# 10 I Parametric, Vector, and Polar

```5128_Ch10_pp530-561 1/13/06 3:50 PM Page 530
Chapter
10
Parametric,
Vector, and Polar
Functions
I
n 1935, air traffic control was conducted with a
system of teletype machines, wall-sized blackboards, large table maps, and movable markers
representing airplanes. Today’s radar data processing includes an automatic display of aircraft
identification, speed, altitude, and velocity vectors.
A DC-10 plane flying due west at 600 mph enters a region with a steady air current coming from
the southwest at 100 mph. How should the pilot
adjust the airplane’s course and speed to maintain
its original velocity vector? This type of problem is
covered in Section 10.2.
530
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Section 10.1 Parametric Functions
531
Chapter 10 Overview
The material in this book is generally described as the calculus of a single variable, since it
deals with functions of one independent variable (usually x or t). In this chapter you will apply
your understanding of single-variable calculus in three kinds of two-variable contexts, enabling
you to analyze some new kinds of curves (parametrically defined and polar) and to analyze
motion in the plane that does not proceed along a straight line. Interestingly enough, this will
not require the tools of multi-variable calculus, which you will probably learn in your next calculus course. We will simply use single-variable calculus in some new and interesting ways.
10.1
• Parametric Curves in the Plane
• Slope and Concavity
• Arc Length
• Cycloids
. . . and why
Parametric equations enable us
to define some interesting and
important curves that would be
difficult or impossible to define in
the form y f (x).
Parametric Functions
Parametric Curves in the Plane
We reviewed parametrically defined functions in Section 1.4. Instead of defining the
points (x, y) on a planar curve by relating y directly to x, we can define both coordinates as
functions of a parameter t. The resulting set of points may or may not define y as a function of x (that is, the parametric curve might fail the vertical line test).
EXAMPLE 1
Reviewing Some Parametric Curves
Sketch the parametric curves and identify those which define y as a function of x. In
each case, eliminate the parameter to find an equation that relates x and y directly.
(a) x cos t and y sin t for t in the interval [0, 2)
(b) x 3 cos t and y 2 sin t for t in the interval [0, 4]
(c) x t and y t 2 for t in the interval [0, 4]
SOLUTION
(a) This is probably the best-known parametrization of all. The curve is the unit circle
(Figure 10.1a), and it does not define y as a function of x. To eliminate the parameter, we
use the identity cos t2 sin t2 1 to write x2 y2 1.
(b) This parametrization stretches the unit circle by a factor of 3 horizontally and by a factor
of 2 vertically. The result is an ellipse (Figure 10.1b), which is traced twice as t covers the interval [0, 4]. (In fact, the point (3, 0) is visited three times.) It does not define y as a
x 2
y 2
function of x. We use the same identity as in part (a) to write 1.
3
2
(c) This parametrization produces a segment of a parabola (Figure 10.1c). It does define
y as a function of x. Since t x2, we write y x2 2.
Now try Exercise 1.
y
y
y
3
3
3
1
x
–2
–4
–1
1
1
4
x
–2
2
–1
–3
(a)
–3
(b)
–3
(c)
Figure 10.1 A collection of parametric curves (Example 1). Each point (x, y) is determined by
parametric functions of t, but only the parametrization in graph (c) determines y as a function of x.
x
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Chapter 10
Parametric, Vector, and Polar Functions
Slope and Concavity
We can analyze the slope and concavity of parametric curves just as we can with explicitlydefined curves. The slope of the curve is still dydx, and the concavity still depends on
d 2ydx 2, so all that is needed is a way of differentiating with respect to x when everything
is given in terms of t. The required parametric differentiation formulas are straightforward
applications of the Chain Rule.
Parametric Differentiation Formulas
If x and y are both differentiable functions of t and if dxdt 0, then
dy
dydt
.
dx
dxdt
If y dydx is also a differentiable function of t, then
d 2y
dydt
d
2 (y) .
dx
dxdt
dx
EXAMPLE 2 Analyzing a Parametric Curve
y
3
1
–6
0
1
6
x
–3
Figure 10.2 The parametric curve
defined in Example 2.
Consider the curve defined parametrically by x t2 5 and y 2 sin t for 0 t .
(a) Sketch a graph of the curve in the viewing window [7, 7] by [4, 4]. Indicate the
direction in which it is traced.
(c) Find all points of inflection on the curve. Justify your answer.
SOLUTION
(a) The curve is shown in Figure 10.2.
(b) We seek to maximize y as a function of t, so we compute dydt 2 cos t. Since dydt
is positive for 0 t 2 and negative for 2 t , the maximum occurs when
t 2. Substituting this t value into the parametrization, we find the highest point to be
approximately (2.533, 2).
(c) First we compute d 2ydx2.
dy dy/dt 2 cos t
cos t
dx dx/dt
2t
t
(sin t)(t) (1)(cos t)
t2
dydt
t sin t cos t
2 dx
dxdt
2t
2t3
d 2y
A graph of
[0, p] by [– 0.1, 0.1]
Figure 10.3 The graph of d 2ydx 2 for the
parametric curve in Example 2 shows a
sign change at t 2.798386 ... , indicating
a point of inflection on the curve.
(Example 2)
t sin t cos t
y on the interval [0, ] (Figure 10.3)
2t3
shows a sign change at t 2.798386... . Substituting this t value into the parametrization, we find the point of inflection to be approximately (2.831, 0.673).
Now try Exercise 19.
Arc Length
In Section 7.4 we derived two different formulas for arc length, each of them based on an
2
approximation of the curve by tiny straight line segments with length x
yk2
k .
(See Figure 10.4.)
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Section 10.1 Parametric Functions
y
533
Here is a third formula based on the same approximation.
(b, d)
d
(xk) 2 (yk) 2
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
√
Q
Arc Length of a Parametrized Curve
yk
y f (x)
P
Let L be the length of a parametric curve that is traversed exactly once as t increases
from t1 to t2.
If dxdt and dydt are continuous functions of t, then
x k
(a, c)
c
0
xk – 1
a
xk
( ) ( )
t2
x
b
L
2
dx
dy
dt
dt
t1
Figure 10.4 The graph of f, approximated by line segments.
2
dt.
EXAMPLE 3 Measuring a Parametric Curve
x=
cos3 t,
y=
sin3 t,
0 ≤ t ≤ 2p
Find the length of the astroid (Figure 10.5)
x cos 3 t,
y sin 3 t,
0 t 2.
SOLUTION
Solve Analytically The curve is traced once as t goes from 0 to 2. Because of the
curve’s symmetry with respect to the coordinate axes, its length is four times the length
of the first quadrant portion. We have
2
( ) (
( ) (
( ) ( )
Figure 10.5 The astroid in Example 3.
dx
dt
2
dy
dt
2
dx
dy
dt
dt
2
2
3 cos 2 tsin t
)
9 cos 4 t sin 2 t
2
3
sin 2
tcos t
)
9 sin 4 t cos 2 t
2
9co
s 2tsin
t
co
s 2t
sin
2t
1
2 t
9co
s 2tsin
3 cos t sin t .
Thus, the length of the first quadrant portion of the curve is
p 2
p 2
3 cos t sin t dt 3
0
cos t sin t dt
3
sin2 t
2
y
cos t sin t 0, 0 t /2
0
]
p2
u sin t, du cos t dt
0
3
.
2
P(x, y)
The length of the astroid is 432 6.
t
O
at
Support Numerically NINT 3 cos t sin t , t, 0, 2 6.
a
C(at, a)
x
Figure 10.6 The position of P(x, y) on
the edge of the wheel when the wheel has
Now try Exercise 29.
Cycloids
Suppose that a wheel of radius a rolls along a horizontal line without slipping (see
Figure 10.6. The path traced by a point P on the wheel’s edge is a cycloid, where P is
originally at the origin.
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Chapter 10
Parametric, Vector, and Polar Functions
EXAMPLE 4 Finding Parametric Equations for a Cycloid
Huygens’s Clock
The problem with a pendulum clock
whose bob swings in a circular arc is
that the frequency of the swing depends on the amplitude of the swing.
The wider the swing, the longer it takes
This does not happen if the bob can
be made to swing in a cycloid. In 1673,
Christiaan Huygens (1629–1695), the
Dutch mathematician, physicist, and
astronomer who discovered the rings of
Saturn, designed a pendulum clock
whose bob would swing in a cycloid.
Driven by a need to make accurate
determinations of longitude at sea, he
hung the bob from a fine wire constrained by guards that caused it to
draw up as it swung away from center.
How were the guards shaped? They
were cycloids, too.
Find parametric equations for the path of the point P in Figure 10.6.
SOLUTION
We suppose that the wheel rolls to the right, P being at the origin when the turn angle t
equals 0. Figure 10.6 shows the wheel after it has turned t radians. The base of the wheel is
at distance at from the origin. The wheel’s center is at (at, a, and the coordinates of P are
x at a cos ,
y a a sin .
To express in terms of t, we observe that t 3 2 2k for some integer k, so
3
t 2k.
2
Thus,
3
cos cos t 2k sin t,
2
(
(
)
)
3
sin sin t 2k cos t.
2
Therefore,
x at a sin t at sin t,
y a a cos t a1 cos t.
Guard
cycloid
Guard
cycloid
Now try Exercise 41.
EXPLORATION 1
Cycloid
Investigating Cycloids
Consider the cycloids with parametric equations
x at sin t,
1.
2.
3.
4.
5.
6.
y a1 cos t,
a 0.
Graph the equations for a 1, 2, and 3.
Find the x-intercepts.
Show that y 0 for all t.
Explain why the arches of a cycloid are congruent.
What is the maximum value of y? Where is it attained?
Describe the graph of a cycloid.
EXAMPLE 5 Finding Length
Find the length of one arch of the cycloid
x at sin t,
y a1 cos t,
a 0.
SOLUTION
[0, 3p] by [–2, 4]
Figure 10.7 shows the first arch of the cycloid and part of the next for a 1. In Exploration 1 you found that the x-intercepts occur at t equal to multiples of 2 and that the
arches are congruent.
The length of the first arch is
( ) ( )
2
Figure 10.7 The graph of the cycloid
x t sin t, y 1 cos t, t 0.
(Example 5)
0
dx
dt
2
dy 2
dt.
dt
continued
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Section 10.1 Parametric Functions
535
We have
( )
( )
( ) ( )
dx
dt
2
dx
dt
2
dy
dt
2
dy
dt
2
a1 cos t 2 a 2 1 2 cos t cos 2 t
a sin t 2 a 2 sin 2 t
a 2
2o
cst.
a 0, sin 2 t cos 2 t 1
Therefore,
( ) ( )
2
0
dx 2
dy 2
dt a
dt
dt
2
2
2cost dt 8a.
The length of one arch of the cycloid is 8a.
Quick Review 10.1
Using NINT
0
Now try Exercise 43.
(For help, go to Appendix A.1.)
Use algebra or a trig identity to write an equation relating x and y.
6. x csc and y cot x2 1 y2
1. x t 1 and y 2t 3 y 2x 1
7. x cos and y cos(2) y 2x2 1
2. x 3t and y 54t3 3 y 2x3 3
8. x sin and y cos(2) y 1 2x2
3. x sin t and y cos t x2 y2 1
9. x cos and y sin 4. x sin t cos t and y sin(2t) y 2x
5. x tan and y sec y2
1
10. x cos and y sin (0 ) y 1
x2
( 2) y 1
x2
x2
Section 10.1 Exercises
In Exercises 1–6, sketch the parametric curves and identify those
which define y as a function of x. In each case, eliminate the parameter to find an equation that relates x and y directly.
1. x 2t 3 and y 4t 3 for t in the interval [0, 3]
t5
2. x t 2 and y for t in the interval [3, 11]
4
3. x tan t and y sec t for t in the interval [0, 4]
In Exercises 17–22,
(a) sketch the curve over the given t-interval, indicating the direction
in which it is traced,
(b) identify the requested point, and
(c) justify that you have found the requested point by analyzing an
appropriate derivative.
17. x t 1,
4. x sin t and y 2 cos t for t in the interval [0, ]
18. x 5. x sin t and y cos(2t) for t in the interval [0, 2]
19. x 2 sin t,
6. x sin 6t and y 2t for t in the interval [0, 2]
20. x tan t,
In Exercises 7–16, find (a) dy dx and (b) d 2y dx2 in terms of t.
7. x 4 sin t,
y 2 cos t
,
1
9. x t
11. x t2
3t,
13. x tan t,
y 3
t
y
t3
y sec t
8. x cos t,
cos t
y 3
10. x 1 t, y 2 ln t
12. x t2
t,
14. x 2cos t,
y
t2
t
y cos(2t)
15. x ln(2t), y ln(3t)4
16. x ln(5t), y e5t
1
7. (a) 2 tan t
8. (a) 3
1
(b) 8 sec3 t
(b) 0
3
3
9. (a) 3 (b) 3/2
t
t
t2
y t 2 t,
2t,
21. x 2 sin t,
22. x ln(5t),
y
y cos t,
y 2 sec t,
y cos(2t),
y
ln(4t2),
Lowest point
2 t 3 Leftmost point
0t
Rightmost point
1 t 1
Lowest point
1.5 t 4.5
0 t 10
Highest point
Rightmost point
In Exercises 23–26, find the points at which the tangent line to the
2
, or curve is (a) horizontal or (b) vertical.
25. (a) At t 3
23. x 2 cos t,
24. x sec t,
y 1 sin t
y tan t
26. x 2 3 cos t,
10. (a) t (b) t2
6t2 18t
3t2
11. (a) (b) 3
(2t 3)
2t 3
2t 1
4
12. (a) (b) 3
2t 1
(2t 1)
2 t 2
2t 3,
t2
13. (a) sin t (b) cos3 t
14. (a) 2 cos t (b) 1
15. (a) 4 (b) 0
16. (a) 5t e5t (b) 25t2 e5t 5t e5t
(0.845, 3.079) and
(3.155, 3.079) (b) Nowhere
25. x 2 t,
y t 3 4t
y 1 3 sin t
23. (a) (2, 0) and (2, 2) (b) (1, 1) and (3, 1)
24. (a) Nowhere (b) (1, 0) and (1, 0)
26. (a) (2, 4) and (2, 2)
(b) (1, 1) and (5, 1)
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Chapter 10
Parametric, Vector, and Polar Functions
In Exercises 27–34, find the length of the curve. (For an algebraic
challenge, try evaluating the integrals without a calculator.)
27. x cos t,
y sin t,
28. x 3 sin t,
0 t 2 2
y 3 cos t,
0 t 3
2a
y 8 sin t 8t cos t, 0 t 2 2
29. x 8 cos t 8t sin t,
30. x 2
y2
0 t 2 12
2t 3 3 2
t2
31. x , y t , 0 t 3 21/2
3
2
(8t 8)32
32. x , y t2 t, 0 t 2 10
12
1
1
22 1
33. x t 3, y t 2, 0 t 1 0.609
3
3
2
34. x lnsec t tan t sin t, y cos t, 0 t 3
cos3 t,
y
sin3 t,
2ap
39. Find the area of the shaded region. (Hint: dx dx dt dt) 3a2
ln 2
35. Length is Independent of Parametrization To illustrate
the fact that the numbers we get for length do not usually
depend on the way we parametrize our curves, calculate the
length of the semicircle y 1
x 2 with these two different
parametrizations.
(a) x cos 2t,
y sin 2t,
(b) x sin t,
y cos t,
0 t 2
1 2 t 1 2
y 4 sin t,
0 t 2.
22.103
37. Cartesian Length Formula The graph of a function y f x
over an interval a, b automatically has the parametrization
x x, y f x,
a x b.
The parameter in this case is x itself. Show that for this
parametrization, the length formula
( ) ( )
b
L
2
dx
dy
dt
dt
a
2
dt
reduces to the Cartesian formula
( )
b
L
a
dy
1 dx
2
dx
Just substitute x for t and note that dx/dx 1.
derived in Section 7.4.
38. (Continuation of Exercise 37) Show that the Cartesian
formula
( )
d
L
c
dx
1 dy
2
dy
for the length of the curve x gy, c y d, from Section 7.4
is a special case of the parametric length formula
( ) ( )
b
L
a
2
dx
dy
dt
dt
2
dt.
Exercises 39 and 40 refer to the region bounded by the x-axis and one
arch of the cycloid
x at sin t,
y a1 cos t
that is shaded in the figure shown at the top of the next column.
38. Use the parametrization x g(y), y y,
c y d, substitute y for t and note dy/dy 1.
40. Find the volume swept out by revolving the region about the
x-axis. (Hint: dV y 2 dx y 2 dx dt dt) 5 2 a3
41. Curtate Cycloid Modify Example 4 slightly to find the
parametric equations for the motion of a point in the interior of
a wheel of radius a as the wheel rolls along the horizontal line
without slipping. Assume that the point is at distance b from the
center of the wheel, where 0 b a. This curve, known as a
curtate cycloid, has been used by artisans in designing the
arches of violins (Source: mathworld.wolfram.com).
x at b sin t and y a b cos t (0 a b)
36. Perimeter of an Ellipse Find the length of the ellipse
x 3 cos t,
x
42. Prolate Cycloid Modify Example 4 slightly to find the
parametric equations for the motion of a point on the exterior of a
wheel of radius a as the wheel rolls along the horizontal line
without slipping. Assume that the point is at distance b from the
center of the wheel, where a b 2a. This curve, known as a
prolate cycloid, is traced out by a point on the outer edge of a
train’s flanged wheel as the train moves along a track. (If you
graph a prolate cycloid, you can see why they say that there is
always part of a forward-moving train that is moving backwards!)
x at b sin t and y a b cos t (a b 2a)
43. Arc Length Find the length of one arch (that is, the curve
over one period) of the curtate cycloid defined parametrically by
x 3t 2 sin t and y 3 2 cos t. 21.010
44. Arc Length Find the length of one arch (that is, the curve
over one period) of the prolate cycloid defined parametrically by
x 2t 3 sin t and y 2 3 cos t. 21.010
Standardized Test Questions
You should solve the following problems without using a graphing
calculator.
45. True or False In a parametrization, if x is a continuous
function of t and y is a continuous function of t, then y is a
46. True or False If f is a function with domain all real numbers,
then the graph of f can be defined parametrically by x t and
47. Multiple Choice For which of the following parametrizations
of the unit circle will the circle be traversed clockwise? B
(A) x cos t,
y sin t,
0 t 2
(B) x sin t,
y cos t,
0 t 2
(C) x cos t,
y sin t, 0 t 2
(D) x sin t,
y cos t,
0 t 2
y cos t,
0 t 2
(E) x sin t,
45. False. Indeed, y may not even be a function of x. (See Example 1.)
46. True. The ordered pairs (x, f(x)) and (t, f(t)) are exactly the same.
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Section 10.1 Parametric Functions
48. Multiple Choice A parametric curve is defined by x sin t
and y csc t for 0 t 2. This curve is C
537
the string is tangent to the circle at Q, and t is the radian measure
of the angle from the positive x-axis to the segment OQ.
(a) Derive parametric equations for the involute by expressing the
coordinates x and y of P in terms of t for t 0. x cos t (A) increasing and concave up.
(B) increasing and concave down.
t sin t, y sin t t cos t
(b) Find the length of the involute for 0 t 2 . 2 2
(C) decreasing and concave up.
(D) decreasing and concave down.
(E) decreasing with a point of inflection.
49. Multiple Choice The parametric curve defined by x ln(t),
y t for t 0 is identical to the graph of the function C
(A) y ln x for all real x.
(B) y lnx
y
for x 0.
String
Q
(C) y e x for all real x.
P(x, y)
(D) y e x for x 0.
t
(E) y ln(e x) for x 0.
O
1
x
(1, 0)
50. Multiple Choice The curve parametrized by
x 6 sin t 3 sin(7t) and y 6 cos t 3 cos(7t),
as shown in the diagram below, is traversed exactly once as t
increases from 0 to 2. The total length of the curve is given by D
(A) 0
2 (6
(6 sin t 3 s
in(7t))
cos t 3 cos
(7t))2 dt
(B) 0
(6 cos t 3 c
os(7t)
)2 (6
sin t 3 sin (
7t))2 dt
(C) 0
(6 cos t 21
cos(7
t))2 (6 sin t 21
sin(7t
))2 dt
(D) 0
(6 cos t 21
cos(7
t))2 (6 si
n t 2
1 sin(7
t))2 dt
2
2
2
2
(6 cos t 3 (6
sin t 3
(E) 0 7
2
cos(7t))2
sin (3t))2 dt
52. (Continuation of Exercise 51) Repeat Exercise 51 using the
circle of radius a centered at the origin, x 2 y 2 a 2.
(a) x a(cos t t sin t), y a(sin t t cos t) (b) 2a 2
In Exercises 53–56, a projectile is launched over horizontal ground at
an angle with the horizontal and with initial velocity v0 ft sec. Its
path is given by the parametric equations
x v0 cos t,
y v0 sin t 16t 2.
(a) Find the length of the path traveled by the projectile.
(b) Estimate the maximum height of the projectile.
53. 20°, v0 150
y
55. 60°,
10
54. 30°,
v0 150
56. 90°,
v0 150
(a) 641.236 ft (b) 5625/64 87.891 ft
(a) 461.749 ft (b) 41.125 ft
v0 150
(a) 840.421 ft (b) 16,875/64 263.672 ft
Extending the Ideas
2
–10
–2
2
10
x
If dx dt and dydt are continuous, the parametric curve defined by
(x(t), y(t)) for a t b is called smooth. If the curve is traversed
exactly once as t increases from a to b, and if y is a positive
function of x, then the curve can be revolved about the x-axis to
form a solid of revolution (see Section 7.3). The surface area of such
a solid is given by
( ) ( )
b
–10
S
2y
a
Explorations
51. Group Activity Involute of a Circle If a string wound
around a fixed circle is unwound while being held taut in the
plane of the circle, its end P traces an involute of the circle as
suggested by the diagram below. In the diagram, the circle is the
unit circle in the xy-plane, and the initial position of the tracing
point is the point 1, 0 on the x-axis. The unwound portion of
(a) 703.125 ft
(b) 5625/16 351.5625 ft
2
dx
dy
dt
dt
2
dt.
Apply this formula in Exercises 57–60 to find the surface area when
the parametric curve is revolved about the x-axis.
57. x cos t,
y 2 sin t,
58. x 2t,
y (2 3)t 3 2,
59. x t2 2,
y t 1,
0 t 2 8 2
0 t 2 14.214
0 t 3 178.561
60. x ln(sec t tan t) sin t,
y cos t,
0 t 3
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Chapter 10
Parametric, Vector, and Polar Functions
10.2
• Two-Dimensional Vectors
• Vector Operations
• Modeling Planar Motion
• Velocity, Acceleration, and
Speed
• Displacement and Distance
Traveled
. . . and why
The jump from one to two dimensions (and eventually higher) is
easier than one might think,
thanks to the mathematics of
vectors.
Vectors in the Plane
Two-Dimensional Vectors
When an object moves along a straight line, its velocity can be determined by a single number that represents both magnitude and direction (forward if the number is positive, backward
if it is negative). The speed of an object moving on a path in a plane can still be represented by
a number, but how can we represent its direction when there are an infinite number of directions possible? Fortunately, we can represent both magnitude and direction with just two
numbers, just as we can represent any point in the plane with just two coordinates (which is
possible essentially for the same reason). This representation is what two-dimensional vectors
were designed to do.
While the pair (a, b) determines a point in the plane, it also determines a directed line
segment (or arrow) with its tail at the origin and its head at (a, b) (Figure 10.8). The length of
this arrow represents magnitude, while the direction in which it points represents direction. In
this way, the ordered pair (a, b) represents a mathematical object with both magnitude and direction, called the position vector of (a, b).
y
y
(a, b)
(a, b)
a, b
O
x
O
x
Figure 10.8 The point represents the ordered pair (a, b). The arrow (directed line segment)
represents the vector a, b.
DEFINITION Two-Dimensional Vector
A two-dimensional vector v is an ordered pair of real numbers, denoted in component
form as a, b. The numbers a and b are the components of the vector v. The standard
representation of the vector a, b is the arrow from the origin to the point (a, b).
The magnitude (or absolute value) of v, denoted v, is the length of the arrow, and the
direction of v is the direction in which the arrow is pointing. The vector 0 0, 0,
called the zero vector, has zero length and no direction.
The distance formula in the plane gives a simple computational formula for magnitude.
Magnitude of a Vector
The magnitude or absolute value of the vector a, b is the nonnegative real
number a, b a2 b2.
Direction can be quantified in several ways; for example, navigators use bearings from
compass points. The simplest choice for us is to measure direction as we do with the
trigonometric functions, using the usual position angle formed with the positive x-axis as
the initial ray and the vector as the terminal ray. In this way, every nonzero vector determines a unique direction angle satisfying (in degrees) 0 360 or (in radians)
0 2. (See Figure 10.10 for an example.)
5128_Ch10_pp530-561 2/3/06 4:41 PM Page 539
Section 10.2 Vectors in the Plane
539
y
R(–1, 6)
Direction Angle of a Vector
P(3, 4)
The direction angle of a nonzero vector v is the smallest nonnegative angle formed with the positive x-axis as the initial ray and the standard representation
of v as the terminal ray.
Q(–4, 2)
1
x
O
1
Figure 10.9 The arrows QR
and OP
both represent the vector 3, 4, as would
any arrow with the same length pointing in
the same direction. Such arrows are called
equivalent.
y
(–1, √3−)
2
v
u
2
–2
x
–2
This textbook uses boldface variables to represent vectors (for example, u and v) to distinguish them from numbers. In handwritten form it is customary to distinguish vector
variables by arrows (for example, u and v).
We also use angled brackets to distinguish a
vector x, y from a point (x, y) in the plane, although it is not uncommon to see (x, y)
used for both, especially in handwritten form.
It is often convenient in applications to represent vectors with arrows that begin at points
other than the origin. The important thing to remember is that any two arrows with the same
length and pointing in the same direction represent the same vector. In Figure 10.9, for ex an arrow with initial point Q and
ample, the vector 3, 4 is shown represented by QR,
terminal point R, as well as by its standard representation OP. Two arrows that represent
the same vector are said to be equivalent.
The quick way to associate arrows with the vectors they represent is to use the following
rule.
Figure 10.10 The vector v in Example 1
is represented by an arrow from the origin
to the point 1, 3.
If an arrow has initial point (x1, y1) and terminal point (x2, y2), it represents the
vector x2 x1, y2 y1.
y
EXAMPLE 1 Finding Magnitude and Direction
Find the magnitude and the direction angle of the vector v 1, 3 (Figure 10.10).
SOLUTION
3
40˚
y = 3 sin 40˚
x
x = 3 cos 40˚
The magnitude of v is v (1 0)2 (3 0)2 2. Using triangle ratios, we see
that the direction angle satisfies cos 1 2 and sin 3 2, so 120º or 2 3
Now try Exercise 5.
3
–1
EXAMPLE 2 Finding Component Form
Figure 10.11 The vector in Example 2
is represented by an arrow from the
origin to the point (3 cos 40º, 3 sin 40º).
Why Not Use Slope for Direction?
Notice that slope is inadequate for determining the direction of a vector, since
two vectors with the same slope could
be pointing in opposite directions. Moreover, vectors are still useful in dimensions higher than 2, while slope is not.
Find the component form of a vector with magnitude 3 and direction angle 40º.
SOLUTION
The components of the vector, found trigonometrically, are x 3 cos 40º and y 3 sin 40º
(Figure 10.11).
The vector is 3 cos 40º, 3 sin 40º 2.298, 1.928.
Now try Exercise 13.
Vector Operations
The algebra of vectors sometimes involves working with vectors and numbers at the same
time. In this context, we refer to the numbers as scalars. The two most basic algebraic operations involving vectors are vector addition (adding a vector to a vector) and scalar multiplication (multiplying a vector by a number). Both operations are easily represented geometrically.
5128_Ch10_pp530-561 1/13/06 3:51 PM Page 540
540
Chapter 10
Parametric, Vector, and Polar Functions
DEFINITION Vector Addition and Scalar Multiplication
Let u u1, u2 and v v1, v2 be vectors and let k be a real number (scalar).
The sum (or resultant) of the vectors u and v is the vector
u v u1 v1, u2 v2.
The product of the scalar k and the vector u is
ku ku1, u2 ku1, ku2.
The opposite of a vector v is v (1)v. We define vector subtraction by
u v u (v).
v
The vector is a vector of magnitude 1, called a unit vector. Its component form is
v
v
cos , sin , where is the direction angle of v. For this reason, is sometimes called
v
the direction vector of v.
The sum of two vectors u and v can be represented geometrically by arrows in two ways.
In the tail-to-head representation, the arrow from the origin to (u1, u2) is the standard representation of u, the arrow from (u1, u2) to (u1 v1, u2 v2,) represents v (as you can verify by
the HMT Rule), and the arrow from the origin to (u1 v1, u2 v2) then is the standard representation of u v (Figure 10.12a).
In the parallelogram representation, the standard representations of u and v determine
a parallelogram whose diagonal is the standard representation of u v (Figure 10.12b).
y
y
v
u
u
u+v
u+v
v
x
x
(a)
(b)
–2 u
Figure 10.12 Two ways to represent vector addition geometrically: (a) tail-to-head and
(b) parallelogram.
2u
u
1_ u
2
Figure 10.13 Representations of u and
several scalar multiples of u.
The product ku of the scalar k and the vector u can be represented by a stretch (or shrink)
of u by a factor of k. If k 0, then ku points in the same direction as u; if k 0, then ku
points in the opposite direction (Figure 10.13).
EXAMPLE 3 Performing Operations on Vectors
Let u 1, 3 and v 4, 7. Find the following.
1
(a) 2u 3v
(b) u v
(c) u
2
SOLUTION
(a) 2u 3v 21, 3 34, 7
21 34, 23 37 10, 27
continued
5128_Ch10_pp530-561 1/13/06 3:51 PM Page 541
Section 10.2 Vectors in the Plane
541
(b) u v 1, 3 4, 7
1 4, 3 7 5, 4
1
1 3
(c) u , 2
2 2
( ) ()
2
1
3
2
2
2
1
10
2
Now try Exercise 21.
Vector operations have many of the properties of their real-number counterparts.
Properties of Vector Operations
Let u, v, w be vectors and a, b be scalars.
1. u v v u
2. u v w u v w
3. u 0 u
4. u u 0
5. 0u 0
6. 1u u
7. abu abu
8. au v au av
9. a bu au bu
Modeling Planar Motion
Although vectors are used in many other physical applications, our primary reason for
introducing them into this course is to model the motion of objects moving in a coordinate plane. You may have seen vector problems of the following type in a physics or
mechanics course.
EXAMPLE 4 Finding Ground Speed and Direction
A Boeing® 727 ® airplane, flying due east at 500 mph in still air, encounters a 70-mph tail
wind acting in the direction 60º north of east. The airplane holds its compass heading due
east but, because of the wind, acquires a new ground speed and direction. What are they?
N
SOLUTION
If u the velocity of the airplane alone and v the velocity of the tail wind, then
u 500 and v 70 (Figure 10.14).
v
30˚
70
u+v
500
u
E
We need to find the magnitude and direction of the resultant vector u v. If we let the
positive x-axis represent east and the positive y-axis represent north, then the component forms of u and v are
u 500, 0
NOT TO SCALE
Figure 10.14 Vectors representing
the velocities of the airplane and tail
wind in Example 4.
and
v 70 cos 60º, 70 sin 60º 35, 35 3 .
Therefore,
u v 535, 35 3 ,
u
and
v 5352
35
3 2 538.4,
35 3
tan1 6.5º.
535
Interpret The new ground speed of the airplane is about 538.4 mph, and its new
direction is about 6.5º north of east.
Now try Exercise 25.
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Chapter 10
Parametric, Vector, and Polar Functions
Recall that if the position x of an object moving along a line is given as a function of time t,
then the velocity of the object is dx dt and the acceleration of the object is d 2x dt2. It is almost as simple to relate position, velocity, and acceleration for an object moving in the
plane, because we can model those functions with vectors and treat the components of the
vectors as separate linear models. Example 5 shows how simple this modeling actually is.
EXAMPLE 5 Doing Calculus Componentwise
A particle moves in the plane so that its position at any time t 0 is given by (sin t, t2 2).
(a) Find the position vector of the particle at time t.
(b) Find the velocity vector of the particle at time t.
(c) Find the acceleration of the particle at time t.
(d) Describe the position and motion of the particle at time t 6.
SOLUTION
6
cos 6
[–2, 2] by [0, 25]
0≤t≤6
Figure 10.15 The path of the particle in
Example 5 from t 0 to t 6. The red
arrow shows the velocity vector at t 6.
(a) The position vector, which has the same components as the position point, is sin t, t22.
In fact, it could also be represented as (sin t, t2 2), since the context would identify it as
a vector.
(b) Differentiate each component of the position vector to get cos t, t.
(c) Differentiate each component of the velocity vector to get sin t, 1.
(d) The particle is at the point (sin 6, 18), with velocity cos 6, 6 and acceleration
sin 6, 1.
You can graph the path of this particle parametrically, letting x sin(t) and y t2 2. In
Figure 10.15 we show the path of the particle from t 0 to t 6. The red arrow at the
point (sin 6, 18) represents the velocity vector (cos 6, 6). It shows both the magnitude and
direction of the velocity at that moment in time.
Now try Exercise 31.
Velocity, Acceleration, and Speed
We are now ready to give some definitions.
DEFINITIONS Velocity, Speed, Acceleration, and Direction
of Motion
Our definitions can be expanded to a
calculus of vectors, in which (for example) d v dt a(t), but it is not our intention to get into that here. We have
therefore finessed the fine point of vector differentiability by requiring the
path of our particle to be “smooth.”
The path can have vertical tangents,
fail the vertical line test, and loop back
on itself, but corners and cusps are still
problematic.
Suppose a particle moves along a smooth curve in the plane so that its position at
any time t is (x(t)), y(t), where x and y are differentiable functions of t.
1. The particle’s position vector is r(t) x(t), y(t).
dx dy
2. The particle’s velocity vector is v(t) , .
dt dt
3. The particle’s speed is the magnitude of v, denoted v. Speed is a scalar, not a
vector.
d 2x d 2y
4. The particle’s acceleration vector is a(t) ,
.
dt 2 dt 2
v
5. The particle’s direction of motion is the direction vector .
v
EXAMPLE 6 Studying Planar Motion
A particle moves in the plane with position vector r (t) sin (3t), cos (5t). Find the
velocity and acceleration vectors and determine the path of the particle.
continued
5128_Ch10_pp530-561 1/13/06 3:51 PM Page 543
Section 10.2 Vectors in the Plane
SOLUTION
543
d
d
Velocity v(t) (sin(3t)), (cos(5t)) 3 cos(3t), 5 sin(5t).
dt
dt
d
d
Acceleration a(t) (3 cos(3t)), (5 sin(5t)) 9 sin(3t), 25 cos(5t).
dt
dt
The path of the particle is found by graphing (in parametric mode) the curve defined by
x sin(3t) and y cos(5t) (Figure 10.16).
Now try Exercise 33.
EXAMPLE 7 Studying Planar Motion
[–1.6, 1.6] by [–1.1, 1.1]
0 ≤ t ≤ 6.3
A particle moves in an elliptical path so that its position at any time t 0 is given by
(4 sin t, 2 cos t).
Figure 10.16 The path of the busy
particle in Example 6.
(a) Find the velocity and acceleration vectors.
(b) Find the velocity, acceleration, speed, and direction of motion at t 4.
(c) Sketch the path of the particle and show the velocity vector at the point (4, 0).
(d) Does the particle travel clockwise or counterclockwise around the origin?
SOLUTION
y
4
(b) Velocity v( 4) 4 cos( 4), 2 sin( 4) 22, 2
Acceleration a( 4) 4 sin( 4), 2 cos( 4) 22, 2
1
–5
d
d
(a) Velocity v(t) (4 sin t), (2 cos t) 4 cos t, 2 sin t
dt
dt
d
d
Acceleration a(t) (4 cos t), (2 sin t) 4 sin t, 2 cos t
dt
dt
1
x
0, –2
–4
Figure 10.17 The ellipse on which the
particle travels in Example 7. The velocity
vector at the point (4, 0) is 0, 2, represented by an arrow tangent to the ellipse at
(4, 0) and pointing down. The direction of
the velocity at that point indicates that the
particle travels clockwise around the origin.
Speed v( 4) 22, 2 (22)2
(
2)2 10
(c) The ellipse defined parametrically by x 4 sin t and y 2 cos t is shown in Figure 10.17.
At the point (4, 0), sin t 1 and cos t 0, so v(t) 4 cos t, 2 sin t 0, 2. The
vector 0, 2 is drawn tangent to the curve at (4, 0).
(d) As the vector in Figure 10.17 shows, the particle travels clockwise around the origin.
Now try Exercise 35.
Displacement and Distance Traveled
Recall that when a particle moves along a line with velocity v(t), the displacement (or
b
net distance traveled) from time t a to time t b is given by a v(t) dt, while the (total)
b
distance traveled in that time interval is given by a v(t) dt. When a particle moves in
the plane with velocity vector v(t), displacement and distance traveled can be found by
applying the same integrals to the vector v, although in slightly different ways.
DEFINITIONS Displacement and Distance Traveled
Suppose a particle moves along a path in the plane so that its velocity at any time t is
v(t) v1(t), v2(t), where v1 and v2 are integrable functions of t.
The displacement from t a to t b is given by the vector
b
a
v1(t) dt,
b
a
v2(t) dt .
continued
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Chapter 10
Parametric, Vector, and Polar Functions
The preceding vector is added to the position at time t a to get the position at
time t b.
The distance traveled from t a to t b is
b
b
v(t)dt a
v1(t)2 v2(t)2 dt.
a
There are two things worth noting about the formula for distance traveled. First of all,
it is a nice example of the integral as an accumulator, since we are summing up bits of
speed multiplied by bits of time, which equals bits of positive distance. Secondly, it is
actually a new look at an old formula. Substitute dx dt for v1(t) and dy dt for v2(t) and
you get the arc length formula for a curve defined parametrically (Section 10.1). This
formula makes sense, since the distance the particle travels is precisely the length of
the path along which it moves.
EXAMPLE 8 Finding Displacement and Distance Traveled
A particle moves in the plane with velocity vector v(t) (t 3 cos t, 2t sin t).
At t 0, the particle is at the point (1, 5).
(a) Find the position of the particle at t 4.
(b) What is the total distance traveled by the particle from t 0 to t 4?
SOLUTION
(a) Displacement (t 3 cos t)dt, (2t sin t)dt 8, 16.
4
4
0
0
The particle is at the point (1 8, 5 16) (9, 21).
(b) Distance traveled 0 (t 3
cos t
)2 (2
t
sin t)2 dt 33.533.
4
Now try Exercise 37.
EXAMPLE 9 Finding the Path of the Particle
Determine the path that the particle in Example 8 travels going from (1, 5) to (9, 21) .
SOLUTION
The velocity vector and the position at t 0 combine to give us the vector equivalent of an
initial value problem. We simply find the components of the position vector separately.
dx
t 3 cos t
dt
t2
Antidifferentiate.
x 3 sin t C
2
t2
x 3 sin t 1
x 1 when t 0.
2
dy
2t sin t
dt
Antidifferentiate.
y t 2 cos t C
[–5, 15] by [0, 23]
0≤t≤4
Figure 10.18 The path traveled by the
particle in Example 8 as it goes from
(1, 5) to (9, 21) in four seconds (Example 9).
y t 2 cos t 4
We then graph the position 2 3 sin t 1,
t 0 to t 4. The path is shown in Figure 10.18.
t2
t2
y 5 when t 0.
cos t 4 parametrically from
Now try Exercise 41.
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Section 10.2 Vectors in the Plane
Quick Review 10.2
545
(For help, go to Sections 1.1, 4.3, and 10.1.)
In Exercises 1–4, let P (1, 2) and Q (5, 3).
7. Find the velocity and acceleration of a particle moving along a
line if its position at time t is given by x(t) t sin t.
17
1. Find the distance between the points P and Q.
v(t) sin t t cos t; a(t) 2 cos t t sin t
2. Find the slope of the line segment PQ. 1/4
3. If R (3, b), determine b so that segments PQ and RQ are
collinear. b 5/2
4. If R (3, b), determine b so that segments PQ and RQ are
perpendicular. b 11
In Exercises 5 and 6, determine the missing coordinate so that the
four points form a parallelogram ABCD.
5. A (0, 0), B (1, 3), C (5, 3), D (a, 0) a 4
8. A particle moves along the x-axis with velocity v(t) 3t2 12t
for t 0. If its position is x 40 when t 0, where is the particle when t 4? x 8
9. A particle moves along the x-axis with velocity v(t) 3t2 12t
for t 0. What is the total distance traveled by the particle from
t 0 to t 4? 32
10. Find the length of the curve defined parametrically by x sin(2t)
and y cos(3t) for 0 t 2. 15.289
6. A (1, 1), B (3, 5), C (8, b), D (6, 2) b 6
Section 10.2 Exercises
In Exercises 1–4, find the component form of the vector.
1. the vector from the origin to the point A (2, 3) 2, 3
2. the vector from the point A (2, 3) to the origin
2, 3
where P (1, 3) and Q (2, 1) 1, 4
3. the vector PQ,
where O is the origin and P is the midpoint of the
4. the vector OP,
segment RS connecting R (2, 1) and S (4, 3). 1, 1
26. A river is flowing due east at 2 mph. A canoeist paddles across
the river at 4 mph with his bow aimed directly northwest (a direction angle of 135º). What is the true direction angle of the
canoeist’s path, and how fast is the canoe going?
In Exercises 27–32, a particle travels in the plane with position vector
r(t). Find (a) the velocity vector v(t) and (b) the acceleration vector a(t).
27. r(t) 3t2, 2t3 See page 547. 28. r(t) sin 2t, 2 cos t
See page 547.
In Exercises 5–10, find the magnitude of the vector and the direction
angle it forms with the positive x-axis (0 360º).
5. 2, 2
6. 2, 2
8, 45°
7. 3, 1
2, 30°
9. 5, 0
5, 180°
8. 2, 23
10. 0, 4
2, 135°
4, 240°
4, 90°
In Exercises 11–16, find the component form of the vector with the
given magnitude that forms the given directional angle with the positive
x-axis.
11. 4, 180º
4, 0
12. 6, 270º
13. 5, 100º
0.868, 4.924
14. 13, 200º
3, 3
0, 6
12.216, 4.446
3, 3
In Exercises 17–24, let u 3, 2 and v 2, 5. Find the
(a) component form and (b) magnitude of the vector.
17. 3u (a) 9, 6 (b) 313
18. 2v (a) 4, 10 (b) 229
19. u v (a) 1, 3 (b) 10
20. u v (a) 5, 7 (b) 74
21. 2u 3v (a) 12, 19
22. 2u 5v (a) 16, 29
(b) 505
(b) 1097
5
3
4
12
23. u v (a) 1/5, 14/5
24. u v (a) 3, 70/13
13
5
5
13
(b) 197
5
13
(b) 6421
25. Navigation An airplane, flying in the direction 20º east of north
at 325 mph in still air, encounters a 40-mph tail wind acting in the
direction 40º west of north. The airplane maintains its compass
heading but, because of the wind, acquires a new ground speed
and direction. What are they? Speed 346.735 mph
direction 14.266° east of north
29. r(t) tet, et See page 547. 30. r(t) 2 cos 3t, 2 sin 4t
31. r(t) t2 sin 2t, t2 cos 2t See page 547.
See page 547.
32. r(t) t sin t, t cos t See page 547.
33. A particle moves in the plane with position vector cos 3t, sin 2t.
Find the velocity and acceleration vectors and determine the path
of the particle.
34. A particle moves in the plane with position vector sin 4t, cos 3t.
Find the velocity and acceleration vectors and determine the path
of the particle.
35. A particle moves in the plane so that its position at any time t 0
is given by x sin 4t cos t and y sin 2t.
(a) Find the velocity and speed of the particle when t 5 4.
(b) Draw the path of the particle and show the velocity vector at
t 5 4.
(c) Is the particle moving to the left or to the right when t 5 4?
36. A particle moves in the plane so that its position at any time
t 0 is given by x et et and y et et.
(a) Find the velocity vector.
dydt
(b) Find lim .
t→ dxdt
(c) Show algebraically that the particle moves on the hyperbola
x2 y2 4.
(d) Sketch the path of the particle, showing the velocity vector
at t 0.
26. The velocity is 2 22, 22, so the true angle is about 106.3° and the
true speed is about 2.95 miles per hour.
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Chapter 10
Parametric, Vector, and Polar Functions
In Exercises 37–40, the velocity v(t) of a particle moving in the plane
is given, along with the position of the particle at time t 0. Find
(a) the position of the particle at time t 3, and (b) the distance the
particle travels from t 0 to t 3.
37. v(t) 3t2 2t, 1 cos t; (2, 6) (a) (20, 9) (b) 19.343
38. v(t) 2 cos 4 t, 4 sin 2 t; (7, 2) (a) (7, 2) (b) 28.523
39. v(t) (t 1)1,
(t 2)2;
(3, 2) (a) (3 ln 4, 1.7) (b) 1.419
40. v(t) et t, et t; (1, 1) (a) (15.586, 24.586) (b) 20.627
41. Sketch the path that the particle travels in Exercise 37.
42. Sketch the path that the particle travels in Exercise 38.
43. A point moves in the plane so that x 5 cos( t 6) and
y 3 sin( t 6).
(a) Find the speed of the point at t 2.
7/12
2.399
(b) Find the acceleration vector at t 2. 52/72, 2324
(c) Eliminate the parameter and find an equation in x and y that
2
2
defines the curve on which the point moves. x y 1
25
9
44. A particle moves with position vector sec t, tan t for
0 t 1 2.
(a) Find the velocity and speed of the particle at t 1 4.
(b) The particle moves along a hyperbola. Eliminate the parameter
to find an equation of the hyperbola in terms of x and y.
(c) Sketch the path of the particle over the time interval
0 t 1 2.
45. A particle moves on the circle x2 y2 1 so that its position vector
(b) Is the particle ever at rest? Justify your answer. See page 547.
(c) Give the coordinates of the point that the particle approaches
as t increases without bound. See page 547.
46. A particle moves in the plane so that its position at any time t,
0 t 2, is given parametrically by x sin t and y cos(2t).
(a) Find the velocity vector for the particle.
(d) Sketch the path of the particle.
47. A particle moves in the plane so that its position at any time t,
0 t 2, is given parametrically by x et sin t and y et cos t.
(a) Find the slope of the path of the particle at time t 2.
3.844
(c) Find the distance traveled by the particle along the path from
t 0 to t 1. 2.430
48. The position of a particle at any time t 0 is given by
2
1088
32.985
(a) Find the magnitude of the velocity vector at t 4.
(b) Find the total distance traveled by the particle from t 0
to t 4. 46.062
(c) Find dy dx as a function of x.
(b) At time t 2, the value of dydt is 6. Write an equation for
the line tangent to the curve at the point (x(2), y(2)).
2
(c) Find the speed of the object at time t 2.
(d) For t 3, the line tangent to the curve at (x(t), y(t)) has a
slope of 2t 1. Find the acceleration vector of the object at time
t 4. 8 cos 16, 2(2 sin 16) 7(8)cos 16 7.661, 50.205
50. For 0 t 3, an object moving along a curve in the xy-plane has
position (x(t), y(t)) with dxdt sin(t3) and dydt 3 cos(t2). At
time t 2, the object is at position (4, 5).
See page 547.
(a) Write an equation for the line tangent to the curve at (4, 5).
(b) Find the speed of the object at time t 2. See page 547.
(c) Find the total distance traveled by the object over the time
interval 0 t 1. 2.741
(d) Find the position of the object at time t 3.
See page 547.
Standardized Test Questions
You may use a graphing calculator to solve the following
problems.
51. True or False A scalar multiple of a vector v has the same
53. Multiple Choice The position of a particle in the xy-plane is
given by x t2 1 and y ln(2t 3) for all t 0. The acceleration vector of the particle is E
2
4
4
(A) 2t, .
(B) 2t, 2 . (C) 2, 2 .
2t 3
(2t 3)
(2t 3)
2
(D) 2, .
(2t 3)
2
(b) For what values of t is the particle at rest?
(c) Write an equation for the path of the particle in terms of
x and y that does not involve trigonometric functions.
x(t) t 2 3 and y(t) 3t 3.
(a) Find the x-coordinate of the position of the object at time
4
t 4. 3 (2 sin(t2)) dt 3.942
52. True or False If a vector with direction angle 0º is added to a
vector with direction angle 90º, the result is a vector with direction
See page 547.
(b) Find the speed of the particle when t 1.
49. An object moving along a curve in the xy-plane has position
(x(t), y(t)) at time t 0 with dxdt 2 sin(t2). The derivative
dydt is not explicitly given. At time t 2, the object is at
position (3, 5).
1(u) have opposite directions.
1 t2
2t
at any time t 0 is ,
.
1 t2 1 t2
(a) Find the velocity vector.
6
(2 s
in 4)2 (6)2 6.127
49. (b) y 5 (x 3) (c) 2 sin 4
dy/dx t 3x
4
(E) 2, .
(2t 3)
2
54. Multiple Choice An object moving along a curve in the
xy-plane has position (x(t), y(t)) with dxdt cos(t2) and
dydt sin(t3). At time t 0, the object is at position (4, 7).
Where is the particle when t 2? D
(A) 0.654, 0.989
(B) 0.461, 0.452
(D) 4.461, 7.452
(E) 5.962, 8.962
(C) 3.346, 7.989
55. Multiple Choice A vector with magnitude 7 and direction
angle 40º is added to a vector with magnitude 4 and direction
angle 140º. The result is a vector with magnitude B
(A) 4.684.
(B) 7.435.
(C) 8.062.
(D) 9.369.
(E) 11.
56. Multiple Choice The path of a particle moving in the plane is
defined parametrically as a function of time t by x sin 2t and
y cos 5t. What is the speed of the particle when t 2? B
(A) 1.130
(B) 3.018
(D) 0.757, 0.839
(E) 1.307, 2.720
(C) 1.307, 2.720
, 1, which has a direction
52. False. For example, 3, 0 0, 1 3
angle of 30°.
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Section 10.2 Vectors in the Plane
Explorations
Two nonzero vectors are said to be orthogonal if they are perpendicular to each other. The zero vector is considered to be orthogonal to
every vector.
57. Orthogonal vectors A particle with coordinates (x, y) moves
along a curve in the first quadrant in such a way that dxdt x
and dydt 1 x2 for every t 0. Find the acceleration
vector in terms of x and show that it is orthogonal to the corresponding velocity vector.
58. Orthogonal vectors A particle moves around the unit circle with
position vector cos t, sin t. Use vectors to show that the particle’s
velocity is always orthogonal to both its position and its acceleration.
59. Colliding particles The paths of two particles for t 0 are
given by the position vectors
r1(t) t 3, (t 3)2
3t
3t
r2(t) 4, 2 .
2
2
(a) Determine the exact time(s) at which the particles collide.
(b) Find the direction of motion of each particle at the time(s) of
collision.
60. A Satellite in Circular Orbit A satellite of mass m is moving at a constant speed v around a planet of mass M in a circular
orbit of radius r0 , as measured from the planet’s center of mass.
Determine the satellite’s orbital period T (the time to complete
one full orbit), as follows:
(a) Coordinatize the orbital plane by placing the origin at the
planet’s center of mass, with the satellite on the x-axis at t 0
and moving counterclockwise, as in the accompanying figure.
y
547
Let rt be the satellite’s position vector at time t. Show that
vt r0 and hence that
vt
vt
rt r0 cos , r0 sin r0
r0
(b) Find the acceleration of the satellite.
(c) According to Newton’s law of gravitation, the gravitational
force exerted on the satellite by the planet is directed toward the
origin and is given by
(
)
Gm M r
F ,
r02 r 0
where G is the universal constant of gravitation. Using Newton’s
second law, F ma, show that v 2 GM r0 .
(d) Show that the orbital period T satisfies vT 2 r0 .
(e) From parts (c) and (d), deduce that
4 2
T 2 r03 ;
GM
that is, the square of the period of a satellite in circular orbit is
proportional to the cube of the radius from the orbital center.
Extending the Ideas
Let u u1, u2 and v v1, v2 be vectors in the plane. The dot
product or inner product u v is a scalar defined by
u v u1, u2 v1, v2 u1v1 u2v2.
61. Using the Dot Product Show that the dot product of two perpendicular vectors is zero.
62. An Alternate Formula for Dot Product Let u u1, u2
and v v1, v2 be vectors in the plane, and let w u v.
(a) Explain why w can be represented by the
arrow in the accompanying diagram.
m
r(t)
t 0
(c) Find the component form of w and use it to
prove that
M
r0
w
(b) Explain why |w|2 |u|2 |v|2 2|u||v| cos , u
where is the angle between vectors u and v.
x
|u|2
|v|2
|w|2
u
v
2(u1v1 u2v2).
(d) Finally, prove that u v |u||v| cos , where is the angle
between vectors u and v.
27. v(t) 6t, 6t2, a(t) 6, 12t
28. v(t) 2 cos 2t, 2sin t, a(t) 4 sin 2t, 2 cos t
29. v(t) et tet, et, a(t) 2et tet, et
30. v(t) 6 sin 3t, 8 cos 4t, a(t) 18 cos 3t, 32 sin 4t
31. v(t) 2t 2 cos 2t, 2t 2 sin 2t, a(t) 2 4 sin 2t, 2 4 cos 2t
32. v(t) sin t t cos t, cos t t sin t,
32. a(t) 2 cos t t sin t, 2 sin t t cos t
4t
2 2t2
45. (a) ,
2
2
(1 t ) (1 t2)2
(b) No. The x-component of velocity is zero only if t 0, while the
y-component of velocity is zero only if t 1. At no time will the
velocity be 0, 0.
1 t2
2t
1, 0
(c) lim ,
t→ 1 t2 1 t2
3 cos 4
(3 cos 4)2 (sin 8)2 2.196
50. (a) y 5 (x 4) (b) sin 8
3
3
50. (d) 4 sin (t3) dt, 5 3 cos(t2) dt (4.004, 5.724)
2
2
1 x2
57. The velocity vector is x, 1 x2, which has slope .
x
d
d
The acceleration vector is (x), (
1 x2)
dt
dt
dx
2x dx
, 2 dt 1 x dt
x2
x
x, 2 , which has slope .
Since the slopes are negative
1
x
1
x2
reciprocals of each other, the vectors are orthogonal.
5128_Ch10_pp530-561 2/3/06 4:41 PM Page 548
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Chapter 10
Parametric, Vector, and Polar Functions
10.3
• Polar Coordinates
• Polar Curves
• Slopes of Polar Curves
• Areas Enclosed by Polar Curves
• A Small Polar Gallery
. . . and why
Polar equations enable us to define some interesting and important curves that would be difficult
or impossible to define in the
form y f (x).
Polar Functions
Polar Coordinates
If you graph the two functions y sin 3x and
y cos 5x on the same pair of axes, you will get
two sinusoids. But if you graph the curve defined
parametrically by x sin 3t and y cos 5t, you
will get the figure shown. Parametric graphing opens
up a whole new world of curves that can be defined
using our familiar basic functions.
Another way to enter that world is to use a different coordinate system. In polar coordinates we identify the origin O as the pole and the positive x-axis as
the initial ray of angles measured in the usual trigonometric way. We can then identify each
point P in the plane by polar coordinates (r, ), where r gives the directed distance from O to
In Figure 10.19 we see that
P and gives the directed angle from the initial ray to the ray OP.
the point P with rectangular (Cartesian) coordinates (2, 2) has polar coordinates (22, 4).
y
y
P(2, 2)
2
P(2 2, p/4)
2
2 2
p/4
x
x
O
O
2
Rectangular coordinates
Polar coordinates
Figure 10.19 Point P has rectangular coordinates (2, 2) and polar coordinates (22, 4).
As you would expect, we can also coordinatize point P with the polar coordinates
Less obvi(22, 94) or (22, 74), since those angles determine the same ray OP.
ously, we can also coordinatize P with polar coordinates (22, 34), since the directed
distance 22 in the 34 direction is the same as the directed distance 22 in the 4
direction (Figure 10.20). So, although each pair (r, ) determines a unique point in the plane,
each point in the plane can be coordinatized by an infinite number of polar ordered pairs.
y
y
–2 2
2 2
p/4
x
x
−3p/4
Figure 10.20 The directed negative distance 22 in the 34 direction is the same as the
directed positive distance 22 in the 4 direction. Thus the polar coordinates (22
, 34)
and (22, 4) determine the same point.
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Section 10.3 Polar Functions
549
EXAMPLE 1 Rectangular and Polar Coordinates
(a) Find rectangular coordinates for the points with given polar coordinates.
(i) (4, 2) (ii) (3, ) (iii) (16, 56) (iv) (2, 4)
(b) Find two different sets of polar coordinates for the points with given rectangular
coordinates.
(i) (1, 0) (ii) (3, 3) (iii) (0, 4) (iv) (1, 3)
SOLUTION
(a) (i) (0, 4)
(ii) (3, 0)
(iii) (83, 8)
(iv) (1, 1)
(b) A point has infinitely many sets of polar coordinates, so here we list just two typical
examples for each given point.
(i) (1, 0), (1, 2)
(ii) (32, 34), (32, 4)
(iii) (4, 2), (4, 32)
(iv) (2, 3), (2, 43)
Now try Exercises 1 and 3.
EXAMPLE 2 Graphing with Polar Coordinates
Graph all points in the plane that satisfy the given polar equation
(a) r 2
(b) r 2
(c) 6
SOLUTION
First, note that we do not label our axes r and . We are graphing polar equations in the
usual xy-plane, not renaming our rectangular variables!
(a) The set of all points with directed distance 2 units from the pole is a circle of radius
2 centered at the origin (Figure 10.21a).
(b) The set of all points with directed distance 2 units from the pole is also a circle of
radius 2 centered at the origin (Figure 10.21b).
(c) The set of all points of positive or negative directed distance from the pole in the 6
direction is a line through the origin with slope tan(6) (Figure 10.21c).
Now try Exercise 7.
y
y
3
y
3
2
3
–2
p/6
x
–3
3
–3
(a)
x
x
–3
3
–3
(b )
–3
3
–3
(c)
Figure 10.21 Polar graphs of (a) r 2, (b) r 2, and (c) 6. (Example 2)
Polar Curves
The curves in Example 2 are a start, but we would not introduce a new coordinate system just to
graph circles and lines; there are far more interesting polar curves to study. In the past it was hard
work to produce reasonable polar graphs by hand, but today, thanks to graphing technology, it is
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Chapter 10
Parametric, Vector, and Polar Functions
just a matter of finding the right window and pushing the right buttons. Our intent in this section is to use the technology to produce the graphs and then concentrate on how calculus can
be used to give us further information.
EXAMPLE 3 Polar Graphing with Technology
Find an appropriate graphing window and produce a graph of the polar curve.
(a) r sin 6
(b) r 1 2 cos (c) r 4 sin SOLUTION
For all these graphs, set your calculator to POLAR mode.
(a) First we find the window. Notice that r
sin 6 1 for all , so points on the graph
are all within 1 unit from the pole. We want a window at least as large as [1, 1] by [1, 1],
but we choose the window [1.5, 1.5] by [1, 1] in order to keep the aspect ratio close to
the screen dimensions, which have a ratio of 3:2. We choose a -range of 0 2 to
get a full rotation around the graph, after which we know that sin 6 will repeat the same
graph periodically. Choose step 0.05. The result is shown in Figure 10.22a.
(b) In this graph we notice that r
12 cos 3, so we choose [3, 3] for our
y-range and, to get the right aspect ratio, [4.5, 4.5] for our x-range. Due to the cosine’s
period, 0 2 again suffices for our -range. The graph is shown in Figure 10.22b.
(c) Since r
4 sin 4, we choose [4, 4] for our y-range and [6, 6] for our x-range.
Due to the sine’s period, 0 2 again suffices for our -range. The graph is shown in
Figure 10.22c.
Now try Exercise 13.
[–1.5, 1.5] by [–1, 1]
0 ≤ u ≤ 2p
[– 4.5, 4.5] by [–3, 3]
0 ≤ u ≤ 2p
[–6, 6] by [– 4, 4]
0 ≤ u ≤ 2p
(a)
(b)
(c)
Figure 10.22 The graphs of the three polar curves in Example 3. The curves are (a) a 12-petaled rose, (b) a limaçon, and
(c) a circle.
A Rose is a Rose
The graph in Figure 10.22a is called a
12-petaled rose, because it looks like a
flower and some flowers are roses. The
graph in Figure 10.22b is called a
limaçon (LEE-ma-sohn) from an old
French word for snail. We will have
more names for you at the end of the
section.
With a little experimentation, it is possible to improve on the “safe” windows we chose in
Example 3 (at least in parts (b) and (c)), but it is always a good idea to keep a 3:2 ratio of
the x-range to the y-range so that shapes do not become distorted. Also, an astute observer
may have noticed that the graph in part (c) was traversed twice as went from 0 to 2, so
a range of 0 would have sufficed to produce the entire graph. From 0 to , the
circle is swept out by positive r values; then from to 2, the same circle is swept out by
negative r values.
Although the graph in Figure 10.22c certainly looks like a circle, how can we tell for sure
that it really is? One way is to convert the polar equation to a Cartesian equation and verify
that it is the equation of a circle. Trigonometry gives us a simple way to convert polar equations to rectangular equations and vice versa.
5128_Ch10_pp530-561 1/13/06 3:51 PM Page 551
Section 10.3 Polar Functions
551
Polar–Rectangular Conversion Formulas
x r cos r2 x2 y2
y r sin y
tan x
EXAMPLE 4 Converting Polar to Rectangular
Use the polar–rectangular conversion formulas to show that the polar graph of r 4 sin is a circle.
SOLUTION
To facilitate the substitutions, multiply both sides of the original equation by r. (This could
introduce extraneous solutions with r 0, but the pole is the only such point, and we notice
that it is already on the graph.)
r 4 sin r2 4r sin x2
y2
4y
Multiply by r.
Polar–rectangular conversion
x2 y2 4y 0
x2 y2 4y 4 4
x2 (y 2)2 22
Completing the square
Circle in standard form
Sure enough, the graph is a circle centered at (0, 2) with radius 2.
Now try Exercise 25.
The polar–rectangular conversion formulas also reveal the calculator’s secret to polar graphing:
It is really just parametric graphing with as the parameter.
Parametric Equations of Polar Curves
The polar graph of r f () is the curve defined parametrically by:
x r cos f () cos y r sin f () sin EXPLORATION 1
Graphing Polar Curves Parametrically
Switch your grapher to parametric mode and enter the equations
x sin (6t) cos t
y sin (6t) sin t.
1. Set an appropriate window and see if you can reproduce the polar graph in
Figure 10.22a.
2. Then produce the graphs in Figures 10.22b and 10.22c in the same way.
5128_Ch10_pp530-561 01/16/06 12:15 PM Page 552
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Chapter 10
Parametric, Vector, and Polar Functions
Slopes of Polar Curves
Since polar curves are drawn in the xy-plane, the slope of a polar curve is still the slope of the
tangent line, which is dydx. The polar–rectangular conversion formulas enable us to write x
and y as functions of , so we can find dydx as we did with parametrically defined functions:
dy
dyd
.
dx
dxd
EXAMPLE 5 Finding Slope of a Polar Curve
Find the slope of the rose curve r 2 sin 3 at the point where 6 and use it to find
the equation of the tangent line (Figure 10.23).
SOLUTION
The slope is
dy
dyd
dx 6
dxd 6
[–3, 3] by [–2, 2]
0≤u≤p
Figure 10.23 The 3-petaled rose curve
r 2 sin 3. Example 5 shows how to find
the tangent line to the curve at 6.
d
(2 sin 3 sin)
d
d
(2 sin 3 cos)
d
.
6
This expression can be computed by hand, but it is an excellent candidate for your calculator’s numerical derivative functionality (Section 3.2). NDERIV quickly gives an answer of
1.732050808, which you might recognize as 3.
When 6,
x 2 sin(2) cos(6) 3 and y 2 sin(2) sin(6) 1.
So the tangent line has equation y 1 3 (x 3).
Now try Exercise 39.
Areas Enclosed by Polar Curves
We would like to be able to use numerical integration to find areas enclosed by polar curves
just as we did with curves defined by their rectangular coordinates. Converting the equations
to rectangular coordinates is not a reasonable option for most polar curves, so we would like to
have a formula involving small changes in rather than small changes in x. While a small
change x produces a thin rectangular strip of area, a small change produces a thin circular
sector of area (Figure 10.24).
y
y
∆u
∆x
x
x
Figure 10.24 A small change in x produces a rectangular strip of area, while a small change
in produces a thin sector of area.
5128_Ch10_pp530-561 1/13/06 3:51 PM Page 553
Section 10.3 Polar Functions
553
1
Recall from geometry that the area of a sector of a circle is 2r 2, where r is the radius and is
the central angle measured in radians. If we replace by the differential d, we get the area
1
differential dA 2r 2d (Figure 10.25), which is exactly the quantity that we need to integrate to get an area in polar coordinates.
y
1
dA – r 2 d
2
P(r, )
r
d
x
O
Figure 10.25 The area differential dA.
Area in Polar Coordinates
The area of the region between the origin and the curve r f for is
1
1
2
A
r 2 d f d.
2
2
EXAMPLE 6 Finding Area
Find the area of the region in the plane enclosed by the cardioid r 21 cos .
SOLUTION
y
We graph the cardioid (Figure 10.26) and determine that the radius OP sweeps out the
region exactly once as runs from 0 to 2 .
Solve Analytically The area is therefore
r 2 (1 cos )
P(r, )
2
r
4
O
0, 2
x
2
0
1
r 2 d 2
(
2p
1
• 41 cos 2 d
2
0
2p
21 2 cos cos 2 d
0
2p
Figure 10.26 The cardioid in
Example 6.
0
)
1 cos 2 2 4 cos 2 d
2
2p
3 4 cos cos 2 d
0
[
sin 2 3 4 sin 2
]
2
6 0 6 .
0
Support Numerically NINT 21 cos 2, , 0, 2 18.84955592, which
agrees with 6 to eight decimal places.
Now try Exercise 43.
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Chapter 10
Parametric, Vector, and Polar Functions
y
EXAMPLE 7 Finding Area
r 2 cos 1
2–
—
3
Find the area inside the smaller loop of the limaçon r 2 cos 1.
SOLUTION
0
x
After watching the grapher generate the curve over the interval 0 2 (Figure
10.27), we see that the smaller loop is traced by the point r, as increases from
2 3 to 4 3 (the values for which r 2 cos 1 0). The area we seek is
4 3
A
4–
—
3
1
1
r 2 d 2
2
2 3
4 3
2 cos 1 2 d.
2 3
Solve Numerically
Figure 10.27 The limaçon in
Example 7.
1
NINT 2 cos 1 2, , 2 3, 4 3 0.544.
2
Now try Exercise 47.
y
r2
To find the area of a region like the one in Figure 10.28, which lies between two polar curves
r1 r1() and r2 r2() from to , we subtract the integral of (1 2)r12 from the
integral of (1 2)r 22. This leads to the following formula.
r1
Area Between Polar Curves
The area of the region between r1 and r2 for is
A
x
O
Figure 10.28 The area of the shaded
region is calculated by subtracting the area
of the region between r1 and the origin
from the area of the region between r2 and
the origin.
1
1
1
r22 d r12 d r22 r 12 d.
2
2
2
EXAMPLE 8 Finding Area Between Curves
Find the area of the region that lies inside the circle r 1 and outside the cardioid
r 1 cos .
SOLUTION
y
r1 1 cos The region is shown in Figure 10.29. The outer curve is r2 1, the inner curve is
r1 1 cos , and runs from 2 to 2. Using the formula for the area between
polar curves, the area is
Upper limit
/2
r2 1
A
2
x
0
2
1
r 22 r12 d
2
2
2
0
1
r 22 r12 d
2
Symmetry
2
Lower limit
– /2
Figure 10.29 The region and limits of
integration in Example 8.
1 1 2 cos cos 2 d
0
2
2 cos cos 2 d 1.215.
Using NINT
0
In case you are interested, the exact value is 2 4.
Now try Exercise 53.
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Section 10.3 Polar Functions
555
A SMALL POLAR GALLERY
Here are a few of the more common polar graphs and the -intervals that can be used to
produce them.
CIRCLES
y
y
y
x
x
x
r a cos 0
r a sin 0
r constant
0 2
ROSE CURVES
y
y
x
x
r a sin n, n even
0 2
2n petals
y-axis symmetry and
x-axis symmetry
r a sin n, n odd
0
n petals
y-axis symmetry
y
y
x
r a cos n, n odd
0
n petals
x-axis symmetry
x
r a cos n, n even
0 2
2n petals
y-axis symmetry and
x-axis symmetry
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Chapter 10
Parametric, Vector, and Polar Functions
LIMAÇON CURVES
r a b sin or r a b cos with a 0 and b 0
(r a b sin has y-axis symmetry; r a b cos has x-axis symmetry.)
y
y
x
x
a
1
b
0 2
Cardioid
a
1
b
0 2
Limaçon with loop
y
y
x
x
a
2
b
0 2
Convex limaçon
a
1 2
b
0
Dimpled limaçon
LEMNISCATE CURVES
y
y
x
x
r2 a2 sin 2
0
r2 a2 cos 2
0
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Section 10.3 Polar Functions
557
SPIRAL OF ARCHIMEDES
y
x
r0
Quick Review 10.3
(For help, go to Sections 10.1 and 10.2.)
1. Find the component form of a vector with magnitude 4 and
direction angle 30º. 23, 2
2. Find the area of a 30º sector of a circle of radius 6.
Exercises 6–10 refer to the parametrized curve
x 3 cos t, y 5 sin t, 0 t 2.
5
3
6. Find dydx. cot t
3
8. Find the points on the curve where the slope is zero. (0, 5) and (0, 5)
4. Find the rectangular equation of a circle of radius 5 centered at
the origin. x2 y2 25
9. Find the points on the curve where the slope is undefined. (3, 0) and
5. Explain how to use your calculator in function mode to graph 2
4x
4 x2 1/2
the curve x2 3y2 4. Graph y and y 3
3
5
3
7. Find the slope of the curve at t 2. cot 2 0.763
3. Find the area of a sector of a circle of radius 8 that has a central
10. Find the length of the curve from t 0 to t .
(3, 0)
12.763
1/2
Section 10.3 Exercises
In Exercises 1 and 2, plot each point with the given polar coordinates
and find the corresponding rectangular coordinates.
1. (a) 2, 4
(c) 0, 2
(b) 1, 0
(d) 2, 4
2. (a) 3, 5 6
(b) 5, tan1 4 3
(c) 1, 7 (d) 2 3, 2 3
In Exercises 3 and 4, plot each point with the given rectangular coordinates and find two sets of corresponding polar coordinates.
3. (a) 1, 1
(c) 0, 3
4. (a) 3, 1
(c) 0, 2
(b) 1, 3 (d) 1, 0
(b) 3, 4
(d) 2, 0
In Exercises 5–10, graph the set of points whose polar coordinates
satisfy the given equation.
5. r 3
7.
r2
4
9. 6
6. r 3
8. 4
10. r2 8 6r
23. x y 1, a line (slope 1, y-intercept 1)
24. x2 y2 1, a circle (center (0, 0), radius 1)
25. y 2x 5, a line (slope 2, y-intercept 5)
In Exercises 11–20, find an appropriate window and use a graphing
calculator to produce the polar curve. Then sketch the complete curve
and identify the type of curve by name.
11. r 1 cos 12. r 2 2 cos 13. r 2 cos 3
14. r 3 sin 2
15. r 1 2 sin 16. r 32 cos 17.
r2
4 cos 2
19. r 4 sin 18. r 2 sin 2
20. r 3 cos In Exercises 21–30, replace the polar equation by an equivalent Cartesian
(rectangular) equation. Then identify or describe the graph.
21. r 4 csc 22. r 3 sec 23. r cos r sin 1
5
25. r sin 2 cos 24. r 2 1
y 4, a horizontal line
27. cos 2 sin 2 x 3, a vertical line
26. r 2 sin 2 2
xy 1, a hyperbola
28. r 2 4r cos 29. r 8 sin 30. r 2 cos 2 sin 27. x2 y2, the union of two lines: y x
28. (x 2)2 y2 4, a circle (center (2, 0), radius 2)
29. x2 (y 4)2 16, a circle (center (0, 4), radius 4)
30. (x 1)2 (y 1)2 2, a circle (center (1, 1), radius 2)
5128_Ch10_pp530-561 2/3/06 4:41 PM Page 558
558
Chapter 10
Parametric, Vector, and Polar Functions
62. False. Integrating from 0 to 2 traverses the curve twice, giving twice the
area. The correct upper limit of integration is .
In Exercises 31–38, find an appropriate window and use a graphing
calculator to produce the polar curve. Then sketch the complete curve
and identify the type of curve by name. (Note: You won’t find these in
the Polar Gallery.)
57. Sketch the polar curves r 3 cos and r 1 cos and find
the area that lies inside the circle and outside the cardioid.
31. r sec tan 1
33. r 1 cos 14
35. r 5 9 cos 1
37. r 1 0.8 cos 32. r csc cot
2
34. r 1 sin 12
36. r 8 6 cos 1
38. r 1 1.3 cos 58. Sketch the polar curves r 2 and r 2(1 sin ) and find the
area that lies inside the circle and outside the cardioid.
59. Sketch the polar curve r 2 sin 3. Find the area enclosed by
the curve and find the slope of the curve at the point where
4.
60. The accompanying figure shows the parts of the graphs of the line
5
x 3 y and the curve x 1 y2 that lie in the first quadrant.
Region R is enclosed by the line, the curve, and the x-axis.
y
In Exercises 39–42, find the slope of the curve at each indicated point.
39. r 1 sin , 0, At 0: 1; At : 1
40. r cos 2 , 0, 2, At 0: undefined; At 2: 0
At 2: 0; At : undefined
41. r 2 3 sin At (2, 0): 2/3
At (1: 2): 0
At (2, ): 2/3
At (5, 32): 0
y
(2, p)
(2, 0)
x
p
–1, 2
(
R
(
O
1
x
5y
1 y dy 0.347
3
3/4
2
0
(a) Set up and evaluate an integral expression with respect to y
that gives the area of R.
(5, 3p2 (
42. r 31 cos At (1.5, 3): undefined
At (4.5, 23): 0
At (6, ): undefined
At (3, 32): 1
1 y2 can be described in polar
(b) Show that the curve x 1
Let x r cos and
2
coordinates by r .
cos2 sin2 y r sin and solve for r 2.
y
(4.5
2p
3
(
p
1.5 3
(
(c) Use the polar equation in part (b) to set up an integral
expression with respect to that gives the area of
R.
1
1
(
Let tan1 (3/5). Then the area is
Standardized Test Questions
x
(6, p)
2
d.
cos sin 0
2
2
You may use a graphing calculator to solve the following
problems.
(3 32p (
61. True or False There is exactly one point in the plane with
coordinates determine a unique point.
In Exercises 43–56, find the area of the region described.
43. inside the convex limaçon r 4 2 cos 44. inside the cardioid r 2 2 sin 2 1
2
18
r sin 3 is 0
6
45. inside one petal of the four-petaled rose r cos 2
46. inside the eight-petaled rose r 2 sin 4
48. inside the six-petaled rose
2 sin 3
8
(C) 2
11
51. shared by the circles r 2 cos and r 2 sin 2
(3 cos) d
(E) 3 cos
d
2
0
2
(B) 0 3 cos
d
(D) 0 (3 cos) d
0
50. inside the inner loop of the limaçon r 2 sin 1 0.544
52. shared by the circles r 1 and r 2 sin 63. Multiple Choice The area of the region enclosed by the polar
graph of r 3
co
s is given by which integral? D
2
4
49. inside the dimpled limaçon r 3 2 cos sin2 3 d. Justify your answer.
(A) 0 3 cos
d
2
47. inside one loop of the lemniscate r2 4 cos 2
r2
62. True or False The total area enclosed by the 3-petaled rose
(2) 1
(23) (3
2)
53. shared by the circle r 2 and the cardioid r 2(1 cos ) 5 8
54. shared by the cardioids r 2(1 cos ) and r 2(1 cos ) 6 16
64. Multiple Choice The area enclosed by one petal of the
3-petaled rose r 4 cos(3) is given by which integral? E
3
(A) 16
3
(C) 8
55. inside the circle r 2 and outside the cardioid r 2(1 sin ) 8 56. inside the four-petaled rose r 4 cos 2 and outside the circle
r 2 43 (83)
3
(E) 8
3
6
6
cos(3) d
cos2(3) d
cos2(3) d
6
(B) 8
6
cos(3) d
6
(D) 16
6
cos2(3) d
5128_Ch10_pp530-561 01/16/06 12:43 PM Page 559
Section 10.3 Polar Functions
65. Multiple Choice If a 0 and 0, all of the following
must necessarily represent the same point in polar coordinates
except which ordered pair? B
(A) (a, )
(B) (a, )
(C) (a, )
(D) (a, )
(E) (a, 2)
66. Multiple Choice Which of the following gives the slope of
the polar curve r f() graphed in the xy-plane? D
dr
(A) d
dy
(B) d
dyd
(D) dxd
dx
(C) d
dy dr
(E) dx d
Explorations
67. Rotating Curves Let r 1 31 cos and
r 2 r 1 .
(a) Graph r 2 for 6, 4, 3, and 2 and compare
with the graph of r 1.
(b) Graph r 2 for 6, 4, 3, and 2 and
compare with the graph of r1.
(c) Based on your observations in parts (a) and (b), describe the
relationship between the graphs of r 1 f and r 2 f .
2
68. Let r .
1 k cos (a) Graph r in a square viewing window for k 0.1, 0.3, 0.5,
0.7, and 0.9. Describe the graphs.
(b) Based on your observations in part (a), conjecture what
happens to the graphs for 0 k 1 and k→0.
2
69. Let r .
1 k cos (a) Graph r in a square viewing window for k 1.1, 1.3, 1.5, 1.7,
and 1.9. Describe the graphs.
(b) Based on your observations in part (a), conjecture what
happens to the graphs for k 1 and k→1.
k
70. Let r .
1 cos 559
(a) Graph r in a square viewing window for k 1, 3, 5, 7, and 9.
Describe the graphs.
(b) Based on your observations in part (a), conjecture what
happens to the graphs for k 0 and k→0.
Extending the Ideas
71. Distance Formula Show that the distance between two points
r1, 1 and r 2, 2 in polar coordinates is
2 2
r 1
r 2
r1
r 2cos
d 2 1
2 .
72. Average Value If f is continuous, the average value of the
polar coordinate r over the curve r f , , with
respect to is
1
rav f d.
Use this formula to find the average value of r with respect to over the following curves a 0.
(a) the cardioid r a1 cos a
(b) the circle r a a
(c) the circle r a cos , 2 2
2a/
73. Length of a Polar Curve The parametric form of the arc
length formula (Section 10.1) gives the length of a polar curve as
dx 2
dy 2
d.
d
d
Assuming that the necessary derivatives are continuous, show
that the substitutions x r cos and y r sin transform this
expression into
L
L
dr 2
r 2 d.
d
74. Length of a Cardioid Use the formula in Exercise 73 to find
the length of the cardioid r 1 cos . 8
Quick Quiz for AP* Preparation: Sections 10.1–10.3
You may use a graphing calculator to solve the following
problems.
1. Multiple Choice Which of the following is equal to the area
of the region inside the polar curve r 2 cos and outside the
polar curve r cos ? A
2
cos2 d
3 (C) cos2 d
2
(A) 3
0
2
0
(B) 30 cos2 d
(D) 3
2
0
cos d
(E) 30 cos d
(A) 0 4t 1 dt
(B) 20 t2 1 dt
(D) 0 4t2 1 dt
(E) 20 4t2 1 dt
4
4
4
(C) 0 2t2 1 dt
4
4
4. Free Response A polar curve is defined by the equation
r sin 2 for 0 .
(a) Find the area bounded by the curve and the x-axis.
2. Multiple Choice For what values of t does the curve given by
the parametric equations x t3 t2 1 and y t 4 2t 2 8t
have a vertical tangent? C
(A) 0 only
(B) 1 only
(C) 0 and 2 3 only
(D) 0, 2 3, and 1
(E) No value
3. Multiple Choice The length of the path described by the
parametric equations x t2 and y t from t 0 to t 4 is
given by which integral? D
(b) Find the angle that corresponds to the point on the curve
where x 2.
2 dr
(c) For , is negative. How can this be seen
3 d
3
from the graph?
(d) At what angle in the interval 0 2 is the curve
5128_Ch10_pp530-561 01/16/06 12:43 PM Page 560
560
Chapter 10
Parametric, Vector, and Polar Functions
Chapter 10 Key Terms
Absolute value of a vector (p. 538)
Acceleration vector (p. 542)
Archimedes spiral (p. 557)
Arc length of a parametrized curve (p. 533)
Arc length of a polar curve (p. 553)
Area between polar curves (p. 554)
Area differential (p. 553)
Area in polar coordinates (p. 553)
Arrow (p. 538)
Cardioid (p. 556)
Cartesian equation of a curve (p. 550)
Component form of a vector (p. 539)
Components of a vector (p. 538)
Convex limaçon (p. 556)
Cycloid (p. 533)
Dimpled limaçon (p. 556)
Directed distance (p. 548)
Directed line segment (p. 538)
Direction angle of a vector (p. 539)
Direction of motion (p. 542)
Direction vector (p. 538)
Displacement (p. 543)
Distance traveled (p. 544)
Pole (p. 548)
Position of a particle (p. 544)
Position vector (p. 538)
Properties of vectors (p. 541)
Rectangular coordinates (p. 548)
Resultant vector (p. 540)
Rose curve (p. 555)
Scalar (p. 539)
Scalar multiple of a vector (p. 540)
Speed (p. 542)
Standard representation of
a vector (p. 538)
Sum of vectors (p. 540)
Terminal point of an arrow (p. 539)
Unit vector (p. 540)
Vector (p. 538)
Velocity vector (p. 542)
Zero vector (p. 538)
Dot product of vectors (p. 547)
Equivalent arrows (p. 539)
Head Minus Tail Rule (p. 539)
Huygens’s clock (p. 534)
Initial point of an arrow (p. 539)
Initial ray of angle of direction (p. 548)
Lemniscate (p. 556)
Limaçon (p. 556)
Limaçon with inner loop (p. 556)
Magnitude of a vector (p. 538)
Opposite of a vector (p. 540)
Orthogonal vectors (p. 547)
Parallelogram representation of vector
Parametric equations of a polar
curve (p. 551)
Parametric formula for dydx (p. 532)
Parametric formula for d 2 ydx2 (p. 532)
Path of a particle (p. 542)
Polar coordinates (p. 548)
Polar equation of a curve (p. 549)
Polar graphing (p. 549)
Polar–rectangular conversion
formulas (p. 551)
Chapter 10 Review Exercises
In Exercises 1–4, let u 3, 4 and v 2, 5. Find
(a) the component form of the vector and (b) its magnitude.
1. 3u 4v
2. u v
3. 2u
4. 5v
In Exercises 5–8, find the component form of the vector.
5. the vector obtained by rotating 0, 1 through an angle
of 2p 3 radians 32, 1/2 [assuming counterclockwise]
6. the unit vector that makes an angle of 6 radian with the
positive x-axis 32, 1/2
7. the vector 2 units long in the direction 4 i j 8/17
, 2/17
8. the vector 5 units long in the direction opposite to the direction
of 3 5, 4 5 3, 4
In Exercises 9 and 10, (a) find an equation for the tangent to the
curve at the point corresponding to the given value of t, and
(b) find the value of d 2 ydx 2 at this point.
1. (a) 17, 32 (b) 1313
3. (a) 6, 8 (b) 10
2. (a) 1, 1 (b) 2
4. (a) 10, 25 (b) 725
529
9. x 1 2 tan t,
10. x 1 1 t 2,
3
1
y 1 2 sec t; t 3 (a) y 2x 4 (b) 1/4
y 1 3 t; t 2 (a) 1, 1 (b) 2
In Exercises 11–14, find the points at which the tangent
to the curve is (a) horizontal; (b) vertical.
11. x 1 2 tan t,
12. x 2 cos t,
13. x cos t,
(a) (0, 1/2) and (0, 1/2)
(b) Nowhere
y 2 sin t (a) (0, 2) and (0, 2) (b) (2, 0) and (2, 0)
y 1 2 sec t
y cos 2 t
14. x 4 cos t,
y 9 sin t
In Exercises 15–20, find an appropriate window and graph the polar
curve on a graphing calculator. Then sketch the curve on paper and
identify the type of curve.
15. r 1 sin 16. r 2 cos 17. r cos 2 18. r cos 1
sin 2 20. r sin 19.
r2
13. (a) (0, 0) (b) Nowhere
14. (a) (0, 9) and (0, 9) (b) (4, 0) and (4, 0)
5128_Ch10_pp530-561 1/13/06 3:51 PM Page 561
23. Horizontal: y 0, y 0.443, y 1.739
Vertical: x 2, x 0.067, x 1.104
Chapter 10 Review Exercises
24. Horizontal: y 1/2, y 4
Vertical: x 0, x 2.598
561
34. (r cos 2)2 (r sin 5)2 16
In Exercises 21 and 22, find the slope of the tangent lines at the point
where 3.
(b) Find the x- and y-components of the acceleration of
the particle at t 3.
21. r cos 2
(c) Find a single equation in x and y for the path of the particle.
22. r 2 cos 2
4.041
0.346
In Exercises 23 and 24, find equations for the horizontal and vertical
tangent lines to the curves.
23. r 1 cos 2,
0 4
24. r 21 sin ,
48. Particle Motion At time t, 0 t 4, the position of
a particle moving along a path in the plane is given by
the parametric equations
x e t cos t,
0 2
y e t sin t.
25. Find equations for the lines that are tangent to the tips of the
petals of the four-petaled rose r sin 2 . y x 2 and y x 2
(a) Find the slope of the path of the particle at time t . 1
26. Find equations for the lines that are tangent to the cardioid
r 1 sin at the points where it crosses the x-axis.
(c) Find the distance traveled by the particle along the path from
t 0 to t 3. (e3 1)2
y x 1 and y x 1
In Exercises 27–30, replace the polar equation by an equivalent
Cartesian equation. Then identify or describe the graph.
27. r cos r sin x y, a line
28. r 3 cos 29. r2 4 tan sec 30. r cos 3 2 3
x 4y, a parabola
(b) Find the speed of the particle when t 3.
e32
49. Particle Motion The position of a particle at any time
t 0 is given by
2
xt t 2 2, yt t 3.
5
(a) Find the magnitude of the velocity vector at t 4. 104/5
In Exercises 31–34, replace the Cartesian equation by an equivalent
polar equation.
(b) Find the total distance traveled by the particle from
t 0 to t 4. 4144/135
31. x 2 y 2 5y 0
32. x 2 y 2 2y 0 r 2 sin (c) Find dydx as a function of x.
33. x 2 4y 2 16
34. x 2 2 y 5 2 16
r 5 sin dy
3
5x
2
dx
36. enclosed by one petal of the three-petaled rose r sin 3 12
50. Navigation An airplane, flying in the direction 80º east of
north at 540 mph in still air, encounters a 55-mph tail wind
acting in the direction 100º east of north. The airplane holds its
compass heading but, because of the wind, acquires a different
ground speed and direction. What are they?
37. inside the “figure eight” r 1 cos 2 and outside the circle
r 1 (4) 2
AP* Examination Preparation
In Exercises 35–38, find the area of the region described.
35. enclosed by the limaçon r 2 cos 92
38. inside the cardioid r 21 sin and outside the
circle r 2 sin 5
In Exercises 39 and 40, r(t) is the position vector of a particle moving
in the plane at time t. Find (a) the velocity and acceleration vectors,
and (b) the speed at the given value of t.
39. r(t) 4 cos t, 2 sin t, t 4
Speed 591.982 mph;
You may use a graphing calculator to solve the following
problems.
51. A particle moves along the graph of y cos x so that its
x-component of acceleration is always 2. At time t 0, the
particle is at the point (, 1) and the velocity of the particle is
0, 0.
40. r(t) 3 sec t, 3 tan t, t 0
(a) Find the position vector of the particle.
41. The position of a particle in the plane at time t is
t
1
. Find the particle’s maximum speed.
r ,
1
t 2 1
t2
Direction 8.179° north of east
(b) Find the speed of the particle when it is at the point (4, cos 4).
52. Two particles move in the xy-plane. For time t 0, the position
of particle A is given by x t 2 and y (t 2)2, and the
3
3
42. Writing to Learning Suppose that r(t) et cos t, et sin t.
Show that the angle between r and the acceleration vector a
never changes. What is the angle?
position of particle B is given by x 2t 4 and y 2t 2.
In Exercises 43–46, find the position vector.
(c) Determine the exact time when the particles collide.
43. v(t) sin t, cos t and r(0) 0, 1 r(t) cos t 1, sin t 1
1
t
44. v(t) , 2 and r(0) 1, 1
1
t 2 1 t
t2 1
r(t) tan1 t 1, 45. a(t) 0, 2 and v(0) 0, 0 and r(0) 1, 0 r(t) 1, t2
(a) Find the velocity vector for each particle at time t 3.
(b) Find the distance traveled by particle A from t 0 to t 3.
53. A region R in the xy-plane is bounded below by the x-axis
4
and above by the polar curve defined by r for
1 sin 0 .
(a) Find the area of R by evaluating an integral in polar
coordinates.
46. a(t) 2, 2 and v(1) 4, 0 and r(1) 3, 3
47. Particle Motion A particle moves in the plane in such a
manner that its coordinates at time t are
x 3 cos t,
4
(b) The curve resembles an arch of the parabola 8y 16 x2.
Convert the polar equation to rectangular coordinates and prove
that the curves are the same.
y 5 sin t.
4
(c) Set up an integral in rectangular coordinates that gives the
area of R.
(a) Find the length of the velocity vector at t 3.
3
3
28. x2 y2 3x, a circle center , 0 , radius 2
2
x
4, a line
30. x 3
y 43 or y 3
46. r(t) t2 6t 2, t2 2t 2
```