and same direction, in which case we write v = w. Geometrically, two vectors are equal if they are translations of one another; thus, the three vectors in Figure 12.2.2a are equal, even though they are in different positions. Because vectors are not affected by translation, the initial point of a vector v can be moved to any convenient point A by making an appropriate translation. If the initial point −→ of v is A and the terminal point is B, then we write v = AB when we want to emphasize the initial and terminal points (Figure 12.2.2b). If the initial and terminal points of a vector coincide, then the vector has length zero; we call this the zero vector and denote it by 0. The zero vector does not have a specific direction, so we will agree that it can be assigned any convenient direction in a specific problem. There are various algebraic operations that are performed on vectors, all of whose definitions originated in physics. We begin with vector addition.

m

••••••••••••••••••••••••••••••••••••••

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w

12.2.1 DEFINITION. If v and w are vectors, then the sum v + w is the vector from the initial point of v to the terminal point of w when the vectors are positioned so the initial point of w is at the terminal point of v (Figure 12.2.3a).

v+w

w v+w w+v

Yo

In Figure 12.2.3b we have constructed two sums, v + w (purple arrows) and w + v (green arrows). It is evident that

(a)

v

799

v+w=w+v v

and that the sum coincides with the diagonal of the parallelogram determined by v and w when these vectors are positioned so they have the same initial point. Since the initial and terminal points of 0 coincide, it follows that

w

(b)

0+v=v+0=v

Figure 12.2.3

az

v

Vectors

us uf i

12.2

Ri

12.2.2 DEFINITION. If v is a nonzero vector and k is a nonzero real number (a scalar), then the scalar multiple kv is defined to be the vector whose length is |k| times the length of v and whose direction is the same as that of v if k > 0 and opposite to that of v if k < 0. We define kv = 0 if k = 0 or v = 0.

2v v 1 v 2

n

Figure 12.2.4 shows the geometric relationship between a vector v and various scalar multiples of it. Observe that if k and v are nonzero, then the vectors v and kv lie on the same line if their initial points coincide and lie on parallel or coincident lines if they do not. Thus, we say that v and kv are parallel vectors. Observe also that the vector (−1)v has the same length as v but is oppositely directed. We call (−1)v the negative of v and denote it by −v (Figure 12.2.5). In particular, −0 = (−1)0 = 0. Vector subtraction is defined in terms of addition and scalar multiplication by

sa

(−1)v

(− 23 )v

Ha

Figure 12.2.4

v − w = v + (−w)

M uh am

m

ad

The difference v − w can be obtained geometrically by first constructing the vector −w and then adding v and −w, say by the parallelogram method (Figure 12.2.6a). However, if v and w are positioned so their initial points coincide, then v − w can be formed more directly, as shown in Figure 12.2.6b, by drawing the vector from the terminal point of w (the second term) to the terminal point of v (the first term). In the special case where v = w the terminal points of the vectors coincide, so their difference is 0; that is,

••••••••••••••••••••••••••••••••••••••

VECTORS IN COORDINATE SYSTEMS

v + (−v) = v − v = 0

v

v–w

v −v

–w

v

w

(a)

v–w

w

(b)

Figure 12.2.6

Figure 12.2.5

Problems involving vectors are often best solved by introducing a rectangular coordinate system. If a vector v is positioned with its initial point at the origin of a rectangular coordinate system, then its terminal point will have coordinates of the form (v1 , v2 ) or (v1 , v2 , v3 ), depending on whether the vector is in 2-space or 3-space (Figure 12.2.7). We call these coordinates the components of v, and we write v = v1 , v2 2-space

or

v = v1 , v2 , v3 3-space

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Three-Dimensional Space; Vectors y

us uf i

800

In particular, the zero vector is

(v1, v2 )

0 = 0, 0

v

and

0 = 0, 0, 0

2-space

x

3-space

Components provide a simple way of identifying equivalent vectors. For example, consider the vectors v = v1 , v2 and w = w1 , w2 in 2-space. If v = w, then the vectors have the same length and same direction, and this means that their terminal points coincide when their initial points are placed at the origin. It follows that v1 = w1 and v2 = w2 , so we have shown that equivalent vectors have the same components. Conversely, if v1 = w1 and v2 = w2 , then the terminal points of the vectors coincide when their initial points are placed at the origin. It follows that the vectors have the same length and same direction, so we have shown that vectors with the same components are equivalent. A similar argument holds for vectors in 3-space, so we have the following result:

Yo

z

(v1, v2, v3) v y

az

x

Figure 12.2.7

Ri

12.2.3 THEOREM. Two vectors are equivalent if and only if their corresponding components are equal. For example,

a, b, c = 1, −4, 2

ARITHMETIC OPERATIONS ON VECTORS

sa

••••••••••••••••••••••••••••••••••••••

n

if and only if a = 1, b = −4, and c = 2. The next theorem shows how to perform arithmetic operations on vectors using components.

Ha

12.2.4 THEOREM. If v = v1 , v2 and w = w1 , w2 are vectors in 2-space and k is any scalar, then v + w = v1 + w1 , v2 + w2 v − w = v1 − w1 , v2 − w2 kv = kv1 , kv2

ad

y

Similarly, if v = v1 , v2 , v3 and w = w1 , w2 , w3 are vectors in 3-space and k is any scalar, then

(v1 + w1, v2 + w2)

v

w

+

w

(v1, v2 )

M uh am

w2

(w1, w 2 )

m

v2

v

v1

(kv1, kv2 )

kv

v2

v

(v1, v2 )

v + w = v1 + w1 , v2 + w2 , v3 + w3 v − w = v1 − w1 , v2 − w2 , v3 − w3 kv = kv1 , kv2 , kv3

(4) (5) (6)

x

w1

y

kv2

(1) (2) (3)

x

v1

kv1

Figure 12.2.8

••••••••••••••••••••••••••••••••••••••

VECTORS WITH INITIAL POINT NOT AT THE ORIGIN

We will not prove this theorem. However, results (1) and (3) should be evident from Figure 12.2.8. Similar figures in 3-space can be used to motivate (4) and (6). Formulas (2) and (5) can be obtained by writing v + w = v + (−1)w. Example 1 If v = −2, 0, 1 and w = 3, 5, −4, then v + w = −2, 0, 1 + 3, 5, −4 = 1, 5, −3 3v = −6, 0, 3 −w = −3, −5, 4 w − 2v = 3, 5, −4 − −4, 0, 2 = 7, 5, −6

Recall that we defined the components of a vector to be the coordinates of its terminal point when its initial point is at the origin. We will now consider the problem of finding

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OP1

P1 P2

OP2 x

O

the components of a vector whose initial point is not at the origin. To be specific, suppose that P1 (x1 , y1 ) and P2 (x2 , y2 ) are points in 2-space and we are interested in finding the −−→ components of the vector P1P2 . As illustrated in Figure 12.2.9, we can write this vector as −−→ −→ −→ P1P2 = OP2 − OP1 = x2 , y2 − x1 , y1 = x2 − x1 , y2 − y1 −−→ Thus, we have shown that the components of the vector P1P2 can be obtained by subtracting the coordinates of its initial point from the coordinates of its terminal point. Similar computations hold in 3-space, so we have established the following result: −−→ 12.2.5 THEOREM. If P1P2 is a vector in 2-space with initial point P1 (x1 , y1 ) and terminal point P2 (x2 , y2 ), then

Figure 12.2.9

−−→ P1P2 = x2 − x1 , y2 − y1

az

P1(x1, y1)

P2 (x2, y2 )

801

Yo

y

Vectors

us uf i

12.2

(7)

Ri

−−→ Similarly, if P1P2 is a vector in 3-space with initial point P1 (x1 , y1 , z1 ) and terminal point P2 (x2 , y2 , z2 ), then (8)

n

−−→ P1P2 = x2 − x1 , y2 − y1 , z2 − z1

sa

Example 2 In 2-space the vector from P1 (1, 3) to P2 (4, −2) is −−→ P1P2 = 4 − 1, −2 − 3 = 3, −5

••••••••••••••••••••••••••••••••••••••

The following theorem shows that many of the familiar rules of ordinary arithmetic also hold for vector arithmetic. 12.2.6 THEOREM. For any vectors u, v, and w and any scalars k and !, the following relationships hold: (a) u + v = v + u (e) k(!u) = (k!)u (b) (u + v) + w = u + (v + w) ( f ) k(u + v) = ku + kv (c) u + 0 = 0 + u = u (g) (k + !)u = ku + !u (d ) u + (−u) = 0 (h) 1u = u

M uh am

m

ad

RULES OF VECTOR ARITHMETIC

Ha

and in 3-space the vector from A(0, −2, 5) to B(3, 4, −1) is −→ AB = 3 − 0, 4 − (−2), −1 − 5 = 3, 6, −6

The results in this theorem can be proved either algebraically by using components or geometrically by treating the vectors as arrows. We will prove part (b) both ways and leave some of the remaining proofs as exercises.

Proof (b) (Algebraic in 2-space). Let u = u1 , u2 , v = v1 , v2 , and w = w1 , w2 . Then (u + v) + w = ( u1 , u2 + v1 , v2 ) + w1 , w2 = u1 + v1 , u2 + v2 + w1 , w2 = (u1 + v1 ) + w1 , (u2 + v2 ) + w2 = u1 + (v1 + w1 ), u2 + (v2 + w2 ) = u1 , u2 + v1 + w1 , v2 + w2 = u + (v + w)

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−→ −→

v u

v+

Figure 12.2.10. Then − → − → v + w = QS and u + (v + w) = PS − → − → u + v = PR and (u + v) + w = PS

R

u+v u + (v + w) (u + v )+w

w

w

Therefore,

S

(u + v) + w = u + (v + w)

Figure 12.2.10

• REMARK.

••••••••••••••••••••••••••••••••••••••

+x

x

v

n

u+

w v+

The distance between the initial and terminal points of a vector v is called the length, the norm, or the magnitude of v and is denoted by v . This distance does not change if the vector is translated, so for purposes of calculating the norm we can assume that the vector is positioned with its initial point at the origin (Figure 12.2.12). This makes it evident that the norm of a vector v = v1 , v2 in 2-space is given by

v = v12 + v22 (9)

Ri

NORM OF A VECTOR

It follows from part (b) of this theorem that the symbol u + v + w is unambiguous since the same vector results no matter how the terms are grouped. Moreover, Figure 12.2.10 shows that if the vectors u, v, and w are placed “tip to tail,” then the sum u + v + w is the vector from the initial point of u to the terminal point of w. This also works for four or more vectors (Figure 12.2.11).

az

• • • • • • • • • • • • • • • • • • • • • • •

u

− →

Proof (b) (Geometric). Let u, v, and w be represented by PQ, QR, and RS as shown in

Q

P

us uf i

Three-Dimensional Space; Vectors

Yo

802

w

and the norm of a vector v = v1 , v2 , v3 in 3-space is given by

v = v12 + v22 + v32

sa

Figure 12.2.11

Example 3 Find the norms of v = −2, 3 and w = 2, 3, 6.

(v1, v2 ) ||v||

(10)

Ha

y

Solution. From (9) and (10)

v2

z

ad

v1

√ (−2)2 + 32 = 13 √ √

w = 22 + 32 + 62 = 49 = 7

v =

x

Recall from Definition 12.2.2 that the length of kv is |k| times the length of v; that is,

||v||

v3

m

(v1, v2, v3)

M uh am v1

x

Figure 12.2.12

••••••••••••••••••••••••••••••••••••••

UNIT VECTORS

(11)

Thus, for example,

y

v2

kv = |k| v

3v = |3| v = 3 v

−2v = |−2| v = 2 v

−v = |−1| v = v

This applies to vectors in 2-space and 3-space. A vector of length 1 is called a unit vector. In an xy-coordinate system the unit vectors along the x- and y-axes are denoted by i and j, respectively; and in an xyz-coordinate system the unit vectors along the x-, y-, and z-axes are denoted by i, j, and k, respectively (Figure 12.2.13). Thus, i = 1, 0,

j = 0, 1

i = 1, 0, 0,

j = 0, 1, 0,

In 2-space

k = 0, 0, 1

In 3-space

Every vector in 2-space is expressible uniquely in terms of i and j, and every vector in

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y

Vectors

803

us uf i

12.2

3-space is expressible uniquely in terms of i, j, and k as follows:

(0, 1)

v = v1 , v2 = v1 , 0 + 0, v2 = v1 1, 0 + v2 0, 1 = v1 i + v2 j

j x

v = v1 , v2 , v3 = v1 1, 0, 0 + v2 0, 1, 0 + v3 0, 0, 1 = v1 i + v2 j + v3 k

(1, 0)

z

• REMARK. • • • • • • •

(0, 0, 1)

The bracket and unit vector notations for vectors are completely interchangeable, the choice being a matter of convenience or personal preference.

Yo

i

k y

i (1, 0, 0)

(0, 1, 0)

Example 4 2-space

x

Figure 12.2.13

3-space

〈2, 3〉 = 2i + 3j

〈2, –3, 4〉 = 2i – 3j + 4k

〈–4, 0〉 = –4i + 0j = –4i

〈0, 3, 0〉 = 3j

az

j

〈0, 0, 0〉 = 0i + 0j + 0k = 0 (3i + 2j – k) – (4i – j + 2k) = –i + 3j – 3k

2(i + j – k) + 4(i – j) = 6i – 2j – 2k

5(6i – 2j) = 30i – 10j

|| 2i – 3j || = √ + || v1i + v2 j || = √v12 + v22

(–3)2

= √13

||i + 2j – 3k|| = √12 + 22 + (–3)2 = √14 ||〈v1, v2, v3〉|| = √v12 + v22 + v32

n

22

Ri

〈0, 0〉 = 0i + 0j = 0 (3i + 2j) + (4i + j) = 7i + 3j

••••••••••••••••••••••••••••••••••••••

sa

A common problem in applications is to find a unit vector u that has the same direction as some given nonzero vector v. This can be done by multiplying v by the reciprocal of its length; that is, 1 v u= v=

v

v

is a unit vector with the same direction as v—the direction is the same because k = 1/ v

is a positive scalar, and the length is 1 because 1

u = kv = |k| v = k v =

v = 1

v

The process of multiplying a vector v by the reciprocal of its length to obtain a unit vector with the same direction is called normalizing v.

M uh am

m

ad

Ha

NORMALIZING A VECTOR

••••••••••••••••••••••••••••••••••••••

VECTORS DETERMINED BY LENGTH AND ANGLE

Example 5 Find the unit vector that has the same direction as v = 2i + 2 j − k.

Solution. The vector v has length

v =

22 + 22 + (−1)2 = 3

so the unit vector u in the same direction as v is

u = 31 v = 23 i + 23 j − 13 k • FOR THE READER. • • • • • • • • • • • •

Many calculating utilities can perform vector operations, and some have built-in norm and normalization operations. If your calculating utility has these capabilities, use it to check the computations in Examples 1, 3, and 5. If v is a nonzero vector with its initial point at the origin of an xy-coordinate system, and if φ is the angle from the positive x-axis to the radial line through v, then the x-component of v can be written as v cos φ and the y-component as v sin φ (Figure 12.2.14); and hence v can be expressed in trigonometric form as v = v

cos φ, sin φ

or

v = v cos φi + v sin φ j

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y

us uf i

Three-Dimensional Space; Vectors

In the special case of a unit vector u this simplifies to u = cos φ, sin φ

||v||

||v|| sin f x f ||v||cos f

or

u = cos φi + sin φ j

(13)

Example 6 (a) (b)

Find the vector of length 2 that makes an angle of π/4 with the positive x-axis. √ Find the angle that the vector v = − 3 i + j makes with the positive x-axis.

Yo

804

Solution (a). From (12)

√ √ π π v = 2 cos i + 2 sin j = 2 i + 2 j 4 4

Figure 12.2.14

az

Solution (b). We will normalize v, then use (13) to find sin φ and cos φ, and then use these

It is a common problem in many applications that a direction in 2-space or 3-space is determined by some known unit vector u, and it is of interest to find the components of a vector v that has the same direction as u and some specified length v . This can be done by expressing v as v = v u

z

n

VECTORS DETERMINED BY LENGTH AND A VECTOR IN THE SAME DIRECTION

sa

••••••••••••••••••••••••••••••••••••••

Ri

values to find φ. Normalizing v yields √ √ v 3 − 3i + j 1 = √ i+ j =−

v

2 2 (− 3 )2 + 12 √ Thus, cos φ = − 3/2 and sin φ = 12 , from which we conclude that φ = 5π/6.

v is equal to its length times a unit vector in the same direction.

and then reading off the components of v u.

A(0, 0, 4)

Ha

||v|| = √5 v

Example 7 Figure 12.2.15 shows a vector v of length through A and B. Find the components of v.

y

√

5 that extends along the line

−→

Solution. First we will find the components of the vector AB, then we will normalize this x

B(2, 5, 0)

vector to obtain a unit vector in the direction of v, and then we will multiply this unit vector by v to obtain the vector v. The computations are as follows: −→ AB = 2, 5, 0 − 0, 0, 4 = 2, 5, −4 √ √ −→

AB = 22 + 52 + (−4)2 = 45 = 3 5 −→ AB 2 5 4 −→ = √ , √ , − √ 3 5 3 5 3 5

AB

−→ √ 4 2 AB 5 4 2 5 = 5 √ , √ ,− √ = , ,− v = v −→ 3 3 3 3 5 3 5 3 5

AB

M uh am

m

ad

Figure 12.2.15

••••••••••••••••••••••••••••••••••••••

RESULTANT OF TWO CONCURRENT FORCES

The effect that a force has on an object depends on the magnitude and direction of the force and the point at which it is applied. Thus, forces are regarded to be vector quantities and, indeed, the algebraic operations on vectors that we have defined in this section have their origin in the study of forces. For example, it is a fact of physics that if two forces F1 and F2 are applied at the same point on an object, then the two forces have the same effect on the object as the single force F1 + F2 applied at the point (Figure 12.2.16). Physicists and engineers call F1 + F2 the resultant of F1 and F2 , and they say that the forces F1 and F2 are concurrent to indicate that they are applied at the same point. In many applications, the magnitudes of two concurrent forces and the angle between them are known, and the problem is to find the magnitude and direction of the resultant. For example, referring to Figure 12.2.17, suppose that we know the magnitudes of the forces F1 and F2 and the angle φ between them, and we are interested in finding the magnitude of

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Vectors

805

us uf i

12.2

F2

Yo

the resultant F1 + F2 and the angle α that the resultant makes with the force F1 . This can be done by trigonometric methods based on the laws of sines and cosines. For this purpose, recall that the law of sines applied to the triangle in Figure 12.2.18 states that a b c = = sin α sin β sin γ and the law of cosines implies that

F1 + F2

c2 = a 2 + b2 − 2ab cos γ

F1

Referring to Figure 12.2.19, and using the fact that cos(π − φ) = − cos φ, it follows from the law of cosines that

The single force F1 + F2 has the same effect as the two forces F1 and F2.

F1 + F2 2 = F1 2 + F2 2 + 2 F1

F2 cos φ

Figure 12.2.16

(14)

F1 + F2 F2

sin α =

Ri

az

Moreover, it follows from the law of sines that

F2

F1 + F2

= sin α sin(π − φ) which, with the help of the identity sin(π − φ) = sin φ, can be expressed as

F2

sin φ

F1 + F2

(15)

f

Example 8 Suppose that two forces are applied to an eye bracket, as shown in Figure 12.2.20. Find the magnitude of the resultant and the angle θ that it makes with the positive x-axis.

n

a

sa

F1

Figure 12.2.17

Solution. We are given that F1 = 200 N and F2 = 300 N and that the angle between

Ha

the vectors F1 and F2 is φ = 40 ◦ . Thus, it follows from (14) that the magnitude of the resultant is

F1 + F2 = F1 2 + F2 2 + 2 F1

F2 cos φ = (200)2 + (300)2 + 2(200)(300) cos 40 ◦

a c b g b

|| F2 ||

1

|| F2 ||

F2

p–f

f

M uh am

|| F

m

||

+

f

a

|| F1||

Figure 12.2.19

≈ 471 N Moreover, it follows from (15) that the angle α between F1 and the resultant is

F2

−1 α = sin sin φ

F1 + F2

300 ≈ sin−1 sin 40 ◦ 471

ad

a Figure 12.2.18

≈ 24.2 ◦ Thus, the angle θ that the resultant makes with the positive x-axis is θ = α + 30 ◦ ≈ 24.2 ◦ + 30 ◦ = 54.2 ◦ (Figure 12.2.21).

y

||F1 + F2|| ≈ 471 N

y

|| F2 || = 300 N

|| F2 || = 300 N

|| F1|| = 200 N

40° 30°

24.2° x

54.2° Figure 12.2.20

|| F1|| = 200 N x

Figure 12.2.21

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Three-Dimensional Space; Vectors

us uf i

806

• REMARK.

Yo

The resultant of three or more concurrent forces can be found by working in pairs. For example, the resultant of three concurrent forces can be found by finding the resultant of any two of the three forces and then finding the resultant of that resultant with the third force.

• • • • • • • • • • • • • • • • •

EXERCISE SET 12.2

10. (a) Find the terminal point of v = 7, 6 if the initial point is (2, −1). (b) Find the terminal point of v = i + 2 j − 3k if the initial point is (−2, 1, 4).

(c) 2, 0 (f ) −6 j

2. (a) −3, 7 (d) 4i + 2 j

(b) 6, −2 (e) −2i − j

(c) 0, −8 (f ) 4i

3. (a) 1, −2, 2 (c) −i + 2 j + 3k

(b) 2, 2, −1 (d) 2i + 3 j − k

4. (a) −1, 3, 2 (c) 2 j − k

(b) 3, 4, 2 (d) i − j + 2k

In Exercises 11 and 12, perform the stated operations on the vectors u, v, and w.

Ri

(b) −5, −4 (e) 3i − 2 j

11. u = 3i − k, v = i − j + 2k, (a) w − v (c) −v − 2w (e) −8(v + w) + 2u

12. u = 2, −1, 3, v = 4, 0, −2, w = 1, 1, 3 (a) u − w (b) 7v + 3w (c) −w + v (d) 3(u − 7v) (e) −3v − 8w (f) 2v − (u + w)

sa

In Exercises 5 and 6, find the components of the vector, and sketch an equivalent vector with its initial point at the origin.

w = 3j (b) 6u + 4w (d) 4(3u + v) (f ) 3w − (v − w)

n

1. (a) 2, 5 (d) −5i + 3 j

az

In Exercises 1–4, sketch the vectors with their initial points at the origin.

5. (a)

y

z

(b) (1, 5)

(0, 0, 4)

Ha

In Exercises 13 and 14, find the norm of v.

y

(4, 1)

ad

6. (a)

y

(2, 3, 0)

(0, 4, 4)

m

(2, 3)

(3, 0, 4)

M uh am

(–3, 3)

z

(b)

14. (a) v = 3, 4 (c) v = 0, −3, 0

(b) v = −i + 7 j (d) v = −3i + 2 j + k √ √ (b) v = 2i − 7 j (d) v = i + j + k

15. Let u = i − 3 j + 2k, v = i + j, and w = 2i + 2 j − 4k. Find

x x

13. (a) v = 1, −1 (c) v = −1, 2, 4

y

x

x

−−→ In Exercises 7 and 8, find the components of the vector P1P2 .

7. (a) P1 (3, 5), P2 (2, 8) (b) P1 (7, −2), P2 (0, 0) (c) P1 (5, −2, 1), P2 (2, 4, 2) 8. (a) P1 (−6, −2), P2 (−4, −1) (b) P1 (0, 0, 0), P2 (−1, 6, 1) (c) P1 (4, 1, −3), P2 (9, 1, −3) 9. (a) Find the terminal point of v = 3i − 2 j if the initial point is (1, −2). (b) Find the initial point of v = −3, 1, 2 if the terminal point is (5, 0, −1).

(a) u + v

(b) u + v

(c) −2u + 2 v

(d) 3u − 5v + w

1 (f ) w w .

1 (e) w

w

16. Is it possible to have u + v = u + v if u and v are nonzero vectors? Justify your conclusion geometrically. In Exercises 17 and 18, find unit vectors that satisfy the stated conditions. 17. (a) Same direction as −i + 4 j. (b) Oppositely directed to 6i − 4 j + 2k. (c) Same direction as the vector from the point A(−1, 0, 2) to the point B(3, 1, 1). 18. (a) Oppositely directed to 3i − 4 j. (b) Same direction as 2i − j − 2k. (c) Same direction as the vector from the point A(−3, 2) to the point B(1, −1). In Exercises 19 and 20, find vectors that satisfy the stated conditions.

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19. (a) Oppositely directed to v = 3, −4 and half the length of v. √ (b) Length 17 and same direction as v = 7, 0, −6.

Vectors

807

us uf i

12.2

34. What do you know about k and v if kv = 0?

21. In each part, find the component form of the vector v in 2-space that has the stated length and makes the stated angle φ with the positive x-axis. (a) v = 3; φ = π/4 (b) v = 2; φ = 90 ◦ ◦ (c) v = 5; φ = 120 (d) v = 1; φ = π

36. In each part, find two unit vectors in 3-space that satisfy the stated condition. (a) Perpendicular to the xy-plane (b) Perpendicular to the xz-plane (c) Perpendicular to the yz-plane

Yo

20. (a) Same direction as v = −2i + 3 j and three times the length of v. (b) Length 2 and oppositely directed to v = −3i + 4 j + k.

35. In each part, find two unit vectors in 2-space that satisfy the stated condition. (a) Parallel to the line y = 3x + 2 (b) Parallel to the line x + y = 4 (c) Perpendicular to the line y = −5x + 1

37. Let r = x, y be an arbitrary vector. In each part, describe the set of all points (x, y) in 2-space that satisfy the stated condition. (a) r = 1 (b) r ≤ 1 (c) r > 1

In Exercises 23 and 24, find the component form of v + w, given that v and w are unit vectors.

38. Let r = x, y and r0 = x0 , y0 . In each part, describe the set of all points (x, y) in 2-space that satisfy the stated condition. (a) r − r0 = 1 (b) r − r0 ≤ 1 (c) r − r0 > 1 39. Let r = x, y, z be an arbitrary vector. In each part, describe the set of all points (x, y, z) in 3-space that satisfy the stated condition. (a) r = 1 (b) r ≤ 1 (c) r > 1

w 135°

sa

w

v

v

120°

x

30°

Ri

y

24.

n

y

23.

x

y

(b)

Ha

25. In each part, sketch the vector u + v + w and express it in component form. (a)

az

22. Find the component forms of v + w and v − w in 2-space, given that v = 1, w = 1, v makes an angle of π/6 with the positive x-axis, and w makes an angle of 3π/4 with the positive x-axis.

y

40. Let r1 = x1 , y1 , r2 = x2 , y2 , and r = x, y. Describe the set of all points (x, y) for which r − r1 + r − r2 = k, assuming that k > r2 − r1 . In Exercises 41–46, find the magnitude of the resultant force and the angle that it makes with the positive x-axis.

v

w

v

w

x

41.

M uh am

26. In each part of Exercise 25, sketch the vector u − v + w and express it in component form.

y

42.

y

100 N

u

m

u

ad

x

30 lb 60 lb x

y

43.

27. Let u = 1, 3, v = 2, 1, w = 4, −1. Find the vector x that satisfies 2u − v + x = 7x + w.

120 N x

60°

y

44.

400 N

2 lb 120°

28. Let u = −1, 1, v = 0, 1, and w = 3, 4. Find the vector x that satisfies u − 2x = x − w + 3v.

50°

x

30°

29. Find u and v if u + 2v = 3i − k and 3u − v = i + j + k.

x

27°

400 N

4 lb

30. Find u and v if u + v = 2, −3 and 3u + 2v = −1, 2.

31. Use vectors to find the lengths of the diagonals of the parallelogram that has i + j and i − 2 j as adjacent sides.

32. Use vectors to find the fourth vertex of a parallelogram, three of whose vertices are (0, 0), (1, 3), and (2, 4). [Note: There is more than one answer.]

33. (a) Given that v = 3, find all values of k such that

kv = 5. (b) Given that k = −2 and kv = 6, find v .

y

45.

y

46.

200 N

150 N 40 N 60°

50 N 75 N x

75° 60°

30°

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100 N

x

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Three-Dimensional Space; Vectors

us uf i

808

51. A vector w is said to be a linear combination of the vectors v1 and v2 if w can be expressed as w = c1 v1 + c2 v2 , where c1 and c2 are scalars. (a) Find scalars c1 and c2 to express the vector 4 j as a linear combination of the vectors v1 = 2i − j and v2 = 4i + 2 j. (b) Show that the vector 3, 5 cannot be expressed as a linear combination of the vectors v1 = 1, −3 and v2 = −2, 6.

y

47.

10 lb

Yo

A particle is said to be in static equilibrium if the resultant of all forces applied to it is zero. In Exercises 47 and 48, find the force F that must be applied to the point to produce static equilibrium. Describe F by specifying its magnitude and the angle that it makes with the positive x-axis.

y

48.

150 N

120 N

52. A vector w is said to be a linear combination of the vectors v1 , v2 , and v3 if w can be expressed as w = c1 v1 + c2 v2 + c3 v3 , where c1 , c2 , and c3 are scalars. (a) Find scalars c1 , c2 , and c3 to express −1, 1, 5 as a linear combination of the vectors v1 = 1, 0, 1, v2 =

3, 2, 0, and v3 = 0, 1, 1. (b) Show that the vector 2i + j − k cannot be expressed as a linear combination of the vectors v1 = i−j, v2 = 3i+k, and v3 = 4i − j + k.

75° 60°

x

100 N

45°

x

az

8 lb

53. Use a theorem from plane geometry to show that if u and v are vectors in 2-space or 3-space, then

n

u + v ≤ u + v

which is called the triangle inequality for vectors. Give some examples to illustrate this inequality.

sa

50. Find the tensions in the cables shown in the accompanying figure if the block is in static equilibrium (see Exercise 49).

Ri

49. The accompanying figure shows a 250-lb traffic light supported by two flexible cables. The magnitudes of the forces that the cables apply to the eye ring are called the cable tensions. Find the tensions in the cables if the traffic light is in static equilibrium (defined above Exercise 47).

54. Prove parts (a), (c), and (e) of Theorem 12.2.6 algebraically in 2-space.

30°

45°

60°

Ha

30°

ad

200 N

Figure Ex-50

56. Prove part ( f ) of Theorem 12.2.6 geometrically. 57. Use vectors to prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. 58. Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.

m

Figure Ex-49

55. Prove parts (d ), (g), and (h) of Theorem 12.2.6 algebraically in 2-space.

M uh am

12.3 DOT PRODUCT; PROJECTIONS

••••••••••••••••••••••••••••••••••••••

DEFINITION OF THE DOT PRODUCT

In the last section we defined three operations on vectors—addition, subtraction, and scalar multiplication. In scalar multiplication a vector is multiplied by a scalar and the result is a vector. In this section we will define a new kind of multiplication in which two vectors are multiplied to produce a scalar. This multiplication operation has many uses, some of which we will also discuss in this section. 12.3.1 DEFINITION. If u = u1 , u2 and v = v1 , v2 are vectors in 2-space, then the dot product of u and v is written as u · v and is defined as u · v = u 1 v1 + u 2 v2 Similarly, if u = u1 , u2 , u3 and v = v1 , v2 , v3 are vectors in 3-space, then their dot product is defined as u · v = u 1 v 1 + u 2 v 2 + u 3 v3

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Dot Product; Projections

809

us uf i

12.3

In words, the dot product of two vectors is formed by multiplying their corresponding components and adding the products. Note that the dot product of two vectors is a scalar. Example 1

2, 3 · −3, 2 = 2(−3) + 3(2) = 0

Yo

3, 5 · −1, 2 = 3(−1) + 5(2) = 7

1, −3, 4 · 1, 5, 2 = 1(1) + (−3)(5) + 4(2) = −6

Here are the same computations expressed another way: (3i + 5j) · (−i + 2 j) = 3(−1) + 5(2) = 7

az

(2i + 3j) · (−3i + 2 j) = 2(−3) + 3(2) = 0 (i − 3 j + 4k) · (i + 5 j + 2k) = 1(1) + (−3)(5) + 4(2) = −6 • FOR THE READER.

••••••••••••••••••••••••••••••••••••••

Many calculating utilities have a built-in dot product operation. If your calculating utility has this capability, use it to check the computations in Example 1.

Ri

• • • • • • •

The following theorem provides some of the basic algebraic properties of the dot product:

n

ALGEBRAIC PROPERTIES OF THE DOT PRODUCT

sa

12.3.2 THEOREM. If u, v, and w are vectors in 2- or 3-space and k is a scalar, then (a) u · v = v · u (b)

u · (v + w) = u · v + u · w

(c)

k(u · v) = (ku) · v = u · (kv)

(e)

Ha

(d ) v · v = v 2 0·v=0

ad

We will prove parts (c) and (d ) for vectors in 3-space and leave some of the others as exercises.

Proof (c). Let u = u1 , u2 , u3 and v = v1 , v2 , v3 . Then

M uh am

m

k(u · v) = k(u1 v1 + u2 v2 + u3 v3 ) = (ku1 )v1 + (ku2 )v2 + (ku3 )v3 = (ku) · v

••••••••••••••••••••••••••••••••••••••

ANGLE BETWEEN VECTORS

Similarly, k(u · v) = u · (kv).

Proof (d). v · v = v1 v1 + v2 v2 + v3 v3 = v12 + v22 + v32 = v 2 .

• REMARK. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Pay particular attention to the two zeros that appear in part (e) of the last theorem—the zero on the left side is the zero vector (boldface), and the zero on the right side is the zero scalar (lightface). It is also worth noting that the result in part (d ) can be written as √

v = v · v (1) which provides a way of expressing the norm of a vector in terms of a dot product.

Suppose that u and v are nonzero vectors in 2-space or 3-space that are positioned so their initial points coincide. We define the angle between u and v to be the angle θ determined by the vectors that satisfies the condition 0 ≤ θ ≤ π (Figure 12.3.1). In 2-space, θ is the smallest counterclockwise angle through which one of the vectors can be rotated until it aligns with the other.

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810

Three-Dimensional Space; Vectors

u

u

u

u

u

v

v

v

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

u

v

u

Yo

u

Figure 12.3.1

az

The next theorem provides a way of calculating the angle between two vectors from their components.

cos θ =

u·v

u

v

(2)

n

Proof. Suppose that the vectors u, v, and v − u are positioned to form three sides of a v–u

triangle, as shown in Figure 12.3.2. It follows from the law of cosines that

sa

v

Ri

12.3.3 THEOREM. If u and v are nonzero vectors in 2-space or 3-space, and if θ is the angle between them, then

v − u 2 = u 2 + v 2 − 2 u

v cos θ

u u Figure 12.3.2

(3)

Ha

Using the properties of the dot product in Theorem 12.3.2, we can rewrite the left side of this equation as

v − u 2 = (v − u) · (v − u) = (v − u) · v − (v − u) · u =v·v−u·v−v·u+u·u = v 2 − 2u · v + u 2

ad

Substituting this back into (3) yields

v 2 − 2u · v + u 2 = u 2 + v 2 − 2 u

v cos θ

m

which we can simplify and rewrite as u · v = u

v cos θ

M uh am

Finally, dividing both sides of this equation by u

v yields (2).

Example 2 Find the angle between the vector u = i − 2 j + 2k and (a) v = −3i + 6 j + 2k

(b) w = 2i + 7 j + 6k

(c) z = −3i + 6 j − 6k

Solution (a). cos θ =

−11 11 u·v = =−

u

v

(3)(7) 21

Thus,

θ = cos−1 − 11 ≈ 2.12 radians ≈ 121.6 ◦ 21

Solution (b). 0 u·w = =0

u

w

u

w

Thus, θ = π/2, which means that the vectors are perpendicular. cos θ =

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Solution (c). cos θ =

−27 u·z = = −1

u

z

(3)(9)

Dot Product; Projections

811

us uf i

12.3

••••••••••••••••••••••••••••••••••••••

It will often be convenient to express Formula (2) as

INTERPRETING THE SIGN OF THE DOT PRODUCT

u · v = u

v cos θ

Yo

Thus, θ = π, which means that the vectors are oppositely directed. In retrospect, we could have seen this without computing θ, since z = −3u.

(4)

Ri

az

which expresses the dot product of u and v in terms of the lengths of these vectors and the angle between them. Since u and v are assumed to be nonzero vectors, this version of the formula makes it clear that the sign of u · v is the same as the sign of cos θ . Thus, we can tell from the dot product whether the angle between two vectors is acute or obtuse or whether the vectors are perpendicular (Figure 12.3.3). v

n

v

v

u

u

u

u.v < 0

u.v = 0

Ha

Figure 12.3.3

u

sa

u

u.v>0

• REMARK.

The terms “perpendicular,” “orthogonal,” and “normal” are all commonly used to describe geometric objects that meet at right angles. For consistency, we will say that two vectors are orthogonal, a vector is normal to a plane, and two planes are perpendicular. Moreover, although the zero vector does not make a well-defined angle with other vectors, we will consider 0 to be orthogonal to all vectors. This convention allows us to say that u and v are orthogonal vectors if and only if u · v = 0, and it makes Formula (4) valid if u or v (or both) is zero.

ad

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

M uh am

DIRECTION ANGLES

m

••••••••••••••••••••••••••••••••••••••

y

j

v

b

a

x

In an xy-coordinate system, the direction of a nonzero vector v is completely determined by the angles α and β between v and the unit vectors i and j (Figure 12.3.4), and in an xyz-coordinate system the direction is completely determined by the angles α, β, and γ between v and the unit vectors i, j, and k (Figure 12.3.5). In both 2-space and 3-space the angles between a nonzero vector v and the vectors i, j, and k are called the direction angles of v, and the cosines of those angles are called the direction cosines of v. Formulas for the direction cosines of a vector can be obtained from Formula (2). For example, if v = v1 i + v2 j + v3 k, then cos α =

i

Figure 12.3.4

v·i v1 = ,

v

i

v

cos β =

v·j v2 = ,

v

j

v

cos γ =

v·k v3 =

v

k

v

Thus, we have the following result:

12.3.4

THEOREM.

v1 cos α = ,

v

The direction cosines of a nonzero vector v = v1 i + v2 j + v3 k are cos β =

v2 ,

v

cos γ =

v3

v

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The direction cosines of a vector v = v1 i + v2 j + v3 k can be computed by normalizing v and reading off the components of v/ v , since v v2 v3 v1 i+ j+ k = (cos α)i + (cos β) j + (cos γ )k =

v

v

v

v

We leave it as an exercise for you to show that the direction cosines of a vector satisfy the equation

z

v

k

g b

y

cos2 α + cos2 β + cos2 γ = 1

j

a

us uf i

Three-Dimensional Space; Vectors

Yo

812

i x

(5)

az

Example 3 Find the direction cosines of the vector v = 2i − 4 j + 4k, and approximate the direction angles to the nearest degree.

Figure 12.3.5

Solution. √ First we will normalize the vector v and then read off the components. We have 4 + 16 + 16 = 6, so that v/ v = 13 i − 23 j + 23 k. Thus,

cos α = 13 ,

cos β = − 23 ,

cos γ =

2 3

Ri

v =

With the help of a calculating utility we obtain α = cos−1 13 ≈ 71 ◦ , β = cos−1 − 23 ≈ 132 ◦ ,

z

γ = cos−1

2 3

≈ 48 ◦

(0, 0, a)

n

Example 4 Find the angle between a diagonal of a cube and one of its edges. k

(a, a, a)

Solution. Assume that the cube has side a, and introduce a coordinate system as shown

sa

d

in Figure 12.3.6. In this coordinate system the vector y

d = ai + a j + ak

(0, a, 0)

j

is a diagonal of the cube and the unit vectors i, j, and k run along the edges. By symmetry, the diagonal makes the same angle with each edge, so it is sufficient to find the angle between d and i (the direction angle α). Thus, d·i 1 a a cos α = =√ = =√ 2

d

i

d

3 3a

Ha

a i (a, 0, 0) x

Figure 12.3.6

and hence

ad

1 α = cos−1 √ ≈ 0.955 radian ≈ 54.7 ◦ 3

In many applications it is desirable to “decompose” a vector into a sum of two orthogonal vectors with convenient specified directions. For example, Figure 12.3.7 shows a block on an inclined plane. The downward force F that gravity exerts on the block can be decomposed into the sum

m

••••••••••••••••••••••••••••••••••••••

M uh am

DECOMPOSING VECTORS INTO ORTHOGONAL COMPONENTS

F1

F2

F

The force of gravity pulls the block against the ramp and down the ramp.

Figure 12.3.7

F = F1 + F2 where the force F1 is parallel to the ramp and the force F2 is perpendicular to the ramp. The forces F1 and F2 are useful because F1 is the force that pulls the block along the ramp, and F2 is the force that the block exerts against the ramp. Thus, our next objective is to develop a computational procedure for decomposing a vector into a sum of orthogonal vectors. For this purpose, suppose that e1 and e2 are two orthogonal unit vectors in 2-space, and suppose that we want to express a given vector v as a sum v = w1 + w2 where w1 is a scalar multiple of e1 and w2 is a scalar multiple of e2 (Figure 12.3.8a); that is, we want to find scalars k1 and k2 such that v = k1 e1 + k2 e2

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Dot Product; Projections

||v|| cos u w2

(||v|| sin u)e2

e2

e2

v

||v||

||v|| sin u

u w1

e1

(a)

(||v|| cos u)e1

(b)

Figure 12.3.8

Yo

e1

813

us uf i

12.3

We can find k1 by taking the dot product of v with e1 . This yields

v · e1 = (k1 e1 + k2 e2 ) · e1 = k1 (e1 · e1 ) + k2 (e2 · e1 ) = k1 e1 2 + 0 = k1

az

and, similarly,

v · e2 = (k1 e1 + k2 e2 ) · e2 = k1 (e1 · e2 ) + k2 (e2 · e2 ) = 0 + k2 e2 2 = k2 v = (v · e1 )e1 + (v · e2 )e2

Ri

Substituting these expressions for k1 and k2 in (6) yields

(7)

and

v · e2 = v sin θ

sa

v · e1 = v cos θ

n

In this formula we call (v · e1 )e1 and (v · e2 )e2 the vector components of v along e1 and e2 , respectively; and we call v · e1 and v · e2 the scalar components of v along e1 and e2 , respectively. If θ denotes the angle between v and e1 , then the scalar components of v can be written in trigonometric form as (8)

(Figure 12.3.8b). Moreover, the vector components of v can be expressed as and

(v · e2 )e2 = ( v sin θ)e2

Ha

(v · e1 )e1 = ( v cos θ)e1 b 0l

and the decomposition (6) can be expressed as

10

v = ( v cos θ)e1 + ( v sin θ)e2 30°

(a)

Example 5 A rope is attached to a 100-lb block on a ramp that is inclined at an angle of 30 ◦ with the ground (Figure 12.3.9a). How much force does the block exert against the ramp, and how much force must be applied to the rope in a direction parallel to the ramp to prevent the block from sliding down the ramp? (Assume that the ramp is smooth, that is, exerts no frictional forces.)

60° F2

(b)

M uh am

Figure 12.3.9

m

30° F

(10)

ad

F1

(9)

••••••••••••••••••••••••••••••••••••••

ORTHOGONAL PROJECTIONS

Solution. Let F denote the downward force of gravity on the block (so F = 100 lb), and let F1 and F2 be the vector components of F parallel and perpendicular to the ramp (as shown in Figure 12.3.9b). The lengths of F1 and F2 are

1

F1 = F cos 60 ◦ = 100 = 50 lb 2 √ 3 ◦

F2 = F sin 60 = 100 ≈ 86.6 lb 2 Thus, the block exerts a force of approximately 86.6 lb against the ramp, and it requires a force of 50 lb to prevent the block from sliding down the ramp. The vector components of v along e1 and e2 in (7) are also called the orthogonal projections of v on e1 and e2 and are commonly denoted by proje1 v = (v · e1 )e1

and

proje2 v = (v · e2 )e2

In general, if e is a unit vector, then we define the orthogonal projection of v on e to be proje v = (v · e)e

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The orthogonal projection of v on an arbitrary nonzero vector b can be obtained by normalizing b and then applying Formula (11); that is,

b b projb v = v ·

b

b

which can be rewritten as

proj bv

Acute angle between v and b

projb v = v

(12)

Geometrically, if b and v have a common initial point, then projb v is the vector that is determined when a perpendicular is dropped from the terminal point of v to the line through b (illustrated in Figure 12.3.10 in two cases). Moreover, it is evident from Figure 12.3.10 that if we subtract projb v from v, then the resulting vector

v – proj bv b

proj bv

v·b b

b 2

Obtuse angle between v and b

az

b

Yo

v

v – proj bv

us uf i

Three-Dimensional Space; Vectors

v − projb v

Figure 12.3.10

will be orthogonal to b; we call this the vector component of v orthogonal to b.

Ri

814

Example 6 Find the orthogonal projection of v = i + j + k on b = 2i + 2 j, and then find the vector component of v orthogonal to b.

Solution. We have

b 2 = 22 + 22 = 8

k

sa

v=i+j+k

Thus, the orthogonal projection of v on b is v·b 4 projb v = b = (2i + 2 j) = i + j 2

b

8 and the vector component of v orthogonal to b is

y

Ha

i+j

x

n

v · b = (i + j + k) · (2i + 2 j) = 2 + 2 + 0 = 4

z

v − projb v = (i + j + k) − (i + j) = k

b = 2i + 2j

Figure 12.3.11

These results are consistent with Figure 12.3.11.

••••••••••••••••••••••••••••••••••••••

In Section 6.6 we discussed the work done by a constant force acting on an object that moves along a line. We defined the work W done on the object by a constant force of magnitude F acting in the direction of motion over a distance d to be

ad

WORK

W = Fd = force × distance

(13)

M uh am

m

If we let F denote a force vector of magnitude F = F acting in the direction of motion, then we can write (13) as W = F d

Furthermore, if we assume that the object moves along a line from point P to point Q, then −→ d = PQ , so that the work can be expressed entirely in vector form as −→ W = F

PQ

−→ (Figure 12.3.12a). The vector PQ is called the displacement vector for the object. In the case where a constant force F is not in the direction of motion, but rather makes an angle θ || F ||

|| F ||

F

u || F || cos u

F

P

Q

|| PQ || (a)

|| PQ || (b)

Figure 12.3.12

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Dot Product; Projections

815

us uf i

12.3

with the displacement vector, then we define the work W done by F to be −→ −→ W = ( F cos θ) PQ = F · PQ

(14)

(Figure 12.3.12b). • REMARK.

Note that in Formula (14) the quantity F cos θ is the scalar component of force along the displacement vector. Thus, in the case where cos θ > 0, a force of magnitude

F acting at an angle θ does the same work as a force of magnitude F cos θ acting in the direction of motion.

Yo

• • • • • • • • • • • • • • • • • •

az

Example 7 A wagon is pulled horizontally by exerting a constant force of 10 lb on the handle at an angle of 60 ◦ with the horizontal. How much work is done in moving the wagon 50 ft?

y

F

Solution. Introduce an xy-coordinate system so that the wagon moves from P (0, 0) to

60° x

P(0, 0)

Q(50, 0)

Q(50, 0) along the x-axis (Figure 12.3.13). In this coordinate system −→ PQ = 50i

Ri

10 lb

and

Figure 12.3.13

n

√ F = (10 cos 60 ◦ )i + (10 sin 60 ◦ ) j = 5i + 5 3 j

so the work done is √ −→ W = F · PQ = (5i + 5 3 j) · (50i) = 250 (foot-pounds)

EXERCISE SET 12.3

Graphing Utility

Ha

sa

C

CAS

m

ad

1. In each part, find the dot product of the vectors and the cosine of the angle between them. (a) u = i + 2 j, v = 6i − 8 j (b) u = −7, −3, v = 0, 1 (c) u = i − 3 j + 7k, v = 8i − 2 j − 2k (d) u = −3, 1, 2, v = 4, 2, −5

M uh am

2. In each part use the given information to find u · v. (a) u = 1, v = 2, the angle between u and v is π/6. (b) u = 2, v = 3, the angle between u and v is 135 ◦ . 3. In each part, determine whether u and v make an acute angle, an obtuse angle, or are orthogonal. (a) u = 7i + 3 j + 5k, v = −8i + 4 j + 2k (b) u = 6i + j + 3k, v = 4i − 6k (c) u = 1, 1, 1, v = −1, 0, 0 (d) u = 4, 1, 6, v = −3, 0, 2

6. The accompanying figure shows six vectors that are equally spaced around a circle of radius 5. Find the dot product of v0 with each of the other five vectors. v2 v3

v2

v1

v4

v0 v5

v3

v0

v7 v6

Figure Ex-5

v1

v4

v5

Figure Ex-6

4. Does the triangle in 3-space with vertices (−1, 2, 3), (2, −2, 0), and (3, 1, −4) have an obtuse angle? Justify your answer.

7. (a) Use vectors to show that A(2, −1, 1), B(3, 2, −1), and C(7, 0, −2) are vertices of a right triangle. At which vertex is the right angle? (b) Use vectors to find the interior angles of the triangle with vertices (−1, 0), (2, −1), and (1, 4). Express your answers to the nearest degree.

5. The accompanying figure shows eight vectors that are equally spaced around a circle of radius 1. Find the dot product of v0 with each of the other seven vectors.

8. Find k so that the vector from the point A(1, −1, 3) to the point B(3, 0, 5) is orthogonal to the vector from A to the point P (k, k, k).

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Three-Dimensional Space; Vectors

us uf i

816

9. (a) Show that if v = ai + b j is a vector in 2-space, then the vectors v1 = âˆ’bi + a j and v2 = bi âˆ’ a j are both orthogonal to v. (b) Use the result in part (a) to find two unit vectors that are orthogonal to the vector v = 3i âˆ’ 2 j. Sketch the vectors v, v1 , and v2 .

Yo

21. In each part, find the vector component of v along b and the vector component of v orthogonal to b. Then sketch the vectors v, projb v, and v âˆ’ projb v. (a) v = 2i âˆ’ j, b = 3i + 4 j (b) v = 4, 5, b = 1, âˆ’2 (c) v = âˆ’3i âˆ’ 2 j, b = 2i + j 22. In each part, find the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i âˆ’ j + 3k, b = i + 2 j + 2k (b) v = 4, âˆ’1, 7, b = 2, 3, âˆ’6

10. Find two unit vectors in 2-space that make an angle of 45 â—Ś with 4i + 3 j. 11. Explain why each of the following expressions makes no sense. (a) u Âˇ (v Âˇ w) (b) (u Âˇ v) + w (d) k Âˇ (u + v) (c) u Âˇ v

az

In Exercises 23 and 24, express the vector v as the sum of a vector parallel to b and a vector orthogonal to b. 23. (a) v = 2i âˆ’ 4 j, b = i + j (b) v = 3i + j âˆ’ 2k, b = 2i âˆ’ k

Ri

12. Verify parts (b) and (c) of Theorem 12.3.2 for the vectors u = 6i âˆ’ j + 2k, v = 2i + 7 j + 4k, w = i + j âˆ’ 3k and k = âˆ’5.

24. (a) v = âˆ’3, 5, b = 1, 1 (b) v = âˆ’2, 1, 6, b = 0, âˆ’2, 1

14. True or False? If a Âˇ b = a Âˇ c and if a = 0, then b = c. Justify your conclusion.

Ha

sa

In Exercises 15 and 16, find the direction cosines of v, and confirm that they satisfy Equation (5). Then use the direction cosines to approximate the direction angles to the nearest degree.

25. If L is a line in 2-space or 3-space that passes through the points A and B, then the distance from a point P to the line âˆ’â†’ L is equal to the length of the component of the vector AP âˆ’ â†’ that is orthogonal to the vector AB (see the accompanying figure). Use this result to find the distance from the point P (1, 0) to the line through A(2, âˆ’3) and B(5, 1).

n

13. Let u = 1, 2, v = 4, âˆ’2, and w = 6, 0. Find (a) u Âˇ (7v + w) (b) (u Âˇ w)w

(d) ( u v) Âˇ w. (c) u (v Âˇ w)

15. (a) v = i + j âˆ’ k

(b) v = 2i âˆ’ 2 j + k

16. (a) v = 3i âˆ’ 2 j âˆ’ 6k

(b) v = 3i âˆ’ 4k

ad

17. Show that the direction cosines of a vector satisfy cos2 Îą + cos2 Î˛ + cos2 Îł = 1

cos Îą = cos Îť cos Î¸ cos Î˛ = cos Îť sin Î¸ cos Îł = sin Îť

m

18. Let Î¸ and Îť be the angles shown in the accompanying figure. Show that the direction cosines of v can be expressed as

M uh am

[Hint: Express v in component form and normalize.]

19. Use the result in Exercise 18 to find the direction angles of the vector shown in the accompanying figure to the nearest degree. z

z

v

l

Figure Ex-18

L A Figure Ex-25

B

26. Use the method of Exercise 25 to find the distance from the point P (âˆ’3, 1, 2) to the line through A(1, 1, 0) and B(âˆ’2, 3, âˆ’4). 27. As shown in the accompanying figure, a block with a mass of 10 kg rests on a smooth (frictionless) ramp that is inclined at an angle of 45 â—Ś with the ground. How much force does the block exert on the ramp, and how much force must be applied in the direction of P to prevent the block from sliding down the ramp? Take the acceleration due to gravity to be 9.8 m/s2 . 28. For the block in Exercise 27, how much force must be applied in the direction of Q (shown in the accompanying figure) to prevent the block from sliding down the ramp?

v P

y

30Â°

u

x

P

y

Q

60Â° 45Â°

45Â°

x

Figure Ex-19

20. Show that two nonzero vectors v1 and v2 are orthogonal if and only if their direction cosines satisfy cos Îą1 cos Îą2 + cos Î˛1 cos Î˛2 + cos Îł1 cos Îł2 = 0

Figure Ex-27

Figure Ex-28

29. A block weighing 300 lb is suspended by cables A and B, as shown in the accompanying figure. Determine the forces that the block exerts along the cables.

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A

Yo

30° B

39. Prove that

u + v 2 + u − v 2 = 2 u 2 + 2 v 2 and interpret the result geometrically by translating it into a theorem about parallelograms. 40. Prove: u · v = 14 u + v 2 − 14 u − v 2 .

20 ft d

41. Show that if v1 , v2 , and v3 are mutually orthogonal nonzero vectors in 3-space, and if a vector v in 3-space is expressed as v = c1 v1 + c2 v2 + c3 v3

B

az

45° A

817

38. Let u and v be adjacent sides of a parallelogram. Use vectors to prove that the parallelogram is a rectangle if the diagonals are equal in length.

30. A block weighing 100 N is suspended by cables A and B, as shown in the accompanying figure. (a) Use a graphing utility to graph the forces that the block exerts along cables A and B as functions of the “sag” d. (b) Does increasing the sag increase or decrease the forces on the cables? (c) How much sag is required if the cables cannot tolerate forces in excess of 150 N? 10 ft

Cross Product

us uf i

12.4

then the scalars c1 , c2 , and c3 are given by the formulas 100 N

300 lb

42. Show that the three vectors v1 = 3i − j + 2k, v2 = i + j − k, v3 = i − 5 j − 4k are mutually orthogonal, and then use the result of Exercise 41 to find scalars c1 , c2 , and c3 so that

Figure Ex-30

31. Find the work done by a force F = −3 j (pounds) applied to a point that moves on a line from (1, 3) to (4, 7). Assume that distance is measured in feet.

n

32. A boat travels 100 meters due north while the wind exerts a force of 500 newtons toward the northeast. How much work does the wind do?

i = 1, 2, 3

Ri

Figure Ex-29

ci = (v · vi )/ vi 2 ,

43. For each x in (−⬁, +⬁), let u(x) be the vector from the origin to the point P (x, y) on the curve y = x 2 + 1, and v(x) the vector from the origin to the point Q(x, y) on the line y = −x − 1. (a) Use a CAS to find, to the nearest degree, the minimum angle between u(x) and v(x) for x in (−⬁, +⬁). (b) Determine whether there are any real values of x for which u(x) and v(x) are orthogonal.

sa

C

c1 v1 + c2 v2 + c3 v3 = i − j + k

Ha

33. A box is dragged along the floor by a rope that applies a force of 50 lb at an angle of 60 ◦ with the floor. How much work is done in moving the box 15 ft?

34. A force of F = 4i − 6 j + k newtons is applied to a point that moves a distance of 15 meters in the direction of the vector i + j + k. How much work is done?

C

ad

35. Find, to the nearest degree, the acute angle formed by two diagonals of a cube.

m

36. Find, to the nearest degree, the angles that a diagonal of a box with dimensions 10 cm by 15 cm by 25 cm makes with the edges of the box.

M uh am

37. Let u and v be adjacent sides of a parallelogram. Use vectors to prove that the diagonals of the parallelogram are perpendicular if the sides are equal in length.

••••••••••••••••••••••••••••••••••••••

DETERMINANTS

44. Let u be a unit vector in the xy-plane of an xyz-coordinate system, and let v be a unit vector in the yz-plane. Let θ1 be the angle between u and i, let θ2 be the angle between v and k, and let θ be the angle between u and v. (a) Show that cos θ = ± sin θ1 sin θ2 . (b) Find θ if θ is acute and θ1 = θ2 = 45 ◦ . (c) Use a CAS to find, to the nearest degree, the maximum and minimum values of θ if θ is acute and θ2 = 2θ1 . 45. Prove parts (b) and (e) of Theorem 12.3.2 for vectors in 3-space.

12.4 CROSS PRODUCT In many applications of vectors in mathematics, physics, and engineering, there is a need to find a vector that is orthogonal to two given vectors. In this section we will discuss a new type of vector multiplication that can be used for this purpose. Some of the concepts that we will develop in this section require basic ideas about determinants, which are functions that assign numerical values to square arrays of numbers. For example, if a1 , a2 , b1 , and b2 are real numbers, then we define a 2 × 2 determinant by

a1 a2 b1 b2 a1b2 − a2b1

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Three-Dimensional Space; Vectors

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

The purpose of the arrows is to help you remember the formula—the determinant is the product of the entries on the rightward arrow minus the product of the entries on the leftward arrow. For example,

3 4

–2 (3)(5) − (−2)(4) 15 + 8 23 5

Yo

A 3 × 3 determinant is defined in terms of 2 × 2 determinants by

a 1 a 2 a 3

b1 b2 b3 = a1 b2 b3 − a2 b1 b3 + a3 b1 b2

c1 c3

c1 c2

c2 c3

c1 c2 c3

(2)

1 −4

− (−2)

0 2

n

For example,

3 −2 −5

1

= 3 4 4 −4

3

0 3 2

Ri

az

The right side of this formula is easily remembered by noting that a1 , a2 , and a3 are the entries in the first “row” of the left side, and the 2 × 2 determinants on the right side arise by deleting the first row and an appropriate column from the left side. The pattern is as follows: a1 a2 a3 a1 a2 a3 a1 a2 a3 a1 a2 a3 b1 b2 b3 a1 b1 b2 b3 − a2 b1 b2 b3 + a3 b1 b2 b3 c1 c2 c3 c1 c2 c3 c1 c2 c3 c1 c2 c3

1 −4

+ (−5)

0 2

4

3

Ha

sa

= 3(20) + 2(2) − 5(3) = 49 There are also definitions of 4 × 4 determinants, 5 × 5 determinants, and higher, but we will not need them in this text. Properties of determinants are studied in a branch of mathematics called linear algebra, but we will only need the two properties stated in the following theorem:

ad

12.4.1 THEOREM. (a) If two rows in the array of a determinant are the same, then the value of the determinant is 0. (b) Interchanging two rows in the array of a determinant multiplies its value by −1.

m

We will give the proofs of parts (a) and (b) for 2 × 2 determinants and leave the proofs for 3 × 3 determinants as exercises.

M uh am

Proof (a).

••••••••••••••••••••••••••••••••••••••

CROSS PRODUCT

a1

a1

a2

= a1 a2 − a2 a1 = 0 a2

Proof (b).

b1

a1

a1 b2

= b a − b a = −(a b − a b ) = − 1 2 2 1 1 2 2 1

b1 a2

a2

b2

We now turn to the main concept in this section. 12.4.2 DEFINITION. If u = u1 , u2 , u3 and v = v1 , v2 , v3 are vectors in 3-space, then the cross product u × v is the vector defined by

u1 u3

u1 u2

u2 u3

k

i−

j+

(3) u×v=

v2 v3

v1 v 3

v1 v 2

or, equivalently, u × v = (u2 v3 − u3 v2 )i − (u1 v3 − u3 v1 ) j + (u1 v2 − u2 v1 )k

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Cross Product

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12.4

Observe that the right side of Formula (3) has the same form as the right side of Formula (2), the difference being notation and the order of the factors in the three terms. Thus, we can rewrite (3) as j u2 v2

k

u3

v3

(5)

Yo

i

u × v =

u1

v 1

However, this is just a mnemonic device and not a true determinant since the entries in a determinant are numbers, not vectors.

(a) u × v

az

Example 1 Let u = 1, 2, −2 and v = 3, 0, 1. Find (b) v × u

1 −2

j +

3 1

n

j k

2 −2

0 1

1 −2

i −

3 1

2

k = 2i − 7 j − 6k 0

sa

i

u × v =

1

3

2 =

0

Ri

Solution (a).

Solution (b). We could use the method of part (a), but it is really not necessary to perform

Ha

any computations. We need only observe that reversing u and v interchanges the second and third rows in (5), which in turn interchanges the rows in the arrays for the 2×2 determinants in (3). But interchanging the rows in the array of a 2 × 2 determinant reverses its sign, so the net effect of reversing the factors in a cross product is to reverse the signs of the components. Thus, by inspection v × u = −(u × v) = −2i + 7 j + 6k

ad

Example 2 Show that u × u = 0 for any vector u in 3-space.

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m

Solution. We could let u = u1 i + u2 j + u3 k and apply the method in part (a) of Example

••••••••••••••••••••••••••••••••••••••

ALGEBRAIC PROPERTIES OF THE CROSS PRODUCT

1 to show that

i

u × u =

u1

u 1

j u2 u2

k

u3

= 0 u3

However, the actual computations are unnecessary. We need only observe that if the two factors in a cross product are the same, then each 2 × 2 determinant in (3) is zero because its array has identical rows. Thus, u × u = 0 by inspection. Our next goal is to establish some of the basic algebraic properties of the cross product. As you read the discussion, keep in mind the essential differences between the cross product and the dot product: • •

The cross product is defined only for vectors in 3-space, whereas the dot product is defined for vectors in 2-space and 3-space. The cross product of two vectors is a vector, whereas the dot product of two vectors is a scalar. The main algebraic properties of the cross product are listed in the next theorem.

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Three-Dimensional Space; Vectors

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820

Yo

12.4.3 THEOREM. If u, v, and w are any vectors in 3-space and k is any scalar, then (a) u Ă— v = âˆ’(v Ă— u) (b) u Ă— (v + w) = (u Ă— v) + (u Ă— w) (c) (u + v) Ă— w = (u Ă— w) + (v Ă— w) (d ) k(u Ă— v) = (ku) Ă— v = u Ă— (kv) (e) u Ă— 0 = 0 Ă— u = 0 (f) u Ă— u = 0

Parts (a) and ( f ) were addressed in Examples 1 and 2. The other proofs are left as exercises. â€˘ WARNING.

az

In ordinary multiplication and in dot products the order of the factors does not matter, but in cross products it does. Part (a) of the last theorem shows that reversing the order of the factors in a cross product reverses the direction of the resulting vector.

Ri

â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘

The following cross products occur so frequently that it is helpful to be familiar with them: iĂ—j=k jĂ—k=i kĂ—i=j (6) j Ă— i = âˆ’k k Ă— j = âˆ’i i Ă— k = âˆ’j

sa

n

These results are easy to obtain; for example,

i j k

1

0 0

1 0

j +

iâˆ’

i Ă— j = 1 0 0 =

0 1 0 0 0

0 1 0

i

0

k=k 1

k

Ha

However, rather than computing these cross products each time you need them, you can use the diagram in Figure 12.4.1. In this diagram, the cross product of two consecutive vectors in the clockwise direction is the next vector around, and the cross product of two consecutive vectors in the counterclockwise direction is the negative of the next vector around. j

â€˘ WARNING.

Figure 12.4.1

We can write a product of three real numbers as uvw because the associative law u(vw) = (uv)w ensures that the same value for the product results no matter where the parentheses are inserted. However, the associative law does not hold for cross products. For example,

ad

â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘

i Ă— ( j Ă— j) = i Ă— 0 = 0

and

(i Ă— j) Ă— j = k Ă— j = âˆ’i

m

so that i Ă— ( j Ă— j) = (i Ă— j) Ă— j. Thus, we cannot write a cross product with three vectors as u Ă— v Ă— w, since this expression is ambiguous without parentheses.

M uh am

GEOMETRIC PROPERTIES OF THE CROSS PRODUCT

The following theorem shows that the cross product of two vectors is orthogonal to both factors. 12.4.4 THEOREM. If u and v are vectors in 3-space, then (a) u Âˇ (u Ă— v) = 0 (u Ă— v is orthogonal to u) (b) v Âˇ (u Ă— v) = 0 (u Ă— v is orthogonal to v) We will prove part (a). The proof of part (b) is similar.

Proof (a). Let u = u1 , u2 , u3 and v = v1 , v2 , v3 . Then from (4) u Ă— v = u2 v3 âˆ’ u3 v2 , u3 v1 âˆ’ u1 v3 , u1 v2 âˆ’ u2 v1 so that u Âˇ (u Ă— v) = u1 (u2 v3 âˆ’ u3 v2 ) + u2 (u3 v1 âˆ’ u1 v3 ) + u3 (u1 v2 âˆ’ u2 v1 ) = 0

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Cross Product

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12.4

Example 3 In Example 1 we showed that the cross product u × v of u = 1, 2, −2 and v = 3, 0, 1 is u × v = 2i − 7 j − 6k = 2, −7, −6

Theorem 12.4.4 guarantees that this vector is orthogonal to both u and v; this is confirmed by the computations

Yo

u · (u × v) = 1, 2, −2 · 2, −7, −6 = (1)(2) + (2)(−7) + (−2)(−6) = 0 v · (u × v) = 3, 0, 1 · 2, −7, −6 = (3)(2) + (0)(−7) + (1)(−6) = 0

u×v

It can be proved that if u and v are nonzero and nonparallel vectors, then the direction ∗ of u × v relative to u and v is determined by a right-hand rule; that is, if the fingers of the right hand are cupped so they curl from u toward v in the direction of rotation that takes u into v in less than 180 ◦ , then the thumb will point (roughly) in the direction of u × v (Figure 12.4.2). For example, we stated in (6) that i × j = k,

u

j × k = i,

k×i=j

az

u

Ri

all of which are consistent with the right-hand rule (verify). The next theorem lists some more important geometric properties of the cross product.

v Figure 12.4.2

sa

n

12.4.5 THEOREM. Let u and v be nonzero vectors in 3-space, and let θ be the angle between these vectors when they are positioned so their initial points coincide. (a) u × v = u

v sin θ (b) The area A of the parallelogram that has u and v as adjacent sides is A = u × v

Ha

(c)

(8)

u × v = 0 if and only if u and v are parallel vectors, that is, if and only if they are scalar multiples of one another.

Proof (a).

M uh am

m

ad

u

v sin θ = u

v 1 − cos2 θ (u · v)2 Theorem 12.3.3 = u

v 1 −

u 2 v 2 = u 2 v 2 − (u · v)2 = (u21 + u22 + u23 )(v12 + v22 + v32 ) − (u1 v1 + u2 v2 + u3 v3 )2 =

= u × v

v

||v|| sin u

||v||

||u||

Figure 12.4.3

See Formula (4).

Proof (b). Referring to Figure 12.4.3, the parallelogram that has u and v as adjacent sides can be viewed as having base u and altitude v sin θ . Thus, its area A is A = (base)(altitude) = u

v sin θ = u × v

u

u

(u2 v3 − u3 v2 )2 + (u1 v3 − u3 v1 )2 + (u1 v2 − u2 v1 )2

Proof (c). Since u and v are assumed to be nonzero vectors, it follows from part (a) that u × v = 0 if and only if sin θ = 0; this is true if and only if θ = 0 or θ = π (since ∗

Recall that we agreed to consider only right-handed coordinate systems in this text. Had we used left-handed systems instead, a “left-hand rule” would apply here.

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Three-Dimensional Space; Vectors

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822

0 ≤ θ ≤ π). Geometrically, this means that u × v = 0 if and only if u and v are parallel vectors.

z

Example 4 Find the area of the triangle that is determined by the points P1 (2, 2, 0), P2 (−1, 0, 2), and P3 (0, 4, 3). P3(0, 4, 3)

Solution. The area A of the triangle is half the area of the parallelogram determined by

Yo

−−→ −−→ −−→ −−→ the vectors P1P2 and P1P3 (Figure 12.4.4). But P1P2 = −3, −2, 2 and P1P3 = −2, 2, 3, so −−→ −−→ P1P2 × P1P3 = −10, 5, −10

x

P1(2, 2, 0)

(verify), and consequently −−→ −−→ A = 12 P1P2 × P1P3 =

Figure 12.4.4

15 2

az

y

If u = u1 , u2 , u3 , v = v1 , v2 , v3 , and w = w1 , w2 , w3 are vectors in 3-space, then the number

••••••••••••••••••••••••••••••••••••••

SCALAR TRIPLE PRODUCTS

Ri

P2(–1, 0, 2)

u · (v × w)

Ha

sa

n

is called the scalar triple product of u, v, and w. It is not necessary to compute the dot product and cross product to evaluate a scalar triple product—the value can be obtained directly from the formula

u1 u2 u3

(9) u · (v × w) =

v1 v2 v3

w 1 w2 w3

m

ad

the validity of which can be seen by writing

v 2 v 3

v1 v3

v1 v2

k u · (v × w) = u ·

i−

j+

w 2 w3

w1 w3

w 1 w2

v

v1 v3

v1 v2

v3

= u1

2 − u + u 2

3

w2 w3

w1 w3

w 1 w2

u1 u2 u3

=

v1 v2 v3

w1 w2 w3

M uh am

Example 5 Calculate the scalar triple product u · (v × w) of the vectors

••••••••••••••••••••••••••••••••••••••

GEOMETRIC PROPERTIES OF THE SCALAR TRIPLE PRODUCT

u = 3i − 2 j − 5k,

v = i + 4 j − 4k,

Solution.

3

u · (v × w) =

1

0

−2 4 3

w = 3 j + 2k

−5

−4

= 49 2

• FOR THE READER. • • • • • • • • • • • •

Many calculating utilities have built-in cross product and determinant operations. If your calculating utility has these capabilities, use it to check the computations in Examples 1 and 5.

If u, v, and w are nonzero vectors in 3-space that are positioned so their initial points coincide, then these vectors form the adjacent sides of a parallelepiped (Figure 12.4.5). The following theorem establishes a relationship between the volume of this parallelepiped and the scalar triple product of the sides.

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Cross Product

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12.4

12.4.6 THEOREM. Let u, v, and w be nonzero vectors in 3-space. (a) The volume V of the parallelepiped that has u, v, and w as adjacent edges is

u

V = |u · (v × w)|

w

u · (v × w) = 0 if and only if u, v, and w lie in the same plane.

Yo

(b)

(10)

v Figure 12.4.5

Proof (a). Referring to Figure 12.4.6, let us regard the base of the parallelepiped with u, v, and w as adjacent sides to be the parallelogram determined by v and w. Thus, the area of the base is v × w , and the altitude h of the parallelepiped (shown in the figure) is the length of the orthogonal projection of u on the vector v × w. Therefore, from Formula (12) of Section 12.3 we have |u · (v × w)| |u · (v × w)| h = projv × w u =

v × w = 2

v × w

v × w

It now follows that the volume of the parallelepiped is

v×w

az

u

Ri

w

V = (area of base)(height) = v × w h = |u · (v × w)| v h = || proj v× w u || Figure 12.4.6

• REMARK.

It follows from Formula (10) that

sa

u · (v × w) = ±V

The + occurs when u makes an acute angle with v × w and the − occurs when it makes an obtuse angle.

Ha

• • • • • • • • • • • • • • • • • • • • •

n

Proof (b). The vectors u, v, and w lie in the same plane if and only if the parallelepiped with these vectors as adjacent sides has volume zero (why?). Thus, from part (a) the vectors lie in the same plane if and only if u · (v × w) = 0.

We observed earlier in this section that the expression u × v × w must be avoided because it is ambiguous without parentheses. However, the expression u · v × w is not ambiguous—it has to mean u · (v × w) and not (u · v) × w because we cannot form the cross product of a scalar and a vector. Similarly, the expression u × v · w must mean (u × v) · w and not u × (v · w). Thus, when you see an expression of the form u · v × w or u × v · w, the cross product is formed first and the dot product second. Since interchanging two rows of a determinant multiplies its value by −1, making two row interchanges in a determinant has no effect on its value. This being the case, it follows that

••••••••••••••••••••••••••••••••••••••

M uh am

m

ad

ALGEBRAIC PROPERTIES OF THE SCALAR TRIPLE PRODUCT

u · (v × w) = w · (u × v) = v · (w × u)

(11)

since the 3 × 3 determinants that are used to compute these scalar triple products can be obtained from one another by two row interchanges (verify).

• REMARK. • • • • • • • • • • • • • • • • •

Observe that the second expression in (11) can be obtained from the first by leaving the dot, the cross, and the parentheses fixed, moving the first two vectors to the right, and bringing the third vector to the first position. The same procedure produces the third expression from the second and the first expression from the third (verify). Another useful formula can be obtained by rewriting the first equality in (11) as u · (v × w) = (u × v) · w

and then omitting the superfluous parentheses to obtain u·v×w=u×v·w

(12)

In words, this formula states that the dot and cross in a scalar triple product can be interchanged (provided the factors are grouped appropriately).

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Three-Dimensional Space; Vectors

••••••••••••••••••••••••••••••••••••••

DOT AND CROSS PRODUCTS ARE COORDINATE INDEPENDENT

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In Definitions 12.3.1 and 12.4.2 we defined the dot product and the cross product of two vectors in terms of the components of those vectors in a coordinate system. Thus, it is theoretically possible that changing the coordinate system might change u · v or u × v, since the components of a vector depend on the coordinate system that is chosen. However, the relationships u · v = u

v cos θ

Yo

(13)

u × v = u

v sin θ

(14)

••••••••••••••••••••••••••••••••••••••

Cross products play an important role in describing rotational motion in 3-space. For example, suppose that an astronaut on a satellite repair mission in space applies a force F at a point Q on the surface of a spherical satellite. If the force is directed along a line that passes through the center P of the satellite, then Newton’s Second Law of Motion implies that the force will accelerate the satellite in the direction of F. However, if the astronaut applies −→ the same force at an angle θ with the vector PQ, then F will tend to cause a rotation, as well as an acceleration in the direction of F. To see why this is so, let us resolve F into a sum of orthogonal components F = F1 + F2 , where F1 is the orthogonal projection of F −→ −→ on the vector PQ and F2 is the component of F orthogonal to PQ (Figure 12.4.7). Since the force F1 acts along the line through the center of the satellite, it contributes to the linear acceleration of the satellite but does not cause any rotation. However, the force F2 is tangent −→ to the circle around the satellite in the plane of F and PQ, so it causes the satellite to rotate about an axis that is perpendicular to that plane.

ad

Ha

sa

n

MOMENTS AND ROTATIONAL MOTION IN 3-SPACE

Ri

az

that were obtained in Theorems 12.3.3 and 12.4.5 show that this is not the case. Formula (13) shows that the value of u · v depends only on the lengths of the vectors and the angle between them—not on the coordinate system. Similarly, Formula (14), in combination with the right-hand rule and Theorem 12.4.4, shows that u × v does not depend on the coordinate system (as long as it is right handed). These facts are important in applications because they allow us to choose any convenient coordinate system for solving a problem with full confidence that the choice will not affect computations that involve dot products or cross products.

M uh am

m

Astronauts use tools that are designed to limit forces that would impart unintended rotational motion to a satellite.

F2 P

Q

F

u F1

Figure 12.4.7

You know from your own experience that the “tendency” for rotation about an axis depends both on the amount of force and how far from the axis it is applied. For example, it is easier to close a door by pushing on its outer edge than applying the same force close to the hinges. In fact, the tendency of rotation of the satellite can be measured by −→ distance from the center × magnitude of the force

PQ

F2

(15) However, F2 = F sin θ, so we can rewrite (15) as −→ −→

PQ

F sin θ = PQ × F

This is called the scalar moment or torque of F about the point P . Scalar moments have units −→ of force times distance—pound-feet or newton-meters, for example. The vector PQ × F is called the vector moment or torque vector of F about P .

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F = 100k

Q(1, 1, 1) y

x

(a) z

F

Example 6 Figure 12.4.8a shows a force F of 100 N applied in the positive z-direction at the point Q(1, 1, 1) of a cube whose sides have a length of 1 m. Assuming that the cube is free to rotate about the point P (0, 0, 0) (the origin), find the scalar moment of the force about P , and describe the direction of rotation. −→ Solution. The force vector is F = 100k, and the vector from P to Q is PQ = i + j + k, so the vector moment of F about P is

i j k

−→ 1

= 100i − 100 j PQ × F =

1 1

0 0 100

√ Thus, the scalar moment of F about P is 100i − 100 j = 100 2 ≈ 141 N·m, and the direction of rotation is counterclockwise looking along the vector 100i − 100 j = 100(i − j) toward its initial point (Figure 12.4.8b).

Ri

y

100(i – j)

−→ Recalling that the direction of PQ × F is determined by the right-hand rule, it follows that the direction of rotation about P that results by applying the force F at the point Q −→ is counterclockwise looking down the axis of PQ × F (Figure 12.4.7). Thus, the vector −→ moment PQ × F captures the essential information about the rotational effect of the force— the magnitude of the cross product provides the scalar moment of the force, and the cross product vector itself provides the axis and direction of rotation.

Yo

P

825

az

z

Cross Product

us uf i

12.4

x

(b)

EXERCISE SET 12.4

C

CAS

sa

n

Figure 12.4.8

z

Ha

1. (a) Use a determinant to find the cross product i × (i + j + k)

(1, 1, 1)

(b) Check your answer in part (a) by rewriting the cross product as

v y

ad

i × (i + j + k) = (i × i) + (i × j) + (i × k) and evaluating each term.

m

2. In each part, use the two methods in Exercise 1 to find (b) k × (i + j + k). (a) j × (i + j + k)

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In Exercises 3–6, find u × v, and check that it is orthogonal to both u and v. 3. u = 1, 2, −3, v = −4, 1, 2 4. u = 3i + 2 j − k, v = −i − 3 j + k

5. u = 0, 1, −2, v = 3, 0, −4 6. u = 4i + k, v = 2i − j

C

u x

Figure Ex-9

10. Find two unit vectors that are normal to both u = −7i + 3 j + k, v = 2i + 4k 11. Find two unit vectors that are perpendicular to the plane determined by the points A(0, −2, 1), B(1, −1, −2), and C(−1, 1, 0). 12. Find two unit vectors that are parallel to the yz-plane and are orthogonal to the vector 3i − j + 2k. In Exercises 13 and 14, find the area of the parallelogram that has u and v as adjacent sides.

7. Let u = 2, −1, 3, v = 0, 1, 7, and w = 1, 4, 5. Find (a) u × (v × w) (b) (u × v) × w (d) (v × w) × (u × v). (c) (u × v) × (v × w)

13. u = i − j + 2k, v = 3 j + k

8. Use a CAS or a calculating utility that can compute determinants or cross products to solve Exercise 7.

14. u = 2i + 3 j, v = −i + 2 j − 2k

9. Find the direction cosines of u × v for the vectors u and v in the accompanying figure.

In Exercises 15 and 16, find the area of the triangle with vertices P , Q, and R.

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Three-Dimensional Space; Vectors

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826

15. P (1, 5, −2), Q(0, 0, 0), R(3, 5, 1)

(b) Use the cross product to find sin θ. (c) Confirm that sin2 θ + cos2 θ = 1.

16. P (2, 0, −3), Q(1, 4, 5), R(7, 2, 9)

31. What can you say about the angle between nonzero vectors u and v if u · v = u × v ?

In Exercises 17–20, find u · (v × w). 17. u = 2i − 3 j + k, v = 4i + j − 3k, w = j + 5k

32. Show that if u and v are vectors in 3-space, then

18. u = 1, −2, 2, v = 0, 3, 2, w = −4, 1, −3

Yo

u × v 2 = u 2 v 2 − (u · v)2 [Note: This result is sometimes called Lagrange’s identity.]

19. u = 2, 1, 0, v = 1, −3, 1, w = 4, 0, 1 20. u = i, v = i + j, w = i + j + k

22. u = 3i + j + 2k, v = 4i + 5 j + k, w = i + 2 j + 4k

33. The accompanying figure shows a force F of 10 lb applied in the positive y-direction to the point Q(1, 1, 1) of a cube whose sides have a length of 1 ft. In each part, find the scalar moment of F about the point P , and describe the direction of rotation, if any, if the cube is free to rotate about P . (a) P is the point (0, 0, 0). (b) P is the point (1, 0, 0). (c) P is the point (1, 0, 1).

23. In each part, use a scalar triple product to determine whether the vectors lie in the same plane. (a) u = 1, −2, 1, v = 3, 0, −2, w = 5, −4, 0 (b) u = 5i − 2 j + k, v = 4i − j + k, w = i − j (c) u = 4, −8, 1, v = 2, 1, −2, w = 3, −4, 12

34. The accompanying figure shows a force F of 1000 N applied to the corner of a box. (a) Find the scalar moment of F about the point P . (b) Find the direction angles of the vector moment of F about the point P to the nearest degree.

az

In Exercises 21 and 22, use a scalar triple product to find the volume of the parallelepiped that has u, v, and w as adjacent edges.

n

Ha

25. Consider the parallelepiped with adjacent edges

sa

24. Suppose that u · (v × w) = 3. Find (a) u · (w × v) (b) (v × w) · u (d) v · (u × w) (c) w · (u × v) (f ) v · (w × w). (e) (u × w) · v

Ri

21. u = 2, −6, 2, v = 0, 4, −2, w = 2, 2, −4

u = 3i + 2 j + k v = i + j + 2k w = i + 3 j + 3k

ad

(a) Find the volume. (b) Find the area of the face determined by u and w. (c) Find the angle between u and the plane containing the face determined by v and w.

z

Q(1, 1, 1)

M uh am

27. Use the result in Exercise 26 to find the distance between the point P and the line through the points A and B. (a) P (−3, 1, 2), A(1, 1, 0), B(−2, 3, −4) (b) P (4, 3), A(2, 1), B(0, 2)

28. It is a theorem of solid geometry that the volume of a tetrahedron is 13 (area of base) · (height). Use this result to prove that the volume of a tetrahedron with adjacent edges given by the vectors u, v, and w is 16 |u · (v × w)|. 29. Use the result of Exercise 28 to find the volume of the tetrahedron with vertices P (−1, 2, 0), Q(2, 1, −3), R(1, 0, 1), S(3, −2, 3)

30. Let θ be the angle between the vectors u = 2i + 3 j − 6k and v = 2i + 3 j + 6k. (a) Use the dot product to find cos θ .

10 lb y

1 ft

1000 N

P

y

1m

1 ft 2m

1 ft

1m Q

x

x

Figure Ex-33

Figure Ex-34

35. As shown in the accompanying figure, a force of 200 N is applied at an angle of 18 ◦ to a point near the end of a monkey wrench. Find the scalar moment of the force about the center of the bolt. [Treat this as a problem in two dimensions.]

m

26. Show that in 3-space the distance d from a point P to the line L through points A and B can be expressed as → −→ −

AP × AB

d= − →

AB

z

18° 200 mm

200 N

30 mm

Figure Ex-35

36. Prove parts (b) and (c) of Theorem 12.4.3. 37. Prove parts (d ) and (e) of Theorem 12.4.3. 38. Prove part (b) of Theorem 12.4.1 for 3 × 3 determinants. [Just give the proof for the first two rows.] Then use (b) to prove (a). 39. Expressions of the form u × (v × w) and (u × v) × w are called vector triple products. It can be proved with some

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effort that u × (v × w) = (u · w)v − (u · v)w (u × v) × w = (w · u)v − (w · v)u

Parametric Equations of Lines

827

us uf i

12.5

41. Prove: If a, b, c, and d lie in the same plane when positioned with a common initial point, then (a × b) × (c × d) = 0

u = i + 3 j − k, v = i + j + 2k, w = 3i − j + 2k 40. (a) Use the result in Exercise 39 to show that: u × (v × w) lies in the same plane as v and w (u × v) × w lies in the same plane as u and v. (b) Use a geometrical argument to justify the results in part (a).

Yo

42. Use a CAS to approximate the minimum area of a triangle if two of its vertices are (2, −1, 0) and (3, 2, 2) and its third vertex is on the curve y = ln x in the xy-plane. 43. If a force F is applied to an object at a point Q, then the line through Q parallel to F is called the line of action of the force. We defined the vector moment of F about a point P to −→ be PQ × F. Show that if Q is any point on the line of action −→ −→ of F, then PQ × F = PQ × F; that is, it is not essential to use the point of application to compute the vector moment— any point on the line of action will do. [Hint: Write −→ −→ −−→ PQ = PQ + QQ and use properties of the cross product.]

az

See if you can figure out what the expressions “outer,” “remote,” and “adjacent” mean in this rule, and then use the rule to find the two vector triple products of the vectors

C

Ri

These expressions can be summarized with the following mnemonic rule: vector triple product = (outer · remote)adjacent − (outer · adjacent)remote

n

12.5 PARAMETRIC EQUATIONS OF LINES

••••••••••••••••••••••••••••••••••••••

LINES DETERMINED BY A POINT AND A VECTOR

Ha

sa

In this section we will discuss parametric equations of lines in 2-space and 3-space. In 3-space, parametric equations of lines are especially important because they generally provide the most convenient form for representing lines algebraically. A line in 2-space or 3-space can be determined uniquely by specifying a point on the line and a nonzero vector parallel to the line (Figure 12.5.1). The following theorem gives parametric equations of the line through a point P0 and parallel to a nonzero vector v:

y

12.5.1 THEOREM. (a) The line in 2-space that passes through the point P0 (x0 ,y0 ) and is parallel to the nonzero vector v = a, b = ai + bj has parametric equations

ad

L

v

x

M uh am

z

m

P0(x0, y0) (a, b)

L

(b)

x = x0 + at,

y = y0 + bt

(1)

The line in 3-space that passes through the point P0 (x0 , y0 , z0 ) and is parallel to the nonzero vector v = a, b, c = ai + bj + ck has parametric equations x = x0 + at,

y = y0 + bt,

z = z0 + ct

(2)

P0(x0, y0, z 0)

(a, b, c)

v

x

A unique line L passes through P0 and is parallel to v.

Figure 12.5.1

y

We will prove part (b). The proof of (a) is similar.

Proof (b). If L is the line in 3-space that passes through the point P0 (x0 , y0 , z0 ) and is par-

allel to the nonzero vector v = a, b, c, then L consists precisely of those points P (x, y, z) −→ for which the vector P0P is parallel to v (Figure 12.5.2). In other words, the point P (x, y, z) −→ is on L if and only if P0P is a scalar multiple of v, say −→ P0P = tv This equation can be written as

x − x0 , y − y0 , z − z0 = ta, tb, tc

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z

which implies that

P(x, y, z)

x − x0 = ta, P0 (x0, y0, z 0)

y − y0 = tb,

z − z0 = tc

from which (2) follows. (a, b, c)

• REMARK.

L v

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Three-Dimensional Space; Vectors

y

• • • • • • •

Although it is not stated explicitly, it is understood in Equations (1) and (2) that −⬁ < t < +⬁, which reflects the fact that lines extend indefinitely.

Example 1 Find parametric equations of the line

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828

Figure 12.5.2

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(a) passing through (4, 2) and parallel to v = −1, 5; (b) passing through (1, 2, −3) and parallel to v = 4i + 5j − 7k; (c) passing through the origin in 3-space and parallel to v = 1, 1, 1.

x

Solution (a). From (1) with x0 = 4, y0 = 2, a = −1, and b = 5 we obtain y = 2 + 5t

Solution (b). From (2) we obtain x = 1 + 4t,

y = 2 + 5t,

Ri

x = 4 − t,

z = −3 − 7t

Solution (c). From (2) with x0 = 0, y0 = 0, z0 = 0, a = 1, b = 1, and c = 1 we obtain y = t,

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Example 2

z=t

n

x = t,

Find parametric equations of the line L passing through the points P1 (2, 4, −1) and P2 (5, 0, 7). (b) Where does the line intersect the xy-plane?

Ha

(a)

−−→

Solution (a). The vector P1P2 = 3, −4, 8 is parallel to L and the point P1 (2, 4, −1) lies on L, so it follows from (2) that L has parametric equations x = 2 + 3t,

y = 4 − 4t,

z = −1 + 8t

(3)

ad

Had we used P2 as the point on L rather than P1 , we would have obtained the equations x = 5 + 3t,

y = −4t,

z = 7 + 8t

M uh am

m

Although these equations look different from those obtained using P1 , the two sets of equations are actually equivalent in that both generate L as t varies from −⬁ to +⬁. To see this, note that if t1 gives a point (x, y, z) = (2 + 3t1 , 4 − 4t1 , −1 + 8t1 )

on L using the first set of equations, then t2 = t1 − 1 gives the same point (x, y, z) = (5 + 3t2 , −4t2 , 7 + 8t2 ) = (5 + 3(t1 − 1), −4(t1 − 1), 7 + 8(t1 − 1))

= (2 + 3t1 , 4 − 4t1 , −1 + 8t1 ) on L using the second set of equations. Conversely, if t2 gives a point on L using the second set of equations, then t1 = t2 + 1 gives the same point using the first set.

Solution (b). It follows from (3) in part (a) that the line intersects the xy-plane at the point where z = −1 + 8t = 0, that is, when t = 18 . Substituting this value of t in (3) yields the , 7, 0 . point of intersection (x, y, z) = 19 8 2 Example 3 Let L1 and L2 be the lines L1 : x = 1 + 4t, y = 5 − 4t, z = −1 + 5t L2 : x = 2 + 8t, y = 4 − 3t, z = 5 + t

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(a)

Are the lines parallel?

(b)

Do the lines intersect?

Parametric Equations of Lines

829

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12.5

Solution (a). The line L1 is parallel to the vector 4i − 4j + 5k, and the line L2 is parallel

Yo

to the vector 8i − 3j + k. These vectors are not parallel since neither is a scalar multiple of the other. Thus, the lines are not parallel.

Solution (b). For L1 and L2 to intersect at some point (x0 , y0 , z0 ) these coordinates would have to satisfy the equations of both lines. In other words, there would have to exist values t1 and t2 for the parameters such that x0 = 1 + 4t1 ,

y0 = 5 − 4t1 ,

z0 = −1 + 5t1

y0 = 4 − 3t2 ,

z0 = 5 + t2

x0 = 2 + 8t2 ,

az

and

Ri

This leads to three conditions on t1 and t2 , 1 + 4t1 = 2 + 8t2 5 − 4t1 = 4 − 3t2

n

−1 + 5t1 = 5 + t2 Thus, the lines intersect if there are values of t1 and t2 that satisfy all three equations, and the lines do not intersect if there are no such values. You should be familiar with methods for solving systems of two linear equations in two unknowns; however, this is a system of three linear equations in two unknowns. To determine whether this system has a solution we will solve the first two equations for t1 and t2 and then check whether these values satisfy the third equation. We will solve the first two equations by the method of elimination. We can eliminate the unknown t1 by adding the equations. This yields the equation

L2

Ha

sa

L1

6 = 6 + 5t2

from which we obtain t2 = 0. We can now find t1 by substituting this value of t2 in either the first or second equation. This yields t1 = 14 . However, the values t1 = 41 and t2 = 0 do not satisfy the third equation in (4), so the lines do not intersect. Two lines in 3-space that are not parallel and do not intersect (such as those in Example 3) are called skew lines. As illustrated in Figure 12.5.3, any two skew lines lie in parallel planes.

ad

Parallel planes containing skew lines L1 and L2 can be determined by translating each line until it intersects the other.

Figure 12.5.3

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m

••••••••••••••••••••••••••••••••••••••

LINE SEGMENTS

(4)

••••••••••••••••••••••••••••••••••••••

VECTOR EQUATIONS OF LINES

Sometimes one is not interested in an entire line, but rather some segment of a line. Parametric equations of a line segment can be obtained by finding parametric equations for the entire line, then restricting the parameter appropriately so that only the desired segment is generated. Example 4 Find parametric equations for the line segment that joins the points P1 (2, 4, −1) and P2 (5, 0, 7).

Solution. From Example 2, the line through the points P1 and P2 has parametric equations

x = 2 + 3t, y = 4 − 4t, z = −1 + 8t. With these equations, the point P1 corresponds to t = 0 and P2 to t = 1. Thus, the line segment that joins P1 and P2 is given by x = 2 + 3t,

y = 4 − 4t,

z = −1 + 8t

(0 ≤ t ≤ 1)

We will now show how vector notation can be used to express the parametric equations of a line more compactly. Because two vectors are equal if and only if their components are equal, (1) and (2) can be written in vector form as

x, y = x0 + at, y0 + bt

x, y, z = x0 + at, y0 + bt, z0 + ct

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830

Three-Dimensional Space; Vectors

or, equivalently, as

x, y = x0 , y0 + t a, b

x, y, z = x0 , y0 , z0 + t a, b, c

(5)

(6)

For the equation in 2-space we define the vectors r, r0 , and v as r0 = x0 , y0 ,

v = a, b

and for the equation in 3-space we define them as r = x, y, z,

r0 = x0 , y0 , z0 ,

v = a, b, c

(7)

Yo

r = x, y,

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

(8)

Substituting (7) and (8) in (5) and (6), respectively, yields the equation r = r0 + tv

P0

L

r tv

r0

(9)

v

n

x

in both cases. We call this the vector equation of a line in 2-space or 3-space. In this equation, v is a nonzero vector parallel to the line, and r0 is a vector whose components are the coordinates of a point on the line. We can interpret Equation (9) geometrically by positioning the vectors r0 and v with their initial points at the origin and the vector tv with its initial point at P0 (Figure 12.5.4). The vector tv is a scalar multiple of v and hence is parallel to v and L. Moreover, since the initial point of tv is at the point P0 on L, this vector actually runs along L; hence, the vector r = r0 + tv can be interpreted as the vector from the origin to a point on L. As the parameter t varies from 0 to +⬁, the terminal point of r traces out the portion of L that extends from P0 in the direction of v, and as t varies from 0 to −⬁, the terminal point of r traces out the portion of L that extends from P0 in the direction that is opposite to v. Thus, the entire line is traced as t varies over the interval (−⬁, +⬁), and it is traced in the direction of v as t increases.

Ri

tv

az

y

r = r0 + tv

Ha

sa

Figure 12.5.4

Example 5 The equation

x, y, z = −1, 0, 2 + t 1, 5, −4 is of form (9) with

r0 = −1, 0, 2

and

v = 1, 5, −4

ad

Thus, the equation represents the line in 3-space that passes through the point (−1, 0, 2) and is parallel to the vector 1, 5, −4.

m

Example 6 Find an equation of the line in 3-space that passes through the points P1 (2, 4, −1) and P2 (5, 0, 7).

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Solution. The vector −−→ P1P2 = 3, −4, 8

is parallel to the line, so it can be used as v in (9). For r0 we can use either the vector from the origin to P1 or the vector from the origin to P2 . Using the former yields r0 = 2, 4, −1 Thus, a vector equation of the line through P1 and P2 is

x, y, z = 2, 4, −1 + t 3, −4, 8 If needed, we can express the line parametrically by equating corresponding components on the two sides of this vector equation, in which case we obtain the parametric equations in Example 2 (verify).

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EXERCISE SET 12.5

C

Graphing Utility

Parametric Equations of Lines

CAS

831

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12.5

y

z

L3

L1

L2

(1, 1, 1)

(1, 1)

az

y x

In Exercises 9 and 10, express the given parametric equations of a line in vector form using bracket notation and also using i, j, k notation. 9. (a) x = −3 + t, y = 4 + 5t (b) x = 2 − t, y = −3 + 5t, z = t

L3

L2

8. (a) x, y = −1, 5 + t 2, 3 (b) xi + y j + zk = (i + j − 2k) + t j

10. (a) x = t, y = −2 + t (b) x = 1 + t, y = −7 + 3t, z = 4 − 5t

L4 x

(a)

(b)

Ri

L1

7. (a) xi + y j = (2i − j) + t (4i − j) (b) x, y, z = −1, 2, 4 + t 5, 7, −8

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1. (a) Find parametric equations for the lines through the corner of the unit square shown in part (a) of the accompanying figure. (b) Find parametric equations for the lines through the corner of the unit cube shown in part (b) of the accompanying figure.

In Exercises 11–18, find parametric equations of the line that satisfies the stated conditions.

Figure Ex-1

11. The line through (−5, 2) that is parallel to 2i − 3 j. 12. The line through (0, 3) that is parallel to the line x = −5+t, y = 1 − 2t.

n

2. (a) Find parametric equations for the line segments on the unit square in part (a) of the accompanying figure. (b) Find parametric equations for the line segments in the unit cube shown in part (b) of the accompanying figure.

sa

z

L1

(1, 1) L3

L3

(1, 1, 1)

L2 x

14. The line that is tangent to the parabola y = x 2 at the point (−2, 4).

L2

L4

L1

x

y

(b)

ad

(a)

Ha

y

Figure Ex-2

m

In Exercises 3 and 4, find parametric equations for the line through P1 and P2 and also for the line segment joining those points.

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3. (a) P1 (3, −2), P2 (5, 1)

13. The line that is tangent to the circle x 2 + y 2 = 25 at the point (3, −4).

(b) P1 (5, −2, 1), P2 (2, 4, 2)

4. (a) P1 (0, 1), P2 (−3, −4) (b) P1 (−1, 3, 5), P2 (−1, 3, 2) In Exercises 5 and 6, find parametric equations for the line whose vector equation is given.

5. (a) x, y = 2, −3 + t 1, −4 (b) xi + y j + zk = k + t (i − j + k)

6. (a) xi + y j = (3i − 4 j) + t (2i + j) (b) x, y, z = −1, 0, 2 + t −1, 3, 0 In Exercises 7 and 8, find a point P on the line and a vector v parallel to the line by inspection.

15. The line through (−1, 2, 4) that is parallel to 3i − 4 j + k. 16. The line through (2, −1, 5) that is parallel to −1, 2, 7. 17. The line through (−2, 0, 5) that is parallel to the line x = 1 + 2t, y = 4 − t, z = 6 + 2t. 18. The line through the origin that is parallel to the line x = t, y = −1 + t, z = 2. 19. Where does the line x = 1 + 3t, y = 2 − t intersect (a) the x-axis (b) the y-axis (c) the parabola y = x 2 ? 20. Where does the line x, y = 4t, 3t intersect the circle x 2 + y 2 = 25? In Exercises 21 and 22, find the intersections of the lines with the xy-plane, the xz-plane, and the yz-plane. 21. x = −2, y = 4 + 2t, z = −3 + t 22. x = −1 + 2t, y = 3 + t, z = 4 − t 23. Where does the line x = 1 + t, y = 3 − t, z = 2t intersect the cylinder x 2 + y 2 = 16? 24. Where does the line x = 2 − t, y = 3t, z = −1 + 2t intersect the plane 2y + 3z = 6? In Exercises 25 and 26, show that the lines L1 and L2 intersect, and find their point of intersection. 25. L1 : x = 2 + t, y = 2 + 3t, z = 3 + t L2 : x = 2 + t, y = 3 + 4t, z = 4 + 2t

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Three-Dimensional Space; Vectors

26. L1 : x + 1 = 4t, y − 3 = t, z − 1 = 0 L2 : x + 13 = 12t, y − 1 = 6t, z − 2 = 3t

40. L1 : x = 2t, y = 3 + 4t, z = 2 − 6t L2 : x = 1 + 3t, y = 6t, z = −9t

41. (a) Find parametric equations for the line through the points (x0 , y0 , z0 ) and (x1 , y1 , z1 ). (b) Find parametric equations for the line through the point (x1 , y1 , z1 ) and parallel to the line

In Exercises 27 and 28, show that the lines L1 and L2 are skew. 27. L1 : x = 1 + 7t, y = 3 + t, z = 5 − 3t L2 : x = 4 − t, y = 6, z = 7 + 2t

x = x0 + at,

28. L1 : x = 2 + 8t, y = 6 − 8t, z = 10t L2 : x = 3 + 8t, y = 5 − 3t, z = 6 + t

az

Ri

30. L1 : x = 5 + 3t, y = 4 − 2t, z = −2 + 3t L2 : x = −1 + 9t, y = 5 − 6t, z = 3 + 8t

43. (a) Describe the line whose symmetric equations are x−1 y+3 =z−5 = 4 2

In Exercises 31 and 32, determine whether the points P1 , P2 , and P3 lie on the same line.

n

[See Exercise 42.] (b) Find parametric equations for the line in part (a).

31. P1 (6, 9, 7), P2 (9, 2, 0), P3 (0, −5, −3)

ad

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In Exercises 33 and 34, show that the lines L1 and L2 are the same.

In Exercises 35 and 36, describe the line segment represented by the vector equation.

m

(0 ≤ t ≤ 2)

36. x, y, z = −2, 1, 4 + t 3, 0, −1

(0 ≤ t ≤ 3)

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In Exercises 37 and 38, use the method in Exercise 25 of Section 12.3 to find the distance from the point P to the line L, and then check your answer using the method in Exercise 26 of Section 12.4.

37. P (−2, 1, 1) L: x = 3 − t, y = t, z = 1 + 2t

38. P (1, 4, −3) L: x = 2 + t, y = −1 − t, z = 3t In Exercises 39 and 40, show that the lines L1 and L2 are parallel, and find the distance between them.

39. L1 : x = 2 − t, y = 2t, z = 1 + t L2 : x = 1 + 2t, y = 3 − 4t, z = 5 − 2t

44. Find the point on the line segment joining P1 (1, 4, −3) and P2 (1, 5, −1) that is 23 of the way from P1 to P2 .

sa

32. P1 (1, 0, 1), P2 (3, −4, −3), P3 (4, −6, −5)

35. x, y = 1, 0 + t −2, 3

z = z0 + ct

x − x0 y − y0 z − z0 = = a b c These equations, which are called the symmetric equations of L, provide a nonparametric representation of L.

29. L1 : x = 3 − 2t, y = 4 + t, z = 6 − t L2 : x = 5 − 4t, y = −2 + 2t, z = 7 − 2t

34. L1 : x = 1 + 3t, y = −2 + t, z = 2t L2 : x = 4 − 6t, y = −1 − 2t, z = 2 − 4t

y = y0 + bt,

42. Let L be the line that passes through the point (x0 , y0 , z0 ) and is parallel to the vector v = a, b, c, where a, b, and c are nonzero. Show that a point (x, y, z) lies on the line L if and only if

In Exercises 29 and 30, determine whether the lines L1 and L2 are parallel.

33. L1 : x = 3 − t, y = 1 + 2t L2 : x = −1 + 3t, y = 9 − 6t

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832

45. Let L1 and L2 be the lines whose parametric equations are L1 : x = 1 + 2t, L2 : x = 9 + t,

y = 2 − t, y = 5 + 3t,

z = 4 − 2t z = −4 − t

(a) Show that L1 and L2 intersect at the point (7, −1, −2). (b) Find, to the nearest degree, the acute angle between L1 and L2 at their intersection. (c) Find parametric equations for the line that is perpendicular to L1 and L2 and passes through their point of intersection. 46. Let L1 and L2 be the lines whose parametric equations are L1 : x = 4t, L2 : x = 1 + t,

y = 1 − 2t, y = 1 − t,

z = 2 + 2t z = −1 + 4t

(a) Show that L1 and L2 intersect at the point (2, 0, 3). (b) Find, to the nearest degree, the acute angle between L1 and L2 at their intersection. (c) Find parametric equations for the line that is perpendicular to L1 and L2 and passes through their point of intersection. In Exercises 47 and 48, find parametric equations of the line that contains the point P and intersects the line L at a right angle. 47. P (0, 2, 1) L: x = 2t, y = 1 − t, z = 2 + t 48. P (3, 1, −2) L: x = −2 + 2t, y = 4 + 2t, z = 2 + t

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833

(b) Use a graphing utility to graph the distance between the bugs as a function of time from t = 0 to t = 5. (c) What does the graph tell you about the distance between the bugs? (d) How close do the bugs get?

49. Two bugs are walking along lines in 3-space. At time t bug 1 is at the point (x, y, z) on the line

50. Suppose that the temperature T at a point (x, y, z) on the line x = t, y = 1 + t, z = 3 − 2t is T = 25x 2 yz. Use a CAS or a calculating utility with a root-finding capability to approximate the maximum temperature on that portion of the line that extends from the xz-plane to the xy-plane.

Yo

C

az

x = 4 − t, y = 1 + 2t, z = 2 + t and at the same time t bug 2 is at the point (x, y, z) on the line x = t, y = 1 + t, z = 1 + 2t Assume that distance is in centimeters and that time is in minutes. (a) Find the distance between the bugs at time t = 0.

Planes in 3-Space

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12.6

12.6 PLANES IN 3-SPACE

••••••••••••••••••••••••••••••••••••••

The graph of the equation x = a in an xyz-coordinate system consists of all points of the form (a, y, z), where y and z are arbitrary. One such point is (a, 0, 0), and all others are in the plane that passes through this point and is parallel to the yz-plane (Figure 12.6.1). Similarly, the graph of y = b is the plane through (0, b, 0) that is parallel to the xz-plane, and the graph of z = c is the plane through (0, 0, c) that is parallel to the xy-plane.

Ha

z

sa

n

PLANES PARALLEL TO THE COORDINATE PLANES

Ri

In this section we will use vectors to derive equations of planes in 3-space, and then we will use these equations to solve various geometric problems.

z

z

y=b (0, 0, c)

x=a

y

y

(0, b, 0)

(a, 0, 0)

x

z=c

y

x

ad

x

Figure 12.6.1

••••••••••••••••••••••••••••••••••••••

m

PLANES DETERMINED BY A POINT AND A NORMAL VECTOR

M uh am

n

P

The colored plane is uniquely determined by the point P and the vector n perpendicular to the plane.

Figure 12.6.2

A plane in 3-space can be determined uniquely by specifying a point in the plane and a vector perpendicular to the plane (Figure 12.6.2). A vector perpendicular to a plane is called a normal to the plane. Suppose that we want to find an equation of the plane passing through P0 (x0 , y0 , z0 ) and perpendicular to the vector n = a, b, c. Define the vectors r0 and r as r0 = x0 , y0 , z0

and

r = x, y, z

It should be evident from Figure 12.6.3 that the plane consists precisely of those points P (x, y, z) for which the vector r − r0 is orthogonal to n; or, expressed as an equation, n · (r − r0 ) = 0

(1)

If preferred, we can express this vector equation in terms of components as

a, b, c · x − x0 , y − y0 , z − z0 = 0

(2)

from which we obtain a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0

(3)

This is called the point-normal form of the equation of a plane. Formulas (1) and (2) are vector versions of this formula.

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Three-Dimensional Space; Vectors â€˘ FOR THE READER.

n P(x, y, z)

r â€“ r0

â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘

What does Equation (1) represent if n = a, b, r0 = x0 , y0 , and r = x, y are vectors in an xy-plane in 2-space? Draw a picture.

Example 1 Find an equation of the plane passing through the point (3, âˆ’1, 7) and perpendicular to the vector n = 4, 2, âˆ’5.

P0 (x0 , y0, z 0) r r0

Solution. From (3), a point-normal form of the equation is

O

Yo

834

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4(x âˆ’ 3) + 2(y + 1) âˆ’ 5(z âˆ’ 7) = 0

Figure 12.6.3

(4)

If preferred, this equation can be written in vector form as

4, 2, âˆ’5 Âˇ x âˆ’ 3, y + 1, z âˆ’ 7 = 0

az

Observe that if we multiply out the terms in (3) and simplify, we obtain an equation of the form ax + by + cz + d = 0

(5)

Ri

For example, Equation (4) in Example 1 can be rewritten as 4x + 2y âˆ’ 5z + 25 = 0

n

The following theorem shows that every equation of form (5) represents a plane in 3-space.

sa

12.6.1 THEOREM. If a, b, c, and d are constants, and a, b, and c are not all zero, then the graph of the equation (6)

Ha

ax + by + cz + d = 0

is a plane that has the vector n = a, b, c as a normal.

Proof. Since a, b, and c are not all zero, there is at least one point (x0 , y0 , z0 ) whose

ad

coordinates satisfy Equation (6). For example, if a = 0, then such a point is (âˆ’d /a, 0, 0), and similarly if b = 0 or c = 0 (verify). Thus, let (x0 , y0 , z0 ) be any point whose coordinates satisfy (6); that is, ax0 + by0 + cz0 + d = 0

m

Subtracting this equation from (6) yields a(x âˆ’ x0 ) + b(y âˆ’ y0 ) + c(z âˆ’ z0 ) = 0

M uh am

which is the point-normal form of a plane with normal n = a, b, c. Equation (6) is called the general form of the equation of a plane.

Example 2 Determine whether the planes 3x âˆ’ 4y + 5z = 0

and

âˆ’ 6x + 8y âˆ’ 10z âˆ’ 4 = 0

are parallel.

Solution. It is clear geometrically that two planes are parallel if and only if their normals are parallel vectors. A normal to the first plane is n1 = 3, âˆ’4, 5 and a normal to the second plane is n2 = âˆ’6, 8, âˆ’10 Since n2 is a scalar multiple of n1 , the normals are parallel, and hence so are the planes.

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P

Planes in 3-Space

835

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12.6

We have seen that a unique plane is determined by a point in the plane and a nonzero vector normal to the plane. In contrast, a unique plane is not determined by a point in the plane and a nonzero vector parallel to the plane (Figure 12.6.4). However, a unique plane is determined by a point in the plane and two nonparallel vectors that are parallel to the plane (Figure 12.6.5). A unique plane is also determined by three noncollinear points that lie in the plane (Figure 12.6.6).

Yo

v

Example 3 Find an equation of the plane through the points P1 (1, 2, −1), P2 (2, 3, 1), and P3 (3, −1, 2).

There are infinitely many planes containing P and parallel to v.

−−→ −−→ and P1P3 = 2, −3, 3 are parallel to the plane. Therefore,

i j k

−−→ −−→

1 2

= 9i + j − 5k P1P2 × P1P3 = 1

2 −3 3

−−→ −−→ is normal to the plane, since it is orthogonal to both P1P2 and P1P3 . By using this normal and the point P1 (1, 2, −1) in the plane, we obtain the point-normal form

Solution. Since the points P1 , P2 , and P3 lie in the plane, the vectors P1P2 = 1, 1, 2

v

Ri

w

az

Figure 12.6.4

P There is a unique plane through P that is parallel to both v and w.

9(x − 1) + (y − 2) − 5(z + 1) = 0 which can be rewritten as

n

Figure 12.6.5

sa

9x + y − 5z − 16 = 0

Example 4 Determine whether the line P3

x = 3 + 8t,

P2

y = 4 + 5t,

z = −3 − t

Ha

is parallel to the plane x − 3y + 5z = 12.

P1

Solution. The vector v = 8, 5, −1 is parallel to the line and the vector n = 1, −3, 5

There is a unique plane through three noncollinear points.

is normal to the plane. For the line and plane to be parallel, the vectors v and n must be orthogonal. But this is not so, since the dot product v · n = (8)(1) + (5)(−3) + (−1)(5) = −12

ad

Figure 12.6.6

is nonzero. Thus, the line and plane are not parallel.

M uh am

m

Example 5 Find the intersection of the line and plane in Example 4.

••••••••••••••••••••••••••••••••••••••

ANGLES BETWEEN PLANES

Solution. If we let (x0 , y0 , z0 ) be the point of intersection, then the coordinates of this point satisfy both the equation of the plane and the parametric equations of the line. Thus, x0 − 3y0 + 5z0 = 12

(7)

and for some value of t, say t = t0 , x0 = 3 + 8t0 ,

y0 = 4 + 5t0 ,

z0 = −3 − t0

(8)

Substituting (8) in (7) yields (3 + 8t0 ) − 3(4 + 5t0 ) + 5(−3 − t0 ) = 12 Solving for t0 yields t0 = −3 and on substituting this value in (8), we obtain (x0 , y0 , z0 ) = (−21, −11, 0)

Two distinct intersecting planes determine two positive angles of intersection—an (acute) angle θ that satisfies the condition 0 ≤ θ ≤ π/2 and the supplement of that angle (Figure 12.6.7a). If n1 and n2 are normals to the planes, then depending on the directions of n1

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Three-Dimensional Space; Vectors

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836

and n2 , the angle θ is either the angle between n1 and n2 or the angle between n1 and −n2 (Figure 12.6.7b). In both cases, Theorem 12.3.3 yields the following formula for the acute angle θ between the planes: cos θ =

180° – u

(a)

(9)

Example 6 Find the acute angle of intersection between the two planes 2x − 4y + 4z = 7

n1

and

6x + 2y − 3z = 2

u

Thus, Formula (9) yields |n1 · n2 | |−8| 4 cos θ = =√ √ =

n1 n2

21 36 49 from which we obtain

4 θ = cos−1 ≈ 79 ◦ 21

Ri

Plane 1

az

Solution. The given equations yield the normals n1 = 2, −4, 4 and n2 = 6, 2, −3.

u

Plane 2

(b) Figure 12.6.7 ••••••••••••••••••••••••••••••••••••••

DISTANCE PROBLEMS INVOLVING PLANES

Next we will consider three basic “distance problems” in 3-space:

n

n2

|n1 · n2 |

n1

n2

Yo

u

•

P0

•

Find the distance between two skew lines.

(a)

The three problems are related. If we can find the distance between a point and a plane, then we can find the distance between parallel planes by computing the distance between one of the planes and an arbitrary point P0 in the other plane (Figure 12.6.8a). Moreover, we can find the distance between two skew lines by computing the distance between parallel planes containing them (Figure 12.6.8b).

ad

Ha

sa

•

Find the distance between a point and a plane. Find the distance between two parallel planes.

M uh am

m

12.6.2 THEOREM. The distance D between a point P0 (x0 , y0 , z0 ) and the plane ax + by + cz + d = 0 is D=

|ax0 + by0 + cz0 + d| √ a 2 + b2 + c2

(10)

(b)

Figure 12.6.8

n

projn QP0

D

P0 (x0, y0, z 0)

D

Q(x1, y1, z 1)

Figure 12.6.9

Proof. Let Q(x1 , y1 , z1 ) be any point in the plane, and position the normal n = a, b, c so that its initial point is at Q. As illustrated in Figure 12.6.9, the distance D is equal to the −−→ length of the orthogonal projection of QP0 on n. Thus, from (12) of Section 12.3, −−→ −→ −−→ QP · n |− QP0 · n| |QP0 · n| −−→ 0 D = projn QP0 = n =

n = n 2

n 2

n

But −−→ QP0 = x0 − x1 , y0 − y1 , z0 − z1 −−→ QP0 · n = a(x0 − x1 ) + b(y0 − y1 ) + c(z0 − z1 ) √

n = a 2 + b2 + c2

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Planes in 3-Space

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12.6

Thus,

|a(x0 − x1 ) + b(y0 − y1 ) + c(z0 − z1 )| (11) √ a 2 + b2 + c2 Since the point Q(x1 , y1 , z1 ) lies in the plane, its coordinates satisfy the equation of the plane; that is, ax1 + by1 + cz1 + d = 0 or d = −ax1 − by1 − cz1 Combining this expression with (11) yields (10).

Yo

D=

az

Example 7 Find the distance D between the point (1, −4, −3) and the plane 2x − 3y + 6z = −1

Ri

Solution. Formula (10) requires the plane to be rewritten in the form ax +by +cz+d = 0. Thus, we rewrite the equation of the given plane as 2x − 3y + 6z + 1 = 0

|(2)(1) + (−3)(−4) + 6(−3) + 1| 3 |−3| = = 2 2 2 7 7 2 + (−3) + 6

• REMARK.

See Exercise 48 for an analog of Formula (10) in 2-space that can be used to compute the distance between a point and a line.

Ha

• • • • • • •

sa

D=

n

from which we obtain a = 2, b = −3, c = 6, and d = 1. Substituting these values and the coordinates of the given point in (10), we obtain

Example 8 The planes x + 2y − 2z = 3

and

2x + 4y − 4z = 7

ad

are parallel since their normals, 1, 2, −2 and 2, 4, −4, are parallel vectors. Find the distance between these planes.

M uh am

m

Solution. To find the distance D between the planes, we can select an arbitrary point in

P2

Q2(2, 4, 5)

L2

D

P1

L1

Q1(1, 5, –1)

Figure 12.6.10

one of the planes and compute its distance to the other plane. By setting y = z = 0 in the equation x + 2y − 2z = 3, we obtain the point P0 (3, 0, 0) in this plane. From (10), the distance from P0 to the plane 2x + 4y − 4z = 7 is D=

|(2)(3) + 4(0) + (−4)(0) − 7| 1 = 2 2 2 6 2 + 4 + (−4)

Example 9 It was shown in Example 3 of Section 12.5 that the lines L1 : x = 1 + 4t,

y = 5 − 4t,

z = −1 + 5t

L2 : x = 2 + 8t,

y = 4 − 3t,

z=5+t

are skew. Find the distance between them.

Solution. Let P1 and P2 denote parallel planes containing L1 and L2 , respectively (Figure 12.6.10). To find the distance D between L1 and L2 , we will calculate the distance from a point in P1 to the plane P2 . Since L1 lies in plane P1 , we can find a point in P1 by finding a point on the line L1 ; we can do this by substituting any convenient value of t in the parametric equations of L1 . The simplest choice is t = 0, which yields the point Q1 (1, 5, −1).

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Three-Dimensional Space; Vectors

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838

Yo

The next step is to find an equation for the plane P2 . For this purpose, observe that the vector u1 = 4, −4, 5 is parallel to line L1 , and therefore also parallel to planes P1 and P2 . Similarly, u2 = 8, −3, 1 is parallel to L2 and hence parallel to P1 and P2 . Therefore, the cross product

i j k

5

= 11i + 36j + 20k n = u1 × u2 =

4 −4

8 −3 1

is normal to both P1 and P2 . Using this normal and the point Q2 (2, 4, 5) found by setting t = 0 in the equations of L2 , we obtain an equation for P2 : 11(x − 2) + 36(y − 4) + 20(z − 5) = 0

az

or 11x + 36y + 20z − 266 = 0

The distance between Q1 (1, 5, −1) and this plane is

Ri

|(11)(1) + (36)(5) + (20)(−1) − 266| 95 =√ √ 1817 112 + 362 + 202 which is also the distance between L1 and L2 .

sa

n

D=

EXERCISE SET 12.6

Ha

1. Find equations of the planes P1 , P2 , and P3 that are parallel to the coordinate planes and pass through the corner (3, 4, 5) of the box shown in the accompanying figure.

z

z

7.

1

z

P1

(3, 4, 5) P2

P3

m

P1

P2

M uh am

y

x

Figure Ex-1

1

y

1

(x0, y0, z 0)

1

y

z

10. 1

y

1

x

Figure Ex-2

1 x

y

1 x

In Exercises 11 and 12, find an equation of the plane that passes through the given points.

3. P (2, 6, 1); n = 1, 4, 2

11. (−2, 1, 1), (0, 2, 3), and (1, 0, −1)

4. P (−1, −1, 2); n = −1, 7, 6

12. (3, 2, 1), (2, 1, −1), and (−1, 3, 2)

6. P (0, 0, 0); n = 2, −3, −4

y

x z

1

In Exercises 3–6, find an equation of the plane that passes through the point P and has the vector n as a normal.

5. P (1, 0, 0); n = 0, 0, 1

1 1

x

9.

P3

z

8.

1

ad

2. Find equations of the planes P1 , P2 , and P3 that are parallel to the coordinate planes and pass through the corner (x0 , y0 , z0 ) of the box shown in the accompanying figure.

In Exercises 7–10, find an equation of the plane indicated in the figure.

In Exercises 13 and 14, determine whether the planes are parallel, perpendicular, or neither.

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13. (a) 2x − 8y − 6z − 2 = 0 (b) 3x − 2y + z = 1 −x + 4y + 3z − 5 = 0 4x + 5y − 2z = 4

25. The plane through (1, 2, −1) that is perpendicular to the line of intersection of the planes 2x + y + z = 2 and x + 2y + z = 3.

(b) y = 4x − 2z + 3 x = 14 y + 12 z

Yo

26. The plane through the points P1 (−2, 1, 4), P2 (1, 0, 3) that is perpendicular to the plane 4x − y + 3z = 2.

(c) x + 4y + 7z = 3 5x − 3y + z = 0

27. The plane through (−1, 2, −5) that is perpendicular to the planes 2x − y + z = 1 and x + y − 2z = 3. 28. The plane that contains the point (2, 0, 3) and the line x = −1 + t, y = t, z = −4 + 2t.

In Exercises 15 and 16, determine whether the line and plane are parallel, perpendicular, or neither.

az

29. The plane whose points are equidistant from (2, −1, 1) and (3, 1, 5).

15. (a) x = 4 + 2t, y = −t, z = −1 − 4t; 3x + 2y + z − 7 = 0 (b) x = t, y = 2t, z = 3t; x − y + 2z = 5 (c) x = −1 + 2t, y = 4 + t, z = 1 − t; 4x + 2y − 2z = 7

Ri

30. The plane that contains the line x = 3t, y = 1 + t, z = 2t and is parallel to the intersection of the planes 2x −y +z = 0 and y + z + 1 = 0. 31. Find parametric equations of the line through the point (5, 0, −2) that is parallel to the planes x − 4y + 2z = 0 and 2x + 3y − z +1 = 0.

n

ad

Ha

In Exercises 17 and 18, determine whether the line and plane intersect; if so, find the coordinates of the intersection. 17. (a) x = t, y = t, z = t; 3x − 2y + z − 5 = 0 (b) x = 2 − t, y = 3 + t, z = t; 2x + y + z = 1

32. Do the points (1, 0, −1), (0, 2, 3), (−2, 1, 1), and (4, 2, 3) lie in the same plane? Justify your answer two different ways.

sa

16. (a) x = 3 − t, y = 2 + t, z = 1 − 3t; 2x + 2y − 5 = 0 (b) x = 1 − 2t, y = t, z = −t; 6x − 3y + 3z = 1 (c) x = t, y = 1 − t, z = 2 + t; x+y+z=1

m

18. (a) x = 3t, y = 5t, z = −t; 2x − y + z + 1 = 0 (b) x = 1 + t, y = −1 + 3t, z = 2 + 4t; x − y + 4z = 7

In Exercises 19 and 20, find the acute angle of intersection of the planes to the nearest degree.

M uh am

839

24. The plane through (−1, 4, −3) that is perpendicular to the line x − 2 = t, y + 3 = 2t, z = −t.

(c) x − y + 3z − 2 = 0 2x + z = 1 14. (a) 3x − 2y + z = 4 6x − 4y + 3z = 7

Planes in 3-Space

us uf i

12.6

19. x = 0 and 2x − y + z − 4 = 0

20. x + 2y − 2z = 5 and 6x − 3y + 2z = 8 In Exercises 21–30, find an equation of the plane that satisfies the stated conditions.

21. The plane through the origin that is parallel to the plane 4x − 2y + 7z + 12 = 0.

22. The plane that contains the line x = −2 + 3t, y = 4 + 2t, z = 3 − t and is perpendicular to the plane x − 2y + z = 5. 23. The plane through the point (−1, 4, 2) that contains the line of intersection of the planes 4x − y + z − 2 = 0 and 2x + y − 2z − 3 = 0.

33. Show that the line x = 0, y = t, z = t (a) lies in the plane 6x + 4y − 4z = 0 (b) is parallel to and below the plane 5x − 3y + 3z = 1 (c) is parallel to and above the plane 6x + 2y − 2z = 3.

34. Show that if a, b, and c are nonzero, then the plane whose intercepts with the coordinate axes are x = a, y = b, and z = c is given by the equation x y z + + =1 a b c 35. Show that the lines x = −2 + t, y = 3 + 2t, z = 4 − t x = 3 − t, y = 4 − 2t, z = t are parallel and find an equation of the plane they determine. 36. Show that the lines L1 : x + 1 = 4t, L2 : x + 13 = 12t,

y − 3 = t, y − 1 = 6t,

z−1=0 z − 2 = 3t

intersect and find an equation of the plane they determine. In Exercises 37 and 38, find parametric equations of the line of intersection of the planes. 37. −2x + 3y + 7z + 2 = 0 x + 2y − 3z + 5 = 0 38. 3x − 5y + 2z = 0 z=0 In Exercises 39 and 40, find the distance between the point and the plane. 39. (1, −2, 3); 2x − 2y + z = 4

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840

Three-Dimensional Space; Vectors

40. (0, 1, 5); 3x + 6y − 2z − 5 = 0

can be expressed as n · (r − r0 ) = 0

In Exercises 41 and 42, find the distance between the given parallel planes.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

where r = x, y and r0 = x0 , y0 . (b) Show that the vector equation in part (a) can be expressed as

41. −2x + y + z = 0 6x − 3y − 3z − 5 = 0

Yo

a(x − x0 ) + b(y − y0 ) = 0

42. x + y + z = 1 x + y + z = −1

This is called the point-normal form of a line. (c) Using the proof of Theorem 12.6.1 as a guide, show that if a and b are not both zero, then the graph of the equation

In Exercises 43 and 44, find the distance between the given skew lines.

ax + by + c = 0

az

43. x = 1 + 7t, y = 3 + t, z = 5 − 3t x = 4 − t, y = 6, z = 7 + 2t

is a line that has n = a, b as a normal. (d) Using the proof of Theorem 12.6.2 as a guide, show that the distance D between a point P (x0 , y0 ) and the line ax + by + c = 0 is |ax0 + by0 + c| D= √ a 2 + b2

Ri

44. x = 3 − t, y = 4 + 4t, z = 1 + 2t x = t, y = 3, z = 2t

46. Locate the point of intersection of the plane 2x + y − z = 0 and the line through (3, 1, 0) that is perpendicular to the plane.

50. (a) Show that the distance D between parallel planes

sa

47. Show that the line x = −1 + t, y = 3 + 2t, z = −t and the plane 2x − 2y − 2z + 3 = 0 are parallel, and find the distance between them.

49. Use the formula in part (d) of Exercise 48 to find the distance between the point P (−3, 5) and the line y = −2x + 1.

n

45. Find an equation of the sphere with center (2, 1, −3) that is tangent to the plane x − 3y + 2z = 4.

ax + by + cz + d2 = 0 is

|d1 − d2 | D= √ a 2 + b2 + c2 (b) Use the formula in part (a) to solve Exercise 41.

ad

Ha

48. Formulas (1), (2), (3), (5), and (10), which apply to planes in 3-space, have analogs for lines in 2-space. (a) Draw an analog of Figure 12.6.3 in 2-space to illustrate that the equation of the line that passes through the point P (x0 , y0 ) and is perpendicular to the vector n = a, b

ax + by + cz + d1 = 0

m

12.7 QUADRIC SURFACES

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In this section we will study an important class of surfaces that are the threedimensional analogs of the conic sections.

••••••••••••••••••••••••••••••••••••••

TRACES OF SURFACES z

x

y

Although the general shape of a curve in 2-space can be obtained by plotting points, this method is not usually helpful for surfaces in 3-space because too many points are required. It is more common to build up the shape of a surface with a network of mesh lines, which are curves obtained by cutting the surface with well-chosen planes. For example, Figure 12.7.1, which was generated by a CAS, shows the graph of z = x 3 − 3xy 2 rendered with a combination of mesh lines and colorization to produce the surface detail. This surface is called a “monkey saddle” because a monkey sitting astride the surface has a place for its two legs and tail. The mesh line that results when a surface is cut by a plane is called the trace of the surface in the plane (Figure 12.7.2). Usually, surfaces are built up from traces in planes that are parallel to the coordinate planes, so we will begin by showing how the equations of such traces can be obtained. For this purpose, we will consider the surface z = x2 + y2

Figure 12.7.1

shown in Figure 12.7.3a.

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(1)

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z

841

z

z=1 o f s urface

x

y

(a)

x

y

(b)

Figure 12.7.3

az

Figure 12.7.2

Yo

Trace

Quadric Surfaces

us uf i

12.7

x2 + y2 = 1

(z = 1)

Ri

The basic procedure for finding the equation of a trace is to substitute the equation of the plane into the equation of the surface. For example, to find the trace of the surface z = x 2 + y 2 in the plane z = 1, we substitute z = 1 in (1), which yields (2)

This is a circle of radius 1 centered at the point (0, 0, 1) (Figure 12.7.3b). • REMARK.

n

The parenthetical part of Equation (2) is a reminder that the z-coordinate of all points on the trace is z = 1. This needs to be stated explicitly because z does not appear in the equation x 2 + y 2 = 1.

sa

• • • • • • • • • • • •

Ha

Figure 12.7.4a suggests that the traces of (1) in planes that are parallel to and above the xy-plane form a family of circles that are centered on the z-axis and whose radii increase with z. To confirm this, let us consider the trace in a general plane z = k that is parallel to the xy-plane. The equation of the trace is x2 + y2 = k

(z = k)

M uh am

m

ad

√ If k ≥ 0, then the trace is a circle of radius k centered at the point (0, 0, k). In particular, if k = 0, then the radius is zero, so the trace in the xy-plane is the single point (0, 0, 0). Thus, for nonnegative values of k the traces parallel to the xy-plane form a family of circles, centered on the z-axis, whose radii start at zero and increase with k. This confirms our conjecture. If k < 0, then the equation x 2 + y 2 = k has no graph, which means that there is no trace. Now let us examine the traces of (1) in planes parallel to the yz-plane. Such planes have equations of the form x = k, so we substitute this in (1) to obtain z = k2 + y 2

(x = k)

which we can rewrite as z − k2 = y 2

(x = k)

(3)

For simplicity, let us start with the case where k = 0 (the trace in the yz-plane), in which case the trace has the equation z = y2

(x = 0)

You should be able to recognize that this is a parabola that has its vertex at the origin, opens in the positive z-direction, and is symmetric about the z-axis (Figure 12.7.4b shows a two-dimensional view). You should also be able to recognize that the −k 2 term in (3) has the effect of translating the parabola z = y 2 in the positive z-direction, so the new vertex falls at (k, 0, k 2 ). Thus, the traces parallel to the yz-plane form a family of parabolas whose vertices move upward as k 2 increases. This is consistent with Figure 12.7.4c. Similarly, the traces in planes parallel to the xz-plane have equations of the form z − k2 = x 2

(y = k)

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Three-Dimensional Space; Vectors z

us uf i

842

z

z z

y x

x

y

(a)

Yo

z = y2

y

(b)

(c)

x

y

(d)

az

Figure 12.7.4

••••••••••••••••••••••••••••••••••••••

THE QUADRIC SURFACES

Ri

which again is a family of parabolas whose vertices move upward as k 2 increases (Figure 12.7.4d ). In the discussion of Formula (2) in Section 11.5 we noted that a second-degree equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0

n

represents a conic section (possibly degenerate). The analog of this equation in an xyzcoordinate system is (4)

sa

Ax 2 + By 2 + Cz2 + Dxy + Exz + F yz + Gx + Hy + I z + J = 0

ad

Ha

which is called a second-degree equation in x, y, and z. The graphs of such equations are called quadric surfaces or sometimes quadrics. The six nondegenerate types of quadric surfaces are shown in Table 12.7.1—ellipsoids, hyperboloids of one sheet, hyperboloids of two sheets, elliptic cones, elliptic paraboloids, and hyperbolic paraboloids. (The constants a, b, and c that appear in the equations in the table are assumed to be positive.) Observe that none of the quadric surfaces in the table have cross-product terms in their equations. This is because of their orientations relative to the coordinate axes. Later in this section we will discuss other possible orientations that produce equations of the quadric surfaces with no cross-product terms. In the special case where the elliptic cross sections of an elliptic cone or an elliptic paraboloid are circles, the terms circular cone and circular paraboloid are used.

••••••••••••••••••••••••••••••••••••••

Accurate graphs of quadric surfaces are best left for graphing utilities. However, the techniques that we will now discuss can be used to generate rough sketches of these surfaces that are useful for various purposes. A rough sketch of an ellipsoid

M uh am

m

TECHNIQUES FOR GRAPHING QUADRIC SURFACES

Rough sketch

Figure 12.7.5

x2 y2 z2 + + =1 a2 b2 c2

(a > 0, b > 0, c > 0)

(5)

can be obtained by first plotting the intersections with the coordinate axes, then sketching the elliptical traces in the coordinate planes, and then sketching the surface itself using the traces as a guide. Example 1 illustrates this technique. Example 1 Sketch the ellipsoid x2 y2 z2 + + =1 4 16 9

(6)

Solution. The x-intercepts can be obtained by setting y = 0 and z = 0 in (6). This yields

x = ±2. Similarly, the y-intercepts are y = ±4, and the z-intercepts are z = ±3. From these intercepts we obtain the elliptical traces and the ellipsoid sketched in Figure 12.7.5.

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Table 12.7.1 surface

equation

surface

ellipsoid

z

a2

y x

+

y2 b2

+

z2 c2

z2 =

=1

The traces in the coordinate planes are ellipses, as are the traces in those planes that are parallel to the coordinate planes and intersect the surface in more than one point.

y

x2 y2 + 2 2 a b

The trace in the xy-plane is a point (the origin), and the traces in planes parallel to the xy-plane are ellipses. The traces in the yzand xz-planes are pairs of lines intersecting at the origin. The traces in planes parallel to these are hyperbolas.

Ri

az

x

hyperboloid of one sheet

Yo

x2

843

equation

elliptic cone

z

Quadric Surfaces

us uf i

12.7

elliptic paraboloid

z

a2

+

y2 b2

–

z2 c2

=1

Ha

y

x

hyperboloid of two sheets

m M uh am

The trace in the xy-plane is a point (the origin), and the traces in planes parallel to and above the xy-plane are ellipses. The traces in the yz- and xz-planes are parabolas, as are the traces in planes parallel to these.

x

hyperbolic paraboloid z

z2 y2 x2 – 2 – 2 =1 c2 a b

y

x2 y2 + 2 2 a b

y

ad

z

z=

sa

The trace in the xy-plane is an ellipse, as are the traces in planes parallel to the xy-plane. The traces in the yz-plane and xz-plane are hyperbolas, as are the traces in those planes that are parallel to these and do not pass through the x- or y-intercepts. At these intercepts the traces are pairs of intersecting lines.

z

n

x2

There is no trace in the xy-plane. In planes parallel to the xy-plane that intersect the surface in more than one point the traces are ellipses. In the yz- and xz-planes, the traces are hyperbolas, as are the traces in those planes that are parallel to these.

z=

y

x

x

x2 y2 – 2 2 b a

The trace in the xy-plane is a pair of lines intersecting at the origin. The traces in planes parallel to the xy-plane are hyperbolas. The hyperbolas above the xy-plane open in the y-direction, and those below in the x-direction. The traces in the yz- and xz-planes are parabolas, as are the traces in planes parallel to these.

A rough sketch of a hyperboloid of one sheet x2 y2 z2 + 2 − 2 =1 2 b c a

(a > 0, b > 0, c > 0)

(7)

can be obtained by first sketching the elliptical trace in the xy-plane, then the elliptical traces in the planes z = ±c, and then the hyperbolic curves that join the endpoints of the axes of these ellipses. The next example illustrates this technique.

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844

Three-Dimensional Space; Vectors

Example 2 Sketch the graph of the hyperboloid of one sheet x2 + y2 −

z2 =1 4

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

(8)

Solution. The trace in the xy-plane, obtained by setting z = 0 in (8), is (z = 0)

Yo

x2 + y2 = 1

which is a circle of radius 1 centered on the z-axis. The traces in the planes z = 2 and z = −2, obtained by setting z = ±2 in (8), are given by (z = ±2) √ which are circles of radius 2 centered on the z-axis. Joining these circles by the hyperbolic traces in the vertical coordinate planes yields the graph in Figure 12.7.6.

az

x2 + y2 = 2

A rough sketch of the hyperboloid of two sheets z2 x2 y2 − − =1 c2 a2 b2

Ri

Rough sketch

(a > 0, b > 0, c > 0)

(9)

can be obtained by first plotting the intersections with the z-axis, then sketching the elliptical traces in the planes z = ±2c, and then sketching the hyperbolic traces that connect the zaxis intersections and the endpoints of the axes of the ellipses. (It is not essential to use the planes z = ±2c, but these are good choices since they simplify the calculations slightly and have the right spacing for a good sketch.) The next example illustrates this technique.

sa

n

Figure 12.7.6

Example 3 Sketch the graph of the hyperboloid of two sheets y2 =1 4

Ha

z2 − x 2 −

(10)

Solution. The z-intercepts, obtained by setting x = 0 and y = 0 in (10), are z = ±1. The traces in the planes z = 2 and z = −2, obtained by setting z = ±2 in (10), are given by

ad

x2 y2 + =1 (z = ±2) 3 12 Sketching these ellipses and the hyperbolic traces in the vertical coordinate planes yields Figure 12.7.7.

Rough sketch

M uh am

Figure 12.7.7

m

A rough sketch of the elliptic cone

Rough sketch

Figure 12.7.8

z2 =

x2 y2 + a2 b2

(a > 0, b > 0)

(11)

can be obtained by first sketching the elliptical traces in the planes z = ±1 and then sketching the linear traces that connect the endpoints of the axes of the ellipses. The next example illustrates this technique. Example 4 Sketch the graph of the elliptic cone z2 = x 2 +

y2 4

(12)

Solution. The traces of (12) in the planes z = ±1 are given by y2 =1 (z = ±1) 4 Sketching these ellipses and the linear traces in the vertical coordinate planes yields the graph in Figure 12.7.8. x2 +

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• REMARK. • • • • • • •

Quadric Surfaces

845

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12.7

Observe that if a = b in (11), then the traces parallel to the xy-plane are circles, in which case we call the surface a circular cone. A rough sketch of the elliptic paraboloid y2 x2 + 2 2 a b

(a > 0, b > 0)

Yo

z=

(13)

can be obtained by first sketching the elliptical trace in the plane z = 1 and then sketching the parabolic traces in the vertical coordinate planes to connect the origin to the ends of the axes of the ellipse. The next example illustrates this technique.

x2 y2 + 9 4

(14)

Ri

z=

az

Example 5 Sketch the graph of the elliptic paraboloid

Solution. The trace of (14) in the plane z = 1 is

n

x2 y2 + =1 (z = 1) 4 9 Sketching this ellipse and the parabolic traces in the vertical coordinate planes yields the graph in Figure 12.7.9.

Rough sketch

x2 y2 − b2 a2

(a > 0, b > 0)

(15)

Ha

z=

sa

A rough sketch of the hyperbolic paraboloid

Figure 12.7.9

ad

can be obtained by first sketching the two parabolic traces that pass through the origin (one in the plane x = 0 and the other in the plane y = 0). After the parabolic traces are drawn, sketch the hyperbolic traces in the planes z = ±1 and then fill in any missing edges. The next example illustrates this technique. Example 6 Sketch the graph of the hyperbolic paraboloid

M uh am

m

z=

y2 x2 − 4 9

(16)

Solution. Setting x = 0 in (16) yields y2 (x = 0) 4 which is a parabola in the yz-plane with vertex at the origin and opening in the positive z-direction (since z ≥ 0), and setting y = 0 yields z=

x2 (y = 0) 9 which is a parabola in the xz-plane with vertex at the origin and opening in the negative z-direction. The trace in the plane z = 1 is z=−

y2 x2 =1 (z = 1) − 9 4 which is a hyperbola that opens along a line parallel to the y-axis (verify), and the trace in

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Three-Dimensional Space; Vectors

us uf i

846

the plane z = −1 is

Yo

x2 y2 − =1 (z = −1) 9 4 which is a hyperbola that opens along a line parallel to the x-axis. Combining all of the above information leads to the sketch in Figure 12.7.10. • REMARK.

Rough sketch

Figure 12.7.10

The hyperbolic paraboloid in Figure 12.7.10 has an interesting behavior at the origin—the trace in the xz-plane has a relative maximum at (0, 0, 0), and the trace in the yz-plane has a relative minimum at (0, 0, 0). Thus, a bug walking on the surface may view the origin as a highest point if traveling along one path, or may view the origin as a lowest point if traveling along a different path. A point with this property is commonly called a saddle point or a minimax point.

az

• • • • • • • • • • • • • • • • • • • • • • • • • • • •

Ri

Figure 12.7.11 shows two computer-generated views of the hyperbolic paraboloid in Example 6. The first view, which is much like our rough sketch in Figure 12.7.10, has cuts at the top and bottom that are hyperbolic traces parallel to the xy-plane. In the second view the top horizontal cut has been omitted; this helps to emphasize the parabolic traces parallel to the xz-plane.

n

z

sa

z

y

Ha

y

x

x

Figure 12.7.11

In Section 11.4 we saw that a conic in an xy-coordinate system can be translated by substituting x − h for x and y − k for y in its equation. To understand why this works, think of the xy-axes as fixed, and think of the plane as a transparent sheet of plastic on which all graphs are drawn. When the coordinates of points are modified by substituting (x −h, y −k) for (x, y), the geometric effect is to translate the sheet of plastic (and hence all curves) so that the point on the plastic that was initially at (0, 0) is moved to the point (h, k) (see Figure 12.7.12a).

M uh am

m

TRANSLATIONS OF QUADRIC SURFACES

ad

••••••••••••••••••••••••••••••••••••••

y

z

(h, k, l ) (h, k)

y

x

(0, 0) x

(a)

(b)

Figure 12.7.12

For the analog in three dimensions, think of the xyz-axes as fixed, and think of 3-space as a transparent block of plastic in which all surfaces are embedded. When the coordinates of

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Quadric Surfaces

847

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12.7

points are modified by substituting (x − h, y − k, z − !) for (x, y, z), the geometric effect is to translate the block of plastic (and hence all surfaces) so that the point in the plastic block that was initially at (0, 0, 0) is moved to the point (h, k, !) (see Figure 12.7.12b). Example 7 Describe the surface z = (x − 1)2 + (y + 2)2 + 3.

Yo

Solution. The equation can be rewritten as z − 3 = (x − 1)2 + (y + 2)2

This surface is the paraboloid that results by translating the paraboloid z = x2 + y2

Example 8 Describe the surface

Rough sketch

4x 2 + 4y 2 + z2 + 8y − 4z = −4

Ri

Figure 12.7.13

az

in Figure 12.7.3 so that the new “vertex” is at the point (1, −2, 3). A rough sketch of this paraboloid is shown in Figure 12.7.13.

Solution. Completing the squares yields

4x 2 + 4(y + 1)2 + (z − 2)2 = −4 + 4 + 4 or

sa

n

(z − 2)2 =1 4 Thus, the surface is the ellipsoid that results when the ellipsoid x 2 + (y + 1)2 +

z2 =1 4 is translated so that the new “center” is at the point (0, −1, 2). A rough sketch of this ellipsoid is shown in Figure 12.7.14.

Rough sketch

Ha

x2 + y2 +

• FOR THE READER. • • • • • • •

The ellipsoid in Figure 12.7.14 was sketched with its cross section in the yz-plane tangent to the y- and z-axes. Confirm that this is correct.

ad

Figure 12.7.14

••••••••••••••••••••••••••••••••••••••

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REFLECTIONS OF SURFACES IN 3-SPACE

Recall that in an xy-coordinate system a point (x, y) is reflected about the x-axis if y is replaced by −y, and it is reflected about the y-axis if x is replaced by −x. In an xyzcoordinate system, a point (x, y, z) is reflected about the xy-plane if z is replaced by −z, it is reflected about the yz-plane if x is replaced by −x, and it is reflected about the xz-plane if y is replaced by −y (Figure 12.7.15). It follows that replacing a variable by its negative in the equation of a surface causes that surface to be reflected about a coordinate plane. Recall also that in an xy-coordinate system a point (x, y) is reflected about the line y = x if x and y are interchanged. However, in an xyz-coordinate system, interchanging x and y reflects the point (x, y, z) about the plane y = x (Figure 12.7.16). Similarly, interchanging z z

(–x, y, z)

(y, x, z)

(x, y, z)

(x, y, z)

(x, –y, z) y

y

Plane x

Figure 12.7.15

(x, y, –z)

x

y=x

Figure 12.7.16

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Three-Dimensional Space; Vectors

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848

x and z reflects the point about the plane x = z, and interchanging y and z reflects it about the plane y = z. Thus, it follows that interchanging two variables in the equation of a surface reflects that surface about a plane that makes a 45 ◦angle with two of the coordinate planes.

(a) y 2 = x 2 + z2

Yo

Example 9 Describe the surfaces (b) z = −(x 2 + y 2 )

Solution (a). The graph of the equation y 2 = x 2 + z2 results from interchanging y and z in

az

the equation z2 = x 2 + y 2 . Thus, the graph of the equation y 2 = x 2 + z2 can be obtained by reflecting the graph of z2 = x 2 + y 2 about the plane y = z. Since the graph of z2 = x 2 + y 2 is a circular cone opening along the z-axis (see Table 12.7.1), it follows that the graph of y 2 = x 2 + z2 is a circular cone opening along the y-axis (Figure 12.7.17).

Solution (b). The graph of the equation z = −(x 2 + y 2 ) can be written as −z = x 2 + y 2 ,

Ha

z

sa

n

Ri

which can be obtained by replacing z with −z in the equation z = x 2 + y 2 . Since the graph of z = x 2 +y 2 is a circular paraboloid opening in the positive z-direction (see Table 12.7.1), it follows that the graph of z = −(x 2 + y 2 ) is a circular paraboloid opening in the negative z-direction (Figure 12.7.18).

z

y

x

y

x

ad

Figure 12.7.17

The equations of the quadric surfaces in Table 12.7.1 have certain characteristics that make it possible to identify quadric surfaces that are derived from these equations by reflections. These identifying characteristics, which are shown in Table 12.7.2, are based on writing the equation of the quadric surface so that all of the variable terms are on the left side of the equation and there is a 1 or a 0 on the right side. When there is a 1 on the right side the surface is an ellipsoid, hyperboloid of one sheet, or a hyperboloid of two sheets, and when there is a 0 on the right side it is an elliptic cone, an elliptic paraboloid, or a hyperbolic paraboloid. Within the group with a 1 on the right side, ellipsoids have no minus signs, hyperboloids of one sheet have one minus sign, and hyperboloids of two sheets have two minus signs. Within the group with a 0 on the right side, elliptic cones have no linear terms, elliptic paraboloids have one linear term and two quadratic terms with the same sign, and hyperbolic paraboloids have one linear term and two quadratic terms with opposite signs. These characteristics do not change when the surface is reflected about a coordinate plane or planes of the form x = y, x = z, or y = z, thereby making it possible to identify the reflected quadric surface from the form of its equation.

m

••••••••••••••••••••••••••••••••••••••

Figure 12.7.18

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A TECHNIQUE FOR IDENTIFYING QUADRIC SURFACES

Example 10 Identify the surfaces (a) 3x 2 − 4y 2 + 12z2 + 12 = 0

(b) 4x 2 − 4y + z2 = 0

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Table 12.7.2

equation

characteristic

classification

No minus signs

Ellipsoid

x 2 y2 z2 + – =1 a 2 b2 c2

One minus sign

Hyperboloid of one sheet

z 2 x 2 y2 – – =1 c 2 a 2 b2

Two minus signs

Hyperboloid of two sheets

No linear terms

az

x 2 y2 – =0 a 2 b2

849

Yo

x 2 y2 z2 + + =1 a 2 b2 c2

z2 –

Quadric Surfaces

us uf i

12.7

Elliptic cone

x 2 y2 – =0 a 2 b2

One linear term; two quadratic terms with the same sign

z–

y2 x 2 + =0 b2 a 2

One linear term; two quadratic terms with opposite signs

Ri

z–

Elliptic paraboloid

n

Hyperbolic paraboloid

sa

Solution (a). The equation can be rewritten as

Ha

y2 x2 − − z2 = 1 3 4 This equation has a 1 on the right side and two negative terms on the left side, so its graph is a hyperboloid of two sheets.

Solution (b). The equation has one linear term and two quadratic terms with the same

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sign, so its graph is an elliptic paraboloid. EXERCISE SET 12.7

M uh am

m

In Exercises 1 and 2, identify the quadric surface as an ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic cone, elliptic paraboloid, or hyperbolic paraboloid by matching the equation with one of the forms given in Table 12.7.1. State the values of a, b, and c in each case. x2 y2 + 4 9 (c) x 2 + y 2 − z2 = 16 (e) 4z = x 2 + 4y 2

1. (a) z =

2. (a) 6x 2 + 3y 2 + 4z2 = 12 (c) 9x 2 + y 2 − 9z2 = 9 (e) 2z − x 2 − 4y 2 = 0

y2 − x2 25 (d) x 2 + y 2 − z2 = 0 (f ) z2 − x 2 − y 2 = 1 (b) z =

(b) y 2 − x 2 − z = 0 (d) 4x 2 + y 2 − 4z2 = −4 (f ) 12z2 − 3x 2 = 4y 2

3. Find an equation for and sketch the surface that results when the circular paraboloid z = x 2 + y 2 is reflected about the plane (a) z = 0 (b) x = 0 (c) y = 0 (d) y = x (e) x = z (f ) y = z.

4. Find an equation for and sketch the surface that results when the hyperboloid of one sheet x 2 + y 2 − z2 = 1 is reflected about the plane (a) z = 0 (d) y = x

(b) x = 0 (e) x = z

(c) y = 0 (f ) y = z.

5. The given equations represent quadric surfaces whose orientations are different from those in Table 12.7.1. In each part, identify the quadric surface, and give a verbal description of its orientation (e.g., an elliptic cone opening along the z-axis or a hyperbolic paraboloid straddling the y-axis). (a)

z2 y2 x2 − 2 + 2 =1 2 c b a

(c) x =

y2 z2 + 2 2 b c

(e) y =

z2 x2 − c2 a2

(b)

x2 y2 z2 − 2 − 2 =1 2 a b c

y2 z2 + 2 2 b c

2 x z2 (f ) y = − + a2 c2 (d) x 2 =

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6. For each of the surfaces in Exercise 5, find the equation of the surface that results if the given surface is reflected about the xz-plane and that surface is then reflected about the plane z = 0.

28. 4x 2 − y 2 + 4z2 = 16

x2 + y2 31. z = x 2 + y 2 − 1

1 − x2 − y2 32. z = 1 + x 2 + y 2 30. z =

In Exercises 33–36, identify the surface, and make a rough sketch that shows its position and orientation.

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34. 4x 2 − y 2 + 16(z − 2)2 = 100

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35. 9x 2 + y 2 + 4z2 − 18x + 2y + 16z = 10 36. z2 = 4x 2 + y 2 + 8x − 2y + 4z Exercises 37 and 38 are concerned with the ellipsoid 4x 2 + 9y 2 + 18z2 = 72.

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9x 2 − y 2 + 4z2 = 9; x = 2 9x 2 − y 2 + 4z2 = 9; y = 4 x 2 + 4y 2 − 9z2 = 0; y = 1 x 2 + 4y 2 − 9z2 = 0; z = 1 z = x 2 − 4y 2 ; x = 1 z = x 2 − 4y 2 ; z = 4

m

ad

10. (a) (b) (c) (d) (e) (f )

In Exercises 11–22, identify and sketch the quadric surface.

M uh am

x2 y2 − 4 9

29. z =

(b) 4x 2 − y 2 + 4z2 = 4

4x 2 + y 2 + z2 = 4; y = 1 4x 2 + y 2 + z2 = 4; x = 12 9x 2 − y 2 − z2 = 16; x = 2 9x 2 − y 2 − z2 = 16; z = 2 z = 9x 2 + 4y 2 ; y = 2 z = 9x 2 + 4y 2 ; z = 4

12. x 2 + 4y 2 + 9z2 = 36 14. x 2 + y 2 − z2 = 9 16. 9x 2 + 4y 2 − 36z2 = 0

19. z = y 2 − x 2

x2 z2 − =1 4 9 20. 16z = y 2 − x 2

21. 4z = x 2 + 2y 2

22. z − 3x 2 − 3y 2 = 0

17. 9z2 − 4y 2 − 9x 2 = 36

26. x 2 − 3y 2 − 3z2 = 9

33. z = (x + 2)2 + (y − 3)2 − 9

9. (a) (b) (c) (d) (e) (f )

y2 z2 + =1 4 9 y2 z2 x2 + − =1 13. 4 9 16 15. 4z2 = x 2 + 4y 2

25. 2y 2 − x 2 + 2z2 = 8

In Exercises 29–32, sketch the surface.

(b) z = x 2 + 4y 2

In Exercises 9 and 10, the traces of the surfaces in the planes are conic sections. In each part, find an equation of the trace, and state whether it is an ellipse, a parabola, or a hyperbola.

11. x 2 +

24. x − y 2 − 4z2 = 0

n

x2 y2 z2 =1 + + 4 9 25 2 2 2 x y z (c) + − =1 9 16 4 8. (a) y 2 + 9z2 = x y2 (c) z2 = x 2 + 4 7. (a)

23. x 2 − 3y 2 − 3z2 = 0

27. z =

In Exercises 7 and 8, find equations of the traces in the coordinate planes, and sketch the traces in an xyz-coordinate system. [Suggestion: If you have trouble sketching a trace directly in three dimensions, start with a sketch in two dimensions by placing the coordinate plane in the plane of the paper; then transfer that sketch to three dimensions.]

us uf i

Three-Dimensional Space; Vectors

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850

18. y 2 −

In Exercises 23–28, the given equations represent quadric surfaces whose orientations are different from those in Table 12.7.1. Identify and sketch the surface.

37. (a) Find√an equation of the elliptical trace in the plane z = 2. (b) Find the lengths of the major and minor axes of the ellipse in part (a). (c) Find the coordinates of the foci of the ellipse in part (a). (d) Describe the orientation of the focal axis of the ellipse in part (a) relative to the coordinate axes. 38. (a) Find an equation of the elliptical trace in the plane x = 3. (b) Find the lengths of the major and minor axes of the ellipse in part (a). (c) Find the coordinates of the foci of the ellipse in part (a). (d) Describe the orientation of the focal axis of the ellipse in part (a) relative to the coordinate axes. Exercises 39–42 refer to the hyperbolic paraboloid z = y2 − x2. 39. (a) Find an equation of the hyperbolic trace in the plane z = 4. (b) Find the vertices of the hyperbola in part (a). (c) Find the foci of the hyperbola in part (a). (d) Describe the orientation of the focal axis of the hyperbola in part (a) relative to the coordinate axes. 40. (a) Find an equation of the hyperbolic trace in the plane z = −4. (b) Find the vertices of the hyperbola in part (a). (c) Find the foci of the hyperbola in part (a). (d) Describe the orientation of the focal axis of the hyperbola in part (a) relative to the coordinate axes.

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41. (a) Find an equation of the parabolic trace in the plane x = 2. (b) Find the vertex of the parabola in part (a). (c) Find the focus of the parabola in part (a). (d) Describe the orientation of the focal axis of the parabola in part (a) relative to the coordinate axes.

43. The paraboloids z = x 2 + y 2 and z = 4 − x 2 − y 2 44. The hyperbolic paraboloid x 2 = y 2 + z and the ellipsoid x 2 = 4 − 2y 2 − 2z

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47. Find an equation of the surface consisting of all points P (x, y, z) that are equidistant from the point (0, 0, 1) and the plane z = −1. Identify the surface. 48. Find an equation of the surface consisting of all points P (x, y, z) that are twice as far from the plane z = −1 as from the point (0, 0, 1). Identify the surface. 2

2

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m

ad

x y z + 2 + 2 =1 a a a2 of radius a is compressed in the z-direction, then the resulting surface, called an oblate spheroid, has an equation of

••••••••••••••••••••••••••••••••••••••

CYLINDRICAL AND SPHERICAL COORDINATE SYSTEMS

Yo

n

North Pole

y

sa

46. y = 2x (z = 0) about the y-axis

2

50. The Earth’s rotation causes a flattening at the poles, so its shape is often modeled as an oblate spheroid rather than a sphere (see Exercise 49 for terminology). One of the models used by global positioning satellites is the World Geodetic System of 1984 (WGS-84), which treats the Earth as an oblate spheroid whose equatorial radius is 6378.1370 km and whose polar radius (the distance from the Earth’s center to the poles) is 6356.5231 km. Use the WGS-84 model to find an equation for the surface of the Earth relative to the coordinate system shown in the accompanying figure. z

In Exercises 45 and 46, find an equation for the surface generated by revolving the curve about the axis.

49. If a sphere

the form y2 z2 x2 + 2 + 2 =1 2 a a c where c < a. Show that the oblate spheroid has a circular trace of radius a in the xy-plane and an elliptical trace in the xz-plane with major axis of length 2a along the x-axis and minor axis of length 2c along the z-axis.

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In Exercises 43 and 44, sketch the region enclosed between the surfaces and describe their curve of intersection.

45. y = 4x 2 (z = 0) about the y-axis

851

az

42. (a) Find an equation of the parabolic trace in the plane y = 2. (b) Find the vertex of the parabola in part (a). (c) Find the focus of the parabola in part (a). (d) Describe the orientation of the focal axis of the parabola in part (a) relative to the coordinate axes.

Cylindrical and Spherical Coordinates

us uf i

12.8

Equator x

Figure Ex-50

51. Use the method of slicing to show that the volume of the ellipsoid x2 y2 z2 + 2 + 2 =1 2 a b c is 43 πabc.

12.8 CYLINDRICAL AND SPHERICAL COORDINATES In this section we will discuss two new types of coordinate systems in 3-space that are often more useful than rectangular coordinate systems for studying surfaces with symmetries. These new coordinate systems also have important applications in navigation, astronomy, and the study of rotational motion about an axis. Three coordinates are required to establish the location of a point in 3-space. We have already done this using rectangular coordinates. However, Figure 12.8.1 shows two other possibilities: part (a) of the figure shows the rectangular coordinates (x, y, z) of a point P , part (b) shows the cylindrical coordinates (r, θ, z) of P , and part (c) shows the spherical coordinates (ρ, θ, φ) of P . In a rectangular coordinate system the coordinates can be any real numbers, but in cylindrical and spherical coordinate systems there are restrictions on the allowable values of the coordinates (as indicated in Figure 12.8.1).

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z

z

z

P( r, u, z)

P( x, y, z) y

z

f

x x

x

x Cylindrical coordinates

Rectangular coordinates

y

u

Yo

u

P(r, u, f)

r

y

z r

y

(x, y, z)

(r, u, z) (r ≥ 0, 0 ≤ u < 2p)

(a)

(b)

Figure 12.8.1

••••••••••••••••••••••••••••••••••••••

us uf i

Three-Dimensional Space; Vectors

Spherical coordinates

(r, u, f) (r ≥ 0, 0 ≤ u < 2p, 0 ≤ f ≤ p)

(c)

az

852

x = x0 ,

y = y0 ,

Ri

In rectangular coordinates the surfaces represented by equations of the form

CONSTANT SURFACES

z = z0

and

θ = θ0 ,

z = z0

and

sa

r = r0 ,

n

where x0 , y0 , and z0 are constants, are planes parallel to the yz-plane, xz-plane, and xyplane, respectively (Figure 12.8.2a). In cylindrical coordinates the surfaces represented by equations of the form where r0 , θ0 , and z0 are constants, are shown in Figure 12.8.2b: The surface r = r0 is a right circular cylinder of radius r0 centered on the z-axis. At each point (r, θ, z) on this cylinder, r has the value r0 , but θ and z are unrestricted except for our general restriction that 0 ≤ θ < 2π.

•

The surface θ = θ0 is a half-plane attached along the z-axis and making an angle θ0 with the positive x-axis. At each point (r, θ, z) on this surface, θ has the value θ0 , but r and z are unrestricted except for our general restriction that r ≥ 0.

•

The surface z = z0 is a horizontal plane. At each point (r, θ, z) on this plane, z has the value z0 , but r and θ are unrestricted except for the general restrictions.

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Ha

•

m

z

M uh am

z0

z

z

f0

r = r0

r0

f = f0

x = x0

z0

z = z0

z = z0 y

y

r = r0

x0

y = y0

u0

y0

u = u0

x

x

y

r0

u0

x

u = u0

(a)

(b)

(c)

Figure 12.8.2

In spherical coordinates the surfaces represented by equations of the form ρ = ρ0 ,

θ = θ0 ,

and

φ = φ0

where ρ0 , θ0 , and φ0 are constants, are shown in Figure 12.8.2c:

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The surface ρ = ρ0 consists of all points whose distance ρ from the origin is ρ0 . Assuming ρ0 to be nonnegative, this is a sphere of radius ρ0 centered at the origin.

•

As in cylindrical coordinates, the surface θ = θ0 is a half-plane attached along the z-axis, making an angle of θ0 with the positive x-axis.

•

The surface φ = φ0 consists of all points from which a line segment to the origin makes an angle of φ0 with the positive z-axis. Depending on whether 0 < φ0 < π/2 or π/2 < φ0 < π, this will be the nappe of a cone opening up or opening down. (If φ0 = π/2, then the cone is flat, and the surface is the xy-plane.)

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CONVERTING COORDINATES

Just as we needed to convert between rectangular and polar coordinates in 2-space, so we will need to be able to convert between rectangular, cylindrical, and spherical coordinates in 3-space. Table 12.8.1 provides formulas for making these conversions. Table 12.8.1

(r, u, z) → (x, y, z) (x, y, z) → (r, u, z)

x = r cos u, y = r sin u, z = z r = √x 2 + y2, tan u = y/ x, z = z

Spherical to cylindrical Cylindrical to spherical

(r, u, f) → (r, u, z) (r, u, z) → (r, u, f)

r = r sin f, u = u, z = r cos f r = √r 2 + z 2, u = u, tan f = r/ z

Spherical to rectangular Rectangular to spherical

(r, u, f) → (x, y, z) (x, y, z) → (r, u, f)

x = r sin f cos u, y = r sin f sin u, z = r cos f r = √x 2 + y2 + z 2, tan u = y/ x, cos f = z /√x 2 + y2 + z 2

(r, u, 0)

(a)

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z

P

r

sa

x = r cos θ,

(r, u, f) (r, u, z)

z

r

y

(b)

z = ρ cos φ

(1)

y = r sin θ,

z=z

(2)

we can obtain direct conversion formulas from spherical coordinates to rectangular coordinates by substituting (1) in (2). This yields x = ρ sin φ cos θ,

y = ρ sin φ sin θ,

z = ρ cos φ

The other conversion formulas in Table 12.8.1 are left as exercises. Example 1 (a)

Find the rectangular coordinates of the point with cylindrical coordinates (r, θ, z) = (4, π/3, −3)

(b)

Find the rectangular coordinates of the point with spherical coordinates (ρ, θ, φ) = (4, π/3, π/4)

x

Figure 12.8.3

θ = θ,

Moreover, since the cylindrical coordinates (r, θ, z) of P can be converted to rectangular coordinates (x, y, z) by the conversion formulas

f

f

u

r = ρ sin φ,

m

y x

r ≥ 0, r ≥ 0 0 ≤ u < 2p 0≤f≤p

The diagrams in Figure 12.8.3 will help you to understand how the formulas in Table 12.8.1 are derived. For example, part (a) of the figure shows that in converting between rectangular coordinates (x, y, z) and cylindrical coordinates (r, θ, z), we can interpret (r, θ ) as polar coordinates of (x, y). Thus, the polar-to-rectangular and rectangular-to-polar conversion formulas (1) and (2) of Section 11.1 provide the conversion formulas between rectangular and cylindrical coordinates in the table. Part (b) of Figure 12.8.3 suggests that the spherical coordinates (ρ, θ, φ) of a point P can be converted to cylindrical coordinates (r, θ, z) by the conversion formulas

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y

r

u

restrictions

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(x, y, z) (r, u, z)

z

n

Cylindrical to rectangular Rectangular to cylindrical

z

x

formulas

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conversion

P

853

•

az

••••••••••••••••••••••••••••••••••••••

Cylindrical and Spherical Coordinates

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12.8

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(3)

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Three-Dimensional Space; Vectors z

Solution (a). Applying the cylindrical-to-rectangular conversion formulas in Table 12.8.1

y

yields

4

p/3

us uf i

854

âˆš Ď€ Ď€ = 2, y = r sin Î¸ = 4 sin = 2 3, z = âˆ’3 3 3 âˆš Thus, the rectangular coordinates of the point are (x, y, z) = (2, 2 3, âˆ’3) (Figure 12.8.4). x = r cos Î¸ = 4 cos

3 (2, 2âˆš3, â€“3) (4, p/3, â€“3)

Solution (b). Applying the spherical-to-rectangular conversion formulas in Table 12.8.1

Figure 12.8.4

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x

yields

Ď€ Ď€ âˆš cos = 2 3 4 Ď€ Ď€ âˆš y = Ď sin Ď† sin Î¸ = 4 sin sin = 6 4 3 âˆš Ď€ z = Ď cos Ď† = 4 cos = 2 2 4 âˆš âˆš âˆš Thus, the rectangular coordinates of the point are (x, y, z) = ( 2, 6, 2 2) (Figure 12.8.5). x = Ď sin Ď† cos Î¸ = 4 sin

z

az

âˆš6

âˆš2

2âˆš2

p/4 4

Ri

y

p/3

Example 2 Find the spherical coordinates of the point that has rectangular coordinates âˆš (x, y, z) = (4, âˆ’4, 4 6)

x

Figure 12.8.5

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Solution. From the rectangular-to-spherical conversion formulas in Table 12.8.1 we obtain

âˆš âˆš âˆš x 2 + y 2 + z2 = 16 + 16 + 96 = 128 = 8 2 y tan Î¸ = = âˆ’1 x âˆš âˆš 3 4 6 z = âˆš = cos Ď† = 2 8 2 x 2 + y 2 + z2 From the restriction 0 â‰¤ Î¸ < 2Ď€ and the computed value of tan Î¸ , the possibilities for Î¸ are Î¸ = 3Ď€/4 and Î¸ = 7Ď€/4. However, the given point has a negative y-coordinate, so we must have Î¸ = 7Ď€/4. Moreover, from the restriction 0 â‰¤ Ď† â‰¤ Ď€ and the computed value of cos Ď†, the only possibility for Ď† is Ď† = Ď€/6. Thus, the spherical coordinates of the point âˆš are (Ď , Î¸, Ď†) = (8 2, 7Ď€/4, Ď€/6) (Figure 12.8.6).

ad

Ha

sa

Ď =

M uh am

z

4

4

8âˆš2

4âˆš6

p/6

7p/4 y

4

x

Figure 12.8.6

Surfaces of revolution about the z-axis of a rectangular coordinate system usually have simpler equations in cylindrical coordinates than in rectangular coordinates, and the equations of surfaces with symmetry about the origin are usually simpler in spherical coordinates than in rectangular coordinates. For example, consider the upper nappe of the circular cone whose equation in rectangular coordinates is z = x2 + y2

m

EQUATIONS OF SURFACES IN CYLINDRICAL AND SPHERICAL COORDINATES

(Table 12.8.2). The corresponding equation in cylindrical coordinates can be obtained from the cylindrical-to-rectangular conversion formulas in Table 12.8.1. This yields âˆš z = (r cos Î¸ )2 + (r sin Î¸)2 = r 2 = |r| = r so the equation of the cone in cylindrical coordinates is z = r. Going a step further, the equation of the cone in spherical coordinates can be obtained from the spherical-tocylindrical conversion formulas from Table 12.8.1. This yields Ď cos Ď† = Ď sin Ď† which, if Ď = 0, can be rewritten as Ď€ tan Ď† = 1 or Ď† = 4 Geometrically, this tells us that the radial line from the origin to any point on the cone makes an angle of Ď€/4 with the z-axis.

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Cylindrical and Spherical Coordinates

Table 12.8.2

cylinder

sphere

paraboloid

hyperboloid

z

z

z

z

z

x

y

y x

x

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cone

y

855

us uf i

12.8

y

x

y

x

z = √x 2 + y 2

x2 + y2 = 1

x2 + y2 + z2 = 1

z = x2 + y2

x2 + y2 – z2 = 1

cylindrical

z=r

r=1

z2 = 1 – r2

z = r2

z2 = r2 – 1

spherical

f = p/4

r = csc f

r=1

r = cos f csc2 f

r 2 = –sec 2f

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Example 3 coordinates.

az

rectangular

Find equations of the paraboloid z = x 2 + y 2 in cylindrical and spherical

Solution. The rectangular-to-cylindrical conversion formulas in Table 12.8.1 yield

n

z = r2

(4)

sa

which is the equation in cylindrical coordinates. Now applying the spherical-to-cylindrical conversion formulas to (4) yields ρ cos φ = ρ 2 sin2 φ

Ha

which we can rewrite as ρ = cos φ csc2 φ

Alternatively, we could have obtained this equation directly from the equation in rectangular coordinates by applying the spherical-to-rectangular conversion formulas (verify). • FOR THE READER.

••••••••••••••••••••••••••••••••••••••

M uh am

m

SPHERICAL COORDINATES IN NAVIGATION

Confirm that the equations for the cylinder and hyperboloid in cylindrical and spherical coordinates given in Table 12.8.2 are correct.

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• • • • • • •

Spherical coordinates are related to longitude and latitude coordinates used in navigation. To see why this is so, let us construct a right-hand rectangular coordinate system with its origin at the center of the Earth, its positive z-axis passing through the North Pole, and its positive x-axis passing through the prime meridian (Figure 12.8.7). If we assume the Earth to be a sphere of radius ρ = 4000 miles, then each point on the Earth has spherical coordinates of the form (4000, θ, φ), where φ and θ determine the latitude and longitude of the point. It is common to specify longitudes in degrees east or west of the prime meridian z

Prime meridian y

East

New Orleans x Equator West

Figure 12.8.7

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Three-Dimensional Space; Vectors

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856

and latitudes in degrees north or south of the equator. However, the next example shows that it is a simple matter to determine φ and θ from such data.

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Example 4 The city of New Orleans is located at 90 ◦ west longitude and 30 ◦ north latitude. Find its spherical and rectangular coordinates relative to the coordinate axes of Figure 12.8.7. (Assume that distance is in miles.)

Solution. A longitude of 90 ◦ west corresponds to θ = 360 ◦ − 90 ◦ = 270 ◦ or θ = 3π/2

EXERCISE SET 12.8

Graphing Utility

C

n

Ri

az

radians; and a latitude of 30 ◦ north corresponds to φ = 90 ◦ − 30 ◦ = 60 ◦ or φ = π/3 radians. Thus, the spherical coordinates (ρ, θ, φ) of New Orleans are (4000, 3π/2, π/3). To find the rectangular coordinates we apply the spherical-to-rectangular conversion formulas in Table 12.8.1. This yields √ π 3 3π x = 4000 sin cos = 4000 (0) = 0 mi 3 2 2 √ √ 3 3π π (−1) = −2000 3 mi = 4000 y = 4000 sin sin 2 3 2

π 1 z = 4000 cos = 4000 = 2000 mi 3 2

CAS

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8. (a) (1, 2π/3, 3π/4) (c) (8, π/6, π/4)

In Exercises 1 and 2, convert from rectangular to cylindrical coordinates.

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√ 1. (a) (4 3, 4, −4) (c) (0, 2, 0) √ √ 2. (a) ( 2, − 2, 1) (c) (−4, 4, −7)

(b) (−5, 5, 6) √ (d) (4, −4 3, 6)

In Exercises 9 and 10, convert from cylindrical to spherical coordinates.

(b) (0, 1, 1) (d) (2, −2, −2)

√ 9. (a) ( 3, π/6, 3) (c) (2, 3π/4, 0)

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In Exercises 3 and 4, convert from cylindrical to rectangular coordinates.

10. (a) (4, 5π/6, 4) (c) (4, π/2, 3)

(b) (8, 3π/4, −2) (d) (7, π, −9)

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4. (a) (6, 5π/3, 7) (c) (3, π/2, 5)

(b) (1, π/2, 0) (d) (4, π/2, −1)

In Exercises 5 and 6, convert from rectangular to spherical coordinates.

√ 5. (a) (1, 3, −2) √ (c) (0, 3 3, 3) √ 6. (a) (4, 4, 4 6) (c) (2, 0, 0)

√ (b) (1, −1, 2) √ (d) (−5 3, 5, 0) √ (b) (1, − 3, −2) √ √ (d) ( 3, 1, 2 3)

In Exercises 7 and 8, convert from spherical to rectangular coordinates.

7. (a) (5, π/6, π/4) (c) (1, π, 0)

(b) (7, 0, π/2) (d) (2, 3π/2, π/2)

(b) (1, π/4, −1) √ (d) (6, 1, −2 3) (b) (2, 0, −2) (d) (6, π, 2)

In Exercises 11 and 12, convert from spherical to cylindrical coordinates.

m

3. (a) (4, π/6, 3) (c) (5, 0, 4)

(b) (3, 7π/4, 5π/6) (d) (4, π/2, π/3)

C

C

11. (a) (5, π/4, 2π/3) (c) (3, 0, 0)

(b) (1, 7π/6, π) (d) (4, π/6, π/2)

12. (a) (5, π/2, 0) √ (c) ( 2, 3π/4, π)

(b) (6, 0, 3π/4) (d) (5, 2π/3, 5π/6)

13. Use a CAS or a programmable calculating utility to set up the conversion formulas in Table 12.8.1, and then use the CAS or calculating utility to solve the problems in Exercises 1, 3, 5, 7, 9, and 11. 14. Use a CAS or a programmable calculating utility to set up the conversion formulas in Table 12.8.1, and then use the CAS or calculating utility to solve the problems in Exercises 2, 4, 6, 8, 10, and 12.

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ure 12.8.7. Take miles as the unit of distance and assume the Earth to be a sphere of radius 4000 miles.

In Exercises 15–22, an equation is given in cylindrical coordinates. Express the equation in rectangular coordinates and sketch the graph. 16. θ = π/4 19. r = 4 sin θ 22. r 2 cos 2θ = z

48. (a) Show that the curve of intersection of the surfaces z = sin θ and r = a (cylindrical coordinates) is an ellipse. (b) Sketch the surface z = sin θ for 0 ≤ θ ≤ π/2.

17. z = r 2 20. r = 2 sec θ

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15. r = 3 18. z = r cos θ 21. r 2 + z2 = 1

49. The accompanying figure shows a right circular cylinder of radius 10 cm spinning at 3 revolutions per minute about the z-axis. At time t = 0 s, a bug at the point (0, 10, 0) begins walking straight up the face of the cylinder at the rate of 0.5 cm/min. (a) Find the cylindrical coordinates of the bug after 2 min. (b) Find the rectangular coordinates of the bug after 2 min. (c) Find the spherical coordinates of the bug after 2 min.

In Exercises 23–30, an equation is given in spherical coordinates. Express the equation in rectangular coordinates and sketch the graph.

26. ρ = 2 sec φ

27. ρ = 4 cos φ

28. ρ sin φ = 1

29. ρ sin φ = 2 cos θ

30. ρ − 2 sin φ cos θ = 0

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24. θ = π/3

25. φ = π/4

z

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23. ρ = 3

In Exercises 31–42, an equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates.

3 rev/min

35. x 2 + y 2 = 4

36. x 2 + y 2 − 6y = 0

37. x + y + z = 9

38. z2 = x 2 − y 2

39. 2x + 3y + 4z = 1

40. x 2 + y 2 − z2 = 1

41. x 2 = 16 − z2

42. x 2 + y 2 + z2 = 2z

2

2

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2

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33. z = 3x 2 + 3y 2

32. y = 2 34. z = 3x 2 + 3y 2

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31. z = 3

In Exercises 43–46, describe the region in 3-space that satisfies the given inequalities.

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43. r 2 ≤ z ≤ 4

44. 0 ≤ r ≤ 2 sin θ, 0 ≤ z ≤ 3 45. 1 ≤ ρ ≤ 3

0≤ρ≤2

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46. 0 ≤ φ ≤ π/6,

47. St. Petersburg (Leningrad), Russia, is located at 30 ◦ east longitude and 60 ◦ north latitude. Find its spherical and rectangular coordinates relative to the coordinate axes of Fig-

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Supplementary Exercises

y

(0, 10, 0) x

Figure Ex-49

50. Referring to Exercise 49, use a graphing utility to graph the bug’s distance from the origin as a function of time. 51. A ship at sea is at point A that is 60 ◦ west longitude and 40 ◦ north latitude. The ship travels to point B that is 40 ◦ west longitude and 20 ◦ north latitude. Assuming that the Earth is a sphere with radius 6370 kilometers, find the shortest distance the ship can travel in going from A to B, given that the shortest distance between two points on a sphere is along the arc of the great circle joining the points. [Suggestion: Introduce an xyz-coordinate system as in Figure 12.8.7, and consider the angle between the vectors from the center of the Earth to the points A and B. If you are not familiar with the term “great circle,” consult a dictionary.]

SUPPLEMENTARY EXERCISES

1. (a) What is the difference between a vector and a scalar? Give a physical example of each. (b) How can you determine whether or not two vectors are orthogonal? (c) How can you determine whether or not two vectors are parallel? (d) How can you determine whether or not three vectors with a common initial point lie in the same plane in 3-space?

2. (a) Sketch vectors u and v for which u + v and u − v are orthogonal. (b) How can you use vectors to determine whether four points in 3-space lie in the same plane? (c) If forces F1 = i and F2 = j are applied at a point in 2-space, what force would you apply at that point to cancel the combined effect of F1 and F2 ? (d) Write an equation of the sphere with center (1, −2, 2) that passes through the origin.

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Three-Dimensional Space; Vectors

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3. (a) Draw a picture that shows the direction angles Îą, Î˛, and Îł of a vector. (b) What are the components of a unit vector in 2-space that makes an angle of 120 â—Ś with the positive x-axis (two answers)? (c) How can you use vectors to determine whether a triangle with known vertices P1 , P2 , and P3 has an obtuse angle? (d) True or false: The cross product of orthogonal unit vectors is a unit vector. Explain your reasoning.

13. Let A, B, C, and D be four distinct points in 3-space. Explain why the line through A and B must intersect the line through â†’ âˆ’â†’ âˆ’â†’ âˆ’ âˆ’ â†’ âˆ’â†’ C and D if AB Ă— CD = 0 and AC Âˇ (AB Ă— CD) = 0.

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14. Let A, B, and C be three distinct noncollinear points in 3space. Describe the set of all points P that satisfy the vector â†’ âˆ’â†’ âˆ’â†’ âˆ’ equation AP Âˇ (AB Ă— AC) = 0. 15. True or false? Explain your reasoning. (a) If u Âˇ v = 0, then u = 0 or v = 0. (b) If u Ă— v = 0, then u = 0 or v = 0. (c) If u Âˇ v = 0 and u Ă— v = 0, then u = 0 or v = 0.

4. (a) Make a table that shows all possible cross products of the vectors i, j, and k. (b) Give a geometric interpretation of u Ă— v . (c) Give a geometric interpretation of |u Âˇ (v Ă— w)|. (d) Write an equation of the plane that passes through the origin and is perpendicular to the line x = t, y = 2t, z = âˆ’t.

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17. Show that if u and v are unit vectors and Î¸ is the angle between them, then u âˆ’ v = 2 sin 12 Î¸ .

n

18. Consider the points A(1, âˆ’1, 2), B(2, âˆ’3, 0), C(âˆ’1, âˆ’2, 0), D(2, 1, âˆ’1) (a) Find the volume of the parallelepiped that has the vecâˆ’ â†’ âˆ’â†’ âˆ’â†’ tors AB, AC, AD as adjacent edges. (b) Find the distance from D to the plane containing A, B, and C.

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5. (a) List the six basic types of quadric surfaces, and describe their traces in planes parallel to the coordinate planes. (b) Give the coordinates of the points that result when the point (x, y, z) is reflected about the plane y = x, the plane y = z, and the plane x = z. (c) Describe the intersection of the surfaces r = 5 and z = 1 in cylindrical coordinates. (d) Describe the intersection of the surfaces Ď† = Ď€/4 and Î¸ = 0 in spherical coordinates.

16. In each part, use the result in Exercise 39 of Section 12.4 to prove the vector identity. (a) (a Ă— b) Ă— (c Ă— d) = (a Ă— b Âˇ d)c âˆ’ (a Ă— b Âˇ c)d (b) (a Ă— b) Ă— c + (b Ă— c) Ă— a + (c Ă— a) Ă— b = 0

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6. In each part, find an equation of the sphere with center (âˆ’3, 5, âˆ’4) and satisfying the given condition. (a) Tangent to the xy-plane (b) Tangent to the xz-plane (c) Tangent to the yz-plane

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7. (a) Find the area of the triangle with vertices A(1, 0, 1), B(0, 2, 3), and C(2, 1, 0). (b) Use the result in part (a) to find the length of the altitude from vertex C to side AB.

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8. Find the largest and smallest distances between the point P (1, 1, 1) and the sphere x 2 + y 2 + z2 âˆ’ 2y + 6z âˆ’ 6 = 0

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9. Let a = ci + j and b = 4i + 3 j. Find c so that (a) a and b are orthogonal (b) the angle between a and b is Ď€/4 (c) the angle between a and b is Ď€/6 (d) a and b are parallel.

19. (a) Find parametric equations for the intersection of the planes 2x + y âˆ’ z = 3 and x + 2y + z = 3. (b) Find the acute angle between the two planes.

20. A diagonal of a box makes angles of 50 â—Ś and 70 â—Ś with two of its edges. Find to the nearest degree the angle that it makes with the third edge. 21. Find the vector with length 5 and direction angles Îą = 60 â—Ś , Î˛ = 120 â—Ś , Îł = 135 â—Ś . 22. The accompanying figure shows a cube. (a) Find the angle between the vectors d and u to the nearest degree. (b) Make a conjecture about the angle between the vectors d and v, and confirm your conjecture by computing the angle. z

v

10. Given the points P (3, 4), Q(1, 1), and R(5, 2), use vector methods to find the coordinates of the fourth vertex of the âˆ’â†’ âˆ’â†’ parallelogram whose adjacent sides are PQ and QR.

11. Let r0 = x0 , y0 , z0 and r = x, y, z. Describe the set of all points (x, y, z) for which (a) r Âˇ r0 = 0 (b) (r âˆ’ r0 ) Âˇ r0 = 0. 12. What condition must the constants satisfy for the planes a1 x + b1 y + c1 z = d1 to be perpendicular?

and

a2 x + b2 y + c2 z = d2

d y

x

u

Figure Ex-22

23. In each part, identify the surface by completing the squares. (a) x 2 + 4y 2 âˆ’ z2 âˆ’ 6x + 8y + 4z = 0 (b) x 2 + y 2 + z2 + 6x âˆ’ 4y + 12z = 0 (c) x 2 + y 2 âˆ’ z2 âˆ’ 2x + 4y + 5 = 0

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34. Assuming that force is in pounds and distance is in feet, find the work done by a constant force F = 3i − 4 j + k acting on a particle that moves on a straight line from P (5, 7, 0) to Q(6, 6, 6). 35. Assuming that force is in newtons and distance is in meters, find the work done by the resultant of the constant forces F1 = i−3 j+k and F2 = i+2 j+2k acting on a particle that moves on a straight line from P (−1, −2, 3) to Q(0, 2, 0). 36. As shown in the accompanying figure, a force of 250 N is applied to a boat at an angle of 38 ◦ with the positive x-axis. What force F should be applied to the boat to produce a resultant force of 1000 N acting in the positive x-direction? State your answer by giving the magnitude of the force and its angle with the positive x-axis to the nearest degree.

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In Exercises 27 and 28, sketch the solid in 3-space that is described in spherical coordinates by the stated inequalities. 27. (a) 0 ≤ ρ ≤ 2 (b) 0 ≤ φ ≤ π/6 (c) 0 ≤ ρ ≤ 2 and 0 ≤ φ ≤ π/6 28. (a) 0 ≤ ρ ≤ 5, 0 ≤ φ ≤ π/2, and 0 ≤ θ ≤ π/2 (b) 0 ≤ φ ≤ π/3 and 0 ≤ ρ ≤ 2 sec φ (c) 0 ≤ ρ ≤ 2 and π/6 ≤ φ ≤ π/3

33. Sketch the surface whose equation in spherical coordinates is ρ = a(1 − cos φ). [Hint: The surface is shaped like a familiar fruit.]

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26. In each part, express the equation in rectangular coordinates. (a) z = r 2 cos 2θ (b) ρ 2 sin φ cos φ cos θ = 1

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25. In each part, express the equation in cylindrical and spherical coordinates. (a) x 2 + y 2 = z (b) x 2 − y 2 − z2 = 0

32. In each part, use the idea in Exercise 31(a) to derive a formula for the stated surface of revolution. (a) The surface generated by revolving the curve x = f(y) in the xy-plane about the y-axis. (b) The surface generated by revolving the curve y = f(z) in the yz-plane about the z-axis. (c) The surface generated by revolving the curve z = f(x) in the xz-plane about the x-axis.

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24. In Exercise 42 of Section 12.5 we defined the symmetric equations of a line in 3-space. Consider the lines L1 and L2 whose symmetric equations are y + 32 z+1 x−1 = L1 : = 1 2 2 x−4 y−3 z+4 = = L2 : 2 −1 −2 (a) Are L1 and L2 parallel? Perpendicular? (b) Find parametric equations for L1 and L2 . (c) Do L1 and L2 intersect? If so, where?

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Supplementary Exercises

In Exercises 29 and 30, sketch the solid in 3-space that is described in cylindrical coordinates by the stated inequalities.

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y

250 N

29. (a) 1 ≤ r ≤ 2 (b) 2 ≤ z ≤ 3 (c) π/6 ≤ θ ≤ π/3 (d) 1 ≤ r ≤ 2, 2 ≤ z ≤ 3, and π/6 ≤ θ ≤ π/3 30. (a) r 2 + z2 ≤ 4

(b) r ≤ 1

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31. (a) The accompanying figure shows a surface of revolution that is generated by revolving the curve y = f(x) in the xy-plane about the x-axis. Show that the equation of this surface is y 2 + z2 = [f(x)]2 . [Hint: Each point on the curve traces a circle as it revolves about the x-axis.] (b) Find an equation of the surface of revolution that is generated by revolving the curve y = ex in the xy-plane about the x-axis. (c) Show that the ellipsoid 3x 2 + 4y 2 + 4z2 = 16 is a surface of revolution about the x-axis by finding a curve y = f(x) in the xy-plane that generates it.

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38°

(c) r 2 + z2 ≤ 4 and r > 1

F

Figure Ex-36

37. Suppose that a force F with a magnitude of 9 lb is applied to the lever–shaft assembly shown in the accompanying figure. (a) Express the force F in component form. (b) Find the vector moment of F about the origin. z

5 in

z

2 in A

y

F

y

3 in

y = f (x)

x

x

x

1 in

B

Figure Ex-31

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Figure Ex-37

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

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V ECTOR- V ALUED F UNCTIONS

n this chapter we will consider functions whose values are vectors. Such functions provide a unified way of studying parametric curves in 2-space and 3-space and are a basic tool for analyzing the motion of particles along curved paths. We will begin by developing the calculus of vector-valued functionsâ€”we will show how to differentiate and integrate such functions, and we will develop some of the basic properties of these operations. We will then apply these calculus tools to define three fundamental vectors that can be used to describe such basic characteristics of curves as curvature and twisting tendencies. Once this is done, we will develop the concepts of velocity and acceleration for such motion, and we will apply these concepts to explain various physical phenomena. Finally, we will use the calculus of vector-valued functions to develop basic principles of gravitational attraction and to derive Keplerâ€™s laws of planetary motion.

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Vector-Valued Functions

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13.1 INTRODUCTION TO VECTOR-VALUED FUNCTIONS

PARAMETRIC CURVES IN 3-SPACE

(0, 3, 2) (t = 1)

x = f(t),

y = g(t)

(1)

generates a curve in 2-space that is traced in a specific direction as the parameter t increases. We defined this direction to be the orientation of the curve or the direction of increasing parameter, and we called the curve together with its orientation the graph of the equations or the parametric curve represented by the equations. Analogously, if f, g, and h are three well-behaved functions, then the parametric equations

y

(1, 0, 0) x (t = 0)

x = f(t),

Figure 13.1.1 z

y = g(t),

z = h(t)

(2)

n

generate a curve in 3-space that is traced in a specific direction as t increases. As in 2-space, this direction is called the orientation or direction of increasing parameter, and the curve together with its orientation is called the graph of the equations or the parametric curve represented by the equations. If no restrictions are stated explicitly or are implied by the equations, then it will be understood that t varies over the interval (−⬁, +⬁).

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(t = c)

(t = i )

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(t = 6 ) y

Example 1 The parametric equations

O

x = 1 − t,

(t = 0) x

x = a cos t, y = a sin t, z = ct

y = 3t,

z = 2t

represent a line in 3-space that passes through the point (1, 0, 0) and is parallel to the vector −1, 3, 2. Since x, y, and z increase as t increases, the line has the orientation shown in Figure 13.1.1.

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Figure 13.1.2

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z

Recall from Section 1.8 that if f and g are well-behaved functions, then the pair of parametric equations

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••••••••••••••••••••••••••••••••••••••

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In Section 12.5 we discussed parametric equations of lines in 3-space. In this section we will discuss more general parametric curves in 3-space, and we will show how vector notation can be used to express parametric equations in 2-space and 3-space in a more compact form. This will lead us to consider a new kind of function—namely, functions that associate vectors with real numbers. Such functions have many important applications in physics and engineering.

Example 2 Describe the parametric curve represented by the equations y = a sin t,

z = ct

m

x = a cos t,

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where a and c are positive constants.

Computer representation of the twin helix DNA molecule (deoxyribonucleic acid). This structure contains all the inherited instructions necessary for the development of a living organism.

••••••••••••••••••••••••••••••••••••••

PARAMETRIC CURVES GENERATED WITH TECHNOLOGY

Solution. As the parameter t increases, the value of z = ct also increases, so the point (x, y, z) moves upward. However, as t increases, the point (x, y, z) also moves in a path directly over the circle x = a cos t,

y = a sin t

in the xy-plane. The combination of these upward and circular motions produces a corkscrew-shaped curve that wraps around a right circular cylinder of radius a centered on the z-axis (Figure 13.1.2). This curve is called a circular helix. Except in the simplest cases, parametric curves in 3-space can be difficult to visualize and draw without the help of a graphing utility. For example, Figure 13.1.3a shows the graph of the parametric curve called a torus knot that was produced by a CAS. However, even this computer rendering is difficult to visualize because it is unclear whether the points of overlap are intersections or whether one portion of the curve is in front of the other. To resolve this visualization problem, some graphing utilities provide the capability of enclosing the curve within a thin tube, as in Figure 13.1.3b. Such graphs are called tube plots.

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Introduction to Vector-Valued Functions

• FOR THE READER.

If you have a CAS, read the documentation on graphing parametric curves in 3-space, and then use it to generate the line in Example 1 and the helix x = 4 cos t,

y = 4 sin t,

z=t

(0 ≤ t ≤ 3π)

shown in Figure 13.1.4. z

z

–4

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• • • • • • • • • • • • • • • • • • • • • •

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13.1

0

y 4

9

y

x

az

z

(b)

Figure 13.1.3

4

0 x

Figure 13.1.4

Curves in 3-space often arise as intersections of surfaces. For example, Figure 13.1.5a shows a portion of the intersection of the cylinders z = x 3 and y = x 2 . One method for finding parametric equations for the curve of intersection is to choose one of the variables as the parameter and use the two equations to express the remaining two variables in terms of that parameter. In particular, if we choose x = t as the parameter and substitute this into the equations z = x 3 and y = x 2 , we obtain the parametric equations

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PARAMETRIC EQUATIONS FOR INTERSECTIONS OF SURFACES

–4

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(a)

••••••••••••••••••••••••••••••••••••••

0

y

x

x = t,

y = t 2,

z = t3

(3)

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This curve is called a twisted cubic. The portion of the twisted cubic shown in Figure 13.1.5a corresponds to t ≥ 0; a computer-generated graph of the twisted cubic for positive and negative values of t is shown in Figure 13.1.5b. Some other examples and techniques for finding intersections of surfaces are discussed in the exercises. y = x2

y

0

z

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4

z = x3

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••••••••••••••••••••••••••••••••••••••

VECTOR-VALUED FUNCTIONS

y

z 0 –2 0 x

–8 x

2 x = t, y = t 2, z = t 3 t≥0

(a)

(b)

Figure 13.1.5

The twisted cubic defined by the equations in (3) is the set of points of the form (t, t 2 , t 3 ) for real values of t. If we view each of these points as a terminal point for a vector r whose initial point is at the origin, r = x, y, z = t, t 2 , t 3 = ti + t 2 j + t 3 k then we obtain r as a function of the parameter t, that is, r = r(t). Since this function produces a vector, we say that r = r(t) defines r as a vector-valued function of a real variable, or more simply, a vector-valued function. The vectors that we will consider in

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Vector-Valued Functions

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this text are either in 2-space or 3-space, so we will say that a vector-valued function is in 2-space or in 3-space according to the kind of vectors that it produces. If r(t) is a vector-valued function in 2-space, then for each allowable value of t, the vector r = r(t) can be represented in terms of components as r = r(t) = x(t), y(t) = x(t)i + y(t)j

r(t) = x(t), y(t), z(t) = x(t)i + y(t)j + z(t)k

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As suggested by this notation, the vector-valued function r(t) defines a pair of real-valued functions, x = x(t) and y = y(t), which we call the component functions or the components of r(t). Similarly, a vector-valued function r(t) in 3-space defines three component functions, x(t), y(t), and z(t), via

r(t) = t, t 2 , t 3 = ti + t 2 j + t 3 k are y(t) = t 2 ,

z(t) = t 3

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x(t) = t,

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For example, the component functions of

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The domain of a vector-valued function r(t) is the set of allowable values of t. If r(t) is defined in terms of component functions and the domain is not specified explicitly, then it will be understood that the domain is the set of all values of t for which every component is defined and yields a real value; we call this the natural domain of r(t). That is, the natural domain of a vector-valued function is the intersection of the natural domains for its component functions. For example, the natural domain for √ √ r(t) = ln |t − 1|, et , t = (ln |t − 1|)i + et j + tk is the set of values of t such that 0 ≤ t < 1 or 1 < t, since

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((−⬁, 1) ∪ (1, +⬁)) ∩ (−⬁, +⬁) ∩ [0, +⬁) = [0, 1) ∪ (1, +⬁) is the intersection of the natural domains of the component functions √ x(t) = ln |t − 1|, y(t) = et , and z(t) = t ••••••••••••••••••••••••••••••••••••••

If r(t) is a vector-valued function in 2-space or 3-space, then we define the graph of r(t) to be the parametric curve described by the component functions for r(t). For example, if

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GRAPHS OF VECTOR-VALUED FUNCTIONS

r(t) = 1 − t, 3t, 2t = (1 − t)i + 3tj + 2tk

(4)

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then the graph of r = r(t) is the graph of the parametric equations x = 1 − t,

y = 3t,

z = 2t

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Thus, the graph of (4) is the line in Figure 13.1.1. Example 3 Describe the graph of the vector-valued function r(t) = cos t, sin t, t = cos ti + sin tj + tk

z

(x, y, z)

r

C

x

As t varies, the tip of the radius vector r traces out the curve C.

Figure 13.1.6

y

Solution. The corresponding parametric equations are x = cos t,

y = sin t,

z=t

Thus, as we saw in Example 2, the graph is a circular helix wrapped around a cylinder of radius 1. Up to now we have considered parametric curves to be paths traced by moving points. However, if a parametric curve is viewed as the graph of a vector-valued function, then we can also imagine the graph to be traced by the tip of a moving vector. For example, if the curve C in 3-space is the graph of r(t) = x(t)i + y(t)j + z(t)k and if we position the vector r = x, y, z with its initial point at the origin, then its terminal point will fall at the point (x, y, z) on the curve C (as shown in Figure 13.1.6). Thus, the

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Introduction to Vector-Valued Functions

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13.1

terminal point of r(t) will trace out the curve C as the parameter t varies. We call r the radius vector or the position vector for C.

(a) r(t) = cos ti + sin t j, 0 ≤ t ≤ 2π (b) r(t) = cos ti + sin t j + 2k, 0 ≤ t ≤ 2π

Yo

Example 4 Sketch the graph and a radius vector of

Solution (a). The corresponding parametric equations are x = cos t,

y = sin t

(0 ≤ t ≤ 2π)

az

so the graph is a circle of radius 1, centered at the origin, and oriented counterclockwise. The graph and a radius vector are shown in Figure 13.1.7.

Solution (b). The corresponding parametric equations are x = cos t,

y = sin t,

z=2

(0 ≤ t ≤ 2π)

Ri

From the third equation, the tip of the radius vector traces a curve in the plane z = 2, and from the first two equations, the curve is a circle of radius 1 centered on the z-axis and traced counterclockwise looking down the z-axis. The graph and a radius vector are shown in Figure 13.1.8. z

2

sa

n

y

x

t

Ha

1

1

x

r = cos t i + sin t j + 2k

r = cos t i + sin t j

M uh am

m

VECTOR FORM OF A LINE SEGMENT

t(r

1

–r ) 0

r

r = r0 + tv In particular, if r0 and r1 are vectors in 2-space or 3-space with their initial points at the origin, then the line that passes through the terminal points of these vectors can be expressed in vector form as or

r = (1 − t)r0 + tr1

(5–6)

as indicated in Figure 13.1.9.

r1

r = (1 – t)r0 + tr1

Recall from Formula (9) of Section 12.5 that if r0 is a vector in 2-space or 3-space with its initial point at the origin, then the line that passes through the terminal point of r0 and is parallel to the vector v can be expressed in vector form as

r = r0 + t (r1 − r0 )

r0

O

Figure 13.1.8

ad

Figure 13.1.7

••••••••••••••••••••••••••••••••••••••

y

• REMARK. • • • • • • • • • • • •

It is common to call either (5) or (6) the two-point vector form of a line and to say, for simplicity, that the line passes through the points r0 and r1 (as opposed to saying that it passes through the terminal points of r0 and r1 ).

Figure 13.1.9

It is understood in (5) and (6) that t varies from −⬁ to +⬁. However, if we restrict t to vary over the interval 0 ≤ t ≤ 1, then r will vary from r0 to r1 . Thus, for example, the equation r = (1 − t)r0 + tr1 (0 ≤ t ≤ 1) (7) represents the line segment in 2-space or 3-space that is traced from r0 to r1 .

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Vector-Valued Functions

EXERCISE SET 13.1

Graphing Utility

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

z

x

Q

In Exercises 5–8, express the parametric equations as a single vector equation of the form r = x(t)i + y(t) j or r = x(t)i + y(t) j + z(t)k.

3

Yo

P

4

4. r(t) = 2e−t , sin−1 t, ln(1 − t); t0 = 0

az

25. r(t) = 2i + tj

6. x = t 2 + 1, y = e−2t

2

x

26. r(t) = 3t − 4, 6t + 2

Ri

28. r(t) = 2 cos t, 5 sin t; 0 ≤ t ≤ 2π 29. r(t) = cosh ti + sinh t j

In Exercises 9–12, find the parametric equations that correspond to the given vector equation.

30. r(t) =

√

t i+(2t +4) j

31. r(t) = 2 cos ti + 2 sin t j + tk

n

32. r(t) = 9 cos ti + 4 sin t j + tk

33. r(t) = ti + t 2 j + 2k

sa

√ 11. r = (2t − 1)i − 3 t j + sin 3tk

Q

27. r(t) = (1 + cos t)i + (3 − sin t) j; 0 ≤ t ≤ 2π

8. x = t sin t, y = ln t, z = cos2 t

10. r = sin2 ti + (1 − cos 2t) j

3 y

In Exercises 25–34, sketch the graph of r(t) and show the direction of increasing t.

7. x = 2t, y = 2 sin 3t, z = 5 cos 3t

9. r = 3t 2 i − 2 j

24.

P 4

1. r(t) = cos ti − 3tj; t0 = π √ 2. r(t) = 3t + 1, t 2 ; t0 = 1 √ 3. r(t) = cos πti − ln t j + t − 2 k; t0 = 3

5. x = 3 cos t, y = t + sin t

y

23.

In Exercises 1–4, find the domain of r(t) and the value of r(t0 ).

34. r(t) = ti + tj + sin tk; 0 ≤ t ≤ 2π

12. r = te−t i − 5t 2 k

Ha

In Exercises 13–18, describe the graph of the equation. 13. r = (2 − 3t)i − 4tj

14. r = 3 sin 2ti + 3 cos 2t j

15. r = 2ti − 3 j + (1 + 3t)k

16. r = 3i+2 cos t j+2 sin tk

17. r = 3 cos ti + 2 sin t j − k

18. r = −2i + tj + (t 2 − 1)k

ad

19. (a) Find the slope of the line in 2-space that is represented by the vector equation r = (1 − 2t)i − (2 − 3t) j. (b) Find the coordinates of the point where the line

m

r = (2 + t)i + (1 − 2t) j + 3tk intersects the xz-plane.

M uh am

20. (a) Find the y-intercept of the line in 2-space that is represented by the vector equation r = (3 + 2t)i + 5tj. (b) Find the coordinates of the point where the line r = ti + (1 + 2t) j − 3tk

intersects the plane 3x − y − z = 2.

In Exercises 21 and 22, sketch the line segment represented by the vector equation.

21. (a) r = (1 − t)i + tj; 0 ≤ t ≤ 1 (b) r = (1 − t)(i + j) + t (i − j); 0 ≤ t ≤ 1

22. (a) r = (1 − t)(i + j) + tk; 0 ≤ t ≤ 1 (b) r = (1 − t)(i + j + k) + t (i + j); 0 ≤ t ≤ 1 In Exercises 23 and 24, write a vector equation for the line segment from P to Q.

In Exercises 35 and 36, sketch the curve of intersection of the surfaces, and find parametric equations for the intersection in terms of parameter x = t. Check your work with a graphing utility by generating the parametric curve over the interval −1 ≤ t ≤ 1.

35. z = x 2 + y 2 , x − y = 0 36. y + x = 0, z = 2 − x 2 − y 2 In Exercises 37 and 38, sketch the curve of intersection of the surfaces, and find a vector equation for the curve in terms of the parameter x = t. 37. 9x 2 + y 2 + 9z2 = 81, y = x 2

(z > 0)

38. y = x, x + y + z = 1 39. Show that the graph of r = t sin ti + t cos t j + t 2 k lies on the paraboloid z = x 2 + y 2 . 40. Show that the graph of 1 − t2 1+t j+ k, t > 0 t t lies in the plane x − y + z + 1 = 0. r = ti +

41. Show that the graph of

√ r = sin ti + 2 cos t j + 3 sin tk is a circle, and find its center and radius. [Hint: Show that the curve lies on both a sphere and a plane.]

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44. How many revolutions will the circular helix r = a cos ti + a sin t j + 0.2tk make in a distance of 10 units measured along the z-axis? 45. Show that the curve r = t cos ti + t sin t j + tk, t â‰Ľ 0, lies on the cone z = x 2 + y 2 . Describe the curve. 46. Describe the curve r = a cos ti + b sin t j + ctk, where a, b, and c are positive constants such that a = b.

49. (a) Find parametric equations for the curve of intersection of the circular cylinder x 2 + y 2 = 9 and the parabolic cylinder z = x 2 in terms of a parameter t for which x = 3 cos t. (b) Use a graphing utility to generate the curve of intersection in part (a). 50. Use a graphing utility to generate the intersection of the cone z = x 2 + y 2 and the plane z = y + 2. Identify the curve and explain your reasoning. 51. (a) Sketch the graph of 2 r(t) = 2t, 1 + t2 (b) Prove that the curve in part (a) is also the graph of the function 8 y= 4 + x2 [The graphs of y = a 3 /(a 2 + x 2 ), where a denotes a constant, were first studied by the French mathematician Pierre de Fermat, and later by the Italian mathematicians Guido Grandi and Maria Agnesi. Any such curve is now known as a â€œwitch of Agnesi.â€? There are a number of theories for the origin of this name. Some suggest there was a mistranslation by either Grandi or Agnesi of some less colorful Latin name into Italian. Others lay the blame on a translation into English of Agnesiâ€™s 1748 treatise, Analytical Institutions.]

z

y x

x

y

II

ad

I

Ha

z

sa

n

47. In each part, match the vector equation with one of the accompanying graphs, âˆš and explain your reasoning. (a) r = ti âˆ’ tj + 2 âˆ’ t 2 k (b) r = sin Ď€ti âˆ’ tj + tk (c) r = sin ti + cos t j + sin 2tk (d) r = 12 ti + cos 3t j + sin 3tk

Yo

43. For the helix r = a cos ti + a sin t j + ctk, find c (c > 0) so that the helix will make one complete turn in a distance of 3 units measured along the z-axis.

48. Check your conclusions in Exercise 47 by generating the curves with a graphing utility. [Note: Your graphing utility may look at the curve from a different viewpoint. Read the documentation for your graphing utility to determine how to control the viewpoint, and see if you can generate a reasonable facsimile of the graphs shown in the figure by adjusting the viewpoint and choosing the interval of t-values appropriately.]

az

r = 3 cos ti + 3 sin t j + 3 sin tk is an ellipse, and find the lengths of the major and minor axes. [Hint: Show that the graph lies on both a circular cylinder and a plane and use the result in Exercise 60 of Section 11.4.]

z

M uh am

m

z

y

x

y

x

III

867

Ri

42. Show that the graph of

Introduction to Vector-Valued Functions

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13.1

IV

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Vector-Valued Functions

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13.2 CALCULUS OF VECTOR-VALUED FUNCTIONS

In this section we will define limits, derivatives, and integrals of vector-valued functions and discuss their properties. LIMITS AND CONTINUITY

Our first goal in this section is to develop a notion of what it means for a vector-valued function r(t) to approach a limiting vector L as t approaches a number a. That is, we want to define (1) lim r(t) = L

Yo

••••••••••••••••••••••••••••••••••••••

t →a

az

In the introduction to Chapter 12 we mentioned that vectors are useful in many physical contexts because they encapsulate both magnitude (or length) and direction. Equation (1) can be interpreted intuitively through this geometric perspective: as t approaches a, the limit of the length of r(t) must match the length of L, and the limit of the direction of r(t) must match the direction of L. L

Ri

y

13.2.1 GEOMETRIC INTERPRETATION OF LIMITS. in 2-space or 3-space, then lim r(t) = L t →a

t→a

Figure 13.2.1

x →a

was defined informally as the assertion that values of f(x) can be made as close as we like to L by taking values of x sufficiently close to a (but not equal to a). This was formalized in Section 2.4 to the assertion that for any given > 0, we can find a number δ > 0 such that |f(x) − L| < if 0 < |x − a| < δ. To adapt the notion of limits of a real-valued function y = f(x) to limits of a vectorvalued function r = r(t), we need to replace the notion of “closeness” of the real numbers f(x) and L by a corresponding notion for the vectors r(t) and L. But how do we measure how close (or how far apart) two vectors r(t) and L are? We can look at the difference between the vectors, r(t) − L (Figure 13.2.2), but this is a vector. What we need is the length of this vector, r(t) − L, which gives the distance between the terminal points of r(t) and L when they are positioned with the same initial point. To say that a vector r(t) is close to the vector L is to say that r(t) − L is small, say less than some positive number . In 2-space, the set of all vectors r satisfying r − L < can be described geometrically as those vectors that, when positioned with the same initial point as L, have terminal points lying within a disk of radius centered at the terminal point for L. In 3-space, this set is those vectors with terminal points lying within a ball of radius centered at the terminal point for L (Figure 13.2.3). We can now transform Definition 2.4.1 into a definition for (1).

ad

r(t) – L

r(t)

m

L

M uh am

x

|| r(t) – L || is the distance between terminal points for vectors r(t) and L when positioned with the same initial points.

Figure 13.2.2

sa

r(t) approaches L in length and direction if lim r(t) = L.

Although saying that r(t) approaches L in both length and direction may be helpful for visualizing a limit, it is difficult to use the statement in 13.2.1 to establish a limit. Instead, let us go back to Chapter 2 and use the definition of the limit of a real-valued function as a guide. Recall from Section 2.1 that the limit lim f(x) = L

Ha

x

n

if and only if the radius vector r = r(t) approaches L in both length and direction as t → a (Figure 13.2.1).

r(t)

y

If r(t) is a vector-valued function

13.2.2 DEFINITION. Let r(t) be a vector-valued function defined for all t in some open interval containing the number a, except that r(t) need not be defined at a. We will write lim r(t) = L t →a

if given any number > 0 we can find a number δ > 0 such that r(t) − L < if 0 < |t − a| < δ

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y

Calculus of Vector-Valued Functions

z

r

r

Yo

L L

869

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13.2

y

x

x

|| r – L || <

az

Figure 13.2.3

Ri

Similarly, we mirror Definition 2.5.1 to define continuity of a vector-valued function. 13.2.3 DEFINITION. A vector-valued function r(t) is continuous at t = c provided the following conditions are satisfied: 1. r(c) is defined. 2. lim r(t) exists.

n

t →c

lim r(t) = r(c).

t →c

sa

3.

Ha

As before, we say that r(t) is continuous on an interval I if it is continuous at each value of t in I (with only the appropriate one-sided limit results required at any endpoints of I that are included in I ). In practice, limits of vector-valued functions are frequently computed using components. For example, if r(t) = x(t), y(t), z(t) = x(t)i + y(t)j + z(t)k then

ad

lim r(t) = lim x(t), lim y(t), lim z(t) t →a

M uh am

m

t →a

••••••••••••••••••••••••••••••••••••••

DERIVATIVES

•

=

t →a

t →a

lim x(t) i + lim y(t) j + lim z(t) k

t →a

t →a

t →a

(2)

provided each of the component limits exists. Furthermore, it follows immediately that r(t) is continuous at t = c if and only if its component functions x(t), y(t), and z(t) are each continuous at t = c. FOR THE READER.

Write the corresponding statement to (2) when r(t) is in 2-space.

Example 1 Let r(t) = t 2 i + et j − (2 cos πt)k. Then lim r(t) = lim t 2 i + lim et j − lim 2 cos πt k = j − 2 k t →0

t →0

t →0

t →0

Alternatively, using the angle bracket notation for vectors, lim r(t) = lim t 2 , et , −2 cos πt = lim t 2 , lim et , lim (−2 cos πt) = 0, 1, −2 t →0

t →0

t →0

t →0

t →0

Following the lead of the discussion above, we consider substituting a vector-valued function for the real-valued function in the definition of the derivative (Definition 3.2.3). Note that the numerator in the resulting difference quotient is now a difference of vectors, which results in a vector, whereas the denominator is a difference of scalars. Thus, the difference quotient is a scalar multiple of a vector, so it is also a vector.

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13.2.4 DEFINITION. If r(t) is a vector-valued function, we define the derivative of r with respect to t to be the vector-valued function r given by r (t) = lim

w→t

r(w) − r(t) w−t

(3)

Yo

The domain of r consists of all values of t in the domain of r(t) for which the limit exists.

ad

Ha

sa

n

Ri

az

The function r(t) is differentiable at t if the limit in (3) exists. All of the standard notations for derivatives continue to apply. For example, the derivative of r(t) can be expressed as d dr [r(t)], , r (t), or r dt dt It is important to remember that for a given value of t the derivative r (t) is a vector, not a number. Since r (t) is a vector, it has both magnitude and [if r (t) is nonzero] direction. Our next goal is to relate the direction of r (t) to the graph of r(t). [We will study the significance of the magnitude of r (t) in the next section.] To do this, consider parts (a) and (b) of Figure 13.2.4. These illustrations show the graph C of r(t) (with its orientation) and the vectors r(w), r(t), and r(w) − r(t) for w > t and for w < t. In both cases, the vector r(w) − r(t) runs along the secant line joining the terminal points of r(t) and r(w), but with opposite directions in the two cases. In the case where w > t, the vector r(w) − r(t) points in the direction of increasing parameter; and in the case where w < t, the vector r(w) − r(t) points in the opposite direction. However, if w < t, the direction is reversed when we multiply by the negative value 1/(w − t), so that in both cases the vector 1 r(w) − r(t) [r(w) − r(t)] = w−t w−t points in the direction of increasing parameter and runs along the secant line. As w → t the secant line approaches the tangent line at the terminal point of r(t), so we can conclude that the limit r(w) − r(t) r (t) = lim w→t w−t (if it exists and is nonzero) is a vector that is tangent to the curve C at the tip of r(t) and points in the direction of increasing parameter (Figure 13.2.4c).

M uh am

m

y

y

r(w) – r(t)

y

r(w) – r(t)

r′(t)

r(t)

r(t)

r(w) r(t) C

r(w)

C

C x

x

w>t

w<t

(a)

(b)

x

(c)

Figure 13.2.4

13.2.5 GEOMETRIC INTERPRETATION OF THE DERIVATIVE. Suppose that C is the graph of a vector-valued function r(t) in 2-space or 3-space and that r (t) exists and is nonzero for a given value of t. If the vector r (t) is positioned with its initial point at the terminal point of the radius vector r(t), then r (t) is tangent to C and points in the direction of increasing parameter.

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Calculus of Vector-Valued Functions

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13.2

Since limits of vector-valued functions can be computed componentwise, it seems reasonable that we should be able to compute derivatives in terms of component functions as well. This is the result of the next theorem. THEOREM.

If r(t) is a vector-valued function, then r(w) − r(t) r (t) = lim w→t w−t exists if and only if each of the component functions for r(t) is differentiable at t, in which case the component functions for r (t) are the derivatives of the component functions for r(t).

Yo

13.2.6

Ri

az

Proof. For simplicity, we give the proof in 2-space; the proof in 3-space is identical, except for the additional component. Assume that r(t) = x(t)i + y(t)j, so r(w) − r(t) [x(w)i + y(w)j] − [x(t)i + y(t)j] r (t) = lim = lim w→t w→t w−t w−t [x(w) − x(t)]i + [y(w) − y(t)]j w−t

y(w) − y(t) x(w) − x(t) = lim i+ j w→t w−t w−t

x(w) − x(t) y(w) − y(t) = lim i + lim j w→t w→t w−t w−t = lim

sa

n

w→t

= x (t)i + y (t)j

Ha

Example 2 Let r(t) = t 2 i + et j − (2 cos πt)k. Then d d d r (t) = (t 2 )i + (et )j − (2 cos πt)k = 2ti + et j + (2π sin πt)k dt dt dt and

ad

r (1) = 2i + ej ••••••••••••••••••••••••••••••••••••••

M uh am

m

DERIVATIVE RULES

Many of the rules for differentiating real-valued functions have analogs in the context of differentiating vector-valued functions. We state some of these in the following theorem. 13.2.7 THEOREM (Rules of Differentiation). Let r(t), r1 (t), and r2 (t) be vector-valued functions that are all in 2-space or all in 3-space, and let f (t) be a real-valued function, k a scalar, and c a constant vector (that is, a vector whose value does not depend on t). Then the following rules of differentiation hold: d (a) [c] = 0 dt d d (b) [kr(t)] = k [r(t)] dt dt d d d (c) [r1 (t) + r2 (t)] = [r1 (t)] + [r2 (t)] dt dt dt d d d (d ) [r1 (t) − r2 (t)] = [r1 (t)] − [r2 (t)] dt dt dt d d d (e) [f (t)r(t)] = f (t) [r(t)] + [f (t)]r(t) dt dt dt

The proofs of most of these rules are immediate consequences of Definition 13.2.4, although the last rule can be seen more easily by application of the product rule for real-

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Vector-Valued Functions

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valued functions to the component functions. The proof of Theorem 13.2.7 is left as an exercise. â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

13.2.8 DEFINITION. Let P be a point on the graph of a vector-valued function r(t), and let r(t0 ) be the radius vector from the origin to P (Figure 13.2.5). If r (t0 ) exists and r (t0 ) = 0, then we call r (t0 ) the tangent vector to the graph of r(t) at r(t0 ), and we call the line through P that is parallel to the tangent vector the tangent line to the graph of r(t) at r(t0 ).

P Tangent line

r(t0)

x

Let r0 = r(t0 ) and v0 = r (t0 ). It follows from Formula (9) of Section 12.5 that the tangent line to the graph of r(t) at r0 is given by the vector equation

az

râ€˛(t0)

r = r0 + tv0

Figure 13.2.5

(4)

Ri

y

Motivated by the discussion of the geometric interpretation of the derivative of a vectorvalued function, we make the following definition.

Yo

TANGENT LINES TO GRAPHS OF VECTOR-VALUED FUNCTIONS

Example 3 Find parametric equations of the tangent line to the circular helix x = cos t,

y = sin t,

z=t

n

where t = t0 , and use that result to find parametric equations for the tangent line at the point where t = Ď€.

sa

Solution. The vector equation of the helix is r(t) = cos ti + sin t j + tk

Ha

so we have r0 = r(t0 ) = cos t0 i + sin t0 j + t0 k v0 = r (t0 ) = (âˆ’ sin t0 )i + cos t0 j + k It follows from (4) that the vector equation of the tangent line at t = t0 is r = cos t0 i + sin t0 j + t0 k + t[(âˆ’ sin t0 )i + cos t0 j + k]

â€“1

ad

= (cos t0 âˆ’ t sin t0 )i + (sin t0 + t cos t0 )j + (t0 + t)k

Thus, the parametric equations of the tangent line at t = t0 are

y

m

0 1

6

M uh am 0

t=0

1

z = t0 + t

x = âˆ’1,

y = âˆ’t,

z = Ď€+t

The graph of the helix and this tangent line are shown in Figure 13.2.6.

Example 4 Let r1 (t) = (tanâˆ’1 t)i + (sin t)j + t 2 k

â€“1

0 x

y = sin t0 + t cos t0 ,

In particular, the tangent line at the point where t = Ď€ has parametric equations

t=p z 3

x = cos t0 âˆ’ t sin t0 ,

and

Figure 13.2.6

r2 (t) = (t 2 âˆ’ t)i + (2t âˆ’ 2)j + (ln t)k The graphs of r1 (t) and r2 (t) intersect at the origin. Find the degree measure of the acute angle between the tangent lines to the graphs of r1 (t) and r2 (t) at the origin.

Solution. The graph of r1 (t) passes through the origin at t = 0, where its tangent vector is r1 (0) =

1

= 1, 1, 0 , cos t, 2t

2 1+t t=0

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Calculus of Vector-Valued Functions

873

us uf i

13.2

The graph of r2 (t) passes through the origin at t = 1 (verify), where its tangent vector is

1

r2 (1) = 2t − 1, 2, = 1, 2, 1 t t=1

Yo

By Theorem 12.3.3, the angle θ between these two tangent vectors satisfies √ 1, 1, 0 · 1, 2, 1 3 3 cos θ = =√ = 1, 1, 0 1, 2, 1 2 12 It follows that θ = π/6 radians, or 30 ◦ . ••••••••••••••••••••••••••••••••••••••

The following rules, which are derived in the exercises, provide a method for differentiating dot products in 2-space and 3-space and cross products in 3-space. dr2 dr1 d · r2 (t) [r1 (t) · r2 (t)] = r1 (t) · + dt dt dt

(5)

dr1 d dr2 [r1 (t) × r2 (t)] = r1 (t) × + × r2 (t) dt dt dt

(6)

• REMARK.

In (5) the order of the factors in each term on the right does not matter, but in

(6) it does.

n

• • • • • • •

Ri

az

DERIVATIVES OF DOT AND CROSS PRODUCTS

Ha

sa

In plane geometry one learns that a tangent line to a circle is perpendicular to the radius at the point of tangency. Consequently, if a point moves along a circle in 2-space that is centered at the origin, then one would expect the radius vector and the tangent vector at any point on the circle to be orthogonal. This is the motivation for the following useful theorem, which is applicable in both 2-space and 3-space. 13.2.9 THEOREM. If r(t) is a vector-valued function in 2-space or 3-space and r(t) is constant for all t, then r(t) · r (t) = 0 that is, r(t) and r (t) are orthogonal vectors for all t.

ad

(7)

M uh am

m

Proof. It follows from (5) with r1 (t) = r2 (t) = r(t) that

z

d dr dr [r(t) · r(t)] = r(t) · + · r(t) dt dt dt or, equivalently, d dr [r(t)2 ] = 2r(t) · dt dt But r(t)2 is constant, so its derivative is zero. Thus dr 2r(t) · =0 dt from which (7) follows.

(8)

r'(t)

r(t)

x

Figure 13.2.7

y

Example 5 Just as a tangent line to a circle in 2-space is perpendicular to the radius at the point of tangency, so a tangent vector to a curve on the surface of a sphere in 3-space that is centered at the origin is orthogonal to the radius vector at the point of tangency (Figure 13.2.7). To see that this is so, suppose that the graph of r(t) lies on the surface of a sphere of positive radius k centered at the origin. For each value of t we have r(t) = k, so by Theorem 13.2.9 r(t) · r (t) = 0 and hence the radius vector r(t) and the tangent vector r (t) are orthogonal.

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INTEGRALS OF VECTOR-VALUED FUNCTIONS

If r(t) = x(t), y(t) = x(t)i + y(t)j is a vector-valued function in 2-space, we can define the definite integral of r(t) from t = a to t = b via a Riemann sum, as was done for realvalued functions in Definition 5.5.1. It follows immediately that a definite integral of r(t) can be expressed as a vector whose components are the definite integrals of the component functions for r(t). b n r(t) dt = lim r(tkâˆ— )tk max tk â†’ 0

a

k=1

lim

max tk â†’ 0

k=1

=

n

lim

max tk â†’ 0

b

=

x(tkâˆ— )tk

i+

x(tkâˆ— )tk

k=1

x(t) dt i +

b

n k=1

i+

y(tkâˆ— )tk n

lim

max tk â†’ 0

y(t) dt j

a

y(tkâˆ— )tk

j

k=1

a

Alternatively, b x(t), y(t) dt = a

b

b

x(t) dt,

a

y(t) dt

a

sa

n

For vector-valued functions in 3-space this becomes b b b x(t), y(t), z(t) dt = x(t) dt, y(t) dt, a

j

Ri

=

n

Yo

us uf i

Vector-Valued Functions

az

874

a

Ha

=

b

a

z(t) dt

a

x(t) dt i +

a

b

b

y(t) dt j +

a

b

z(t) dt k

a

Example 6 Let r(t) = t 2 i + et j âˆ’ (2 cos Ď€t)k. Then 1

1

1

1 2 t r(t) dt = t dt i + e dt j âˆ’ 2 cos Ď€t dt k

ad

0

0

=

0

3 1

t 3

0

1 i + et

jâˆ’ 0

2 sin Ď€t Ď€

0

1 k= 0

1 i + (e âˆ’ 1)j 3

M uh am

m

An antiderivative for a vector-valued function r(t) is a vector-valued function R(t) such that R (t) = r(t)

(9)

As in Chapter 5, we recast Equation (9) using integral notation as r(t) dt = R(t) + C

(10)

where C is understood to represent an arbitrary constant vector. Note that since differentiation of vector-valued functions can be done componentwise, antidifferentiation can also be done componentwise. This is illustrated in the next example. Example 7

(2ti + 3t 2 j) dt = 2t dt i + 3t 2 dt j = (t 2 + C1 )i + (t 3 + C2 )j = (t 2 i + t 3 j) + (C1 i + C2 j) = (t 2 i + t 3 j) + C where C = C1 i + C2 j is an arbitrary vector constant of integration.

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Calculus of Vector-Valued Functions

875

us uf i

13.2

and

Yo

Most of the familiar integration properties have vector counterparts. For example, vector differentiation and integration are inverse operations in the sense that d r(t) dt = r(t) (11) dt

r (t) dt = r(t) + C

(12)

If R(t) is an antiderivative of r(t) on an interval containing a and b, then b

b

r(t) dt = R(t)

= R(b) − R(a)

a

az

a

2

(2ti + 3t 2 j) dt.

Ri

Example 8 Evaluate the definite integral

(13)

0

Solution. Integrating the components yields

2

2

(2ti + 3t 2 j) dt = t 2

0

2

i + t3 0

j = 4i + 8j

n

0

Alternative Solution. The function R(t) = t 2 i + t 3 j is an antiderivative of the integrand

sa

since R (t) = 2ti + 3t 2 j. Thus, it follows from (13) that 2 2 2 (2ti + 3t 2 j) dt = R(t) = t 2 i + t 3 j = (4i + 8j) − (0i + 0j) = 4i + 8j 0

Ha

0

0

Example 9 Find r(t) given that r (t) = 3, 2t and r(1) = 2, 5.

Solution. Integrating r (t) to obtain r(t) yields

r(t) =

r (t) dt =

3, 2t dt = 3t, t 2 + C

m

ad

where C is a vector constant of integration. To find the value of C we substitute t = 1 and use the given value of r(1) to obtain r(1) = 3, 1 + C = 2, 5 so that C = −1, 4. Thus, r(t) = 3t, t 2 + −1, 4 = 3t − 1, t 2 + 4

M uh am

••••••••••••••••••••••••••••••••••••••

INTEGRAL RULES

As with differentiation, many of the rules for integrating real-valued functions have analogs in the context of integrating vector-valued functions. 13.2.10 THEOREM (Rules of Integration). Let k be a scalar and let r(t), r1 (t), r2 (t), R(t), R1 (t), and R2 (t) be vector-valued functions, all in 2-space or all in 3-space, such that R, R1 , and R2 are antiderivatives of r, r1 , and r2 , respectively; that is, R (t) = r(t), R1 (t) = r1 (t), and R2 (t) = r2 (t). Then (a) kr(t) dt = kR(t) + C (b) [r1 (t) + r2 (t)] dt = R1 (t) + R2 (t) + C (c) [r1 (t) − r2 (t)] dt = R1 (t) − R2 (t) + C The proofs of these rules are left as an exercise.

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Vector-Valued Functions

EXERCISE SET 13.2

us uf i

876

Graphing Utility

In Exercises 1–6, find the limit.

In Exercises 23–26, find parametric equations of the line tangent to the graph of r(t) at the point where t = t0 . 2. lim cos t, sin t t → π/4

√ sin t ti+ j 4. lim+ t →0 t 3 ln t 6. lim 2 , 2 , sin 2t t →1 t t −1

5. lim (ti − 3j + t 2 k) t →2

23. r(t) = t 2 i + (2 − ln t) j; t0 = 1

Yo

1. lim (t 2 i + 2tj) t →3 2 t +1 1 3. lim , t → +⬁ 3t 2 + 2 t

24. r(t) = e2t i − 2 cos 3t j; t0 = 0

25. r(t) = 2 cos πti + 2 sin πtj + 3tk; t0 = 26. r(t) = ln ti + e−t j + t 3 k; t0 = 2

In Exercises 27–30, find a vector equation of the line tangent to the graph of r(t) at the point P0 on the curve.

8. (a) r(t) = et i + j + csc tk √ (b) r(t) = 5i − 3t + 1 j + e2t k

1 j + tk t

27. r(t) = (2t − 1)i +

ad

12. r(t) = 4i − cos tj 1 13. r(t) = i + tan tj + e2t k t √ 14. r(t) = (tan−1 t)i + t cos t j − t k

m

In Exercises 15–18, find the vector r (t0 ); then sketch the graph of r(t) in 2-space and draw the tangent vector r (t0 ). 15. r(t) = t, t 2 ; t0 = 2

n

31. Let r(t) = cos ti + sin tj + k. Find (a) lim (r(t) − r (t)) (b) lim (r(t) × r (t)) t →0

16. r(t) = t 3 i + t 2 j; t0 = 1

M uh am

17. r(t) = sec ti + tan tj; t0 = 0

19. r(t) = 2 sin ti + j + 2 cos tk; t0 = π/2

20. r(t) = cos ti + sin t j + tk; t0 = π/4 In Exercises 21 and 22, use a graphing utility to generate the graph of r(t) and the graph of the tangent line at t0 on the same screen.

21. r(t) = sin πti + t 2 j; t0 =

lim r(t) · (r (t) × r (t))

t →1

In Exercises 33 and 34, calculate d d [r1 (t) · r2 (t)] and [r1 (t) × r2 (t)] dt dt first by differentiating the product directly and then by applying Formulas (5) and (6). 33. r1 (t) = 2ti + 3t 2 j + t 3 k, r2 (t) = t 4 k 34. r1 (t) = cos ti + sin tj + tk, r2 (t) = i + tk In Exercises 35–40, evaluate the indefinite integral.

(3i + 4tj) dt

35.

(t sin i + j) dt

1 39. t 2 i − 2tj + k dt t

37.

(cos ti + sin tj) dt

36.

tet , ln t dt

38.

e−t , et , 3t 2 dt

40.

In Exercises 41– 46, evaluate the definite integral.

π/3

41.

0

0

cos 3t, − sin 3t dt

42.

1

(t 2 i + t 3 j) dt

0 2

43.

1 2

22. r(t) = 3 sin ti + 4 cos tj; t0 = π/4

t →0

32. Let r(t) = ti + t 2 j + t 3 k. Find

18. r(t) = 2 sin ti + 3 cos tj; t0 = π/6 In Exercises 19 and 20, find the vector r (t0 ); then sketch the graph of r(t) in 3-space and draw the tangent vector r (t0 ).

t →0

(c) lim (r(t) · r (t)).

sa

11. r(t) = (4 + 5t)i + (t − t 2 ) j

Ha

In Exercises 11–14, find r (t).

3t + 4 j; P0 (−1, 2)

28. r(t) = 4 cos ti − 3tj; P0 (2, −π) 1 j + (4 − t 2 )k; P0 (4, 1, 0) 29. r(t) = t 2 i − t +1 30. r(t) = sin ti + cosh tj + (tan−1 t)k; P0 (0, 1, 0)

9. Sketch the circle r(t) = cos ti + sin tj, and in each part draw the vector with its correct length. (a) r (π/4) (b) r (π) (c) r(2π) − r(3π/2) 10. Sketch the circle r(t) = cos ti − sin tj, and in each part draw the vector with its correct length. (a) r (π/4) (b) r (π) (c) r(2π) − r(3π/2)

√

Ri

(b) r(t) = t 2 i +

az

In Exercises 7 and 8, determine whether r(t) is continuous at t = 0. Explain your reasoning. 7. (a) r(t) = 3 sin ti − 2tj

1 3

3

44. −3

ti + t 2 j dt (3 − t)3/2 , (3 + t)3/2 , 1 dt

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9

45.

(t 1/2 i + t âˆ’1/2 j) dt

1

46.

1

(e2t i + eâˆ’t j + tk) dt

Yo

55. r1 (t) = t 2 i + tj + 3t 3 k r2 (t) = (t âˆ’ 1)i + 14 t 2 j + (5 âˆ’ t)k; P (1, 1, 3)

47. y (t) = t i + 2tj, y(0) = i + j 2

56. r1 (t) = 2eâˆ’t i + cos tj + (t 2 + 3)k r2 (t) = (1 âˆ’ t)i + t 2 j + (t 3 + 4)k; P (2, 1, 3)

48. y (t) = cos ti + sin tj, y(0) = i âˆ’ j 49. y (t) = i + et j, y(0) = 2i, y (0) = j

57. Use Formula (6) to derive the differentiation formula

50. y (t) = 12t 2 i âˆ’ 2tj, y(0) = 2i âˆ’ 4j, y (0) = 0

d [r(t) Ă— r (t)] = r(t) Ă— r (t) dt

az

In Exercises 51 and 52, let Î¸(t) be the angle between r(t) and r (t). Use a graphing calculator to generate the graph of Î¸ versus t, and make rough estimates of the t-values at which t-intercepts or relative extrema occur. What do these values tell you about the vectors r(t) and r (t)?

Ri

58. Let u = u(t), v = v(t), and w = w(t) be differentiable vector-valued functions. Use Formulas (5) and (6) to show that d [u Âˇ (v Ă— w)] dt du dv dw = Ă—w +uÂˇ vĂ— Âˇ [v Ă— w] + u Âˇ dt dt dt

51. r(t) = 4 cos ti + 3 sin tj; 0 â‰¤ t â‰¤ 2Ď€ 52. r(t) = t 2 i + t 3 j; 0 â‰¤ t â‰¤ 1

r = ti + t j âˆ’ 3tk

sa

2

Ha

intersects the plane 2x âˆ’ y + z = âˆ’2. (b) For the curve and plane in part (a), find, to the nearest degree, the acute angle that the tangent line to the curve makes with a line normal to the plane at each point of intersection. 54. Find where the tangent line to the curve

ad

m M uh am

SMOOTH PARAMETRIZATIONS

60. Prove Theorem 13.2.7 for 2-space. 61. Derive Formulas (5) and (6) for 3-space.

at the point (1, 1, 0) intersects the yz-plane.

59. Let u1 , u2 , u3 , v1 , v2 , v3 , w1 , w2 , and w3 be differentiable functions of t. Use Exercise 58 to show that

u u2 u 3

d

1 v 1 v 2 v3

dt

w w w

1 2 3

u u u u 1 u2 u3 u 1 u2 u3

2 3

1 =

v1 v2 v3

+

v1 v2 v3

+

v1 v2 v3

w1 w2 w3 w1 w2 w3 w w w

1 2 3

n

53. (a) Find the points where the curve

r = eâˆ’2t i + cos tj + 3 sin tk

877

In Exercises 55 and 56, show that the graphs of r1 (t) and r2 (t) intersect at the point P . Find, to the nearest degree, the acute angle between the tangent lines to the graphs of r1 (t) and r2 (t) at the point P .

0

In Exercises 47â€“50, solve the vector initial-value problem for y(t) by integrating and using the initial conditions to find the constants of integration.

Change of Parameter; Arc Length

us uf i

13.3

62. Prove Theorem 13.2.10 for 2-space.

13.3 CHANGE OF PARAMETER; ARC LENGTH

We observed in earlier sections that a curve in 2-space or 3-space can be represented parametrically in more than one way. For example, in Section 1.8 we gave two parametric representations of a circleâ€”one in which the circle was traced clockwise and the other in which it was traced counterclockwise. Sometimes it will be desirable to change the parameter for a parametric curve to a different parameter that is better suited for the problem at hand. In this section we will investigate issues associated with changes of parameter, and we will show that arc length plays a special role in parametric representations of curves. Graphs of vector-valued functions range from continuous and smooth to discontinuous and wildly erratic. In this text we will not be concerned with graphs of the latter type, so we will need to impose restrictions to eliminate the unwanted behavior. We will say that r(t) is smoothly parametrized or that r(t) is a smooth function of t if r (t) is continuous and r (t) = 0 for any allowable value of t. Algebraically, smoothness implies that the components of r(t) have continuous derivatives that are not all zero for the same value

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Vector-Valued Functions

us uf i

878

of t, and geometrically, it implies that the tangent vector r (t) varies continuously along the curve. For this reason a smoothly parametrized function is said to have a continuously turning tangent vector.

(a) r(t) = a cos ti + a sin tj + ctk (b) r(t) = t 2 i + t 3 j

y

(a > 0, c > 0)

Solution (a). We have

Yo

Example 1 Determine whether the following vector-valued functions have continuously turning tangent vectors.

az

r (t) = −a sin ti + a cos tj + ck

x

Ri

The components are continuous functions, and there is no value of t for which all three of them are zero (verify), so r(t) has a continuously turning tangent vector. The graph of r(t) is the circular helix in Figure 13.1.2.

Solution (b). We have r (t) = 2ti + 3t 2 j

n

Although the components are continuous functions, they are both equal to zero if t = 0, so r(t) does not have a continuously turning tangent vector. The graph of r(t), which is shown in Figure 13.3.1, is a semicubical parabola traced in the upward direction (see Example 3 of Section 11.2). Observe that for values of t slightly less than zero the angle between r (t) and i is near π, and for values of t slightly larger than zero the angle is near 0; hence there is a sudden reversal in the direction of the tangent vector as t increases through t = 0.

Figure 13.3.1

••••••••••••••••••••••••••••••••••••••

ARC LENGTH FROM THE VECTOR VIEWPOINT

Ha

sa

r(t) = t 2 i + t 3j

Recall from Theorem 6.4.3 that the arc length L of a parametric curve x = x(t),

y = y(t)

(a ≤ t ≤ b)

(1)

is given by the formula b

ad

L=

a

dx dt

2 +

dy dt

2 (2)

dt

m

Analogously, the arc length L of a parametric curve x = x(t),

y = y(t),

z = z(t)

(a ≤ t ≤ b)

(3)

M uh am

in 3-space is given by the formula L= a

b

dx dt

2

+

dy dt

2

+

dz dt

2 (4)

dt

Formulas (2) and (4) have vector forms that we can obtain by letting r(t) = x(t)i + y(t) j or 2-space

It follows that dr dy dx i+ j or = dt dt dt 2-space

r(t) = x(t)i + y(t) j + z(t)k 3-space

dx dy dz dr = i+ j+ k dt dt dt dt 3-space

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and hence 2 2 dr dx dy = + dt dt dt

or

Change of Parameter; Arc Length

879

us uf i

13.3

2 2 2 dr dx dy dz = + + dt dt dt dt

2-space

3-space

Yo

Substituting these expressions in (2) and (4) leads us to the following theorem.

13.3.1 THEOREM. If C is the graph in 2-space or 3-space of a smooth vector-valued function r(t), then its arc length L from t = a to t = b is L= a

b

dr dt dt

(5)

az

x = cos t,

Ri

Example 2 Find the arc length of that portion of the circular helix y = sin t,

z=t

from t = 0 to t = π.

Solution. Set r(t) = (cos t)i + (sin t)j + tk = cos t, sin t, t. Then

n

r (t) = − sin t, cos t, 1

and

r (t) =

√ (− sin t)2 + (cos t)2 + 1 = 2

••••••••••••••••••••••••••••••••••••••

ARC LENGTH AS A PARAMETER

For many purposes the best parameter to use for representing a curve in 2-space or 3-space parametrically is the length of arc measured along the curve from some fixed reference point. This can be done as follows:

ad

Step 1. s=3

s = –2

s = –3

C

Figure 13.3.2

Step 3.

If P is a point on the curve, let s be the “signed” arc length along C from the reference point to P , where s is positive if P is in the positive direction from the reference point, and s is negative if P is in the negative direction. Figure 13.3.2 illustrates this idea.

ti o

re c

cti on

–d

ire

s = –1

Starting from the reference point, choose one direction along the curve to be the positive direction and the other to be the negative direction.

+

M uh am

di

Reference point

Select an arbitrary point on the curve C to serve as a reference point.

Step 2.

n

s=1

m

s=2

0

Ha

0

sa

From Theorem 13.3.1 the arc length of the helix is π π√ √ dr dt = L= 2 dt = 2π dt

By this procedure, a unique point P on the curve is determined when a value for s is given. For example, s = 2 determines the point that is 2 units along the curve in the positive direction from the reference point, and s = − 32 determines the point that is 32 units along the curve in the negative direction from the reference point. Let us now treat s as a variable. As the value of s changes, the corresponding point P moves along C and the coordinates of P become functions of s. Thus, in 2-space the coordinates of P are (x(s), y(s)), and in 3-space they are (x(s), y(s), z(s)). Therefore, in 2-space or 3-space the curve C is given by the parametric equations x = x(s),

y = y(s)

or

x = x(s),

y = y(s),

z = z(s)

A parametric representation of a curve with arc length as the parameter is called an arc length parametrization of the curve. Note that a given curve will generally have infinitely

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Vector-Valued Functions

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880

many different arc length parametrizations, since the reference point and orientation can be chosen arbitrarily.

Example 3 Find the arc length parametrization of the circle x 2 + y 2 = a 2 with counterclockwise orientation and (a, 0) as the reference point. y

Yo

Solution. The circle with counterclockwise orientation can be represented by the parametric equations P(x, y)

x = a cos t,

s

(0 ≤ t ≤ 2π)

(6)

in which t can be interpreted as the angle in radian measure from the positive x-axis to the radius from the origin to the point P (x, y) (Figure 13.3.3). If we take the positive direction for measuring the arc length to be counterclockwise, and we take (a, 0) to be the reference point, then s and t are related by

x

(a, 0)

s = at

or

az

t

y = a sin t

t = s /a

x = a cos(s /a), ••••••••••••••••••••••••••••••••••••••

y = a sin(s /a)

(0 ≤ s ≤ 2πa)

In many situations the solution of a problem can be simplified by choosing the parameter in a vector-valued function or a parametric curve in the right way. The two most common parameters for curves in 2-space or 3-space are time and arc length. However, there are other useful possibilities as well. For example, in analyzing the motion of a particle in 2space, it is often desirable to parametrize its trajectory in terms of the angle φ between the tangent vector and the positive x-axis (Figure 13.3.4). Thus, our next objective is to develop methods for changing the parameter in a vector-valued function or parametric curve. This will allow us to move freely between different possible parametrizations.

Ha

sa

n

CHANGE OF PARAMETER

Ri

Making this change of variable in (6) and noting that s increases from 0 to 2πa as t increases from 0 to 2π yields the following arc length parametrization of the circle:

Figure 13.3.3

y

y

y

s x

m

ad

r(f)

r(s)

r(t)

Time as parameter

f x

Arc length as parameter

x

f as parameter

M uh am

Figure 13.3.4

A change of parameter in a vector-valued function r(t) is a substitution t = g(τ ) that produces a new vector-valued function r(g(τ )) having the same graph as r(t), but possibly traced differently as the parameter τ increases. Example 4 Find a change of parameter t = g(τ ) for the circle r(t) = cos ti + sin tj

(0 ≤ t ≤ 2π)

such that (a) (b)

the circle is traced counterclockwise as τ increases over the interval [0, 1]; the circle is traced clockwise as τ increases over the interval [0, 1].

Solution (a). The given circle is traced counterclockwise as t increases. Thus, if we choose g to be an increasing function, then it will follow from the relationship t = g(τ ) that

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t

t

2p

Change of Parameter; Arc Length

881

us uf i

13.3

t increases when Ď„ increases, thereby ensuring that the circle will be traced counterclockwise as Ď„ increases. We also want to choose g so that t increases from 0 to 2Ď€ as Ď„ increases from 0 to 1. A simple choice of g that satisfies all of the required criteria is the linear function graphed in Figure 13.3.5a. The equation of this line is

2p

t = g(Ď„ ) = 2Ď€Ď„ t

1

1

t = 2pt

t = 2p(1 â€“ t)

(a)

(b)

which is the desired change of parameter. The resulting representation of the circle in terms of the parameter Ď„ is r(g(Ď„ )) = cos 2Ď€Ď„ i + sin 2Ď€Ď„ j

(0 â‰¤ Ď„ â‰¤ 1)

Yo

t

(7)

Solution (b). To ensure that the circle is traced clockwise, we will choose g to be a

az

decreasing function such that t decreases from 2Ď€ to 0 as Ď„ increases from 0 to 1. A simple choice of g that achieves this is the linear function

Figure 13.3.5

t = g(Ď„ ) = 2Ď€(1 âˆ’ Ď„ )

(8)

Ri

graphed in Figure 13.3.5b. The resulting representation of the circle in terms of the parameter Ď„ is r(g(Ď„ )) = cos(2Ď€(1 âˆ’ Ď„ ))i + sin(2Ď€(1 âˆ’ Ď„ )) j which simplifies to (verify)

n

r(g(Ď„ )) = cos 2Ď€Ď„ i âˆ’ sin 2Ď€Ď„ j

(0 â‰¤ Ď„ â‰¤ 1)

(0 â‰¤ Ď„ â‰¤ 1)

Ha

sa

When making a change of parameter t = g(Ď„ ) in a vector-valued function r(t), it will be important to ensure that the new vector-valued function r(g(Ď„ )) is smooth if r(t) is smooth. To establish conditions under which this happens, we will need the following version of the chain rule for vector-valued functions. The proof is left as an exercise.

ad

13.3.2 THEOREM (Chain Rule). Let r(t) be a vector-valued function in 2-space or 3-space that is differentiable with respect to t. If t = g(Ď„ ) is a change of parameter in which g is differentiable with respect to Ď„, then r(g(Ď„ )) is differentiable with respect to Ď„ and

M uh am

m

dr dr dt = dĎ„ dt dĎ„

FINDING ARC LENGTH PARAMETRIZATIONS

(9)

A change of parameter t = g(Ď„ ) in which r(g(Ď„ )) is smooth if r(t) is smooth is called a smooth change of parameter. It follows from (9) that t = g(Ď„ ) will be a smooth change of parameter if dt /dĎ„ is continuous and dt /dĎ„ = 0 for all values of Ď„, since these conditions imply that dr/dĎ„ is continuous and nonzero if dr/dt is continuous and nonzero. Smooth changes of parameter fall into two categoriesâ€”those for which dt /dĎ„ > 0 for all Ď„ (called positive changes of parameter) and those for which dt /dĎ„ < 0 for all Ď„ (called negative changes of parameter). A positive change of parameter preserves the orientation of a parametric curve, and a negative change of parameter reverses it. Example 5 In Example 4 the change of parameter in (7) is positive since dt /dĎ„ = 2Ď€ > 0, and the change of parameter given by (8) is negative since dt /dĎ„ = âˆ’2Ď€ < 0. The positive change of parameter preserved the orientation of the circle, and the negative change of parameter reversed it. Next we will consider the problem of finding an arc length parametrization of a vectorvalued function that is expressed initially in terms of some other parameter t. The following theorem will provide a general method for doing this.

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Vector-Valued Functions

us uf i

882 y

s r(t0)

C

Yo

13.3.3 THEOREM. Let C be the graph of a smooth vector-valued function r(t) in 2-space or 3-space, and let r(t0 ) be any point on C. Then the following formula defines a positive change of parameter from t to s, where s is an arc length parameter having r(t0 ) as its reference point (Figure 13.3.6): t dr du (10) s= du t0

r(t) x

Figure 13.3.6

az

Proof. From (5) with u as the variable of integration instead of t, the integral represents the arc length of that portion of C between r(t0 ) and r(t) if t > t0 and the negative of that arc length if t < t0 . Thus, s is the arc length parameter with r(t0 ) as its reference point and its positive direction in the direction of increasing t. t

s= t0

t

s=

dx du

2 +

2 +

dy du

dy du

2 du

2

2-space

dz du

+

(11)

2 du

3-space

sa

t0

dx du

n

Ri

When needed, Formula (10) can be expressed in component form as

(12)

Example 6 Find the arc length parametrization of the circular helix

Ha

r = cos ti + sin tj + tk

(13)

that has reference point r(0) = (1, 0, 0) and the same orientation as the given helix.

Solution. Replacing t by u in r for integration purposes and taking t0 = 0 in Formula (10), we obtain

ad

r = cos ui + sin uj + uk

M uh am

m

dr = (− sin u)i + cos uj + k du √ dr = (− sin u)2 + cos2 u + 1 = 2 du t t t√ √ dr √ s= 2 du = 2u = 2t du du = 0 0 0 √ Thus, t = s / 2, so (13) can be reparametrized in terms of s as

s s s r = cos √ i + sin √ j + √ k 2 2 2 We are guaranteed that this reparametrization preserves the orientation of the helix since Formula (10) produces a positive change of parameter.

Example 7 A bug, starting at the reference point (1, 0, 0) of the helix in Example 6, walks up the helix for a distance of 10 units. What are the bug’s final coordinates?

Solution. From Example 6, the arc length parametrization of the helix relative to the reference point (1, 0, 0) is

s s s j+ √ k r = cos √ i + sin √ 2 2 2

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Change of Parameter; Arc Length

Yo

or, expressed parametrically,

s s s x = cos √ , y = sin √ , z = √ 2 2 2 Thus, at s = 10 the coordinates are

10 10 10 cos √ , sin √ , √ ≈ (0.705, 0.709, 7.07) 2 2 2

883

us uf i

13.3

Example 8 Recall from Formula (9) of Section 12.5 that the equation r = r0 + tv

(14)

az

is the vector form of the line that passes through the terminal point of r0 and is parallel to the vector v. Find the arc length parametrization of the line that has reference point r0 and the same orientation as the given line.

Ri

Solution. Replacing t by u in r for integration purposes and taking t0 = 0 in Formula (10), we obtain r = r0 + uv

sa

n

dr Since r0 is constant =v du dr = v du t t t dr s= v du = vu = tv du du = 0 0 0

Ha

Thus, t = s /v, so (14) can be reparametrized in terms of s as

v r = r0 + s v

(15)

• REMARK.

Comparing Formulas (14) and (15) shows that the vector equation of the line through the terminal point of r0 that is parallel to v can be reparametrized in terms of arc length with reference point r0 by normalizing v and then replacing t by s.

ad

• • • • • • • • • • • •

M uh am

m

Example 9 Find the arc length parametrization of the line

••••••••••••••••••••••••••••••••••••••

PROPERTIES OF ARC LENGTH PARAMETRIZATIONS

x = 2t + 1,

y = 3t − 2

that has the same orientation as the given line and uses (1, −2) as the reference point.

Solution. The line passes through the point (1, −2) and is parallel to the vector v = 2i+3j. To find the arc length parametrization of the line, we need only rewrite the given equations using v/v rather than v to determine the direction and replace t by s. Since v 2i + 3j 2 3 = √ = √ i+ √ j v 13 13 13 it follows that the parametric equations for the line in terms of s are 2 3 x = √ s + 1, y = √ s − 2 13 13

Because arc length parameters for a curve C are intimately related to the geometric characteristics of C, arc length parametrizations have properties that are not enjoyed by other parametrizations. For example, the following theorem shows that if a smooth curve is represented parametrically using an arc length parameter, then the tangent vectors all have length 1.

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Vector-Valued Functions

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Yo

13.3.4 THEOREM. (a) If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, where t is a general parameter, and if s is the arc length parameter for C defined by Formula (10), then for every value of t the tangent vector has length dr ds = (16) dt dt

If C is the graph of a smooth vector-valued function r(s) in 2-space or 3-space, where s is an arc length parameter, then for every value of s the tangent vector to C has length dr =1 (17) ds

(c)

If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, and if dr =1 dt

Ri

az

(b)

sa

n

for every value of t, then for any value of t0 in the domain of r, the parameter s = t âˆ’ t0 is an arc length parameter that has its reference point at the point on C where t = t0 .

Proof (a). This result follows by applying the Fundamental Theorem of Calculus (Theorem

Ha

5.6.3) to Formula (10).

Proof (b). Let t = s in part (a). Proof (c). It follows from Theorem 13.3.3 that the formula

ad

t dr du s= du t0

M uh am

m

defines an arc length parameter for C with reference point r(0). However, dr/du = 1 by hypothesis, so we can rewrite the formula for s as t t s= du = u = t âˆ’ t0 t0

t0

The component forms of Formulas (16) and (17) will be of sufficient interest in later sections that we provide them here for reference: 2 2 dr ds dx dy = + = dt dt dt dt

2 2 2 dr ds dx dy dz + + = dt = dt dt dt dt 2 2 dr dx dy = + =1 ds ds ds

(18)

2-space

3-space

2-space

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(19)

(20)

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Change of Parameter; Arc Length

2 2 2 dr dz dx dy = + + =1 ds ds ds ds

3-space

• REMARK.

885

us uf i

13.3

(21)

Note that Formulas (18) and (19) do not involve t0 , and hence do not depend on where the reference point for s is chosen. This is to be expected, since changing the reference point shifts s by a constant (the arc length between the two reference points), and this constant drops out on differentiating.

Yo

• • • • • • • • • • • • • • • • •

EXERCISE SET 13.3

13. r(t) = 3 cos ti + 3 sin tj + tk; 0 ≤ t ≤ 2π 14. r(t) = t 2 i+(cos t +t sin t) j+(sin t −t cos t)k; 0 ≤ t ≤ π In Exercises 15–18, calculate dr/dτ by the chain rule, and then check your result by expressing r in terms of τ and differentiating. 15. r = ti + t 2 j; t = 4τ + 1

n

2. The accompanying figure shows the graph of the vectorvalued function r(t) = sin ti + sin2 tj (0 ≤ t ≤ 2π). Show that this parametric curve is not smooth, even though it has no corners. Give an informal explanation of what causes the lack of smoothness.

12. r(t) = (4 + 3t)i + (2 − 2t) j + (5 + t)k; 3 ≤ t ≤ 4

Ri

1. The accompanying figure shows the graph of the fourcusped hypocycloid r(t) = cos3 ti + sin3 tj (0 ≤ t ≤ 2π). (a) Give an informal explanation of why r(t) is not smooth. (b) Confirm that r(t) is not smooth by examining r (t).

az

16. r = 3 cos t, 3 sin t; t = πτ

sa

y

17. r = et i + 4e−t j; t = τ 2

Ha

18. r = i + 3t 3/2 j + tk; t = 1/τ

x

y

x

Figure Ex-2

ad

Figure Ex-1

m

In Exercises 3–6, determine whether r(t) is a smooth function of the parameter t. 3. r(t) = t 3 i + (3t 2 − 2t) j + t 2 k

M uh am

4. r(t) = cos t 2 i + sin t 2 j + e−t k

5. r(t) = te−t i + (t 2 − 2t) j + cos πtk

6. r(t) = sin πti + (2t − ln t) j + (t 2 − t)k In Exercises 7–10, find the arc length of the parametric curve.

7. x = cos3 t, y = sin3 t, z = 2; 0 ≤ t ≤ π/2 8. x = 3 cos t, y = 3 sin t, z = 4t; 0 ≤ t ≤ π √ 9. x = et , y = e−t , z = 2t; 0 ≤ t ≤ 1

10. x = 12 t, y = 13 (1 − t)3/2 , z = 13 (1 + t)3/2 ; −1 ≤ t ≤ 1 In Exercises 11–14, find the arc length of the graph of r(t).

11. r(t) = t 3 i + tj +

1 2

√ 2 6t k; 1 ≤ t ≤ 3

19. (a) Find the arc length parametrization of the line x = t,

y=t

that has the same orientation as the given line and has reference point (0, 0). (b) Find the arc length parametrization of the line x = t,

y = t,

z=t

that has the same orientation as the given line and has reference point (0, 0, 0). 20. Find arc length parametrizations of the lines in Exercise 19 that have the stated reference points but are oriented opposite to the given lines. 21. (a) Find the arc length parametrization of the line x = 1 + t,

y = 3 − 2t,

z = 4 + 2t

that has the same direction as the given line and has reference point (1, 3, 4). (b) Use the parametric equations obtained in part (a) to find the point on the line that is 25 units from the reference point in the direction of increasing parameter. 22. (a) Find the arc length parametrization of the line x = −5 + 3t,

y = 2t,

z=5+t

that has the same direction as the given line and has reference point (−5, 0, 5). (b) Use the parametric equations obtained in part (a) to find the point on the line that is 10 units from the reference point in the direction of increasing parameter.

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Vector-Valued Functions

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36. (a) Show that r(t) = ti + t 2 j (−1 ≤ t ≤ 1) is a smooth vector-valued function, but the change of parameter t = τ 3 produces a vector-valued function that is not smooth, yet has the same graph as r(t). (b) Examine how the two vector-valued functions are traced and see if you can explain what causes the problem.

In Exercises 23–28, find an arc length parametrization of the curve that has the same orientation as the given curve and has t = 0 as the reference point. 23. r(t) = (3 + cos t)i + (2 + sin t) j; 0 ≤ t ≤ 2π 24. r(t) = cos3 ti + sin3 t j; 0 ≤ t ≤ π/2

Yo

37. Find a change of parameter t = g(τ ) for the semicircle

25. r(t) = 13 t 3 i + 12 t 2 j; t ≥ 0

r(t) = cos ti + sin tj (0 ≤ t ≤ π) such that (a) the semicircle is traced counterclockwise as τ varies over the interval [0, 1] (b) the semicircle is traced clockwise as τ varies over the interval [0, 1].

26. r(t) = (1 + t)2 i + (1 + t)3 j; 0 ≤ t ≤ 1 27. r(t) = et cos ti + et sin t j; 0 ≤ t ≤ π/2 √ 28. r(t) = sin et i + cos et j + 3et k; t ≥ 0

az

29. Show that the arc length of the circular √ helix x = a cos t, y = a sin t, z = ct for 0 ≤ t ≤ t0 is t0 a 2 + c2 .

38. What change of parameter t = g(τ ) would you make if you wanted to trace the graph of r(t) (0 ≤ t ≤ 1) in the opposite direction with τ varying from 0 to 1?

y = a − a cos t

(0 ≤ t ≤ 2π)

with (0, 0) as the reference point.

Ha

x = at − a sin t

sa

31. Find an arc length parametrization of the cycloid

39. As illustrated in the accompanying figure, copper cable with a diameter of 12 inch is to be wrapped in a circular helix around a cylinder that has a 12-inch diameter. What length of cable (measured along its centerline) will make one complete turn around the cylinder in a distance of 20 inches (between centerlines) measured parallel to the axis of the cylinder?

n

r = a cos ti + a sin t j + ctk can be expressed as s cs s i + a sin j+ k r = a cos w w w √ 2 2 where w = a + c and s is an arc length parameter with reference point at (a, 0, 0).

Ri

30. Use the result in Exercise 29 to show the circular helix

ad

32. Show that in cylindrical coordinates a curve given by the parametric equations r = r(t), θ = θ(t), z = z(t) for a ≤ t ≤ b has arc length 2 2 b 2 dr dθ dz L= + r2 + dt dt dt dt a

12 in

Enlarged cross section

20 in

[Hint: Use the relationships x = r cos θ, y = r sin θ.]

m

33. In each part, use the formula in Exercise 32 to find the arc length of the curve. (a) r = e2t , θ = t, z = e2t ; 0 ≤ t ≤ ln 2 (b) r = t 2 , θ = ln t, z = 13 t 3 ; 1 ≤ t ≤ 2

M uh am

34. Show that in spherical coordinates a curve given by the parametric equations ρ = ρ(t), θ = θ(t), φ = φ(t) for a ≤ t ≤ b has arc length 2 2 b 2 dρ dφ dθ L= + ρ 2 sin2 φ + ρ2 dt dt dt dt a [Hint: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.]

35. In each part, use the formula in Exercise 34 to find the arc length of the curve. (a) ρ = e−t , θ = 2t, φ = π/4; 0 ≤ t ≤ 2 (b) ρ = 2t, θ = ln t, φ = π/6; 1 ≤ t ≤ 5

1 2

in

Figure Ex-39

40. Let x = cos t, y = sin t, z = t 3/2 . Find 2 ds (b) r (t) dt. (c) (a) r (t) dt 0 41. Let r(t) = ln ti + 2tj + t 2 k. Find ds (b) (a) r (t) dt

3

(c)

r (t) dt.

1

42. Prove: If r(t) is a smoothly parametrized function, then the angles between r (t) and the vectors i, j, and k are continuous functions of t. 43. Prove the vector form of the chain rule for 2-space (Theorem 13.3.2) by expressing r(t) in terms of components.

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Unit Tangent, Normal, and Binormal Vectors

887

us uf i

13.4

13.4 UNIT TANGENT, NORMAL, AND BINORMAL VECTORS

Yo

In this section we will discuss some of the fundamental geometric properties of vectorvalued functions. Our work here will have important applications to the study of motion along a curved path in 2-space or 3-space and to the study of the geometric properties of curves and surfaces. â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

Recall that if C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, then the vector r (t) is nonzero, tangent to C, and points in the direction of increasing parameter. Thus, by normalizing r (t) we obtain a unit vector T(t) =

that is tangent to C and points in the direction of increasing parameter. We call T(t) the unit tangent vector to C at t.

T(t)

C

(1)

Ri

y

r (t) r (t)

az

UNIT TANGENT VECTORS

â€˘ REMARK.

r(t) x

Unless stated otherwise, we will assume that T(t) is positioned with its initial point at the terminal point of r(t) as in Figure 13.4.1. This will ensure that T(t) is actually tangent to the graph of r(t) and not simply parallel to the tangent line. Example 1 where t = 2.

sa

Figure 13.4.1

Find the unit tangent vector to the graph of r(t) = t 2 i + t 3 j at the point

n

â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘

Solution. Since

Ha

r (t) = 2ti + 3t 2 j we obtain

r (2) 3 4i + 12 j 1 4i + 12 j = âˆš = âˆš i+ âˆš j = âˆš r (2) 4 10 160 10 10 The graph of r(t) and the vector T(2) are shown in Figure 13.4.2. T(2) =

UNIT NORMAL VECTORS y 1

âˆš10

i+

3

âˆš10

M uh am

T(2) =

m

10

(4, 8)

j

N(t) =

T (t) T (t)

(2)

that is normal to C and points in the same direction as T (t). We call N(t) the principal unit normal vector to C at t or more simply the unit normal vector. Observe that the unit normal vector is only defined at points where T (t) = 0. Unless stated otherwise, we will assume that this condition is satisfied. In particular, this excludes straight lines.

r(t) = t 2 i + t 3j

x

8

Figure 13.4.2

Recall from Theorem 13.2.9 that if a vector-valued function r(t) has constant norm, then r(t) and r (t) are orthogonal vectors. In particular, T(t) has constant norm 1, so T(t) and T (t) are orthogonal vectors. This implies that T (t) is perpendicular to the tangent line to C at t, so we say that T (t) is normal to C at t. It follows that if T (t) = 0, and if we normalize T (t), then we obtain a unit vector

ad

â€˘ REMARK. â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘ â€˘

In 2-space there are two unit vectors that are orthogonal to T(t), and in 3-space there are infinitely many such vectors (Figure 13.4.3). In both cases the principal unit normal is that particular normal that points in the direction of T (t). After the next example we will show that for a nonlinear parametric curve in 2-space the principal unit normal is the one that points â€œinwardâ€? toward the concave side of the curve. Example 2 Find T(t) and N(t) for the circular helix x = a cos t,

y = a sin t,

z = ct

where a > 0.

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Vector-Valued Functions y

us uf i

888

z

C T(t)

C

T(t)

There are two unit vectors orthogonal to T(t).

x

Yo

y

x

There are infinitely many unit vectors orthogonal to T(t).

az

Figure 13.4.3

Solution. The radius vector for the helix is r(t) = a cos ti + a sin t j + ctk

Ri

Thus, r (t) = (−a sin t)i + a cos t j + ck √ r (t) = (−a sin t)2 + (a cos t)2 + c2 = a 2 + c2

a cos t c a sin t r (t) i+ j+ k = − r (t) a 2 + c2 a 2 + c2 a 2 + c2 a cos t a sin t T (t) = − i− j a 2 + c2 a 2 + c2

2

2 a cos t a2 a sin t a T (t) = − + − = = 2 2 a +c a 2 + c2 a 2 + c2 a 2 + c2 T (t) = (− cos t)i − (sin t)j N(t) = T (t) T(t) =

n

z

Ha

sa

t = 3c

y

t=0

• FOR THE READER.

(a, 0, 0)

• • • • • • • • • • • •

Figure 13.4.4

••••••••••••••••••••••••••••••••••••••

M uh am

C

n(t)

T(t)

f

Our next objective is to show that for a nonlinear parametric curve C in 2-space the unit normal vector always points toward the concave side of C. For this purpose, let φ(t) be the angle from the positive x-axis to T(t), and let n(t) be the unit vector that results when T(t) is rotated counterclockwise through an angle of π/2 (Figure 13.4.5). Since T(t) and n(t) are unit vectors, it follows from Formula (12) of Section 12.2 that these vectors can be expressed as

m

INWARD UNIT NORMAL VECTORS IN 2-SPACE

Because the k component of N(t) is zero, this vector lies in a horizontal plane for every value of t. Show that N(t) actually points directly toward the z-axis for all t (Figure 13.4.4).

ad

x

Figure 13.4.5

T(t) = cos φ(t)i + sin φ(t) j and n(t) = cos[φ(t) + π/2]i + sin[φ(t) + π/2] j = − sin φ(t)i + cos φ(t) j

(3) (4)

Observe that on intervals where φ(t) is increasing the vector n(t) points toward the concave side of C, and on intervals where φ(t) is decreasing it points away from the concave side (Figure 13.4.6). Now let us differentiate T(t) by using Formula (3) and applying the chain rule. This yields dT dφ dφ dT = = [(− sin φ)i + (cos φ)j] dt dφ dt dt and thus from (4) dT dφ (5) = n(t) dt dt

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y

Unit Tangent, Normal, and Binormal Vectors

y

C n(t)

T(t)

n(t)

T(t)

C f(t)

889

us uf i

13.4

Yo

f(t)

x

x

f(t) increases as t increases.

f(t) decreases as t increases.

az

Figure 13.4.6

Ri

But dφ /dt > 0 on intervals where φ(t) is increasing and dφ /dt < 0 on intervals where φ(t) is decreasing. Thus, it follows from (5) that dT/dt has the same direction as n(t) on intervals where φ(t) is increasing and the opposite direction on intervals where φ(t) is decreasing. Therefore, T (t) = dT/dt points “inward” toward the concave side of the curve in all cases, and hence so does N(t). For this reason, N(t) is also called the inward unit normal when applied to curves in 2-space. In the case where r(s) is parametrized by arc length, the procedures for computing the unit tangent vector T(s) and the unit normal vector N(s) are simpler than in the general case. For example, we showed in Theorem 13.3.4 that if s is an arc length parameter, then r (s) = 1. Thus, Formula (1) for the unit tangent vector simplifies to

sa

COMPUTING T AND N FOR CURVES PARAMETRIZED BY ARC LENGTH

n

••••••••••••••••••••••••••••••••••••••

T(s) = r (s)

(6)

Ha

and consequently Formula (2) for the unit normal vector simplifies to N(s) =

y

(x, y) t

x

M uh am

m

(a, 0)

Figure 13.4.7

T(s)

s

Figure 13.4.8

r = a cos ti + a sin tj (0 ≤ t ≤ 2π)

(8)

In this representation we can interpret t as the angle in radian measure from the positive x-axis to the radius vector (Figure 13.4.7). This angle subtends an arc of length s = at on the circle, so we can reparametrize the circle in terms of s by substituting s /a for t in (8). This yields r(s) = a cos(s /a)i + a sin(s /a) j (0 ≤ s ≤ 2πa)

To find T(s) and N(s) from Formulas (6) and (7), we must compute r (s), r (s), and r (s). Doing so, we obtain r (s) = − sin(s /a)i + cos(s /a) j

y

N(s)

(7)

Example 3 The circle of radius a with counterclockwise orientation and centered at the origin can be represented by the vector-valued function

ad

s = at

r (s) r (s)

x

r (s) = −(1/a) cos(s /a)i − (1/a) sin(s /a)j r (s) = (−1/a)2 cos2 (s /a) + (−1/a)2 sin2 (s /a) = 1/a Thus, T(s) = r (s) = − sin(s /a)i + cos(s /a) j N(s) = r (s)/r (s) = − cos(s /a)i − sin(s /a) j so N(s) points toward the center of the circle for all s (Figure 13.4.8). This makes sense geometrically and is also consistent with our earlier observation that in 2-space the unit normal vector is the inward normal.

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Vector-Valued Functions

••••••••••••••••••••••••••••••••••••••

BINORMAL VECTORS IN 3-SPACE

us uf i

890

If C is the graph of a vector-valued function r(t) in 3-space, then we define the binormal vector to C at t to be B(t) = T(t) × N(t)

r(t)

Rectifying plane

It follows from properties of the cross product that B(t) is orthogonal to both T(t) and N(t) and is oriented relative to T(t) and N(t) by the right-hand rule. Moreover, T(t) × N(t) is a unit vector since T(t) × N(t) = T(t)N(t) sin(π/2) = 1

B

Thus, {T(t), N(t), B(t)} is a set of three mutually orthogonal unit vectors. Just as the vectors i, j, and k determine a right-handed coordinate system in 3-space, so do the vectors T(t), N(t), and B(t). At each point on a smooth parametric curve C in 3-space, these vectors determine three mutually perpendicular planes that pass through the point— the TB-plane (called the rectifying plane), the TN-plane (called the osculating plane), and the NB-plane (called the normal plane) (Figure 13.4.9). Moreover, one can show that a coordinate system determined by T(t), N(t), and B(t) is right-handed in the sense that each of these vectors is related to the other two by the right-hand rule (Figure 13.4.10):

N

Osculating plane

Figure 13.4.9

T

B(t) = T(t) × N(t),

Ri

az

T

N(t) = B(t) × T(t),

T(t) = N(t) × B(t)

(10)

n

The coordinate system determined by T(t), N(t), and B(t) is called the TNB-frame or sometimes the Frenet frame in honor of the French mathematician Jean Fr´ed´eric Frenet (1816– 1900) who pioneered its application to the study of space curves. Typically, the xyzcoordinate system determined by the unit vectors i, j, and k remains fixed, whereas the TNB-frame changes as its origin moves along the curve C (Figure 13.4.11).

N

sa

B

(9)

Yo

Normal plane

Ha

Each vector is the cross product of the other two taken in clockwise order.

Figure 13.4.10

r(t) B

N

T

B

B

N

T

T

N

ad

B

N

T

M uh am

m

Figure 13.4.11

Formula (9) expresses B(t) in terms of T(t) and N(t). Alternatively, the binormal B(t) can be expressed directly in terms of r(t) as B(t) =

r (t) × r (t) r (t) × r (t)

(11)

and in the case where the parameter is arc length it can be expressed in terms of r(s) as B(s) =

r (s) × r (s) r (s)

We omit the proof.

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Unit Tangent, Normal, and Binormal Vectors

891

us uf i

13.4

EXERCISE SET 13.4

â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘â€˘

1. In each part, sketch the unit tangent and normal vectors at the points P , Q, and R, taking into account the orientation of the curve C. (b)

y

In Exercises 15â€“18, use the formula B(t) = T(t) Ă— N(t) to find B(t), and then check your answer by using Formula (11) to find B(t) directly from r(t).

y

C

C P

P

R

x

R

Q

x

Q

Yo

(a)

13. r(t) = sin ti + cos tj + 12 t 2 k; t0 = 0 âˆš 14. r(t) = ti + tj + 9 âˆ’ t 2 k; t0 = 1

15. r(t) = 3 sin ti + 3 cos t j + 4tk

16. r(t) = et sin ti + et cos t j + 3k

17. r(t) = (sin t âˆ’ t cos t)i + (cos t + t sin t) j + k

Figure Ex-1

In Exercises 3â€“10, find T(t) and N(t) at the given point.

20. r(t) = et i + et cos t j + et sin t k; t = 0

4. r(t) = 12 t 2 i + 13 t 3 j; t = 1

n

21. (a) Use the formula N(t) = B(t) Ă— T(t) and Formulas (1) and (11) to show that N(t) can be expressed in terms of r(t) as

6. r(t) = ln ti + tj; t = e

Ha

7. r(t) = 4 cos ti + 4 sin tj + tk; t = Ď€/2

sa

5. r(t) = 5 cos ti + 5 sin t j; t = Ď€/3

9. x = et cos t, y = et sin t, z = et ; t = 0 10. x = cosh t, y = sinh t, z = t; t = ln 2

ad

11. In the remark following Example 8 of Section 13.3, we observed that a line r = r0 + tv can be parametrized in terms of an arc length parameter s with reference point r0 by normalizing v. Use this result to show that the tangent line to the graph of r(t) at the point t0 can be expressed as

m

r = r(t0 ) + sT(t0 ) where s is an arc length parameter with reference point r(t0 ). 12. Use the result in Exercise 11 to show that the tangent line to the parabola

M uh am

In Exercises 19 and 20, find T(t), N(t), and B(t) for the given value of t. Then find equations for the osculating, normal, and rectifying planes at the point that corresponds to that value of t. 19. r(t) = cos ti + sin tj + k; t = Ď€/4

3. r(t) = (t 2 âˆ’ 1)i + tj; t = 1

8. r(t) = ti + 12 t 2 j + 13 t 3 k; t = 0

(a = 0, c = 0)

Ri

r(t) = 3 cos ti + 2 sin tj for 0 â‰¤ t â‰¤ 2Ď€ and the unit tangent and normal vectors at the points t = 0, t = Ď€/4, t = Ď€/2, and t = Ď€.

az

18. r(t) = a cos ti + a sin t j + ctk

2. Make a rough sketch that shows the ellipse

x = t, y = t at the point (1, 1) can be expressed parametrically as s 2s x =1+ âˆš , y =1+ âˆš 5 5 2

In Exercises 13 and 14, use the result in Exercise 11 to find parametric equations for the tangent line to the graph of r(t) at t0 in terms of an arc length parameter s.

r (t) Ă— r (t) r (t) Ă— r (t) Ă— r (t) r (t) (b) Use properties of cross products to show that the formula in part (a) can be expressed as N(t) =

(r (t) Ă— r (t)) Ă— r (t) (r (t) Ă— r (t)) Ă— r (t) (c) Use the result in part (b) and Exercise 39 of Section 12.4 to show that N(t) can be expressed directly in terms of r(t) as N(t) =

N(t) =

u(t) u(t)

where u(t) = r (t)2 r (t) âˆ’ (r (t) Âˇ r (t))r (t) 22. Use the result in part (b) of Exercise 21 to find the unit normal vector requested in (a) Exercise 3 (b) Exercise 7. In Exercises 23 and 24, use the result in part (c) of Exercise 21 to find N(t). 23. r(t) = sin ti + cos tj + tk

24. r(t) = ti + t 2 j + t 3 k

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Vector-Valued Functions

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892

13.5 CURVATURE

••••••••••••••••••••••••••••••••••••••

Suppose that C is the graph of a smooth vector-valued function in 2-space or 3-space that is parametrized in terms of arc length. Figure 13.5.1 suggests that for a curve in 2-space the “sharpness” of the bend in C is closely related to dT/ds, which is the rate of change of the unit tangent vector T with respect to s. (Keep in mind that T has constant length, so only its direction changes.) If C is a straight line (no bend), then the direction of T remains constant (Figure 13.5.1a); if C bends slightly, then T undergoes a gradual change of direction (Figure 13.5.1b); and if C bends sharply, then T undergoes a rapid change of direction (Figure 13.5.1c). T

T

C

T T

Ri

az

DEFINITION OF CURVATURE

Yo

In this section we will consider the problem of obtaining a numerical measure of how sharply a curve in 2-space or 3-space bends. Our results will have applications in geometry and in the study of motion along a curved path.

C

T

T

T

sa

T

n

C

(b)

(c)

Ha

(a)

T

Figure 13.5.1

M uh am

m

ad

The situation in 3-space is more complicated because bends in a curve are not limited to a single plane—they can occur in all directions, as illustrated by the complicated tube plot in Figure 13.1.3. To describe the bending characteristics of a curve in 3-space completely, one must take into account dT/ds, dN/ds, and dB/ds. A complete study of this topic would take us too far afield, so we will limit our discussion to dT/ds, which is the most important of these derivatives in applications.

13.5.1 DEFINITION. If C is a smooth curve in 2-space or 3-space that is parametrized by arc length, then the curvature of C, denoted by κ = κ(s) (κ = Greek “kappa”), is defined by dT κ(s) = (1) ds = r (s)

Observe that κ(s) is a real-valued function of s, since it is the length of dT/ds that measures the curvature. In general, the curvature will vary from point to point along a curve; however, the following example shows that the curvature is constant for circles in 2-space, as you might expect. Example 1 In Example 3 of Section 13.4 we showed that the circle of radius a, centered at the origin, can be parametrized in terms of arc length as s s r(s) = a cos i + a sin j (0 ≤ s ≤ 2πa) a a

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Curvature

893

us uf i

13.5

Thus,

so the circle has constant curvature 1/a.

Yo

s s 1 1 r (s) = − cos i − sin j a a a a and hence from (1) s 2 s 2 1 1 1 κ(s) = r (s) = − cos + − sin = a a a a a

The next example shows that lines have zero curvature, which is consistent with the fact that they do not bend.

az

Example 2 Recall from the remark following Example 8 of Section 13.3 that a line in 2-space or 3-space can be parametrized in terms of arc length as r = r0 + su

Ri

where the terminal point of r0 is a point on the line and u is a unit vector parallel to the line. Thus, dr d r (s) = = [r0 + su] = 0 + u = u ds ds

n

r0 is constant

and hence

d dr = [u] = 0 ds ds

sa

r (s) =

Thus,

Ha

u is constant

κ(s) = r (s) = 0

••••••••••••••••••••••••••••••••••••••

Formula (1) is only applicable if the curve is parametrized in terms of arc length. The following theorem provides two formulas for curvature in terms of a general parameter t.

M uh am

m

ad

FORMULAS FOR CURVATURE

13.5.2 THEOREM. If r(t) is a smooth vector-valued function in 2-space or 3-space, then for each value of t at which T (t) and r (t) exist, the curvature κ can be expressed as (a)

κ(t) =

T (t) r (t)

(2)

(b)

κ(t) =

r (t) × r (t) r (t)3

(3)

Proof (a). It follows from Formula (1) and Formulas (16) and (17) of Section 13.3 that dT dT/dt dT/dt T (t) = = κ(t) = = ds ds /dt dr/dt r (t)

Proof (b). It follows from Formula (1) of Section 13.4 that r (t) = r (t)T(t)

(4)

r (t) = r (t) T(t) + r (t)T (t)

(5)

so

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Vector-Valued Functions

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894

But from Formula (2) of Section 13.4 and part (a) of this theorem we have T (t) = T (t)N(t)

and

T (t) = κ(t)r (t)

so T (t) = κ(t)r (t)N(t)

Yo

Substituting this into (5) yields r (t) = r (t) T(t) + κ(t)r (t)2 N(t) Thus, from (4) and (6)

(6)

r (t) × r (t) = r (t)r (t) (T(t) × T(t)) + κ(t)r (t)3 (T(t) × N(t))

az

But the cross product of a vector with itself is zero, so this equation simplifies to r (t) × r (t) = κ(t)r (t)3 (T(t) × N(t)) = κ(t)r (t)3 B(t) r (t) × r (t) = κ(t)r (t)3 Formula (3) now follows. • REMARKS.

n

Formula (2) is useful if T(t) is known or is easy to obtain; however, Formula (3) will usually be easier to apply, since it involves only r(t) and its derivatives. We also note that cross products were defined only for vectors in 3-space, so to use Formula (3) in 2-space we must first write the 2-space function r(t) = x(t)i+y(t) j as the 3-space function r(t) = x(t)i + y(t) j + 0k with a zero k component.

sa

• • • • • • • • • • • • • • • • • • • • • • •

Ri

It follows from this equation and the fact that B(t) is a unit vector that

Ha

Example 3 Find κ(t) for the circular helix x = a cos t,

y = a sin t,

z = ct

where a > 0.

Solution. The radius vector for the helix is

ad

r(t) = a cos ti + a sin t j + ctk Thus,

m

r (t) = (−a sin t)i + a cos t j + ck

M uh am

so

r (t) = (−a cos t)i + (−a sin t) j

i

r (t) × r (t) = −a sin t

−a cos t

Therefore, r (t) =

j a cos t −a sin t

k

2 c = (ac sin t)i − (ac cos t) j + a k

0

(−a sin t)2 + (a cos t)2 + c2 = a 2 + c2

and r (t) × r (t) =

(ac sin t)2 + (−ac cos t)2 + a 4 √ √ = a 2 c2 + a 4 = a a 2 + c2

so

a a 2 + c2 a r (t) × r (t) = = 2 κ(t) = 3 3 r (t) a + c2 ( a 2 + c2 )

Note that κ does not depend on t, which tells us that the helix has constant curvature.

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y

Example 4 The graph of the vector equation

3

r = 2 cos ti + 3 sin t j

895

(0 ≤ t ≤ 2π)

is the ellipse in Figure 13.5.2. Find the curvature of the ellipse at the endpoints of the major and minor axes, and use a graphing utility to generate the graph of κ(t).

x

Solution. To apply Formula (3), we must treat the ellipse as a curve in the xy-plane of an

2

Yo

-2

Curvature

us uf i

13.5

xyz-coordinate system by adding a zero k component and writing its equation as r = 2 cos ti + 3 sin t j + 0k -3

r = 2 cos ti + 3 sin tj

r (t) = (−2 sin t)i + 3 cos t j

k 1 0.8

Therefore, r (t) =

0.4 t i

o

so

r (t) × r (t) 6 = (7) 2 3 r (t) [4 sin t + 9 cos2 t]3/2 The endpoints of the minor axis are (2, 0) and (−2, 0), which correspond to t = 0 and t = π, respectively. Substituting these values in (7) yields the same curvature at both points, namely κ(t) =

6 [4 sin2 t + 9 cos2 t]3/ 2

Ha

k(t) =

c

n

r (t) × r (t) = 6

0.2 6

(−2 sin t)2 + (3 cos t)2 = 4 sin2 t + 9 cos2 t

sa

0.6

Figure 13.5.3

M uh am Figure 13.5.4

••••••••••••••••••••••••••••••••••••••

RADIUS OF CURVATURE

93/2

=

ad

m

x

k 0.2 0.3 0.4 0.5 0.6 0.7

6

2 6 = 27 9 The endpoints of the major axis are (0, 3) and (0, −3), which correspond to t = π/2 and t = 3π/2, respectively; from (7) the curvature at these points is

π 3 6 3π κ=κ =κ = 3/2 = 2 2 4 4 κ = κ(0) = κ(π) =

y

k

2 0 = [(6 sin t) + (6 cos2 t)]k = 6k

0

Ri

r (t) = (−2 cos t)i + (−3 sin t) j

i j

r (t) × r (t) = −2 sin t 3 cos t

−2 cos t −3 sin t

Figure 13.5.2

az

It is not essential to write the zero k component explicitly as long as you assume it to be there when you calculate a cross product. Thus,

Observe that the curvature is greater at the ends of the major axis than at the ends of the minor axis, as you might expect. Figure 13.5.3 shows the graph κ versus t. This graph illustrates clearly that the curvature is minimum at t = 0 (the right end of the minor axis), increases to a maximum at t = π/2 (the top of the major axis), decreases to a minimum again at t = π (the left end of the minor axis), and continues cyclically in this manner. Figure 13.5.4 provides another way of picturing the curvature.

In the last example we found the curvature at the ends of the minor axis to be 29 and the curvature at the ends of the major axis to be 34 . To obtain a better understanding of the meaning of these numbers, recall from Example 1 that a circle of radius a has a constant curvature of 1/a; thus, the curvature of the ellipse at the ends of the minor axis is the same as that of a circle of radius 92 , and the curvature at the ends of the major axis is the same as that of a circle of radius 43 (Figure 13.5.5). In general, if a curve C in 2-space has nonzero curvature κ at a point P , then the circle of radius ρ = 1/κ sharing a common tangent with C at P , and centered on the concave side of the curve at P , is called the circle of curvature or osculating circle at P (Figure 13.5.6). The osculating circle and the curve C not only touch at P but they have equal curvatures at that point. In this sense, the osculating circle is the circle that best approximates the curve

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Vector-Valued Functions y

Radius

4 3

Osculating circle x

C

Center of curvature

Yo

9 Radius 2

us uf i

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1 r= k

P Figure 13.5.6

az

Figure 13.5.5

••••••••••••••••••••••••••••••••••••••

AN INTERPRETATION OF CURVATURE IN 2-SPACE

Ri

C near P . The radius ρ of the osculating circle at P is called the radius of curvature at P , and the center of the circle is called the center of curvature at P (Figure 13.5.6). A useful geometric interpretation of curvature in 2-space can be obtained by considering the angle φ measured counterclockwise from the direction of the positive x-axis to the unit tangent vector T (Figure 13.5.7). By Formula (12) of Section 12.2, we can express T in terms of φ as T(φ) = cos φi + sin φ j

T

Thus, dT = (− sin φ)i + cos φ j dφ

sa

f

n

C

Figure 13.5.7

Ha

dT dT dφ = ds dφ ds from which we obtain

dT dφ dT dφ

dφ

2 2

κ(s) = = = (− sin φ) + cos φ =

ds ds dφ ds ds

ad

In summary, we have shown that

dφ

κ(s) =

ds

M uh am

m

(8)

which tells us that curvature in 2-space can be interpreted as the magnitude of the rate of change of φ with respect to s—the greater the curvature, the more rapidly φ changes with s (Figure 13.5.8). In the case of a straight line, the angle φ is constant (Figure 13.5.9) and consequently κ(s) = |dφ /ds| = 0, which is consistent with the fact that a straight line has zero curvature at every point.

T f

f T

f

f f

T f

In 2-space, k(s) is the magnitude of the rate of change of f with respect to s.

Figure 13.5.8

f is constant, so the line has zero curvature. Figure 13.5.9

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FORMULA SUMMARY

Curvature

897

us uf i

13.5

We conclude this section with a summary of formulas for T, N, and B. These formulas have either been derived in the text or are easily derivable from formulas we have already established. T(s) = r (s) r (s) r (s) 1 dT = = κ(s) κ(s) ds r (s)

B(s) =

r (s) × r (s) r (s) × r (s) = κ(s) r (s)

T(t) =

r (t) r (t)

B(t) =

r (t) × r (t) r (t) × r (t)

az

N(s) =

Yo

(9)

(10)

(11)

Ri

(12)

(13)

N(t) = B(t) × T(t)

EXERCISE SET 13.5

C

Graphing Utility

sa

n

(14)

CAS

Ha

In Exercises 1 and 2, use the osculating circle shown in the figure to estimate the curvature at the indicated point. y

1.

ad

x

M uh am

x

-3

m

1

14. x = sin t, y = cos t, z = 12 t 2 ; t = 0 In Exercises 15 and 16, confirm that s is an arc length parameter by showing that dr/ds = 1, and then apply Formula (1) to find κ(s).

3

1

-1

13. x = et cos t, y = et sin t, z = et ; t = 0

y

2.

0.5

12. r(t) = et i + e−t j + tk; t = 0

3

-3

s i + cos 1 + 15. r = sin 1 + 2 3/2 /2 3 i + 23 s j 16. r = 1 − 23 s

17. (a) Use Formula (3) to show that in 2-space the curvature of a smooth parametric curve

In Exercises 3–10, use Formula (3) to find κ(t).

3. r(t) = t i + t j 2

3

−t

5. r(t) = e i + e j 3t

x = x(t),

4. r(t) = 4 cos ti + sin t j 6. x = 1−t , y = t −t 3

2

7. r(t) = 4 cos ti + 4 sin tj + tk

8. r(t) = ti + 12 t 2 j + 13 t 3 k

9. x = cosh t, y = sinh t, z = t

10. r(t) = i + tj + t 2 k

In Exercises 11–14, find the curvature and the radius of curvature at the stated point.

11. r(t) = 3 cos ti + 4 sin tj + tk; t = π/2

√ s s j+ 3 1+ k 2 2 0 ≤ s ≤ 32

is κ(t) =

y = y(t)

|x y − y x | (x 2 + y 2 )3/2

where primes denote differentiation with respect to t. (b) Use the result in part (a) to show that in 2-space the curvature of the plane curve given by y = f(x) is κ(x) =

|d 2 y /dx 2 | [1 + (dy/dx)2 ]3/2

[Hint: Express y = f(x) parametrically with x = t as the parameter.]

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Vector-Valued Functions

18. Use part (b) of Exercise 17 to show that the curvature of y = f(x) can be expressed in terms of the angle of inclination of the tangent line as

2

d y

3

Îş(Ď†) = 2 cos Ď†

dx

In Exercises 35 and 36, use a graphing utility to generate the graph of y = f(x), and then make a conjecture about the shape of the graph of y = Îş(x). Check your conjecture by generating the graph of y = Îş(x). 35. f(x) = xeâˆ’x

In Exercises 19â€“24, use the result in Exercise 17(b) to find the curvature at the stated point. 20. y = x 3 /3; x = 0

21. y = 1/x; x = 1

22. y = eâˆ’x ; x = 1

23. y = tan x; x = Ď€/4

24. y 2 âˆ’ 4x 2 = 9; (2, 5)

C

In Exercises 25â€“30, use the result in Exercise 17(a) to find the curvature at the stated point. 1 2

C

26. x = 4 cos t, y = sin t; t = Ď€/2 28. x = 1 âˆ’ t 3 , y = t âˆ’ t 2 ; t = 1 30. x = 2 sin 2t, y = 3 sin t; t = Ď€/2

Ha

31. In each part, use the formulas in Exercise 17 to help find the radius of curvature at the stated points. Then sketch the graph together with the osculating circles at those points. (a) y = cos x at x = 0 and x = Ď€ (b) x = 2 cos t, y = sin t (0 â‰¤ t â‰¤ 2Ď€) at t = 0 and t = Ď€/2

ad

32. Use the formula in Exercise 17(a) to find Îş(t) for the curve x = eâˆ’t cos t, y = eâˆ’t sin t. Then sketch the graph of Îş(t).

m

M uh am II

x

II

I x

34. (a)

y

II

I x

[Hint: Let Î¸ be the parameter and use the relationships x = r cos Î¸, y = r sin Î¸.]

In Exercises 41â€“44, use the formula of Exercise 39 to find the curvature at the indicated point.

y

I

39. Use the formula in Exercise 17(a) to show that for a curve in polar coordinates described by r = f(Î¸ ) the curvature is

2

dr d 2 r

2 âˆ’r 2

r + 2

dÎ¸ dÎ¸

Îş(Î¸ ) = 2 3/2 dr r2 + dÎ¸

40. Use the result in Exercise 39 to show that a circle has constant curvature.

In each part of Exercises 33 and 34, the graphs of f(x) and the associated curvature function Îş(x) are shown. Determine which is which, and explain your reasoning. (b)

38. (a) Use a CAS to graph the parametric curve x = t cos t, y = t sin t for t â‰Ľ 0. (b) Make a conjecture about the behavior of Îş(t) as t â†’ +âŹ . (c) Use the CAS and part (a) of Exercise 17 to find Îş(t). (d) Check your conjecture by finding the limit of Îş(t) as t â†’ +âŹ .

sa

29. x = t, y = 1/t; t = 1

y

37. (a) If you have a CAS, read the documentation on calculating higher-order derivatives. Then use the CAS and part (b) of Exercise 17 to find Îş(x) for f(x) = x 4 âˆ’ 2x 2 . (b) Use the CAS to generate the graphs of f(x) = x 4 âˆ’ 2x 2 and Îş(x) on the same screen for âˆ’2 â‰¤ x â‰¤ 2. (c) Find the radius of curvature at each relative extremum. (d) Make a reasonably accurate hand-drawn sketch that shows the graph of f(x) = x 4 âˆ’ 2x 2 and the osculating circles in their correct proportions at the relative extrema.

n

27. x = e3t , y = eâˆ’t ; t = 0

33. (a)

for âˆ’1 â‰¤ x â‰¤ 1

az

19. y = sin x; x = Ď€/2

25. x = t 2 , y = t 3 ; t =

for 0 â‰¤ x â‰¤ 5

36. f(x) = x 3 âˆ’ x

Yo

[Hint: tan Ď† = dy/dx.]

Ri

898

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

41. r = 1 + cos Î¸; Î¸ = Ď€/2

42. r = e2Î¸ ; Î¸ = 1

43. r = sin 3Î¸; Î¸ = 0

44. r = Î¸ ; Î¸ = 1

45. The accompanying figure is the graph of the radius of curvature versus Î¸ in rectangular coordinates for the cardioid r = 1 + cos Î¸. In words, explain what the graph tells you about the cardioid. y 1.5

(b)

y

1

II

0.5

I

u

x 6

c

i

y = 1/k(u)

o

Figure Ex-45

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49. At what point(s) does 4x 2 +9y 2 = 36 have minimum radius of curvature? 50. Find the value of x, x > 0, where y = x 3 has maximum curvature. 51. Find the maximum and minimum values of the radius of curvature for the curve x = cos t, y = sin t, z = cos t. 52. Find the minimum value of √ the radius of curvature for the curve x = et , y = e−t , z = 2t. 53. Use the formula in Exercise 39 to show that the curvature of the polar curve r = eaθ is inversely proportional to r.

57. Show that the transition at x = 0 from the horizontal line y = 0 for x ≤ 0 to the parabola y = x 2 for x > 0 is not smooth, whereas the transition to y = x 3 for x > 0 is smooth. 58. (a) Sketch the graph of the curve defined piecewise by y = x 2 for x < 0, y = x 4 for x ≥ 0. (b) Show that for the curve in part (a) the transition at x = 0 is not smooth. 59. The accompanying figure shows the arc of a circle of radius r with center at (0, r). Find the value of a so that there is a smooth transition from the circle to the parabola y = ax 2 at the point where x = 0. y

54. Use the formula in Exercise 39 and √ a CAS to show that the curvature of the lemniscate r = a cos 2θ is directly proportional to r.

n

C

Yo

48. At what point(s) does y = ex have maximum curvature?

In Exercises 57–60, we will be concerned with the problem of creating a single smooth curve by piecing together two separate smooth curves. If two smooth curves C1 and C2 are joined at a point P to form a curve C, then we will say that C1 and C2 make a smooth transition at P if the curvature of C is continuous at P .

az

47. Find the radius of curvature of the parabola y 2 = 4px at (0, 0).

sa

55. (a) Use the result in Exercise 18 to show that for the parabola y = x 2 the curvature κ(φ) at points where the tangent line has an angle of inclination of φ is

Ha

κ(φ) = |2 cos3 φ|

ad

(b) Use the result in part (a) to find the radius of curvature of the parabola at the point on the parabola where the tangent line has slope 1. (c) Make a sketch with reasonably accurate proportions that shows the osculating circle at the point on the parabola where the tangent line has slope 1.

M uh am

m

56. The evolute of a smooth parametric curve C in 2-space is the curve formed from the centers of curvature of C. The accompanying figure shows the ellipse x = 3 cos t, y = 2 sin t (0 ≤ t ≤ 2π) and its evolute graphed together. (a) Which points on the evolute correspond to t = 0 and t = π/2? (b) In what direction is the evolute traced as t increases from 0 to 2π? (c) What does the evolute of a circle look like? Explain your reasoning. y

3

-3

-3

Figure Ex-56

x 3

899

Ri

46. Use the formula in Exercise 39 and a graphing utility to generate the graph in Exercise 45.

Curvature

us uf i

13.5

Arc of circle (0, r)

y = ax 2

x

Figure Ex-59

60. Find a, b, and c so that there is a smooth transition at x = 0 from the curve y = ex for x ≤ 0 to the parabola y = ax 2 + bx + c for x > 0. [Hint: The curvature is continuous at those points where y is continuous.] In Exercises 61–64, we assume that s is an arc length parameter for a smooth vector-valued function r(s) in 3-space and that dT/ds and dN/ds exist at each point on the curve. This implies that dB/ds exists as well, since B = T × N. 61. Show that dT = κ(s)N(s) ds and use this result to obtain the formulas in (10). 62. (a) Show that dB/ds is perpendicular to B(s). (b) Show that dB/ds is perpendicular to T(s). [Hint: Use the fact that B(s) is perpendicular to both T(s) and N(s), and differentiate B · T with respect to s.] (c) Use the results in parts (a) and (b) to show that dB/ds is a scalar multiple of N(s). The negative of this scalar is called the torsion of r(s) and is denoted by τ (s). Thus, dB = −τ (s)N(s) ds (d) Show that τ (s) = 0 for all s if the graph of r(s) lies in a plane. [Note: For reasons that we cannot discuss here, the torsion is related to the “twisting” properties of the curve, and τ (s) is regarded as a numerical measure of the tendency for the curve to twist out of the osculating plane.]

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Vector-Valued Functions

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900

63. Let κ be the curvature of C and τ the torsion (defined in Exercise 62). By differentiating N = B × T with respect to s, show that dN/ds = −κT + τ B.

66. (a) Use the chain rule and the first two Frenet–Serret formulas in Exercise 64 to show that

T = κs N and N = −κs T + τ s B where primes denote differentiation with respect to t. (b) Show that Formulas (4) and (6) can be written in the form

64. The following derivatives, known as the Frenet–Serret formulas, are fundamental in the theory of curves in 3-space: [Exercise 61] [Exercise 63] [Exercise 62(c)]

Yo

dT/ds = κN dN/ds = −κT + τ B dB/ds = −τ N

r (t) = s T and r (t) = s T + κ(s )2 N (c) Use the results in parts (a) and (b) to show that r (t) = [s − κ 2 (s )3 ]T

Use the first two Frenet–Serret formulas and the fact that r (s) = T if r = r(s) to show that [r (s) × r (s)] · r (s) r (s) × r (s) τ= and B = r (s)2 r (s)

+ [3κs s + κ (s )2 ]N + κτ (s )3 B

az

(d) Use the results in parts (b) and (c) to show that τ (t) =

65. Use the results in Exercise 64 and the results in Exercise 30 of Section 13.3 to show that for the circular helix

Ri

In Exercises 67–70, use the formula in Exercise 66(d) to find the torsion τ = τ (t). 67. The twisted cubic r(t) = 2ti + t 2 j + 13 t 3 k

n

68. The circular helix r(t) = a cos ti + a sin t j + ctk √ 69. r(t) = et i + e−t j + 2tk 70. r(t) = (t − sin t)i + (1 − cos t) j + tk

Ha

sa

r = a cos ti + a sin t j + ctk with a > 0 the torsion and the binormal vector are c τ= 2 w and c c a s s B= i− cos j+ k sin w w w w w √ where w = a 2 + c2 and s has reference point (a, 0, 0).

[r (t) × r (t)] · r (t) r (t) × r (t)2

13.6 MOTION ALONG A CURVE

m

ad

In earlier sections we considered the motion of a particle along a line. In that situation there are only two directions in which the particle can move—the positive direction or the negative direction. Motion in 2-space or 3-space is more complicated because there are infinitely many directions in which a particle can move. In this section we will show how vectors can be used to analyze motion along curves in 2-space or 3-space.

••••••••••••••••••••••••••••••••••••••

M uh am

VELOCITY, ACCELERATION, AND SPEED

Let us assume that the motion of a particle in 2-space or 3-space is described by a smooth vector-valued function r(t) in which the parameter t denotes time; we will call this the position function or trajectory of the particle. As the particle moves along its trajectory, its direction of motion and its speed can vary from instant to instant. Thus, before we can undertake any analysis of such motion, we must have clear answers to the following questions: •

What is the direction of motion of the particle at an instant of time?

•

What is the speed of the particle at an instant of time?

We will define the direction of motion at time t to be the direction of the unit tangent vector T(t), and we will define the speed to be ds /dt—the instantaneous rate of change of the arc length traveled by the particle from an arbitrary reference point. Taking this a step further, we will combine the speed and the direction of motion to form the vector ds v(t) = T(t) (1) dt

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T(t) v(t) =

ds T(t) dt

The length of the velocity vector is the speed of the particle, and the direction of the velocity vector is the direction of motion.

901

which we call the velocity of the particle at time t. Thus, at each instant of time the velocity vector v(t) points in the direction of motion and has a magnitude that is equal to the speed of the particle (Figure 13.6.1). Recall that for motion along a coordinate line the velocity function is the derivative of the position function. The same is true for motion along a curve, since dr dr ds ds = = T(t) = v(t) dt ds dt dt For motion along a coordinate line, the acceleration function was defined to be the derivative of the velocity function. The definition is the same for motion along a curve.

Yo

r(t)

Motion Along a Curve

us uf i

13.6

dr dt

d 2r dv = 2 dt dt

(3)

n

acceleration = a(t) =

(2)

Ri

velocity = v(t) =

az

13.6.1 DEFINITION. If r(t) is the position function of a particle moving along a curve in 2-space or 3-space, then the instantaneous velocity, instantaneous acceleration, and instantaneous speed of the particle at time t are defined by

Figure 13.6.1

ds dt

(4)

sa

speed = v(t) =

Ha

As shown in Table 13.6.1, the position, velocity, acceleration, and speed can also be expressed in component form:

position

2-space

3-space

r(t) = x(t)i + y(t)j

r(t) = x(t)i + y(t)j + z(t)k

velocity

dy dx v(t) = i+ j dt dt

v(t) =

dx dy dz i+ j+ k dt dt dt

acceleration

a(t) =

a(t) =

d 2x d 2y d 2z i+ 2 j+ 2k 2 dt dt dt

speed

|| v(t) || =

ad m M uh am

Table 13.6.1

d 2x d 2y i+ 2 j 2 dt dt

√(dxdt ) + (dydt ) 2

2

√(dxdt ) + (dydt ) + ( dzdt ) 2

|| v(t) || =

2

2

Example 1 A particle moves along a circular path in such a way that its x- and y-coordinates at time t are x = 2 cos t, (a) (b)

(c)

y = 2 sin t

Find the instantaneous velocity and speed of the particle at time t. Sketch the path of the particle, and show the position and velocity vectors at time t = π/4 with the velocity vector drawn so that its initial point is at the tip of the position vector. Show that at each instant the acceleration vector is perpendicular to the velocity vector.

Solution (a). At time t, the position vector is r(t) = 2 cos ti + 2 sin tj

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Vector-Valued Functions

us uf i

902

Yo

so the instantaneous velocity and speed are dr v(t) = = −2 sin ti + 2 cos tj dt v(t) = (−2 sin t)2 + (2 cos t)2 = 2

Solution (b). The graph of the parametric equations is a circle of radius 2 centered at the v = –√2 i + √2 j r = √2 i + √2 j p/4 –2

x

origin. At time t = π/4 the position and velocity vectors of the particles are √ √ r(π/4) = 2 cos(π/4)i + 2 sin(π/4) j = 2i + 2 j √ √ v(π/4) = −2 sin(π/4)i + 2 cos(π/4) j = − 2 i + 2 j These vectors and the circle are shown in Figure 13.6.2.

az

y

2

Solution (c). At time t, the acceleration vector is

dv = −2 cos ti − 2 sin tj dt One way of showing that v(t) and a(t) are perpendicular is to show that their dot product is zero (try it). However, it is easier to observe that a(t) is the negative of r(t), which implies that v(t) and a(t) are perpendicular, since at each point on a circle the radius and tangent line are perpendicular.

n

Figure 13.6.2

Ri

a(t) =

Ha

sa

Since v(t) can be obtained by differentiating r(t), and since a(t) can be obtained by differentiating v(t), it follows that r(t) can be obtained by integrating v(t), and v(t) can be obtained by integrating a(t). However, such integrations do not produce unique functions because constants of integration occur. Typically, initial conditions are required to determine these constants. Example 2 A particle moves through 3-space in such a way that its velocity is v(t) = i + tj + t 2 k

ad

Find the coordinates of the particle at time t = 1 given that the particle is at the point (−1, 2, 4) at time t = 0.

M uh am

m

Solution. Integrating the velocity function to obtain the position function yields

t3 t2 j+ k+C (5) 2 3 where C is a vector constant of integration. Since the coordinates of the particle at time t = 0 are (−1, 2, 4), the position vector at time t = 0 is r(t) =

v(t) dt =

(i + tj + t 2 k) dt = ti +

r(0) = −i + 2j + 4k

(6)

It follows on substituting t = 0 in (5) and equating the result with (6) that C = −i + 2 j + 4k Substituting this value of C in (5) and simplifying yields 2

3

t t r(t) = (t − 1)i + +2 j+ +4 k 2 3 Thus, at time t = 1 the position vector of the particle is 5 13 j+ k 2 3 so its coordinates at that instant are 0, 52 , 13 . 3 r(1) = 0i +

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••••••••••••••••••••••••••••••••••••••

DISPLACEMENT AND DISTANCE TRAVELED

Motion Along a Curve

903

us uf i

13.6

If a particle travels along a curve C in 2-space or 3-space, the displacement of the particle over the time interval t1 ≤ t ≤ t2 is commonly denoted by r and is defined as r = r(t2 ) − r(t1 )

(7)

r =

t2

v(t) dt = t1

t1

t2

t2

dr dt = r(t) dt

Yo

(Figure 13.6.3). The displacement vector, which describes the change in position of the particle during the time interval, can be obtained by integrating the velocity function from t1 to t2 : = r(t2 ) − r(t1 ) t1

Displacement

(8)

y

Ri

az

It follows from Theorem 13.3.1 that we can find the distance s traveled by a particle over a time interval t1 ≤ t ≤ t2 by integrating the speed over that interval, since t2 t2 dr Distance traveled s= v(t) dt (9) dt dt = t1 t1

L r(t1)

C

n

r(t2)

sa

∆r

Displacement = ∆r = r(t2) – r(t1)

Ha

x

Figure 13.6.3

Example 3 Suppose that a particle moves along a circular helix in 3-space so that its position vector at time t is

ad

r(t) = (4 cos πt)i + (4 sin πt) j + tk

M uh am

m

Find the distance traveled and the displacement of the particle during the time interval 1 ≤ t ≤ 5.

Solution. We have dr = (−4π sin πt)i + (4π cos πt)j + k dt √ v(t) = (−4π sin πt)2 + (4π cos πt)2 + 1 = 16π2 + 1

v(t) =

Thus, it follows from (9) that the distance traveled by the particle from time t = 1 to t = 5 is 5 s= 16π2 + 1 dt = 4 16π2 + 1 1

Moreover, it follows from (8) that the displacement over the time interval is r = r(5) − r(1) = (4 cos 5πi + 4 sin 5πj + 5k) − (4 cos πi + 4 sin πj + k) = (−4i + 5k) − (−4i + k) = 4k which tells us that the change in the position of the particle over the time interval was 4 units straight up.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Vector-Valued Functions

••••••••••••••••••••••••••••••••••••••

You know from your experience as an automobile passenger that if a car speeds up rapidly, then your body is thrown back against the backrest of the seat. You also know that if the car rounds a turn in the road, then your body is thrown toward the outside of the curve—the greater the curvature in the road, the greater the force with which you are thrown. The explanation of these effects can be understood by resolving the velocity and acceleration components of the motion into vector components that are parallel to the unit tangent and unit normal vectors. The following theorem explains how this can be done.

Yo

NORMAL AND TANGENTIAL COMPONENTS OF ACCELERATION

us uf i

904

13.6.2 THEOREM. If a particle moves along a smooth curve C in 2-space or 3-space, then at each point on the curve velocity and acceleration vectors can be written as

d 2s a = 2T+κ dt

ds dt

az

ds T dt

2 N

Ri

v=

2

(dsdt ) N

k

a

d 2s T dt 2

where s is an arc length parameter for the curve, and T, N, and κ denote the unit tangent vector, unit normal vector, and curvature at the point (Figure 13.6.4).

ds T dt

n

C

v= T

(11)

Proof. Formula (10) is just a restatement of (1). To obtain (11), we differentiate both sides of (10) with respect to t; this yields

d ds ds dT d 2s a= T = 2T+ dt dt dt dt dt

sa

N

(10)

Ha

Figure 13.6.4

d 2s ds dT ds T+ 2 dt dt ds dt 2 2 d s ds dT = 2T+ dt ds dt

2 d 2s ds = 2T+ κN dt dt

ad

=

Formula (10) of Section 13.5

M uh am

m

from which (11) follows. The coefficients of T and N in (11) are commonly denoted by aT =

d 2s dt 2

aN = κ

ds dt

2 (12–13)

in which case Formula (11) is expressed as a = aT T + aN N

(14)

In this formula the scalars aT and aN are called the tangential scalar component of acceleration and the normal scalar component of acceleration, and the vectors aT T and aN N are called the tangential vector component of acceleration and the normal vector component of acceleration. The scalar components of acceleration explain the effect that you experience when a car speeds up rapidly or rounds a turn. The rapid increase in speed produces a large value for d 2 s /dt 2 , which results in a large tangential scalar component of acceleration; and by Newton’s second law this corresponds to a large tangential force on the car in the direction

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of motion. To understand the effect of rounding a turn, observe that the normal scalar component of acceleration has the curvature κ and the square of speed ds /dt as factors. Thus, sharp turns or turns taken at high speed both produce large normal forces on the car.

a

aNN

aTT

y

x

905

• REMARK. • • • • • • • • • • • • • • • • • •

Formula (14) applies to motion in both 2-space and 3-space. What is interesting is that the 3-space formula does not involve the binormal vector B, so the acceleration vector always lies in the plane of T and N (the osculating plane), even for the most twisting paths of motion (Figure 13.6.5).

Yo

z

Motion Along a Curve

us uf i

13.6

Although Formulas (12) and (13) provide useful insight into the behavior of particles moving along curved paths, they are not always the best formulas for computations. The following theorem provides some more useful formulas that relate aT , aN , and κ to the velocity v and acceleration a.

az

a = aTT + aNN

Figure 13.6.5

||a||

aTT

u

aN

aN =

v × a v

κ=

v × a v3

(15–17)

aT

sa

aN N

v·a v

n

aT =

Ri

13.6.3 THEOREM. If a particle moves along a smooth curve C in 2-space or 3-space, then at each point on the curve the velocity v and the acceleration a are related to aT , aN , and κ by the formulas

Proof. As illustrated in Figure 13.6.6, let θ be the angle between the vector a and the vector aT T. Thus, aT = a cos θ

and

Ha

aT = ||a|| cos u aN = ||a|| sin u

aN = a sin θ

from which we obtain v·a va cos θ aT = a cos θ = = v v

Figure 13.6.6

ad

aN = a sin θ = κ=

v × a va sin θ = v v

aN v × a aN 1 v × a = = = 2 2 2 / (ds dt) v v v v3

M uh am

m

• REMARK. • • • • • • • • • • • • • • • • • • • • • • •

Theorem 13.6.3 applies to motion in 2-space and 3-space, but for motion in 2-space you will have to add a zero k component to v and a to calculate the cross product. Also, recall that for nonlinear smooth curves in 2-space the unit normal vector N is the inward normal; that is, it points toward the concave side of the curve. Thus, the same is true for aN N, since aN is a nonnegative scalar.

Example 4 time t is

Suppose that a particle moves through 3-space so that its position vector at

r(t) = ti + t 2 j + t 3 k (The path is the twisted cubic shown in Figure 13.1.5.) (a)

Find the scalar tangential and normal components of acceleration at time t.

(b)

Find the scalar tangential and normal components of acceleration at time t = 1.

(c)

Find the vector tangential and normal components of acceleration at time t = 1.

(d)

Find the curvature of the path at the point where the particle is located at time t = 1.

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Vector-Valued Functions

us uf i

906

Solution (a). We have v(t) = r (t) = i + 2tj + 3t 2 k

v(t) · a(t) = 4t + 18t 3

i j k

v(t) × a(t) = 1 2t 3t 2 = 6t 2 i − 6tj + 2k

0 2 6t

Yo

a(t) = v (t) = 2 j + 6tk √ v(t) = 1 + 4t 2 + 9t 4

Thus, from (15) and (16)

az

4t + 18t 3 v·a = v 1 + 4t 2 + 9t 4 36t 4 + 36t 2 + 4 9t 4 + 9t 2 + 1 v × a =2 aN = = 9t 4 + 4t 2 + 1 v 1 + 4t 2 + 9t 4

Ri

aT =

Solution (b). At time t = 1, the components aT and aN in part (a) are

and

aN = 2

n

22 aT = √ ≈ 5.88 14

19 ≈ 2.33 14

sa

Solution (c). Since T and v have the same direction, T can be obtained by normalizing v, that is, v(t) v(t)

Ha

T(t) =

ad

At time t = 1 we have v(1) i + 2 j + 3k 1 T(1) = = = √ (i + 2 j + 3k) v(1) i + 2 j + 3k 14 From this and part (b) we obtain the vector tangential component of acceleration:

m

11 22 33 11 22 aT (1)T(1) = √ T(1) = (i + 2 j + 3k) = i+ j+ k 7 7 7 7 14 To find the normal vector component of acceleration, we rewrite a = aT T + aN N as aN N = a − aT T

M uh am

Thus, at time t = 1 the normal vector component of acceleration is aN (1)N(1) = a(1) − aT (1)T(1)

11 22 33 = (2 j + 6k) − i+ j+ k 7 7 7 =−

11 8 9 i− j+ k 7 7 7

Solution (d ). We will apply Formula (17) with t = 1. From part (a) v(1) =

√ 14

and

v(1) × a(1) = 6i − 6 j + 2k

Thus, at time t = 1

√ v × a 76 1 38 κ= = √ 3 = ≈ 0.17 v3 14 7 ( 14)

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• FOR THE READER.

A MODEL OF PROJECTILE MOTION

aN =

a2 − aT2

(18)

Confirm that this is so in Example 4.

Yo

••••••••••••••••••••••••••••••••••••••

907

It follows from Figure 13.6.6 and the Theorem of Pythagoras that aN can be expressed in terms of v and aT as

Earlier in this text we examined various problems concerned with objects moving vertically in the Earth’s gravitational field (see Free-Fall Model 4.4.4, Example 4 of Section 5.7, and the subsection of Section 9.1 entitled A Model of Free-Fall Motion Retarded by Air Resistance). Now we will consider the motion of a projectile launched along a curved path in the Earth’s gravitational field. For this purpose we will need the vector version of Newton’s Second Law of Motion (9.1.1)

az

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Motion Along a Curve

us uf i

13.6

(19)

Ri

F = ma

and we will need to make three modeling assumptions:

•

The object remains sufficiently close to the Earth that we can assume the force of gravity to be constant.

n

The only force acting on the object after it is launched is the force of the Earth’s gravity. (Thus, air resistance and the gravitational effect of other planets and celestial objects are ignored.)

sa

v0

The mass m of the object is constant.

•

Let us assume that at time t = 0 an object of mass m is launched from a height of s0 above the Earth with an initial velocity vector of v0 . Furthermore, let us introduce an xy-coordinate system as shown in Figure 13.6.7. In this coordinate system the positive y-direction is up, the origin is at the surface of the Earth, and the initial coordinate of the object is (0, s0 ). Our objective is to use basic principles of physics to derive the velocity function v(t) and the position function r(t) from the acceleration function a(t) of the object. Our starting point is the physical observation that the downward force F of the Earth’s gravity on an object of mass m is

Ha

y

•

s0 x Earth

ad

Figure 13.6.7

M uh am

m

F = −mgj

where g is the acceleration due to gravity (see 9.4.3). It follows from this fact and Newton’s second law (19) that ma = −mgj or on canceling m from both sides a = −gj

(20)

Observe that this acceleration function does not involve t and hence is constant. We can now obtain the velocity function v(t) by integrating this acceleration function and using the initial condition v(0) = v0 to find the constant of integration. Integrating (20) with respect to t and keeping in mind that −gj is constant yields v(t) = −gj dt = −gtj + c1 where c1 is a vector constant of integration. Substituting t = 0 in this equation and using the initial condition v(0) = v0 yields v0 = c1

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Vector-Valued Functions

us uf i

908

Thus, the velocity function of the object is v(t) = −gtj + v0

(21)

Yo

To obtain the position function r(t) of the object, we will integrate the velocity function and use the known initial position of the object to find the constant of integration. For this purpose observe that the object has coordinates (0, s0 ) at time t = 0, so the position vector at that time is r(0) = 0i + s0 j = s0 j

(22)

az

This is the initial condition that we will need to find the constant of integration. Integrating (21) with respect to t yields r(t) = (−gtj + v0 ) dt = − 12 gt 2 j + tv0 + c2 (23)

s0 j = c2 so that (23) can be written as r(t) = − 12 gt 2 + s0 j + tv0

Ri

where c2 is another vector constant of integration. Substituting t = 0 in (23) and using initial condition (22) yields

(24)

sa

n

This formula expresses the position function of the object in terms of its known initial position and velocity. • REMARK.

m

PARAMETRIC EQUATIONS OF PROJECTILE MOTION

y

M uh am

v0

v0

(v0 sin a)j

a

(v0 cos a)i

Figure 13.6.8

Formulas (21) and (24) can be used to obtain parametric equations for the position and velocity in terms of the initial speed of the object and the angle that the initial velocity vector makes with the positive x-axis. For this purpose, let v0 = v0 be the initial speed, let α be the angle that the initial velocity vector v0 makes with the positive x-axis, let vx and vy be the horizontal and vertical scalar components of v(t) at time t, and let x and y be the horizontal and vertical components of r(t) at time t. As illustrated in Figure 13.6.8, the initial velocity vector can be expressed as

ad

••••••••••••••••••••••••••••••••••••••

Observe that the mass of the object does not enter into the final formulas for velocity and position. Physically, this means that the mass has no influence on the trajectory or the velocity of the object—these are completely determined by the initial position and velocity. This explains the famous observation of Galileo that two objects of different mass, released from the same height, will reach the ground at the same time if air resistance is neglected.

Ha

• • • • • • • • • • • • • • • • • • • • • • • • • • • •

x

v0 = (v0 cos α)i + (v0 sin α) j

Substituting this expression in (24) and combining like components yields (verify) r(t) = (v0 cos α)ti + s0 + (v0 sin α)t − 12 gt 2 j

(25)

(26)

which is equivalent to the parametric equations x = (v0 cos α)t,

y = s0 + (v0 sin α)t − 12 gt 2

(27)

Similarly, substituting (25) in (21) and combining like components yields v(t) = (v0 cos α)i + (v0 sin α − gt) j which is equivalent to the parametric equations vx = v0 cos α,

vy = v0 sin α − gt

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(28)

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Motion Along a Curve

909

us uf i

13.6

The parameter t can be eliminated in (27) by solving the first equation for t and substituting in the second equation. We leave it for you to show that this yields

g y = s0 + (tan α)x − x2 (29) 2v02 cos2 α

Yo

which is the equation of a parabola, since the right side is a quadratic polynomial in x. Thus, we have shown that the trajectory of the projectile is a parabolic arc. Example 5 A shell, fired from a cannon, has a muzzle speed (the speed as it leaves the barrel) of 800 ft/s. The barrel makes an angle of 45 ◦ with the horizontal and, for simplicity, the barrel opening is assumed to be at ground level.

Figure 13.6.9

Find parametric equations for the shell’s trajectory relative to the coordinate system in Figure 13.6.9.

(b)

How high does the shell rise?

(c)

How far does the shell travel horizontally?

(d)

What is the speed of the shell at its point of impact with the ground?

az

x

45°

(a)

Ri

y

Solution (a). From (27) with v0 = 800 ft/s, α = 45 ◦ , s0 = 0 ft (since the shell starts at ground level), and g = 32 ft/s2 , we obtain the parametric equations y = (800 sin 45 ◦ )t − 16t 2

n

x = (800 cos 45 ◦ )t,

sa

which simplify to √ √ x = 400 2t, y = 400 2t − 16t 2

(t ≥ 0)

(t ≥ 0)

(30)

Solution (b). The maximum height of the shell is the maximum value of y in (30), which

Ha

occurs when dy/dt = 0, that is, when √ √ 25 2 400 2 − 32t = 0 or t = 2 Substituting this value of t in (30) yields y = 5000 ft

ad

as the maximum height of the shell.

Solution (c). The shell will hit the ground when y = 0. From (30), this occurs when

M uh am

m

√ 400 2t − 16t 2 = 0

or

√ t (400 2 − 16t) = 0

The solution t = 0 corresponds to the initial position of the shell and the solution √ t = 25 2 to the time of impact. Substituting the latter value in the equation for x in (30) yields x = 20,000 ft as the horizontal distance traveled by the shell.

Solution (d ). From (30), the position function of the shell is √ √ r(t) = 400 2ti + (400 2t − 16t 2 ) j

so that the velocity function is √ √ v(t) = r (t) = 400 2i + (400 2 − 32t) j √ From part (c), impact occurs when t = 25 2, so the velocity vector at this point is √ √ √ √ √ √ v(25 2) = 400 2i + [400 2 − 32(25 2)] j = 400 2 i − 400 2 j Thus, the speed at impact is √ √ √ v(25 2) = (400 2)2 + (−400 2)2 = 800 ft/s

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Vector-Valued Functions

EXERCISE SET 13.6

Graphing Utility

C

us uf i

910

CAS

(b) What are the maximum and minimum speeds of the particle? (c) Use the graph to estimate the time at which the maximum speed first occurs. (d) Find the exact time at which the maximum speed first occurs.

Yo

In Exercises 1–4, r(t) is the position vector of a particle moving in the plane. Find the velocity, acceleration, and speed at an arbitrary time t. Then sketch the path of the particle together with the velocity and acceleration vectors at the indicated time t.

6. x = 1 + 3t, y = 2 − 4t, z = 7 + t; t = 2

14. Suppose that the position function of a particle moving in 3-space is r = 3 cos 2ti + sin 2tj + 4tk. (a) Use a graphing utility to graph the speed of the particle versus time from t = 0 to t = π. (b) Use the graph to estimate the maximum and minimum speeds of the particle. (c) Use the graph to estimate the time at which the maximum speed first occurs. (d) Find the exact values of the maximum and minimum speeds and the exact time at which the maximum speed first occurs.

7. x = 2 cos t, y = 2 sin t, z = t; t = π/4 8. r(t) = et sin ti + et cos tj + tk; t = π/2

In Exercises 15–18, use the given information to find the position and velocity vectors of the particle.

1. r(t) = 3 cos ti + 3 sin tj; t = π/3 2. r(t) = ti + t 2 j; t = 2 3. r(t) = et i + e−t j; t = 0

az

4. r(t) = (2 + 4t)i + (1 − t) j; t = 1

Ri

In Exercises 5–8, find the velocity, speed, and acceleration at the given time t of a particle moving along the given curve.

15. a(t) = − cos ti − sin tj; v(0) = i; r(0) = j

Ha

sa

9. As illustrated in the accompanying figure, suppose that the equations of motion of a particle moving along an elliptic path are x = a cos ωt, y = b sin ωt. (a) Show that the acceleration is directed toward the origin. (b) Show that the magnitude of the acceleration is proportional to the distance from the particle to the origin.

n

5. r(t) = ti + 12 t 2 j + 13 t 3 k; t = 1

y

b x

ad

a

m

Figure Ex-9

M uh am

10. Suppose that a particle vibrates in such a way that its position function is r(t) = 16 sin πti + 4 cos 2πt j, where distance is in millimeters and t is in seconds. (a) Find the velocity and acceleration at time t = 1 s. (b) Show that the particle moves along a parabolic curve. (c) Show that the particle moves back and forth along the curve.

11. Suppose that the position vector of a particle moving in the √ plane is r = 12 t i + t 3/2 j, t > 0. Find the minimum speed of the particle and its location when it has this speed. 12. Suppose that the motion of a particle is described by the position vector r = (t − t 2 )i − t 2 j. Find the minimum speed of the particle and its location when it has this speed. 13. Suppose that the position function of a particle moving in 2-space is r = sin 3ti − 2 cos 3tj. (a) Use a graphing utility to graph the speed of the particle versus time from t = 0 to t = 2π/3.

16. a(t) = i + e−t j; v(0) = 2i + j; r(0) = i − j

17. a(t) = sin ti + cos tj + et k; v(0) = k; r(0) = −i + k 18. a(t) = (t + 1)−2 j − e−2t k; v(0) = 3i − j; r(0) = 2k 19. What can you say about the trajectory of a particle that moves in 2-space or 3-space with zero acceleration? Justify your answer. 20. Recall from Theorem 13.2.9 that if r(t) is a vector-valued function in 2-space or 3-space, and if r(t) is constant for all t, then r(t) · r (t) = 0. (a) Translate this theorem into a statement about the motion of a particle in 2-space or 3-space. (b) Replace r(t) by r (t) in the theorem, and translate the result into a statement about the motion of a particle in 2-space or 3-space. 21. Find, to the nearest degree, the angle between v and a for r = t 3 i + t 2 j when t = 1. 22. Show that the angle between v and a is constant for the position vector r = et cos ti + et sin tj. Find the angle. 23. (a) Suppose that at time t = t0 an electron has a position vector of r = 3.5i − 1.7 j + k, and at a later time t = t1 it has a position vector of r = 4.2i + j − 2.4k. What is the displacement of the electron during the time interval from t0 to t1 ? (b) Suppose that during a certain time interval a proton has a displacement of r = 0.7i + 2.9 j − 1.2k and its final position vector is known to be r = 3.6k. What was the initial position vector of the proton? 24. Suppose that the position function of a particle moving along a circle in the xy-plane is r = 5 cos 2πti + 5 sin 2πtj.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

In Exercises 25–28, find the displacement and the distance traveled over the indicated time interval. 25. r = t 2 i + 13 t 3 j; 1 ≤ t ≤ 3

In Exercises 29 and 30, the position vectors of two particles are given. Show that the particles move along the same path but the speed of the first is constant and the speed of the second is not.

47. The nuclear accelerator at the Enrico Fermi Laboratory is circular with a radius of 1 km. Find the scalar normal component of acceleration of a proton moving around the accelerator with a constant speed of 2.9 × 105 km/s.

48. Suppose that a particle moves with nonzero acceleration along the curve y = f(x). Use part (b) of Exercise 17 in Section 13.5 to show that the acceleration vector is tangent to the curve at each point where f (x) = 0. In Exercises 49 and 50, use the given information and Exercise 17 of Section 13.5 to find the normal scalar component of acceleration as a function of x. 49. A particle moves along the parabola y = x 2 with a constant speed of 3 units per second.

Ri

29. r1 = 2 cos 3ti + 2 sin 3tj r2 = 2 cos(t 2 )i + 2 sin(t 2 ) j

1 4

az

26. r = (1 − 3 sin t)i + 3 cos tj; 0 ≤ t ≤ 3π/2 √ 27. r = et i + e−t j + 2tk; 0 ≤ t ≤ ln 3 28. r = cos 2ti + (1 − cos 2t) j + 3 + 21 cos 2t k; 0 ≤ t ≤ π

(4t − 1)2 + cos2 πt; t = √ 46. v = t 4 + 5t 2 + 3; t = 1 45. v =

911

Yo

(a) Sketch some typical displacement vectors over the time interval from t = 0 to t = 1. (b) What is the distance traveled by the particle during the time interval?

Motion Along a Curve

us uf i

13.6

50. A particle moves along the curve x = ln y with a constant speed of 2 units per second.

(t ≥ 0)

30. r1 = (3 + 2t)i + tj + (1 − t)k r2 = (5 − 2t 3 )i + (1 − t 3 ) j + t 3 k

51. a(1) = i + 2 j − 2k; aT (1) = 3

Ha

sa

In Exercises 31–38, the position function of a particle is given. Use Theorem 13.6.3 to find (a) the scalar tangential and normal components of acceleration at the stated time t; (b) the vector tangential and normal components of acceleration at the stated time t; (c) the curvature of the path at the point where the particle is located at the stated time t.

n

In Exercises 51 and 52, use the given information to find the normal scalar component of acceleration at time t = 1.

31. r = e−t i + et j; t = 0

√ π/2

ad

32. r = cos(t 2 )i + sin(t 2 ) j; t =

33. r = (t 3 − 2t)i + (t 2 − 4) j; t = 1 34. r = et cos ti + et sin tj; t = π/4

m

35. r = (1/t)i + t 2 j + t 3 k; t = 1 36. r = et i + e−2t j + tk; t = 0

37. r = 3 sin ti + 2 cos tj − sin 2tk; t = π/2

M uh am

38. r = 2i + t 3 j − 16 ln tk; t = 1

In Exercises 39–42, v and a are given at a certain instant of time. Find aT , aN , T, and N at this instant.

39. v = −4 j, a = 2i + 3 j

40. v = i + 2 j, a = 3i

41. v = 2i + 2 j + k, a = i + 2k

42. v = 3i − 4k, a = i − j + 2k In Exercises 43–46, the speed v of a particle at an arbitrary time t is given. Find the scalar tangential component of acceleration at the indicated time.

43. v = 44. v =

√

3t 2 + 4; t = 2

t 2 + e−3t ; t = 0

52. a(1) = 9; aT (1)T(1) = 2i − 2 j + k

53. An automobile travels at a constant speed around a curve whose radius of curvature is 1000 m. What is the maximum allowable speed if the maximum acceptable value for the normal scalar component of acceleration is 1.5 m/s2 ? 54. If an automobile of mass m rounds a curve, then its inward vector component of acceleration aN N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal scalar component of acceleration are related by the equation F = maN N. Thus, 2 ds F = mκ dt Use this result to find the magnitude of the frictional force in newtons exerted by the road on a 500-kg go-cart driven at a speed of 10 km/h around a circular track of radius 15 m. [Note: 1 N = 1 kg·m/s2 ] 55. A shell is fired from ground level with a muzzle speed of 320 ft/s and elevation angle of 60 ◦ . Find (a) parametric equations for the shell’s trajectory (b) the maximum height reached by the shell (c) the horizontal distance traveled by the shell (d) the speed of the shell at impact. 56. Solve Exercise 55 assuming that the muzzle speed is 980 m/s and the elevation angle is 45 ◦ . 57. A rock is thrown downward from the top of a building, 168 ft high, at an angle of 60 ◦ with the horizontal. How far from the base of the building will the rock land if its initial speed is 80 ft/s?

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132 ft/s

58. Solve Exercise 57 assuming that the rock is thrown horizontally at a speed of 80 ft/s. 59. A shell is to be fired from ground level at an elevation angle of 30 ◦ . What should the muzzle speed be in order for the maximum height of the shell to be 2500 ft?

3000 ft

60. A shell, fired from ground level at an elevation angle of 45 ◦ , hits the ground 24,500 m away. Calculate the muzzle speed of the shell.

Figure Ex-66

61. Find two elevation angles that will enable a shell, fired from ground level with a muzzle speed of 800 ft/s, to hit a groundlevel target 10,000 ft away.

67. A shell is fired from ground level at an elevation angle of α and a muzzle speed of v0 . (a) Show that the maximum height reached by the shell is

62. A ball rolls off a table 4 ft high while moving at a constant speed of 5 ft/s. (a) How long does it take for the ball to hit the floor after it leaves the table? (b) At what speed does the ball hit the floor? (c) If a ball were dropped from rest at table height just as the rolling ball leaves the table, which ball would hit the ground first? Justify your answer.

az

maximum height =

(v0 sin α)2 2g

Ri

(b) The horizontal range R of the shell is the horizontal distance traveled when the shell returns to ground level. Show that R = (v02 sin 2α)/g. For what elevation angle will the range be maximum? What is the maximum range?

n

68. A shell is fired from ground level with an elevation angle α and a muzzle speed of v0 . Find two angles that can be used to hit a target at ground level that is a distance of three-fourths the maximum range of the shell. Express your answer to the nearest tenth of a degree. [Hint: See Exercise 67(b).]

sa

63. As illustrated in the accompanying figure, a fire hose sprays water with an initial velocity of 40 ft/s at an angle of 60 ◦ with the horizontal. (a) Confirm that the water will clear corner point A. (b) Confirm that the water will hit the roof. (c) How far from corner point A will the water hit the roof?

us uf i

Vector-Valued Functions

Yo

912

Ha

64. What is the minimum initial velocity that will allow the water in Exercise 63 to hit the roof?

m

ad

65. As shown in the accompanying figure, water is sprayed from a hose with an initial velocity of 35 m/s at an angle of 45 ◦ with the horizontal. (a) What is the radius of curvature of the stream at the point where it leaves the hose? (b) What is the maximum height of the stream above the nozzle of the hose?

A

20 ft

M uh am

60° 4 ft

15 ft

Figure Ex-63

35 m /s 45°

25 ft

Figure Ex-65

66. As illustrated in the accompanying figure, a train is traveling on a curved track. At a point where the train is traveling at a speed of 132 ft/s and the radius of curvature of the track is 3000 ft, the engineer hits the brakes to make the train slow down at a constant rate of 7.5 ft/s2 . (a) Find the magnitude of the acceleration vector at the instant the engineer hits the brakes. (b) Approximate the angle between the acceleration vector and the unit tangent vector T at the instant the engineer hits the brakes.

69. At time t = 0 a baseball that is 5 ft above the ground is hit with a bat. The ball leaves the bat with a speed of 80 ft/s at an angle of 30 ◦ above the horizontal. (a) How long will it take for the baseball to hit the ground? Express your answer to the nearest hundredth of a second. (b) Use the result in part (a) to find the horizontal distance traveled by the ball. Express your answer to the nearest tenth of a foot. 70. At time t = 0 a projectile is fired from a height h above level ground at an elevation angle of α with a speed v. Let R be the horizontal distance to the point where the projectile hits the ground. (a) Show that α and R must satisfy the equation g(sec2 α)R 2 − 2v 2 (tan α)R − 2v 2 h = 0 (b) If g, h, and v are constant, then the equation in part (a) defines R implicitly as a function of α. Let R0 be the maximum value of R and α0 the value of α when R = R0 . Use implicit differentiation to find dR /dα and show that v2 tan α0 = gR0 [Hint: Assume that dR /dα = 0 when R is maximum.] (c) Use the results in parts (a) and (b) to show that v 2 R0 = v + 2gh g and v α0 = tan−1 v 2 + 2gh

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71. At time t = 0 a skier leaves the end of a ski jump with a speed of v0 ft/s at an angle α with the horizontal (see the accompanying figure). The skier lands 259 ft down the incline 2.9 s later. (a) Approximate v0 to the nearest ft/s and α to the nearest degree. (b) Use a CAS or a calculating utility with a numerical integration capability to approximate the distance traveled by the skier. (Use g = 32 ft/s2 as the acceleration due to gravity.)

y v0

913

x

a

23°

259

ft

az

Figure Ex-71

Yo

C

Kepler’s Laws of Planetary Motion

us uf i

13.7

Ri

13.7 KEPLER’S LAWS OF PLANETARY MOTION

KEPLER’S LAWS

In Section 11.6 we stated the following laws of planetary motion that were published by Johannes Kepler in 1609 in his book known as Astronomia Nova.

Ha

••••••••••••••••••••••••••••••••••••••

sa

n

One of the great advances in the history of astronomy occurred in the early 1600s when Johannes Kepler∗ deduced from empirical data that all planets in our solar system move in elliptical orbits with the Sun at a focus. Subsequently, Isaac Newton showed mathematically that such planetary motion is the consequence of an inversesquare law of gravitational attraction. In this section we will use the concepts developed in the preceding sections of this chapter to derive three basic laws of planetary motion, known as Kepler’s laws.

13.7.1 •

KEPLER’S LAWS.

ad

First law (Law of Orbits). Each planet moves in an elliptical orbit with the Sun at a focus. • Second law (Law of Areas). Equal areas are swept out in equal times by the line from the Sun to a planet.

m

•

M uh am

••••••••••••••••••••••••••••••••••••••

CENTRAL FORCES

Third law (Law of Periods). The square of a planet’s period (the time it takes the planet to complete one orbit about the Sun) is proportional to the cube of the semimajor axis of its orbit.

If a particle moves under the influence of a single force that is always directed toward a fixed point O, then the particle is said to be moving in a central force field. The force is called a central force, and the point O is called the center of force. For example, in the simplest model of planetary motion, it is assumed that the only force acting on a planet is the force of the Sun’s gravity, directed toward the center of the Sun. This model, which produces Kepler’s laws, ignores the forces that other celestial objects exert on the planet as well as the minor effect that the planet’s gravity has on the Sun. Central force models are also used to study the motion of comets, asteroids, planetary moons, and artificial satellites. They also have important applications in electromagnetics. Our objective in this section is to develop some basic principles about central force fields and then use those results to derive Kepler’s laws. Suppose that a particle P of mass m moves in a central force field due to a force F that is directed toward a fixed point O, and let r = r(t) be the position vector from O to P ∗

See biography on p. 779.

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Vector-Valued Functions

P

us uf i

914

(Figure 13.7.1). Let v = v(t) and a = a(t) be the velocity and acceleration functions of the particle, and assume that F and a are related by Newton’s second law (F = ma). Our first objective is to show that the particle P moves in a plane containing the point O. For this purpose observe that a has the same direction as F by Newton’s second law, and this implies that a and r are oppositely directed vectors. Thus, it follows from part (c) of Theorem 12.4.5 that

P r

F

Since the velocity and acceleration of the particle are given by v = dr/dt and a = dv/dt, respectively, we have d dv dr + × v = (r × a) + (v × v) = 0 + 0 = 0 (1) (r × v) = r × dt dt dt Integrating the left and right sides of this equation with respect to t yields

O

az

O Figure 13.7.1

Yo

r×a=0

r×v=b

(2)

Ri

where b is a constant (independent of t). However, b is orthogonal to both r and v, so we can conclude that r = r(t) and v = v(t) lie in a fixed plane containing the point O. • REMARK. • • • • • • •

Our next objective is to derive the position function of a particle moving under a central force in a polar coordinate system. For this purpose we will need the following result, known as Newton’s Law of Universal Gravitation.

sa

NEWTON’S LAW OF UNIVERSAL GRAVITATION

n

••••••••••••••••••••••••••••••••••••••

The preceding discussion shows that each planet moves in a plane through the center of the Sun. Astronomers call this plane the ecliptic of the planet.

Ha

13.7.2 NEWTON’S LAW OF UNIVERSAL GRAVITATION. Every particle of matter in the Universe attracts every other particle of matter in the Universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. Specifically, if a particle of mass M and a particle of mass m are at a distance r from one another, then they attract each other with equal and opposite forces, F and −F, of magnitude GMm r2

ad

F =

(3)

m

where G is a constant called the universal gravitational constant.

M uh am

m

r

F

–F

To obtain a formula for the vector force F that mass M exerts on mass m, we will let r be the radius vector from mass M to mass m (Figure 13.7.2). Thus, the distance r between the masses is r, and the force F can be expressed in terms of r as

r r F = F − = F − r r which from (3) can be expressed as

M

M exerts force F on m, and m exerts force –F on M.

Figure 13.7.2

F=−

GMm r r3

(4)

We start by finding a formula for the acceleration function. To do this we use Formula (4) and Newton’s second law to obtain GMm ma = − 3 r r from which we obtain a=−

GM r r3

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Kepler’s Laws of Planetary Motion

• REMARK. • • • • • • • • • • • •

915

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13.7

Observe that the acceleration a depends on the mass M but not on the mass m. Thus, for example, the acceleration of a planet is affected by the mass of the Sun but not by its own mass.

• •

b y

•

A polar coordinate system is introduced with its pole at mass M and oriented so θ = 0 at time t = 0.

•

The vector v0 is perpendicular to the polar axis at time t = 0.

Moreover, to ensure that the polar angle θ increases with t, let us agree to observe this polar coordinate system looking toward the pole from the terminal point of the vector b = r0 × v0 . We will also find it useful to superimpose an xyz-coordinate system on the polar coordinate system with the positive z-axis in the direction of b (Figure 13.7.3). For computational purposes, it will be helpful to denote r0 by r0 and v0 by v0 , in which case we can express the vectors r0 and v0 in xyz-coordinates as

v0

Ri

x

u=0

Figure 13.7.3

r0 = r0 i

y

and

v0 = v0 j

sa

and the vector b as

n

u r0

The distance r from m to M is minimum at time t = 0. The mass m has nonzero position and velocity vectors r0 and v0 at time t = 0.

az

z

Yo

To obtain a formula for the position function of the mass m, we will need to introduce a coordinate system and make some assumptions about the initial conditions. Let us assume:

b = r0 × v0 = r0 i × v0 j = r0 v0 k v0 j

(6)

(Figure 13.7.4). It will also be useful to introduce the unit vector

x

u = cos θi + sin θ j

Ha

r0 i

(7)

which will allow us to express the polar form of the position vector r as r = r cos θ i + r sin θ j = r(cos θ i + sin θ j) = ru

and to express the acceleration vector a in terms of u by rewriting (5) as GM a=− 2 u (9) r We are now ready to derive the position function of the mass m in polar coordinates. For this purpose, recall from (2) that the vector b = r × v is constant, so it follows from (6) that the relationship

M uh am

m

ad

Figure 13.7.4

(8)

b = r × v = r0 v0 k

(10)

holds for all values of t. Now let us examine b from another point of view. It follows from (8) that dr du dr d v= + u = (ru) = r dt dt dt dt and hence

du dr dr du du b = r × v = (ru) × r + u = r 2u × + r u × u = r 2u × (11) dt dt dt dt dt But (7) implies that du du dθ dθ = = (− sin θi + cos θ j) dt dθ dt dt so du dθ u× = k (12) dt dt Substituting (12) in (11) yields dθ b = r2 k (13) dt

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Vector-Valued Functions

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916

Thus, it follows from (7), (9), and (13) that

dθ GM a × b = − 2 (cos θ i + sin θ j) × r 2 k dt r

dθ du = GM dt dt / From this formula and the fact that db dt = 0 (since b is constant), we obtain d du db dv + × b = a × b = GM (v × b) = v × dt dt dt dt Integrating both sides of this equation with respect to t yields

(14)

Yo

= GM(− sin θ i + cos θ j)

v × b = GMu + C

(15)

az

where C is a vector constant of integration. This constant can be obtained by evaluating both sides of the equation at t = 0. We leave it as an exercise to show that from which it follows that v × b = GMu + (r0 v02 − GM)i

Ri

C = (r0 v02 − GM)i

(16) (17)

n

We can now obtain the position function by computing the scalar triple product r · (v × b) in two ways. First we use (10) and property (11) of Section 12.4 to obtain r · (v × b) = (r × v) · b = b · b = r02 v02

(18)

sa

and next we use (17) to obtain

Ha

r · (v × b) = r · (GMu) + r · (r0 v02 − GM)i r = r · GM + ru · (r0 v02 − GM)i r = GMr + r(r0 v02 − GM) cos θ

If we now equate this to (18), we obtain r02 v02 = GMr + r(r0 v02 − GM) cos θ

ad

which when solved for r gives

M uh am

m

r=

Parabola

e=1

Hyperbola

e>1

Ellipse

0 < e<1

Circle

e=0

Figure 13.7.5

GM +

r02 v02 (r0 v02 − GM) cos θ

r02 v02 = GM

r0 v02 1+ − 1 cos θ GM

(19)

or more simply k r= (20) 1 + e cos θ where r 2v2 r0 v02 k = 0 0 and e = −1 (21–22) GM GM We will leave it as an exercise to show that e ≥ 0. Accepting this to be so, it follows by comparing (20) to Formula (3) of Section 11.6 that the trajectory is a conic section with eccentricity e, the focus at the pole, and d = k /e. Thus, depending on whether e < 1, e = 1, or e > 1, the trajectory will be, respectively, an ellipse, a parabola, or a hyperbola (Figure 13.7.5). Note from Formula (22) that e depends on r0 and v0 , so the exact form of the trajectory is determined by the mass M and the initial conditions. If the initial conditions are such that e < 1, then the mass m becomes trapped in an elliptical orbit; otherwise the mass m “escapes” and never returns to its initial position. Accordingly, the initial velocity that

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Kepler’s Laws of Planetary Motion

917

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13.7

Yo

produces an eccentricity of e = 1 is called the escape speed and is denoted by vesc . Thus, it follows from (22) that 2GM vesc = (23) r0 (verify). ••••••••••••••••••••••••••••••••••••••

It follows from our general discussion of central force fields that the planets have elliptical orbits with the Sun at the focus, which is Kepler’s first law. To derive Kepler’s second law, we begin by equating (10) and (13) to obtain dθ r2 (24) = r0 v0 dt To prove that the radial line from the center of the Sun to the center of a planet sweeps out equal areas in equal times, let r = f(θ) denote the polar equation of the planet, and let A denote the area swept out by the radial line as it varies from any fixed angle θ0 to an angle θ. It follows from Theorem 11.3.2 that A can be expressed as θ 1 A= [f(φ)]2 dφ θ0 2

Ri

az

KEPLER’S FIRST AND SECOND LAWS

Ha

sa

n

where the dummy variable φ is introduced for the integration to reserve θ for the upper limit. It now follows from Part 2 of the Fundamental Theorem of Calculus and the chain rule that dA dθ dA dθ 1 1 dθ = = [f(θ)]2 = r2 dt dθ dt 2 dt 2 dt Thus, it follows from (24) that dA 1 (25) = r0 v0 dt 2 which shows that A changes at a constant rate. This implies that equal areas are swept out in equal times. To derive Kepler’s third law, we let a and b be the semimajor and semiminor axes of the elliptical orbit, and we recall that the area of this ellipse is πab. It follows by integrating (25) that in t units of time the radial line will sweep out an area of A = 12 r0 v0 t. Thus, if T denotes the time required for the planet to make one revolution around the Sun (the period), then the radial line will sweep out the area of the entire ellipse during that time and hence

ad

••••••••••••••••••••••••••••••••••••••

M uh am

m

KEPLER’S THIRD LAW

πab = 12 r0 v0 T from which we obtain 4π2 a 2 b2 T2 = r02 v02

(26)

However, it follows from Formula (1) of Section 11.6 and the relationship c2 = a 2 − b2 for an ellipse that c a 2 − b2 e= = a a Thus, b2 = a 2 (1 − e2 ) and hence (26) can be written as T2 =

4π2 a 4 (1 − e2 ) r02 v02

But comparing Equation (20) to Equation (17) of Section 11.6 shows that k = a(1 − e2 )

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Vector-Valued Functions

Finally, substituting this expression and (21) in (27) yields T2 =

4π2 a 3 4π2 a 3 r02 v02 4π2 3 a k = = GM r02 v02 r02 v02 GM

us uf i

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

(28)

Yo

Thus, we have proved that T 2 is proportional to a 3 , which is Kepler’s third law. When convenient, Formula (28) can also be expressed as 2π 3/2 a T =√ GM ••••••••••••••••••••••••••••••••••••••

Kepler’s second and third laws and Formula (23) also apply to satellites that orbit a celestial body; we need only interpret M to be the mass of the body exerting the force and m to be the mass of the satellite. Values of GM that are required in many of the formulas in this section have been determined experimentally for various attracting bodies (Table 13.7.1).

az

ARTIFICIAL SATELLITES

(29)

international system

british engineering system

Earth

GM = 3.99 × 1014 m3/s2 GM = 3.99 × 105 km3/s2

GM = 1.41 × 1016 ft 3/s2 GM = 1.24 × 1012 mi3/ h2

Sun

GM = 1.33 × 1020 m3/s2 GM = 1.33 × 1011 km3/s2

GM = 4.69 × 1021 ft 3/s2 GM = 4.13 × 1017 mi3/ h2

GM = 4.90 × 1012 m3/s2 GM = 4.90 × 103 km3/s2

GM = 1.73 × 1014 ft 3/s2 GM = 1.53 × 1010 mi3/ h2

sa

Ha

u Perigee

Recall that for orbits of planets around the Sun, the point at which the distance between the center of the planet and the center of the Sun is maximum is called the aphelion and the point at which it is minimum the perihelion. For satellites around the Earth the point at which the maximum distance occurs is called the apogee and the point at which the minimum distance occurs is called the perigee (Figure 13.7.6). The actual distances between the centers at apogee and perigee are called the apogee distance and the perigee distance.

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Apogee

n

attracting body

Moon

r

Ri

Table 13.7.1

Example 1 A geosynchronous orbit for a satellite is a circular orbit about the equator of the Earth in which the satellite stays fixed over a point on the equator. Use the fact that the Earth makes one revolution about its axis every 24 hours to find the altitude in miles of a communications satellite in geosynchronous orbit. Assume the Earth to be a sphere of radius 4000 mi.

M uh am

m

Figure 13.7.6

Solution. To remain fixed over a point on the equator, the satellite must have a period of T = 24 h. It follows from (28) or (29) and the Earth value of GM = 1.24 × 1012 mi3 /h2 from Table 13.7.1 that 2 12 2 3 GMT 3 (1.24 × 10 )(24) a= = ≈ 26,250 mi 2 2 4π 4π and hence the altitude h of the satellite is h = 26,250 − 4000 = 22,250 mi

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Kepler’s Laws of Planetary Motion

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13.7

EXERCISE SET 13.7

radius r0 is constant and is given by GM v= r0

10. The universal gravitational constant is approximately G = 6.67 × 10−11 m3 /kg·s2 and the semimajor axis of the Earth’s orbit is approximately

a = 149.6 × 106 km Estimate the mass of the Sun in kg.

n

2. (a) Use the results in Exercise 1 to show that rmax − rmin e= rmax + rmin (b) Show that 1+e rmax = rmin 1−e

Ri

1. Suppose that a particle is in an elliptical orbit in a central force field in which the center of force is at a focus, and let rmin and rmax denote the minimum and maximum distances from the particle to the center of force. Review the discussion of ellipses in polar coordinates in Section 11.6, and show that if the ellipse has eccentricity e and semimajor axis a, then rmin = a(1 − e) and rmax = a(1 + e).

7. Use the result in Exercise 6 to find the speed in km/s of a satellite in a circular orbit that is 200 km above the surface of the Earth. 8. Use the result in Exercise 6 to find the speed in mi/h of a communications satellite that is in geosynchronous orbit around the Earth. [See Example 1.] 9. Find the escape speed in km/s for a space probe in a circular orbit that is 300 km above the surface of the Earth.

az

radius of Earth = 4000 mi = 6440 km radius of Moon = 1080 mi = 1740 km 1 year (Earth year) = 365 days

Yo

In Exercises that require numerical values, use Table 13.7.1 and the following values, where needed:

11. (a) The eccentricity of the Moon’s orbit around the Earth is 0.055, and its semimajor axis is a = 238,900 mi. Find the maximum and minimum distances between the surface of the Earth and the surface of the Moon. (b) Find the period of the Moon’s orbit in days.

sa

3. (a) Obtain the value of C given in Formula (16) by setting t = 0 in (15). (b) Use Formulas (7), (17), and (22) to show that

Ha

v × b = GM[(e + cos θ)i + sin θ j] (c) Show that v × b = vb. (d) Use the results in parts (b) and (c) to show that the speed of a particle in an elliptical orbit is v0 2 v= e + 2e cos θ + 1 1+e

m

ad

4. Use the result in Exercise 3(d) to show that when a particle in an elliptical orbit with eccentricity e reaches an end of the minor axis, its speed is 1−e v = v0 1+e

M uh am

5. Use the result in Exercise 3(d) to show that for a particle in an elliptical orbit with eccentricity e, the maximum and minimum speeds are related by 1+e vmax = vmin 1−e 6. Use Formula (22) and the result in part (d) of Exercise 3 to show that the speed v of a particle in a circular orbit of

12. (a) Vanguard 1 was launched in March 1958 with perigee and apogee altitudes above the Earth of 649 km and 4340 km, respectively. Find the length of the semimajor axis of its orbit. (b) Use the result in part (a) of Exercise 2 to find the eccentricity of its orbit. (c) Find the period of Vanguard I in minutes. 13. (a) Suppose that a space probe is in a circular orbit at an altitude of 180 mi above the surface of the Earth. Use the result in Exercise 6 to find its speed. (b) During a very short period of time, a thruster rocket on the space probe is fired to increase the speed of the probe by 600 mi/h in its direction of motion. Find the eccentricity of the resulting elliptical orbit, and use the result in part (b) of Exercise 2 to find the apogee altitude. 14. Show that the quantity e defined by Formula (22) is nonnegative. [Hint: The polar axis was chosen so that r is minimum when θ = 0.]

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Vector-Valued Functions

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920

SUPPLEMENTARY EXERCISES 1. In words, what is meant by the graph of a vector-valued function r(t)?

(b) What does Theorem 13.2.9 tell you about the acceleration vector of a particle that moves with constant speed? (c) Show that the particle with position function r(t) =

1 − 14 cos2 t cos ti + 1 − 14 cos2 t sin tj + 12 cos tk

Yo

2. Describe the graph of the vector-valued function. (a) r = r0 + t (r1 − r0 ) (b) r = r0 + t (r1 − r0 ) (0 ≤ t ≤ 1) (c) r = r0 + tr (t0 )

moves over a sphere.

3. In words, describe what happens geometrically to r(t) if lim r(t) = L.

9. As illustrated in the accompanying figure, suppose that a particle moves counterclockwise around a circle of radius R centered at the origin at a constant rate of ω radians per second. This is called uniform circular motion. If we assume that the particle is at the point (R, 0) at time t = 0, then its position function will be

t →a

az

4. Suppose that r(t) is the position function of a particle moving in 2-space or 3-space. In each part, explain what the given quantity represents physically. t1 dr dr dt (a) (b) (c) r(t) dt dt

Ri

r(t) = R cos ωti + R sin ωtj

t0

(a) Show that the velocity vector v(t) is always tangent to the circle and that the particle has constant speed v given by

5. Suppose that r(t) is a smooth vector-valued function. State the definitions of T(t), N(t), and B(t).

v = Rω (b) Show that the acceleration vector a(t) is always directed toward the center of the circle and has constant magnitude a given by

sa

7. In Supplementary Exercise 34 of Chapter 11, we defined the Cornu spiral parametrically as 2

2

t t πu πu x= cos sin du, y = du 2 2 0 0

n

6. State the definition of “curvature” and explain what it means geometrically.

T =

2π 2πR = ω v y

v(t)

a(t) vt

M uh am

x

(R, 0)

m

ad

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This curve, which is graphed in the accompanying figure, is used in highway design to create a gradual transition from a straight road (zero curvature) to an exit ramp with positive curvature. (a) Express the Cornu spiral as a vector-valued function r(t), and then use Theorem 13.3.4 to show that s = t is the arc length parameter with reference point (0, 0). (b) Replace t by s and use Formula (1) of Section 13.5 to show that κ(s) = π|s|. [Note: If s ≥ 0, then the curvature κ(s) = πs increases from 0 at a constant rate with respect to s. This makes the spiral ideal for joining a curved road to a straight road.] (c) What happens to the curvature of the Cornu spiral as s → +⬁? In words, explain why this is consistent with the graph.

a = Rω2 (c) Show that the time T required for the particle to make one complete revolution is

Figure Ex-9

y

x

Figure Ex-7

8. (a) What does Theorem 13.2.9 tell you about the velocity vector of a particle that moves over a sphere?

10. If a particle of mass m has uniform circular motion (see Exercise 9), then the acceleration vector a(t) is called the centripetal acceleration. According to Newton’s second law, this acceleration must be produced by some force F(t), called the centripetal force, that is related to a(t) by the equation F(t) = ma(t). If this force is not present, then the particle cannot undergo uniform circular motion. (a) Show that the direction of the centripetal force varies with time but that it has constant magnitude F given by F =

mv 2 R

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(b) An astronaut with a mass of m = 70 kg orbits the Earth at an altitude of h = 3200 km with a constant speed of v = 6.43 km/s. Find her centripetal acceleration assuming that the radius of the Earth is 6440 km. (c) What centripetal gravitational force in newtons does the Earth exert on the astronaut?

y

Trajectory

eu er r

2x 2 + y 2 + 6z2 = 24

and sketch the intersection.

n

12. Sketch the graph of the vector-valued function that is defined piecewise by ďŁą 3ti, 0 â‰¤ t â‰¤ 13 ďŁ´ ďŁ´ ďŁ˛ r(t) = (2 âˆ’ 3t)i + (3t âˆ’ 1) j, 13 â‰¤ t â‰¤ 23 ďŁ´ ďŁ´ ďŁł 2 3(1 âˆ’ t) j, â‰¤t â‰¤1 3

az

and

14. As illustrated in the accompanying figure, the polar coordinates of a rocket are tracked by radar from a point that is b units from the launching pad. Show that the speed v of the rocket can be expressed in terms b, Î¸, and dÎ¸ /dt as dÎ¸ v = b sec2 Î¸ dt

Ri

y = x2

Figure Ex-13

u

sa

13. Suppose that the position function of a point moving in the xy-plane is

x(t) = r(t) cos Î¸(t), This yields

Ha

r = x(t)i + y(t) j This equation can be expressed in polar coordinates by making the substitution y(t) = r(t) sin Î¸(t)

ad

r = r(t) cos Î¸(t)i + r(t) sin Î¸(t) j which can be expressed as

M uh am

m

r = r(t)er (t) where er (t) = cos Î¸(t)i + sin Î¸(t) j. (a) Show that er (t) is a unit vector that has the same direction as the radius vector r if r(t) > 0 and that eÎ¸ (t) = âˆ’ sin Î¸(t)i + cos Î¸(t) j is the unit vector that results when er (t) is rotated counterclockwise through an angle of Ď€/2. The vector er (t) is called the radial unit vector, and the vector eÎ¸ (t) is called the transverse unit vector (see the accompanying figure). (b) Show that the velocity function v = v(t) can be expressed in terms of radial and transverse components as dr dÎ¸ v= er + r eÎ¸ dt dt

(c) Show that the acceleration function a = a(t) can be expressed in terms of radial and transverse components as 2 2 d 2r dÎ¸ d Î¸ dr dÎ¸ a= âˆ’ r + r + 2 e eÎ¸ r dt 2 dt dt 2 dt dt

x

Yo

u

11. (a) Show that the graph of the vector-valued function r(t) = t sin Ď€ti + tj + t cos Ď€tk lies on the surface of a cone, and sketch the cone. (b) Find parametric equations for the intersection of the surfaces

921

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Supplementary Exercises

b Figure Ex-14

15. Find the arc length parametrization of the line through P (âˆ’1, 4, 3) and Q(0, 2, 5) that has reference point P and orients the line in the direction from P to Q. 16. A player throws a ball with an initial speed of 60 ft/s at an unknown angle Îą with the horizontal from a point that is 4 ft above the floor of a gymnasium. Given that the ceiling of the gymnasium is 25 ft high, determine the maximum height h at which the ball can hit a wall that is 60 ft away (see the accompanying figure).

25 ft

60 ft/s

h

a

4 ft 60 ft Figure Ex-16

17. Find all points on the graph of r(t) = t 3 i + 10tj + 5t 2 k at which the tangent line is perpendicular to the tangent line at t = 1. 18. Solve the vector initial-value problem dr = r, r(0) = r0 dt for the unknown vector-valued function r(t).

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Vector-Valued Functions

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922

20. Let v = v(t) and a = a(t) be the velocity and acceleration vectors for a particle moving in 2-space or 3-space. Show that the rate of change of its speed can be expressed as d 1 (v) = (v Âˇ a) dt v

M uh am

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ad

Ha

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Ri

az

Yo

19. At time t = 0 a particle at the origin of an xyz-coordinate system has a velocity vector of v0 = i + 2 j âˆ’ k. The acceleration function of the particle is a(t) = 2t 2 i + j + cos 2tk. (a) Find the position function of the particle. (b) Find the speed of the particle at time t = 1.

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Yo

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

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ad

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P ARTIAL D ERIVATIVES

n this chapter we will extend many of the basic concepts of calculus to functions of two or more variables, commonly called functions of â€œseveral variables.â€? We will begin by discussing limits and continuity for functions of two and three variables, then we will define derivatives of such functions, and then we will use these derivatives to study tangent planes, rates of change, slopes of surfaces, and maximization and minimization problems. Although many of the basic ideas that we developed for functions of one variable will carry over in a natural way, functions of several variables are intrinsically more complicated than functions of one variable, so we will need to develop new tools and new ideas to deal with such functions.

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Partial Derivatives

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14.1 FUNCTIONS OF TWO OR MORE VARIABLES

••••••••••••••••••••••••••••••••••••••

There are many familiar formulas in which a given variable depends on two or more other variables. For example, the area A of a triangle depends on the base length b and height h by the formula A = 12 bh; the volume V of a rectangular box depends on the length l, the width w, and the height h by the formula V = lwh; and the arithmetic average x¯ of n real numbers, x1 , x2 , . . . , xn , depends on those numbers by the formula x¯ = (x1 + x2 + · · · + xn )/n. Thus, we say that

az

NOTATION AND TERMINOLOGY

Yo

In previous sections we studied real-valued functions of a real variable and vectorvalued functions of a real variable. In this section we will consider real-valued functions of two or more real variables.

A is a function of the two variables b and h;

Ri

V is a function of the three variables l, w, and h; x¯ is a function of the n variables x1 , x2 , . . . , xn .

n

The terminology and notation for functions of two or more variables is similar to that for functions of one variable. For example, the expression z = f(x, y)

Ha

w = f(x, y, z)

sa

means that z is a function of x and y in the sense that a unique value of the dependent variable z is determined by specifying values for the independent variables x and y. Similarly, expresses w as a function of x, y, and z, and u = f(x1 , x2 , . . . , xn )

to the nearest integer. Clearly, Table 14.1.1 provides us with more information “at a glance” than does Equation (1). For example, if the temperature is 30 ◦ F and the speed of the wind is 10 mi/h, then it feels as if the temperature is 16 ◦ F. We can also use Table 14.1.1 to obtain convenient estimates of windchill values that are not explicitly displayed. Table 14.1.1 temperature T (°F )

wind speed v (mi/h)

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ad

expresses u as a function of x1 , x2 , . . . , xn . It is sometimes more convenient (or even necessary) to describe a function z = f(x, y) using a table of values instead of an explicit formula. For example, recall from Example 3 of Section 1.2 that the windchill index is the temperature at a wind speed of 4 mi/h that would produce the same sensation on exposed skin as the current temperature and wind speed combination. Table 14.1.1 displays the windchill index (WCI) as a function of the actual air temperature T and the speed v of the wind. The entries in Table 14.1.1 were obtained by rounding the values obtained by the formula √ WCI = 91.4 + (91.4 − T )(0.0203v − 0.304 v − 0.474) (1)

20

25

30

35

5

16

22

27

32

10

3

10

16

22

15

–5

2

9

15

20

–11

–3

4

11

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Functions of Two or More Variables

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14.1

Example 1 Use Table 14.1.1 to estimate the windchill index if the air temperature is 30 ◦ F and the wind speed is 12 mi/h.

Solution. Although there is no entry in Table 14.1.1 that corresponds to T = 30 and

az

Yo

v = 12, we can estimate the corresponding windchill by a process known as linear interpolation. For T = 30 and v = 10 we have WCI = 16, and for T = 30 and v = 15 we have WCI = 9. That is, an increase of 5 mi/h in the value of v is reflected by a decrease of 7 ◦ F in WCI. If WCI were a linear function of v when T is held fixed at 30 ◦ F, then an increase = 2.8 ◦ F in the value of 2 mi/h in the wind speed would produce a decrease of 25 · 7 = 14 5 of WCI. If we assume that WCI is a linear function of v for T = 30 ◦ F and v between 10 and 15 mi/h, then it follows that at a temperature of 30 ◦ F and a wind speed of 12 mi/h the windchill index will be 16 − 2.8 = 13.2 ◦ F. Note that this value compares reasonably well with the value WCI = 12.5939 ◦ F obtained using Equation (1).

Ri

We will find it useful to think of functions of two or three independent variables in geometric terms. For example, if z = f(x, y), then we can view (x, y) as a point in the xy-plane and think of f as a rule that associates a unique numerical value z with the point (x, y); similarly, we can think of w = f(x, y, z) as a rule that associates a unique numerical value w with a point (x, y, z) in an xyz-coordinate system (Figure 14.1.1). z

x

f

(x, y, z)

sa

f

(x, y)

n

y

y

w

z

Figure 14.1.1

Ha

z = f (x, y)

x

w = f (x, y, z)

M uh am

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ad

As with functions of one variable, the independent variables of a function of two or more variables may be restricted to lie in some set D, which we call the domain of f . Sometimes the domain will be determined by physical restrictions on the variables. If the function is defined by a formula and if there are no physical restrictions or other restrictions stated explicitly, then it is understood that the domain consists of all points for which the formula yields a real value for the dependent variable. We call this the natural domain of the function. The following definitions summarize this discussion. 14.1.1 DEFINITION. A function f of two variables, x and y, is a rule that assigns a unique real number f(x, y) to each point (x, y) in some set D in the xy-plane. 14.1.2 DEFINITION. A function f of three variables, x, y, and z, is a rule that assigns a unique real number f(x, y, z) to each point (x, y, z) in some set D in three-dimensional space.

• REMARK. • • • • • • • • • • • • • • • • •

In more advanced courses the notion of “n-dimensional space” for n > 3 is defined, and a function f of n real variables, x1 , x2 , . . . , xn , is regarded as a rule that assigns a unique real number f(x1 , x2 , . . . , xn ) to each “point” (x1 , x2 , . . . , xn ) in some set in n-dimensional space. However, we will not pursue that idea in this text. Example 2 Let √ f(x, y) = 3x 2 y − 1 Find f(1, 4), f(0, 9), f(t 2 , t), f(ab, 9b), and the natural domain of f.

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Partial Derivatives y

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926

Solution. By substitution

y≥0

Yo

x

√ f(1, 4) = 3(1)2 4 − 1 = 5 √ f(0, 9) = 3(0)2 9 − 1 = −1 √ √ f(t 2 , t) = 3(t 2 )2 t − 1 = 3t 4 t − 1 √ √ f(ab, 9b) = 3(ab)2 9b − 1 = 9a 2 b2 b − 1 √ Because of the radical y in the formula for f , we must have y ≥ 0 to avoid imaginary values for f(x, y). Thus, the natural domain of f consists of all points in the xy-plane that are on or above the x-axis. (See Figure 14.1.2.)

The solid boundary line is included in the domain.

Figure 14.1.2

Example 3 Sketch the natural domain of the function f(x, y) = ln(x 2 − y).

Solution. ln(x 2 − y) is defined only when 0 < x 2 − y or y < x 2 . We first sketch the

az

y

parabola y = x 2 as a “dashed” curve. The region y < x 2 then consists of all points below this curve (Figure 14.1.3).

y = x2

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Example 4 Let f(x, y, z) = 1 − x 2 − y 2 − z2 Find f 0, 12 , − 12 and the natural domain of f. x

1 − (0)2 −

1 2 2

sa

f 0, 12 , − 12 =

n

Solution. By substitution,

2 − − 12 = 12

Because of the square root sign, we must have 0 ≤ 1 − x 2 − y 2 − z2 in order to have a real value for f(x, y, z). Rewriting this inequality in the form

The dashed boundary does not belong to the domain.

Ha

x 2 + y 2 + z2 ≤ 1

Figure 14.1.3

we see that the natural domain of f consists of all points on or within the sphere x 2 + y 2 + z2 = 1

Recall that for a function f of one variable, the graph of f(x) in the xy-plane was defined to be the graph of the equation y = f(x). Similarly, if f is a function of two variables, we define the graph of f(x, y) in xyz-space to be the graph of the equation z = f(x, y). In general, such a graph will be a surface in 3-space. Example 5 In each part, describe the graph of the function in an xyz-coordinate system. (b) f(x, y) = 1 − x 2 − y 2 (a) f(x, y) = 1 − x − 12 y (c) f(x, y) = − x 2 + y 2

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GRAPHS OF FUNCTIONS OF TWO VARIABLES

ad

••••••••••••••••••••••••••••••••••••••

Solution (a). By definition, the graph of the given function is the graph of the equation z = 1 − x − 12 y which is a plane. A triangular portion of the plane can be sketched by plotting the intersections with the coordinate axes and joining them with line segments (Figure 14.1.4a).

Solution (b). By definition, the graph of the given function is the graph of the equation z=

1 − x2 − y2

(2)

After squaring both sides, this can be rewritten as x 2 + y 2 + z2 = 1 which represents a sphere of radius 1, centered at the origin. Since (2) imposes the added condition that z ≥ 0, the graph is just the upper hemisphere (Figure 14.1.4b).

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z

z

√1 – x 2 – y 2

z=

z = 1 – x – 12 y

(0, 0, 1)

Functions of Two or More Variables

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14.1

z

1

y

y

z = – √x 2 + y 2

y 1

-1

(1, 0, 0)

x

x

(a)

1

(b)

Figure 14.1.4

Yo

(0, 2, 0)

x

(c)

Solution (c). The graph of the given function is the graph of the equation

az

z = − x2 + y2

After squaring, we obtain

Ri

z2 = x 2 + y 2

(3)

which is the equation of a circular cone (see Table 12.7.1). Since (3) imposes the condition that z ≤ 0, the graph is just the lower nappe of the cone (Figure 14.1.4c). ••••••••••••••••••••••••••••••••••••••

We are all familiar with the topographic (or contour) maps in which a three-dimensional landscape, such as a mountain range, is represented by two-dimensional contour lines or curves of constant elevation. Consider, for example, the model hill and its contour map shown in Figure 14.1.5. The contour map is constructed by passing planes of constant elevation through the hill, projecting the resulting contours onto a flat surface, and labeling the contours with their elevations. In Figure 14.1.5, note how the two gullies appear as indentations in the contour lines and how the curves are close together on the contour map where the hill has a steep slope and become more widely spaced where the slope is gradual.

Ha

6 5 4 3 2 1

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Hundreds of feet

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LEVEL CURVES

z

x

Figure 14.1.6

f (x, y) = k

2

3 45 6

A contour map of the model hill

Figure 14.1.5

z = f (x, y) z=k

Level curve of height k

A perspective view of a model hill with two gullies

1

y

Contour maps are also useful for studying functions of two variables. If the surface z = f(x, y) is cut by the horizontal plane z = k, then at all points on the intersection we have f(x, y) = k. The projection of this intersection onto the xy-plane is called the level curve of height k or the level curve with constant k (Figure 14.1.6). A set of level curves for z = f(x, y) is called a contour plot or contour map of f. Example 6 The graph of the function f(x, y) = y 2 − x 2 in xyz-space is the hyperbolic paraboloid (saddle surface) shown in Figure 14.1.7a. The level curves have equations of the form y 2 − x 2 = k. For k > 0 these curves are hyperbolas opening along lines parallel to the y-axis; for k < 0 they are hyperbolas opening along lines parallel to the x-axis; and for k = 0 the level curve consists of the intersecting lines y + x = 0 and y − x = 0 (Figure 14.1.7b).

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Partial Derivatives

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z

40 30 20

5.0 4.0 4.0 3.0 2.0

3.0

10

0

2.0

– 10

1.0

Yo

–1.0 –2.0 –3.0

0 –1.0 –2.0

x

–40 – 30 – 20 – 10

y

– 20 – 30 –40

928

0

–3.0 –4.0

10

–5.0

az

x

(a)

(b)

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Figure 14.1.7

20 30 40

Example 7

Sketch the contour plot of f(x, y) = 4x 2 + y 2 using level curves of height k = 0, 1, 2, 3, 4, 5.

(b)

Sketch the contour plot of f(x, y) = 2 − x − y using level curves of height k = −6, −4, −2, 0, 2, 4, 6.

sa

n

(a)

Solution (a). The graph of the surface z = 4x 2 + y 2 is the paraboloid shown in Fig-

Ha

ure 14.1.8, so we can reasonably expect the contour plot to be a family of ellipses centered at the origin. The level curve of height k has the equation 4x 2 + y 2 = k. If k = 0, then the graph is the single point (0, 0). For k > 0 we can rewrite the equation as x2 y2 + =1 k /4 k

ad

√ √ which represents a family of ellipses with x-intercepts ± k /2 and y-intercepts ± k. The contour plot for the specified values of k is shown in Figure 14.1.9.

Solution (b). The graph of the surface z = 2 − x − y is the plane shown in Figure 14.1.10,

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so we can reasonably expect the contour plot to be a family of parallel lines. The level curve of height k has the equation 2 − x − y = k, which we can rewrite as y = −x + (2 − k)

This represents a family of parallel lines of slope −1. The contour plot for the specified values of k is shown in Figure 14.1.11. z y

k k k k

z = 4x 2 + y 2

z=k

5 4 3 2 x

y

k=1 k=0

x

Figure 14.1.8

= = = =

Figure 14.1.9

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Functions of Two or More Variables y

z z = 2–x–y

z=k

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14.1

Yo

x

k = –6 k = –4 k = –2 k=0 k=2 k=4 k=6

y

x

Figure 14.1.11

az

Figure 14.1.10 ••••••••••••••••••••••••••••••••••••••

Except in the simplest cases, contour plots can be difficult to produce without the help of a graphing utility. Figure 14.1.12 illustrates how graphing technology can be used to display level curves. The table shows two graphical representations of the level curves of the function f(x, y) = |sin x sin y| produced with a CAS over the domain 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2π.

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CONTOUR PLOTS USING TECHNOLOGY

z

6

n

5

sa

4 3 y

Ha

2

x

1 0

0

Figure 14.1.12

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LEVEL SURFACES

z

y

x

Level surfaces of

f (x, y, z) = x2 + y2 + z2 Figure 14.1.13

2

3

4

5

z

x

0.928 0.833 0.737 0.642 0.547 0.451 0.356 0.261 0.165 0.07

y

6

Observe that the graph of y = f(x) is a curve in 2-space, and the graph of z = f(x, y) is a surface in 3-space, so the number of dimensions required for these graphs is one greater than the number of independent variables. Accordingly, there is no “direct” way to graph a function of three variables since four dimensions are required. However, if k is a constant, then the graph of the equation f(x, y, z) = k will generally be a surface in 3-space (e.g., the graph of x 2 + y 2 + z2 = 1 is a sphere), which we call the level surface with constant k. Some geometric insight into the behavior of the function f can sometimes be obtained by graphing these level surfaces for various values of k.

ad

••••••••••••••••••••••••••••••••••••••

1

z > 0.928 < 0.928 < 0.833 < 0.737 < 0.642 < 0.547 < 0.451 < 0.356 < 0.261 < 0.165 < 0.07

• REMARK. • • • • • • • • • • • •

The term “level surface” is standard but confusing, since a level surface need not be level in the sense of being horizontal; it is simply a surface on which all values of f are the same. Example 8 Describe the level surfaces of (a) f(x, y, z) = x 2 + y 2 + z2

(b) f(x, y, z) = z2 − x 2 − y 2

Solution (a). The level surfaces have equations of the form x 2 + y 2 + z2 = k

√ For k > 0 the graph of this equation is a sphere of radius k, centered at the origin; for k = 0 the graph is the single point (0, 0, 0); and for k < 0 there is no level surface (Figure 14.1.13).

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Partial Derivatives

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Solution (b). The level surfaces have equations of the form z2 − x 2 − y 2 = k

As discussed in Section 12.7, this equation represents a cone if k = 0, a hyperboloid of two sheets if k > 0, and a hyperboloid of one sheet if k < 0 (Figure 14.1.14). ••••••••••••••••••••••••••••••••••••••

Yo

Generating surfaces with a graphing utility is more complicated than generating plane curves because there are more factors that must be taken into account. We can only touch on the ideas here, so if you want to use a graphing utility, its documentation will be your main source of information. Graphing utilities can only show a portion of xyz-space in a viewing screen, so the first step in graphing a surface is to determine which portion of xyz-space you want to display. This region is called the viewing window or viewing box. For example, the first row of Table 14.1.2 shows the effect of graphing the paraboloid z = x 2 + y 2 in three different viewing windows. However, within a fixed viewing window, the appearance of the surface is also affected by the viewpoint, that is, the direction from which the surface is viewed, and the distance from the viewer to the surface. For example, the second row of Table 14.1.2 shows the graph of the paraboloid z = x 2 + y 2 from three different viewpoints using the viewing window in the first part of the table.

GRAPHING FUNCTIONS OF TWO VARIABLES USING TECHNOLOGY

k<0 k=0 k>0

Ri

az

z

y

• FOR THE READER. x

If you have a graphing utility that can generate surfaces in 3-space, read the documentation and try to duplicate Table 14.1.2.

n

• • • • • • •

• FOR THE READER.

ad

Figure 14.1.14

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Level surfaces of

f (x, y, z) = z2 – x2 – y2

Table 14.1.3 shows six surfaces in 3-space. Examine each surface and convince yourself that the contour plot describes its level curves. This will take a little thought because the mesh lines on the surface are traces in vertical planes, whereas the level curves are traces in horizontal planes. In these contour plots the color gradation is from dark to light as z increases. If you have a graphing utility that can generate surfaces in 3-space, try to duplicate some of these figures. You need not match the colors or generate the coordinate axes.

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• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Table 14.1.2 y

–2

M uh am

y

–2 0

2

m

0

0

2

8

10

20

z 4

z 5

z 10

0

0 0

2

y

–2

2

0

–2 x

2

0

–2 x

0

2

–2 x

-2

y 0

8

2 x 0

z 4

-2

0

2 8 z 4

-2

0 y

2

-2

0 y

2

0 -2 0 2

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x

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Functions of Two or More Variables

Table 14.1.3

surface

contour plot

surface

contour plot

z = e x sin y

z = cos y z

z

6

6

4

Yo

5

2

4 y 3

y 0 x

-2

2

-4 y

-6 -3 -2 -1

y

0 x

1

2

3

az

x

z = sin(√x2 + y2 )

1

z = xye – 2(x

2 1 y 0 -1

n

Ha

3

m

-3 -3 -2 -1

Graphing Utility

M uh am

EXERCISE SET 14.1

0

1

3 1 y 0

y

-3 -3 -2 -1

0 x

1

2

3

0 x

1

2

3

z = |xy| 3

z

2 1 y 0 -1 -2

ad

y

x

2

-2

x

-1

-2

x

z = cos(xy) z

0 -2

-1

-6 -4 -2 0 2 4 6 x

y

1

+ y2 )

z

sa

x

2

Ri

6 4 2 y 0 -2 -4 -6

z

931

us uf i

14.1

C

0 x

1

2

3

x

y

-3 -3 -2 -1

CAS

Exercises 1–8 are concerned with functions of two variables.

1. Let f(x, y) = x 2 y + 1. Find (a) f(2, 1) (b) f(1, 2) (d) f(1, −3) (e) f(3a, a) √ 2. Let f(x, y) = x + 3 xy. Find (a) f(t, t 2 ) (b) f(x, x 2 )

(c) f(0, 0) (f ) f(ab, a − b).

5. Find F(g(x), h(y)) if F(x, y) = xexy , g(x) = x 3 , and h(y) = 3y + 1. 6. Find g(u(x, y), v(x, y)) if g(x, y) = y sin(x 2 y), u(x, y) = x 2 y 3 , and v(x, y) = πxy.

3. Let f(x, y) = xy + 3. Find (a) f(x + y, x − y) (b) f(xy, 3x 2 y 3 ).

7. Let f(x, y) = x + 3x 2 y 2 , x(t) = t 2 , and y(t) = t 3 . Find (a) f(x(t), y(t)) (b) f(x(0), y(0)) (c) f(x(2), y(2)). √ 8. Let g(x, y) = ye−3x , x(t) = ln(t 2 + 1), and y(t) = t. Find g(x(t), y(t)).

4. Let g(x) = x sin x. Find (a) g(x /y) (b) g(xy)

9. Refer to Table 14.1.1 and use the method of Example 1 to estimate the windchill index when

(c) f(2y 2 , 4y).

(c) g(x − y).

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Partial Derivatives

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(a) the temperature is 25 ◦ F and the wind speed is 7 mi/h (b) the temperature is 28 ◦ F and the wind speed is 5 mi/h.

Exercises 17 and 18 are concerned with functions of four or more variables.

10. Refer to Table 14.1.1 and use the method of Example 1 to estimate the windchill index when (a) the temperature is 35 ◦ F and the wind speed is 14 mi/h (b) the temperature is 32 ◦ F and the wind speed is 10 mi/h.

20

25

30

3

71

74

77

79

4

62

66

70

73

5

53

59

63

67

Table Ex-11

M uh am

m

ad

12. Use the table in Exercise 11 to complete parts (a)–(c). (a) What is the wet-bulb depression if the air temperature is 30 ◦ C and the relative humidity is 73%? (b) Use the method of Example 1 to estimate the relative humidity if the air temperature is 15 ◦ C and the wet-bulb depression is 4.25 ◦ C. (c) Use the method of Example 1 to estimate the relative humidity if the air temperature is 26 ◦ C and the wet-bulb depression is 3 ◦ C. Exercises 13–16 involve functions of three variables.

13. Let f(x, y, z) = xy 2 z3 + 3. Find (a) f(2, 1, 2) (b) f(−3, 2, 1) (c) f(0, 0, 0) (d) f(a, a, a) (f ) f(a + b, a − b, b). (e) f(t, t 2 , −t)

14. Let f(x, y, z) = zxy + x. Find (a) f(x + y, x − y, x 2 ) (b) f(xy, y /x, xz). 15. Find F(f(x), g(y), h(z)) if F(x, y, z) = yexyz , f(x) = x 2 , g(y) = y + 1, and h(z) = z2 .

16. Find g(u(x, y, z), v(x, y, z), w(x, y, z)) if g(x, y, z) = z sin xy, u(x, y, z) = x 2 z3 , v(x, y, z) = πxyz, and w(x, y, z) = xy /z.

Yo k=1

az

18. (a) Let f(u, v, λ, φ) = eu+v cos λ tan φ. Find f(−2, 2, 0, π/4). (b) Let f(x1 , x2 , . . . , xn ) = x12 + x22 + · · · + xn2 . Find f(1, 2, . . . , n).

Ri

In Exercises 19–22, sketch the domain of f. Use solid lines for portions of the boundary included in the domain and dashed lines for portions not included. 19. f(x, y) = ln(1 − x 2 − y 2 ) 20. f(x, y) = 1 x − y2

n

21. f(x, y) =

Ha

wet-bulb depression (°C )

air temperature (°C ) 15

Find f(1, 1, . . . , 1).

x2 + y2 − 4

22. f(x, y) = ln xy

In Exercises 23 and 24, describe the domain of f in words. √

− y+2 23. (a) f(x, y) = xe (b) f(x, y, z) = 25 − x 2 − y 2 − z2 (c) f(x, y, z) = exyz √ 4 − x2 24. (a) f(x, y) = 2 (b) f(x, y) = ln(y − 2x) y +3 xyz (c) f(x, y, z) = x+y+z

sa

11. One method for determining relative humidity is to wet the bulb of a thermometer, whirl it through the air, and then compare the thermometer reading with the actual air temperature. If the relative humidity is less than 100%, the reading on the thermometer will be less than the temperature of the air. This difference in temperature is known as the wet-bulb depression. The accompanying table gives the relative humidity as a function of the air temperature and the wet-bulb depression. Use the table to complete parts (a)–(c). (a) What is the relative humidity if the air temperature is 20 ◦ C and the wet-bulb thermometer reads 16 ◦ C? (b) Use the method of Example 1 to estimate the relative humidity if the air temperature is 25 ◦ C and the wet-bulb depression is 3.5 ◦ C. (c) Use the method of Example 1 to estimate the relative humidity if the air temperature is 22 ◦ C and the wet-bulb depression is 5 ◦ C.

√ 17. (a) Let f(x,√y, z, t) = x 2 y 3 z + t. Find f( 5, 2, π, 3π). n kxk . (b) Let f(x1 , x2 , . . . , xn ) =

In Exercises 25–34, sketch the graph of f.

9 − x2 − y2

25. f(x, y) = 3 27. f(x, y) = x 2 + y 2

26. f(x, y) =

29. f(x, y) = x 2 − y 2 31. f(x, y) = x 2 + y 2 + 1

30. f(x, y) = 4 − x 2 − y 2 32. f(x, y) = x 2 + y 2 − 1

33. f(x, y) = y + 1

34. f(x, y) = x 2

28. f(x, y) = x 2 + y 2

35. In each part, match the contour plot with one of the functions f(x, y) = x 2 + y 2 , f(x, y) = x 2 + y 2 , f(x, y) = 1 − x 2 − y 2 by inspection, and explain your reasoning. The larger the value of z, the lighter the color in the contour plot, and the contours correspond to equally spaced values of z. (a) 2

(b) 2

(c) 2

y 0

y 0

y 0

-2 -2

0

x

2

-2 -2

0

x

2

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-2 -2

0

x

2

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y

(b)

6

y 0

x

(d)

y 0

2

0

6

x

38. A curve connecting points of equal atmospheric pressure on a weather map is called an isobar. On a typical weather map the isobars refer to pressure at mean sea level and are given in units of millibars (mb). Mathematically, isobars are level curves for the pressure function p(x, y) defined at the geographic points (x, y) represented on the map. Tightly packed isobars correspond to steep slopes on the graph of the pressure function, and these are usually associated with strong winds—the steeper the slope, the greater the speed of the wind. (a) Referring to the accompanying weather map, is the wind speed greater in Medicine Hat, Alberta or in Chicago? Explain your reasoning. (b) Estimate the average rate of change in atmospheric pressure from Medicine Hat to Chicago, given that the distance between the cities is approximately 1400 mi.

z

x

y

(IV)

z

Ri

x

(II)

x

x

n

x z

(III)

0

6

-6 -6

2

0

Figure Ex-37

y 0

0

2 1

sa

(I)

3

-2 -2

6

B

az

0

(c) 2

-2 -2

2

y 0

Elevations in hundreds of feet

A 4

1012

Ha

-6 -6

2 3

y

1010 1008

Medicine Hat

1006

z

ad

1004 1002

x

m

Figure Ex-36

y

x

Chicago

1000 996

y

1000

994

998

37. In each part, the questions refer to the contour map in the accompanying figure. (a) Is A or B the higher point? Explain your reasoning. (b) Is A or B on the steeper slope? Explain your reasoning. (c) Starting at A and moving so that y remains constant and x increases, will the elevation begin to increase or decrease? (d) Starting at B and moving so that y remains constant and x increases, will the elevation begin to increase or decrease? (e) Starting at A and moving so that x remains constant and y decreases, will the elevation begin to increase or decrease? (f ) Starting at B and moving so that x remains constant and y decreases, will the elevation begin to increase or decrease?

M uh am

933

Yo

36. In each part, match the contour plot with one of the surfaces in the accompanying figure by inspection, and explain your reasoning. The larger the value of z, the lighter the color in the contour plot. (a)

Functions of Two or More Variables

us uf i

14.1

998

1006 Pressure in millibars (mb)

Figure Ex-38

In Exercises 39–44, sketch the level curve z = k for the specified values of k. 39. z = x 2 + y 2 ; k = 0, 1, 2, 3, 4 40. z = y /x; k = −2, −1, 0, 1, 2 41. z = x 2 + y; k = −2, −1, 0, 1, 2 42. z = x 2 + 9y 2 ; k = 0, 1, 2, 3, 4 43. z = x 2 − y 2 ; k = −2, −1, 0, 1, 2 44. z = y csc x; k = −2, −1, 0, 1, 2 In Exercises 45–48, sketch the level surface f(x, y, z) = k. 45. f(x, y, z) = 4x 2 + y 2 + 4z2 ; k = 16 46. f(x, y, z) = x 2 + y 2 − z2 ; k = 0

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Partial Derivatives

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47. f(x, y, z) = z − x 2 − y 2 + 4; k = 7

sketch the equipotential curves at which V = 2.0, V = 1.0, and V = 0.5.

48. f(x, y, z) = 4x − 2y + z; k = 1

59. Let f(x, y) = x 2 + y 3 . (a) Use a graphing utility to generate the level curve that passes through the point (2, −1). (b) Generate the level curve of height 1. √ 60. Let f(x, y) = 2 xy. (a) Use a graphing utility to generate the level curve that passes through the point (2, 2). (b) Generate the level curve of height 8.

In Exercises 49–52, describe the level surfaces in words. 49. f(x, y, z) = (x − 2)2 + y 2 + z2 51. f(x, y, z) = x 2 + z2

Yo

50. f(x, y, z) = 3x − y + 2z 52. f(x, y, z) = z − x 2 − y 2

53. Let f(x, y) = x 2 − 2x 3 + 3xy. Find an equation of the level curve that passes through the point (a) (−1, 1) (b) (0, 0) (c) (2, −1). 54. Let f(x, y) = ye . Find an equation of the level curve that passes through the point (a) (ln 2, 1) (b) (0, 3) (c) (1, −2).

az

62. Let f(x, y) =

1 x e 10

sin y.

(a) Use a CAS to generate the graph of f for 0 ≤ x ≤ 4 and 0 ≤ y ≤ 2π. (b) Generate a contour plot for the surface, and confirm visually that it is consistent with the surface obtained in part (a). (c) Read the appropriate documentation and explore the effect of generating the graph of f from various viewpoints.

n

C

Ha

sa

57. If T (x, y) is the temperature at a point (x, y) on a thin metal plate in the xy-plane, then the level curves of T are called isothermal curves. All points on such a curve are at the same temperature. Suppose that a plate occupies the first quadrant and T (x, y) = xy. (a) Sketch the isothermal curves on which T = 1, T = 2, and T = 3. (b) An ant, initially at (1, 4), wants to walk on the plate so that the temperature along its path remains constant. What path should the ant take and what is the temperature along that path?

2

Ri

55. Let f(x, y, z) = x 2 + y 2 − z. Find an equation of the level surface that passes through the point (a) (1, −2, 0) (b) (1, 0, 3) (c) (0, 0, 0). 56. Let f(x, y, z) = xyz + 3. Find an equation of the level surface that passes through the point (a) (1, 0, 2) (b) (−2, 4, 1) (c) (0, 0, 0).

61. Let f(x, y) = xe−(x +y ) . (a) Use a CAS to generate the graph of f for −2 ≤ x ≤ 2 and −2 ≤ y ≤ 2. (b) Generate a contour plot for the surface, and confirm visually that it is consistent with the surface obtained in part (a). (c) Read the appropriate documentation and explore the effect of generating the graph of f from various viewpoints. 2

C

x

(a) g(x, y) = f(x − 1, y) (b) g(x, y) = 1 + f(x, y) (c) g(x, y) = −f(x, y + 1)

ad

58. If V (x, y) is the voltage or potential at a point (x, y) in the xy-plane, then the level curves of V are called equipotential curves. Along such a curve, the voltage remains constant. Given that 8 V (x, y) = 16 + x 2 + y 2

64. (a) Sketch the graph of f(x, y) = e−(x +y ) . (b) In this part, describe in words how the graph of the 2 2 function g(x, y) = e−a(x +y ) is related to the graph of f for positive values of a. 2

m

M uh am

••••••••••••••••••••••••••••••••••••••

LIMITS ALONG CURVES

63. In each part, describe in words how the graph of g is related to the graph of f.

2

14.2 LIMITS AND CONTINUITY In this section we will introduce the notions of limit and continuity for functions of two or more variables. We will not go into great detail—our objective is to develop the basic concepts accurately and to obtain results needed in later sections. A more extensive study of these topics is usually given in advanced calculus. For a function of one variable there are two one-sided limits at a number x0 , namely lim f(x) and

x → x0 +

lim f(x)

x → x0 −

reflecting the fact that there are only two directions from which x can approach x0 , the right or the left. For functions of two or three variables the situation is more complicated

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y

Limits and Continuity

935

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14.2

because there are infinitely many different curves along which one point can approach another (Figure 14.2.1). Our first objective in this section is to define the limit of f(x, y) as (x, y) approaches a point (x0 , y0 ) along a curve C (and similarly for functions of three variables). If C is a smooth parametric curve in 2-space or 3-space that is represented by the equations

(x, y)

x

x = x(t),

y = y(t)

x = x(t),

or

y = y(t),

z = z(t)

Yo

(x 0, y0)

and if x0 = x(t0 ), y0 = y(t0 ), and z0 = z(t0 ), then the limits

Figure 14.2.1

lim

(x, y) → (x0 , y0 ) (along C)

f(x, y)

and

lim

(x, y, z) → (x0 , y0 , z0 ) (along C)

f(x, y, z)

are defined by

lim

(x, y, z) → (x0 , y0 , z0 ) (along C)

(x(t), y(t), f (x(t), y(t)))

az

z

f(x, y) = lim f(x(t), y(t))

(1)

f(x, y, z) = lim f(x(t), y(t), z(t))

(2)

t → t0

Ri

lim

(x, y) → (x0 , y0 ) (along C)

t → t0

L

Simply stated, limits along parametric curves are obtained by substituting the parametric equations into the formula for the function f and computing the appropriate limit of the resulting function of one variable. A geometric interpretation of the limit along a curve for a function of two variables is shown in Figure 14.2.2: As the point (x(t), y(t)) moves along the curve C in the xy-plane toward (x0 , y0 ), the point (x(t), y(t), f(x(t), y(t))) moves directly above it along the graph of z = f(x, y) with f(x(t), y(t)) approaching the limiting value L. In the figure we followed a common practice of omitting the zero z-coordinate for points in the xy-plane.

n

z = f (x, y)

C (x(t), y(t))

x lim

f (x, y) = L

(x, y)→(x 0, y0 ) (along C)

Ha

sa

y

(x 0 , y0 )

• REMARK. • • • • • • •

In both (1) and (2), the limit of the function of t has to be treated as a one-sided limit if (x0 , y0 ) or (x0 , y0 , z0 ) is an endpoint of C. Example 1 Figure 14.2.3a shows a computer-generated graph of the function xy f(x, y) = − 2 x + y2 The graph reveals that the surface has a ridge above the line y = −x, which is to be expected since f(x, y) has a constant value of 12 for y = −x, except at (0, 0) where f is undefined (verify). Moreover, the graph suggests that the limit of f(x, y) as (x, y) → (0, 0) along a line through the origin varies with the direction of the line. Find this limit along

M uh am

m

ad

Figure 14.2.2

(a) the x-axis (d) the line y = −x

(b) the y-axis (e) the parabola y = x 2

(c) the line y = x

Solution (a). The x-axis has parametric equations x = t, y = 0, with (0, 0) corresponding to t = 0, so lim

(x, y) → (0, 0) (along y = 0)

f(x, y) = lim f(t, 0) = lim t →0

t →0

−

0 t2

= lim 0 = 0 t →0

which is consistent with Figure 14.2.3b.

Solution (b). The y-axis has parametric equations x = 0, y = t, with (0, 0) corresponding to t = 0, so lim

(x, y) → (0, 0) (along x = 0)

f(x, y) = lim f(0, t) = lim t →0

t →0

0 − 2 t

= lim 0 = 0 t →0

which is consistent with Figure 14.2.3b.

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