User manual | Chapter 8- Polar Coordinates + Parametric

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CHAPTER
8
POLAR COORDINATES AND PARAMETRIC EQUATIONS
8.1 Polar Coordinates
8.2 Graphs of Polar Equations
8.3 Polar Form of Complex
Numbers; De Moivre’s Theorem
8.4 Plane Curves and Parametric
Equations
FOCUS ON MODELING
The Path of a Projectile
In Section 1.8 we learned how to graph points in rectangular coordinates. In
this chapter we study a different way of locating points in the plane, called polar coordinates. Using rectangular coordinates is like describing a location in a
city by saying that it's at the corner of 2nd Street and 4th Avenue; these directions would help a taxi driver find the location. But we may also describe this
same location “as the crow flies”; we can say that it's 1.5 miles northeast of
City Hall. So instead of specifying the location with respect to a grid of streets
and avenues, we specify it by giving its distance and direction from a fixed
reference point. That's what we do in the polar coordinate system. In polar
coordinates the location of a point is given by an ordered pair of numbers: the
distance of the point from the origin (or pole) and the angle from the positive
x-axis.
Why do we study different coordinate systems? It’s because certain curves
are more naturally described in one coordinate system rather than another. For
example, in rectangular coordinates lines and parabolas have simple equations,
but equations of circles are rather complicated. We'll see that in polar coordinates circles have very simple equations.
541
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542
CHAPTER 8
| Polar Coordinates and Parametric Equations
8.1 P OLAR C OORDINATES
Definition of Polar Coordinates Relationship Between Polar
and Rectangular Coordinates Polar Equations
P
r
▼ Definition of Polar Coordinates
¨
O
In this section we define polar coordinates, and we learn how polar coordinates are related
to rectangular coordinates.
Polar axis
FIGURE 1
The polar coordinate system uses distances and directions to specify the location of a
point in the plane. To set up this system, we choose a fixed point O in the plane called the
pole (or origin) and draw from O a ray (half-line) called the polar axis as in Figure 1.
Then each point P can be assigned polar coordinates P1r, u2 where
r is the distance from O to P
u is the angle between the polar axis and the segment OP
¨
O
|r|
P(r, ¨)
r<0
We use the convention that u is positive if measured in a counterclockwise direction from
the polar axis or negative if measured in a clockwise direction. If r is negative, then P1r, u2
is defined to be the point that lies 0 r 0 units from the pole in the direction opposite to that
given by u (see Figure 2).
Plotting Points in Polar Coordinates
EXAMPLE 1
FIGURE 2
Plot the points whose polar coordinates are given.
(a) 11, 3p/42
(b) 13, p/62
(c) 13, 3p2
(d) 14, p/42
S O L U T I O N The points are plotted in Figure 3. Note that the point in part (d) lies 4 units
from the origin along the angle 5p/4, because the given value of r is negative.
P !1,
3π
4 @
3π
3π
4
O
O
π
4
O
_ π6
P (3, 3π)
P !3,
O
π
_6 @
P !_4, π4 @
(a)
(b)
(c)
(d)
FIGURE 3
■
NOW TRY EXERCISES 3 AND 5
Note that the coordinates 1r, u2 and 1r, u p2 represent the same point, as shown in
Figure 4. Moreover, because the angles u 2np (where n is any integer) all have the same
terminal side as the angle u, each point in the plane has infinitely many representations in
polar coordinates. In fact, any point P1r, u2 can also be represented by
P1r, u 2np2
and
P1r, u 12n 12p2
for any integer n.
P ( r, ¨)
P ( _r, ¨+π)
¨+π
¨
FIGURE 4
O
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SECTION 8.1
| Polar Coordinates 543
Different Polar Coordinates for the Same Point
EXAMPLE 2
(a) Graph the point with polar coordinates P12, p/32 .
(b) Find two other polar coordinate representations of P with r 0 and two with
r 0.
SOLUTION
(a) The graph is shown in Figure 5(a).
(b) Other representations with r 0 are
a 2,
p
7p
2p b a 2,
b
3
3
Add 2p to u
a 2,
p
5p
2p b a 2, b
3
3
Add 2p to u
Other representations with r 0 are
a 2,
p
4p
p b a 2,
b
3
3
Replace r by r and add p to u
a 2,
p
2p
p b a 2, b
3
3
Replace r by r and add p to u
The graphs in Figure 5 explain why these coordinates represent the same point.
π
P !2, 3 @
P !2,
2
7π
3 @
P !2, _
2
5π
3 @
2
O
O
O
(b)
(a)
4π
3 @
P !_2, _
2
2π
3 @
2
4π
3
7π
3
π
3
P !_2,
O
_ 5π
3
(c)
O
(d)
_ 2π
3
(e)
FIGURE 5
■
NOW TRY EXERCISE 9
▼ Relationship Between Polar and Rectangular Coordinates
y
P(r, ¨)
P(x, y)
r
y
¨
0
FIGURE 6
x
x
Situations often arise in which we need to consider polar and rectangular coordinates simultaneously. The connection between the two systems is illustrated in Figure 6, where
the polar axis coincides with the positive x-axis. The formulas in the following box are
obtained from the figure using the definitions of the trigonometric functions and the
Pythagorean Theorem. (Although we have pictured the case where r 0 and u is acute,
the formulas hold for any angle u and for any value of r.)
RELATIONSHIP BETWEEN POLAR AND RECTANGULAR COORDINATES
1. To change from polar to rectangular coordinates, use the formulas
x r cos u
and
y r sin u
2. To change from rectangular to polar coordinates, use the formulas
r2 x2 y2
and
tan u y
x
1x 02
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544
| Polar Coordinates and Parametric Equations
CHAPTER 8
EXAMPLE 3
Converting Polar Coordinates
to Rectangular Coordinates
Find rectangular coordinates for the point that has polar coordinates 14, 2p/32 .
SOLUTION
Since r 4 and u 2p/3, we have
x r cos u 4 cos
y r sin u 4 sin
2p
1
4 # a b 2
3
2
2p
13
4#
2 13
3
2
Thus the point has rectangular coordinates 12, 2 132 .
NOW TRY EXERCISE 27
EXAMPLE 4
y
Converting Rectangular Coordinates
to Polar Coordinates
Find polar coordinates for the point that has rectangular coordinates 12, 22 .
3π
4
0
SOLUTION
π
Using x 2, y 2, we get
r 2 x 2 y 2 22 122 2 8
x
_4
so r 2 12 or 2 12. Also
(2, _2)
π
!2 Ϸ2, _ 4 @
!_2 Ϸ
2,
FIGURE 7
■
3π
4 @
tan u y
2
1
x
2
so u 3p/4 or p/4. Since the point 12, 22 lies in Quadrant IV (see Figure 7), we can
represent it in polar coordinates as 12 12, p/42 or 12 12, 3p/42 .
■
NOW TRY EXERCISE 35
Note that the equations relating polar and rectangular coordinates do not uniquely determine r or u. When we use these equations to find the polar coordinates of a point, we
must be careful that the values we choose for r and u give us a point in the correct quadrant, as we did in Example 4.
▼ Polar Equations
In Examples 3 and 4 we converted points from one coordinate system to the other. Now
we consider the same problem for equations.
EXAMPLE 5
Converting an Equation from Rectangular
to Polar Coordinates
Express the equation x 2 4y in polar coordinates.
SOLUTION
We use the formulas x r cos u and y r sin u:
x 2 4y
1r cos u 2 41r sin u 2
2
r 2 cos2 u 4r sin u
r4
sin u
cos2 u
r 4 sec u tan u
NOW TRY EXERCISE 45
Rectangular equation
Substitute x r cos u, y r sin u
Expand
Divide by r cos2 u
Simplify
■
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SECTION 8.1
| Polar Coordinates 545
As Example 5 shows, converting from rectangular to polar coordinates is straightforward: Just replace x by r cos u and y by r sin u, and then simplify. But converting polar
equations to rectangular form often requires more thought.
Converting Equations from Polar
to Rectangular Coordinates
EXAMPLE 6
Express the polar equation in rectangular coordinates. If possible, determine the graph of
the equation from its rectangular form.
(a) r 5 sec u
(b) r 2 sin u
(c) r 2 2 cos u
SOLUTION
(a) Since sec u 1/cos u, we multiply both sides by cos u:
r 5 sec u
Polar equation
r cos u 5
Multiply by cos u
x5
Substitute x r cos u
The graph of x 5 is the vertical line in Figure 8.
(b) We multiply both sides of the equation by r, because then we can use the
formulas r 2 x 2 y 2 and r sin u y:
r 2 sin u
Polar equation
r 2 2r sin u
Multiply by r
x y 2y
2
2
r 2 x 2 y 2 and r sin u y
x 2 y 2 2y 0
Subtract 2y
x 1y 12 1
2
2
Complete the square in y
This is the equation of a circle of radius 1 centered at the point 10, 12 . It is graphed
in Figure 9.
y
y
x=5
1
0
x
0
FIGURE 8
1
x
FIGURE 9
(c) We first multiply both sides of the equation by r:
r 2 2r 2r cos u
Using r 2 x 2 y 2 and x r cos u, we can convert two terms in the equation into
rectangular coordinates, but eliminating the remaining r requires more work:
x 2 y 2 2r 2x
r 2 x 2 y 2 and r cos u x
x 2 y 2 2x 2r
1x y 2x2 4r
2
2
2
Subtract 2x
2
1x y 2x2 41x y 2
2
2
2
2
2
Square both sides
r2 x2 y2
In this case the rectangular equation looks more complicated than the polar equation.
Although we cannot easily determine the graph of the equation from its rectangular
form, we will see in the next section how to graph it using the polar equation.
NOW TRY EXERCISES 53, 55, AND 57
■
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546
CHAPTER 8
| Polar Coordinates and Parametric Equations
8.1 EXERCISES
19. 14, 23p/42
CONCEPTS
1. We can describe the location of a point in the plane using
different
systems. The point P shown in
the figure has rectangular coordinates 1 ,
2 and polar
coordinates 1 ,
2.
y
1
20. 14, 23p/4 2
21. 14, 101p/42
22. 14, 103p/42
23–24 ■ A point is graphed in rectangular form. Find polar
coordinates for the point, with r 0 and 0 u 2p.
23.
y
24.
y
P
P
1
1
0
1
coordinates 1x, y2 where x tan u 1
and
.
Plot the point that has the given polar coordinates.
5. 16, 7p/6 2
4. 11, 0 2
7. 12, 4p/3 2
8. 15, 17p/6 2
9–14 ■ Plot the point that has the given polar coordinates. Then
give two other polar coordinate representations of the point, one
with r 0 and the other with r 0.
9. 13, p/22
11. 11, 7p/6 2
10. 12, 3p/4 2
12. 12, p/32
13. 15, 0 2
14. 13, 1 2
15–22 ■ Determine which point in the figure, P, Q, R, or S, has
the given polar coordinates.
Q
3
4 P
2
1
π
4
π
4
O
R
17. 14, p/4 2
2π
3
5π
6
1
O
R
27–34 ■ Find the rectangular coordinates for the point whose
polar coordinates are given.
SKILLS
6. 13, 2p/32
26. S
O
_
polar coordinates 1r, u2 where r 2 15. 14, 3p/4 2
25.
.
3. 14, p/42
x
25–26 ■ A point is graphed in polar form. Find its rectangular
coordinates.
and
(b) If P has rectangular coordinates 1x, y2 then it has
■
1
x
2. Let P be a point in the plane.
(a) If P has polar coordinates 1r, u2 then it has rectangular
3–8
0
x
Q
0
y
1
27. 14, p/62
28. 16, 2p/32
31. 15, 5p 2
32. 10, 13p2
29. 1 12, p/42
33. 16 12, 11p/62
30. 11, 5p/22
34. 1 13, 5p/32
35–42 ■ Convert the rectangular coordinates to polar
coordinates with r 0 and 0 u 2p.
35. 11, 1 2
36. 13 13, 32
39. 13, 4 2
40. 11, 2 2
37. 1 18, 182
41. 16, 0 2
43–48
■
38. 1 16, 122
42. 10, 132
Convert the equation to polar form.
43. x y
44. x 2 y 2 9
45. y x 2
46. y 5
47. x 4
48. x 2 y 2 1
49–68 ■ Convert the polar equation to rectangular
coordinates.
S
16. 14, 3p/4 2
18. 14, 13p/4 2
50. r 3
49. r 7
51. u p
2
53. r cos u 6
52. u p
54. r 2 csc u
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SECTION 8.2
55. r 4 sin u
DISCOVERY
56. r 6 cos u
58. r 311 sin u 2
4
1 2 sin u
■
WRITING
d 2r 21 r 22 2r1r2 cos1u2 u1 2
59. r 1 2 sin u
63. r DISCUSSION
69. The Distance Formula in Polar Coordinates
(a) Use the Law of Cosines to prove that the distance
between the polar points 1r1, u1 2 and 1r2, u2 2 is
57. r 1 cos u
60. r 2 cos u
1
61. r sin u cos u
■
| Graphs of Polar Equations 547
(b) Find the distance between the points whose polar coordinates are 13, 3p/42 and 11, 7p/6 2 , using the formula from
part (a).
(c) Now convert the points in part (b) to rectangular coordinates. Find the distance between them using the usual
Distance Formula. Do you get the same answer?
1
62. r 1 sin u
64. r 2
1 cos u
65. r 2 tan u
66. r 2 sin 2u
67. sec u 2
68. cos 2u 1
8.2 G RAPHS OF P OLAR E QUATIONS
Graphing Polar Equations Symmetry Graphing Polar Equations
with Graphing Devices
The graph of a polar equation r f 1u2 consists of all points P that have at least one polar representation 1r, u 2 whose coordinates satisfy the equation. Many curves that arise in
mathematics and its applications are more easily and naturally represented by polar equations than by rectangular equations.
▼ Graphing Polar Equations
A rectangular grid is helpful for plotting points in rectangular coordinates (see Figure
1(a)). To plot points in polar coordinates, it is convenient to use a grid consisting of circles centered at the pole and rays emanating from the pole, as in Figure 1(b). We will use
such grids to help us sketch polar graphs.
π
2
π
3
y
P(_2, 3)
5
4
3
2
1
3π
4
π
4
A!6,
Q(4, 2)
_5 _4 _3 _2 _1 0 1 2 3 4 5
_1
_2
_3
_4
_5 R(3, _5)
x
π
5π
6 @
B!4,
π
[email protected]
O 1 2 3 4 5 6
C !3,
π
6
0
4π
3 @
5π
4
7π
4
3π
2
FIGURE 1
(a) Grid for rectangular coordinates
(b) Grid for polar coordinates
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548
| Polar Coordinates and Parametric Equations
CHAPTER 8
In Examples 1 and 2 we see that circles centered at the origin and lines that pass
through the origin have particularly simple equations in polar coordinates.
3π
4
π
4
r=3
EXAMPLE 1
Sketching the Graph of a Polar Equation
Sketch a graph of the equation r 3, and express the equation in rectangular coordinates.
S O L U T I O N The graph consists of all points whose r-coordinate is 3, that is, all points
that are 3 units away from the origin. So the graph is a circle of radius 3 centered at the
origin, as shown in Figure 2.
Squaring both sides of the equation, we get
O
5π
4
r 2 32
7π
4
x y 9
2
2
Square both sides
Substitute r 2 x 2 y 2
So the equivalent equation in rectangular coordinates is x 2 y 2 9.
FIGURE 2
■
NOW TRY EXERCISE 17
2π
3
π
3
π
¨= 3
π
3
FIGURE 3
EXAMPLE 2
Sketching the Graph of a Polar Equation
Sketch a graph of the equation u p/3, and express the equation in rectangular coordinates.
S O L U T I O N The graph consists of all points whose u-coordinate is p/3. This is the
straight line that passes through the origin and makes an angle of p/3 with the polar axis
(see Figure 3). Note that the points 1r, p/32 on the line with r 0 lie in Quadrant I,
whereas those with r 0 lie in Quadrant III. If the point 1x, y2 lies on this line, then
O
4π
3
In general, the graph of the equation r a is a circle of radius 0 a 0 centered at the origin. Squaring both sides of this equation, we see that the equivalent equation in rectangular coordinates is x 2 y 2 a 2.
y
p
tan u tan 13
x
3
5π
3
Thus, the rectangular equation of this line is y 13x.
■
NOW TRY EXERCISE 19
To sketch a polar curve whose graph isn’t as obvious as the ones in the preceding examples, we plot points calculated for sufficiently many values of u and then join them in
a continuous curve. (This is what we did when we first learned to graph functions in rectangular coordinates.)
EXAMPLE 3
Sketching the Graph of a Polar Equation
Sketch a graph of the polar equation r 2 sin u.
S O L U T I O N We first use the equation to determine the polar coordinates of several
points on the curve. The results are shown in the following table.
u
0
p /6
p /4
p /3
p /2
2p /3
3p /4
5p /6
p
r 2 sin u
0
1
12
13
2
13
12
1
0
We plot these points in Figure 4 and then join them to sketch the curve. The graph appears to be a circle. We have used values of u only between 0 and p, since the same points
(this time expressed with negative r-coordinates) would be obtained if we allowed u to
range from p to 2p.
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SECTION 8.2
| Graphs of Polar Equations 549
The polar equation r 2 sin u in
rectangular coordinates is
x 2 1 y 12 2 1
(see Section 8.1, Example 6(b)). From
the rectangular form of the equation we
see that the graph is a circle of radius 1
centered at 10, 12 .
F I G U R E 4 r 2 sin u
■
NOW TRY EXERCISE 21
In general, the graphs of equations of the form
r 2a sin u
r 2a cos u
and
are circles with radius 0 a 0 centered at the points with polar coordinates 1a, p/22 and 1a, 02 ,
respectively.
Sketching the Graph of a Cardioid
EXAMPLE 4
r
Sketch a graph of r 2 2 cos u.
0
π
2
π
3π
2
S O L U T I O N Instead of plotting points as in Example 3, we first sketch the graph of
r 2 2 cos u in rectangular coordinates in Figure 5. We can think of this graph as a
table of values that enables us to read at a glance the values of r that correspond to increasing values of u. For instance, we see that as u increases from 0 to p/2, r (the distance
from O) decreases from 4 to 2, so we sketch the corresponding part of the polar graph in
Figure 6(a). As u increases from p/2 to p, Figure 5 shows that r decreases from 2 to 0,
so we sketch the next part of the graph as in Figure 6(b). As u increases from p to 3p/2,
r increases from 0 to 2, as shown in part (c). Finally, as u increases from 3p/2 to 2p, r
increases from 2 to 4, as shown in part (d). If we let u increase beyond 2p or decrease
beyond 0, we would simply retrace our path. Combining the portions of the graph from
parts (a) through (d) of Figure 6, we sketch the complete graph in part (e).
¨
2π
F I G U R E 5 r 2 2 cos u
¨=
π
2
¨=
O
¨=0
π
2
¨=π
O
¨=
(a)
O
¨=π
(b)
3π
2
O
¨=
¨=2π
O
3π
2
(c)
(d)
(e)
F I G U R E 6 Steps in sketching r 2 2 cos u
The polar equation r 2 2 cos u in
rectangular coordinates is
1x 2 y 2 2x2 2 41x 2 y 2 2
(see Section 8.1, Example 6(c)). The
simpler form of the polar equation
shows that it is more natural to describe
cardioids using polar coordinates.
■
NOW TRY EXERCISE 25
The curve in Figure 6 is called a cardioid because it is heart-shaped. In general, the
graph of any equation of the form
r a11 cos u 2
or
r a11 sin u 2
is a cardioid.
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550
CHAPTER 8
| Polar Coordinates and Parametric Equations
Sketching the Graph of a Four-Leaved Rose
EXAMPLE 5
Sketch the curve r cos 2u.
S O L U T I O N As in Example 4, we first sketch the graph of r cos 2u in rectangular
coordinates, as shown in Figure 7. As u increases from 0 to p/4, Figure 7 shows that r decreases from 1 to 0, so we draw the corresponding portion of the polar curve in Figure 8
(indicated by ). As u increases from p/4 to p/2, the value of r goes from 0 to 1. This
means that the distance from the origin increases from 0 to 1, but instead of being in Quadrant I, this portion of the polar curve (indicated by ) lies on the opposite side of the origin in Quadrant III. The remainder of the curve is drawn in a similar fashion, with the arrows and numbers indicating the order in which the portions are traced out. The resulting
curve has four petals and is called a four-leaved rose.
π
¨= 2
r
0
π
¨=3π
4
1
π
4
π
2
3π
4
5π
4
π
3π
2
7π
4
2π
¨= 4
¨=π
¨
¨=0
_1
F I G U R E 7 Graph of r cos 2u sketched in rectangular coordinates
F I G U R E 8 Four-leaved rose r cos 2u
sketched in polar coordinates
■
NOW TRY EXERCISE 29
In general, the graph of an equation of the form
r a cos nu
or
r a sin nu
is an n-leaved rose if n is odd or a 2n-leaved rose if n is even (as in Example 5).
▼ Symmetry
In graphing a polar equation, it’s often helpful to take advantage of symmetry. We list
three tests for symmetry; Figure 9 shows why these tests work.
TESTS FOR SYMMETRY
1. If a polar equation is unchanged when we replace u by u, then the graph is
symmetric about the polar axis (Figure 9(a)).
2. If the equation is unchanged when we replace r by r, then the graph is symmetric about the pole (Figure 9(b)).
3. If the equation is unchanged when we replace u by p u, the graph is symmetric about the vertical line u p/2 (the y-axis) (Figure 9(c)).
¨=π2
(r, π _ ¨)
(r, ¨)
(r, ¨)
O
(r, ¨)
π-¨
¨
¨
_¨
O
O
(_r, ¨)
(r, _¨ )
FIGURE 9
(a) Symmetry about the polar axis
(b) Symmetry about the pole
(c) Symmetry about the line ¨= π2
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SECTION 8.2
The graphs in Figures 2, 6(e), and 8 are symmetric about the polar axis. The graph in Figure 8 is also symmetric about the pole. Figures 4 and 8 show graphs that are symmetric about
u p/2. Note that the four-leaved rose in Figure 8 meets all three tests for symmetry.
In rectangular coordinates, the zeros of the function y f1x 2 correspond to the
x-intercepts of the graph. In polar coordinates, the zeros of the function r f 1u2 are the
angles u at which the curve crosses the pole. The zeros help us sketch the graph, as is
illustrated in the next example.
r
3
EXAMPLE 6
0
_1
| Graphs of Polar Equations 551
π
3
2π
3
π
2π
¨
Using Symmetry to Sketch a Limaçon
Sketch a graph of the equation r 1 2 cos u.
SOLUTION
We use the following as aids in sketching the graph:
Symmetry: Since the equation is unchanged when u is replaced by u, the graph is
symmetric about the polar axis.
FIGURE 10
Zeros:
To find the zeros, we solve
0 1 2 cos u
1
cos u 2
2p 4p
u
,
3 3
¨=2π
3
Table of values: As in Example 4, we sketch the graph of r 1 2 cos u in rectangular coordinates to serve as a table of values (Figure 10).
Now we sketch the polar graph of r 1 2 cos u from u 0 to u p, and then use
symmetry to complete the graph in Figure 11.
¨=4π
3
F I G U R E 1 1 r 1 2 cos u
■
NOW TRY EXERCISE 35
The curve in Figure 11 is called a limaçon, after the Middle French word for snail. In
general, the graph of an equation of the form
r a b cos u
or
r a b sin u
is a limaçon. The shape of the limaçon depends on the relative size of a and b (see the
table on the next page).
▼ Graphing Polar Equations with Graphing Devices
F I G U R E 1 2 r sin u sin3 15 u/2 2
Although it’s useful to be able to sketch simple polar graphs by hand, we need a graphing calculator or computer when the graph is as complicated as the one in Figure 12. Fortunately, most graphing calculators are capable of graphing polar equations directly.
EXAMPLE 7
Drawing the Graph of a Polar Equation
Graph the equation r cos12u/32 .
S O L U T I O N We need to determine the domain for u. So we ask ourselves: How many
times must u go through a complete rotation (2p radians) before the graph starts to repeat
itself? The graph repeats itself when the same value of r is obtained at u and u 2np.
Thus we need to find an integer n, so that
1
_1
cos
1
_1
F I G U R E 1 3 r cos12u/3 2
21u 2np2
3
cos
2u
3
For this equality to hold, 4np/3 must be a multiple of 2p, and this first happens when
n 3. Therefore, we obtain the entire graph if we choose values of u between u 0 and
u 0 2132p 6p. The graph is shown in Figure 13.
NOW TRY EXERCISE 43
■
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552
CHAPTER 8
| Polar Coordinates and Parametric Equations
EXAMPLE 8
A Family of Polar Equations
Graph the family of polar equations r 1 c sin u for c 3, 2.5, 2, 1.5, 1. How does
the shape of the graph change as c changes?
S O L U T I O N Figure 14 shows computer-drawn graphs for the given values of c. When
c 1, the graph has an inner loop; the loop decreases in size as c decreases. When
c 1, the loop disappears, and the graph becomes a cardioid (see Example 4).
c=3.0
c=2.5
c=2.0
c=1.5
c=1.0
F I G U R E 1 4 A family of limaçons r 1 c sin u in the viewing rectangle 32.5, 2.5 4 by 30.5, 4.5 4
■
NOW TRY EXERCISE 47
The box below gives a summary of some of the basic polar graphs used in calculus.
SOME COMMON POLAR CURVES
Circles and Spiral
r=a
circle
r=a ß ¨
circle
r=a ç ¨
circle
r=a¨
spiral
a<b
limaçon with
inner loop
a=b
cardioid
a>b
dimpled limaçon
a≥2b
convex limaçon
r=a ç 2¨
4-leaved rose
r=a ç 3¨
3-leaved rose
r=a ç 4¨
8-leaved rose
r=a ç 5¨
5-leaved rose
r™=a™ ß 2¨
lemniscate
r™=a™ ç 2¨
lemniscate
Limaçons
r a b sin u
r a b cos u
1a 0, b 02
Orientation depends on
the trigonometric function
(sine or cosine) and the sign of b.
Roses
r a sin nu
r a cos nu
n-leaved if n is odd
2n-leaved if n is even
Lemniscates
Figure-eight-shaped
curves
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SECTION 8.2
| Graphs of Polar Equations 553
8.2 EXERCISES
CONCEPTS
V
VI
1. To plot points in polar coordinates, we use a grid consisting of
centered at the pole and
emanating from
the pole.
2. (a) To graph a polar equation r f 1u2 , we plot all the points
the equation.
1r, u2 that
(b) The simplest polar equations are obtained by setting
r or u equal to a constant. The graph of the polar equation
r 3 is a
with radius
centered at
. The graph of the polar equation u p/4
the
is a
passing through the
with slope
. Graph these polar equations below.
1
1
3
9–16 ■ Test the polar equation for symmetry with respect to the
polar axis, the pole, and the line u p/2.
9. r 2 sin u
10. r 4 8 cos u
11. r 3 sec u
13. r 12. r 5 cos u csc u
4
3 2 sin u
14. r 15. r 2 4 cos 2u
5
1 3 cos u
16. r 2 9 sin u
17–22 ■ Sketch a graph of the polar equation, and express the
equation in rectangular coordinates.
O
O
2
2
17. r 2
18. r 1
19. u p/2
20. u 5p/6
21. r 6 sin u
22. r cos u
23–42
SKILLS
3–8 ■ Match the polar equation with the graphs labeled I–VI. Use
the table on page 552 to help you.
■
Sketch a graph of the polar equation.
23. r 2 cos u
24. r 2 sin u 2 cos u
25. r 2 2 cos u
26. r 1 sin u
27. r 311 sin u 2
28. r cos u 1
29. r sin 2u
30. r 2 cos 3u
31. r cos 5u
32. r sin 4u
33. r 23 2 sin u
34. r 2 sin u
3. r 3 cos u
4. r 3
5. r 2 2 sin u
6. r 1 2 cos u
35. r 23 cos u
36. r 1 2 cos u
8. r sin 4u
37. r cos 2u
38. r 2 4 sin 2u
7. r sin 3u
2
39. r u, u 0 (spiral)
I
40. r u 1, u 0 (reciprocal spiral)
II
41. r 2 sec u (conchoid)
42. r sin u tan u (cissoid)
43–46 ■ Use a graphing device to graph the polar equation.
Choose the domain of u to make sure you produce the entire
graph.
1
1
III
43. r cos1u/2 2
45. r 1 2 sin1u/22
IV
44. r sin18u/52
(nephroid)
46. r 21 0.8 sin2 u (hippopede)
3
1
3
47. Graph the family of polar equations r 1 sin nu for
n 1, 2, 3, 4, and 5. How is the number of loops related to n?
48. Graph the family of polar equations r 1 c sin 2u for
c 0.3, 0.6, 1, 1.5, and 2. How does the graph change as
c increases?
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554
CHAPTER 8
| Polar Coordinates and Parametric Equations
49–52 ■ Match the polar equation with the graphs labeled I–IV.
Give reasons for your answers.
49. r sin1u/2 2
50. r 1/ 1u
(b) For what angle u is the satellite closest to the earth? Find
the height of the satellite above the earth’s surface for this
value of u.
52. r 1 3 cos13u 2
51. r u sin u
I
II
r
1
1
III
¨
IV
10
60. An Unstable Orbit The orbit described in Exercise 59 is
stable because the satellite traverses the same path over and
over as u increases. Suppose that a meteor strikes the
satellite and changes its orbit to
1
r
53–56 ■ Sketch a graph of the rectangular equation. [Hint: First
convert the equation to polar coordinates.]
53. 1x 2 y 2 2 3 4x 2y 2
54. 1x 2 y 2 2 3 1x 2 y 2 2 2
55. 1x 2 y 2 2 2 x 2 y 2
56. x 2 y 2 1x 2 y 2 x2 2
57. Show that the graph of r a cos u b sin u is a circle, and
find its center and radius.
58. (a) Graph the polar equation r tan u sec u in the viewing
rectangle 33, 34 by 31, 94.
(b) Note that your graph in part (a) looks like a parabola (see
Section 2.5). Confirm this by converting the equation to
rectangular coordinates.
A P P L I C AT I O N S
59. Orbit of a Satellite Scientists and engineers often use
polar equations to model the motion of satellites in earth orbit.
Let’s consider a satellite whose orbit is modeled by the equation r 22500/14 cos u 2 , where r is the distance in miles
between the satellite and the center of the earth and u is the
angle shown in the following figure.
(a) On the same viewing screen, graph the circle r 3960
(to represent the earth, which we will assume to be a
sphere of radius 3960 mi) and the polar equation of the
satellite’s orbit. Describe the motion of the satellite as u
increases from 0 to 2p.
22500 a 1 u
b
40
4 cos u
(a) On the same viewing screen, graph the circle r 3960
and the new orbit equation, with u increasing from 0 to
3p. Describe the new motion of the satellite.
(b) Use the TRACE feature on your graphing calculator to
find the value of u at the moment the satellite crashes into
the earth.
DISCOVERY
■
DISCUSSION
■
61. A Transformation of Polar Graphs
graphs of
and
r 1 sin a u p
b
6
r 1 sin a u p
b
3
WRITING
How are the
related to the graph of r 1 sin u? In general, how is the
graph of r f 1u a2 related to the graph of r f 1u 2 ?
62. Choosing a Convenient Coordinate System Compare the polar equation of the circle r 2 with its equation in
rectangular coordinates. In which coordinate system is the
equation simpler? Do the same for the equation of the fourleaved rose r sin 2u. Which coordinate system would you
choose to study these curves?
63. Choosing a Convenient Coordinate System Compare the rectangular equation of the line y 2 with its polar
equation. In which coordinate system is the equation simpler?
Which coordinate system would you choose to study lines?
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SECTION 8.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 555
8.3 P OLAR F ORM OF C OMPLEX N UMBERS ; D E M OIVRE ’ S T HEOREM
Graphing Complex Numbers Polar Form of Complex Numbers De Moivre’s Theorem nth Roots of Complex Numbers
In this section we represent complex numbers in polar (or trigonometric) form. This enables us to find the nth roots of complex numbers. To describe the polar form of complex
numbers, we must first learn to work with complex numbers graphically.
Imaginary
axis
a+bi
bi
0
Real
axis
a
FIGURE 1
Im
z⁄=2+3i
_i
z⁄+z¤=5+i
2
To graph real numbers or sets of real numbers, we have been using the number line, which
has just one dimension. Complex numbers, however, have two components: a real part
and an imaginary part. This suggests that we need two axes to graph complex numbers:
one for the real part and one for the imaginary part. We call these the real axis and the
imaginary axis, respectively. The plane determined by these two axes is called the complex plane. To graph the complex number a bi, we plot the ordered pair of numbers
1a, b2 in this plane, as indicated in Figure 1.
EXAMPLE 1
3i
2i
i
▼ Graphing Complex Numbers
4
Re
Graphing Complex Numbers
Graph the complex numbers z1 2 3i, z2 3 2i, and z1 z2.
SOLUTION
Figure 2.
We have z1 z2 12 3i 2 13 2i2 5 i. The graph is shown in
■
NOW TRY EXERCISE 19
_2i
z¤=3-2i
EXAMPLE 2
FIGURE 2
Graphing Sets of Complex Numbers
Graph each set of complex numbers.
(a) S 5a bi 0 a 06
(b) T 5a bi 0 a 1, b 06
SOLUTION
(a) S is the set of complex numbers whose real part is nonnegative. The graph is shown
in Figure 3(a).
(b) T is the set of complex numbers for which the real part is less than 1 and the imaginary part is nonnegative. The graph is shown in Figure 3(b).
Im
0
FIGURE 3
(a)
NOW TRY EXERCISE 21
Im
Re
0
1
Re
(b)
■
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556
CHAPTER 8
| Polar Coordinates and Parametric Equations
Recall that the absolute value of a real number can be thought of as its distance from
the origin on the real number line (see Section 1.1). We define absolute value for complex
numbers in a similar fashion. Using the Pythagorean Theorem, we can see from Figure 4
that the distance between a bi and the origin in the complex plane is 2a 2 b 2. This
leads to the following definition.
Im
a+bi
bi
a™+b™
œ∑∑∑∑∑∑
b
0
Re
a
MODULUS OF A COMPLEX NUMBER
The modulus (or absolute value) of the complex number z a bi is
0 z 0 2a 2 b 2
FIGURE 4
Calculating the Modulus
EXAMPLE 3
Find the moduli of the complex numbers 3 4i and 8 5i.
The plural of modulus is moduli.
SOLUTION
0 3 4i 0 232 42 125 5
0 8 5i 0 282 152 2 189
■
NOW TRY EXERCISE 9
Absolute Value of Complex Numbers
EXAMPLE 4
Graph each set of complex numbers.
(a) C 5z @ 0 z 0 16
(b) D 5z @ 0 z 0 16
SOLUTION
(a) C is the set of complex numbers whose distance from the origin is 1. Thus, C is a
circle of radius 1 with center at the origin, as shown in Figure 5.
(b) D is the set of complex numbers whose distance from the origin is less than or
equal to 1. Thus, D is the disk that consists of all complex numbers on and inside
the circle C of part (a), as shown in Figure 6.
Im
_1
i
C
| z |=1
0
1
Im
Re
_1
_i
FIGURE 5
i
D
| z |≤1
0
1
Re
_i
FIGURE 6
Im
a+bi
NOW TRY EXERCISES 23 AND 25
bi
■
r
▼ Polar Form of Complex Numbers
¨
0
FIGURE 7
a
Re
Let z a bi be a complex number, and in the complex plane let’s draw the line segment joining the origin to the point a bi (see Figure 7). The length of this line segment is r 0 z 0 2a 2 b 2. If u is an angle in standard position whose terminal side
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SECTION 8.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 557
coincides with this line segment, then by the definitions of sine and cosine (see Section 6.2)
a r cos u
b r sin u
and
so z r cos u ir sin u r(cos u i sin u). We have shown the following.
POLAR FORM OF COMPLEX NUMBERS
A complex number z a bi has the polar form (or trigonometric form)
z r 1cos u i sin u 2
where r 0 z 0 2a b and tan u b/a. The number r is the modulus of
z, and u is an argument of z.
2
2
The argument of z is not unique, but any two arguments of z differ by a multiple of 2p.
When determining the argument, we must consider the quadrant in which z lies, as we see
in the next example.
EXAMPLE 5
Writing Complex Numbers in Polar Form
Write each complex number in polar form.
(a) 1 i
(b) 1 13i
(c) 413 4i
These complex numbers are graphed in Figure 8, which helps us find their
SOLUTION
arguments.
Im
Im
_1+Ϸ3 i
(d) 3 4i
Im
4i
Im
Ϸ3 i
3+4i
1+i
i
¨
¨
¨
¨
0
1
Re
_1
0
Re
0
_4 Ϸ
3
_4 Ϸ
3-4i
(a)
(b)
Re
0
3
Re
_4i
(c)
(d)
FIGURE 8
tan u 11 1
u
tan u p
4
13
13
1
u 2p
3
tan u 4
1
4 13
13
u 7p
6
1 i 12 a cos
p
p
i sin b
4
4
(b) An argument is u 2p/3 and r 11 3 2. Thus
1 13 i 2 a cos
tan1 43
2p
2p
i sin
b
3
3
(c) An argument is u 7p/6 (or we could use u 5p/6), and r 148 16 8.
Thus
4 13 4i 8 a cos
tan u 43
u
(a) An argument is u p/4 and r 11 1 12. Thus
7p
7p
i sin
b
6
6
(d) An argument is u tan1 43 and r 232 42 5. So
3 4i 53cosAtan1 43 B i sinAtan1 43 B 4
NOW TRY EXERCISES 29, 31, AND 33
■
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558
CHAPTER 8
| Polar Coordinates and Parametric Equations
The Addition Formulas for Sine and Cosine that we discussed in Section 7.2 greatly
simplify the multiplication and division of complex numbers in polar form. The following theorem shows how.
MULTIPLICATION AND DIVISION OF COMPLEX NUMBERS
If the two complex numbers z1 and z2 have the polar forms
z1 r1 1cos u1 i sin u1 2
z2 r2 1cos u2 i sin u2 2
and
then
z1z2 r1r2 3cos1u1 u2 2 i sin1u1 u2 2 4
z1
r1
3cos1u1 u2 2 i sin1u1 u2 2 4
z2
r2
Multiplication
1z2 02
Division
This theorem says:
To multiply two complex numbers, multiply the moduli and add the arguments.
To divide two complex numbers, divide the moduli and subtract the arguments.
P R O O F To prove the Multiplication Formula, we simply multiply the two complex
numbers:
z1z2 r1r2 1cos u1 i sin u1 2 1cos u2 i sin u2 2
r1r2 3cos u1 cos u2 sin u1 sin u2 i1sin u1 cos u2 cos u1 sin u2 2 4
r1r2 3cos1u1 u2 2 i sin1u1 u2 2 4
In the last step we used the Addition Formulas for Sine and Cosine.
The proof of the Division Formula is left as an exercise.
EXAMPLE 6
■
Multiplying and Dividing Complex Numbers
Let
z1 2 a cos
p
p
i sin b
4
4
and
z2 5 a cos
p
p
i sin b
3
3
Find (a) z1z2 and (b) z1/z2.
SOLUTION
(a) By the Multiplication Formula
z1z2 122 152 c cos a
10 a cos
p
p
p
p
b i sin a b d
4
3
4
3
7p
7p
i sin
b
12
12
To approximate the answer, we use a calculator in radian mode and get
z1z2 1010.2588 0.9659i2
2.588 9.659i
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SECTION 8.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 559
(b) By the Division Formula
z1
2
p
p
p
p
c cos a b i sin a b d
z2
5
4
3
4
3
2
p
p
c cos a b i sin a b d
5
12
12
2
p
p
a cos
i sin b
5
12
12
Using a calculator in radian mode, we get the approximate answer:
z1 2
10.9659 0.2588i 2 0.3864 0.1035i
z2 5
■
NOW TRY EXERCISE 55
▼ De Moivre’s Theorem
Repeated use of the Multiplication Formula gives the following useful formula for raising
a complex number to a power n for any positive integer n.
DE MOIVRE’S THEOREM
If z r 1cos u i sin u2 , then for any integer n
z n r n 1cos nu i sin nu 2
This theorem says: To take the nth power of a complex number, we take the nth power of
the modulus and multiply the argument by n.
PROOF
By the Multiplication Formula
z 2 zz r 2 3cos1u u2 i sin1u u2 4
r 2 1cos 2u i sin 2u 2
Now we multiply z 2 by z to get
z 3 z 2z r 3 3cos12u u2 i sin12u u 2 4
r 3 1cos 3u i sin 3u 2
Repeating this argument, we see that for any positive integer n
z n r n 1cos nu i sin nu 2
A similar argument using the Division Formula shows that this also holds for negative
integers.
■
EXAMPLE 7
Find
A 12
Finding a Power Using De Moivre’s Theorem
1 10
2 iB .
SOLUTION
Since 12 12 i 12 11 i2 , it follows from Example 5(a) that
12
p
1
p
1
a cos i sin b
i
2
2
2
4
4
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560
CHAPTER 8
| Polar Coordinates and Parametric Equations
So by De Moivre’s Theorem
a
10p
10p
1
1 10
12 10
b a cos
i sin
b
ib a
2
4
4
2
2
25
5p
5p
1
a cos
i sin
b i
2
2
32
210
■
NOW TRY EXERCISE 69
▼ nth Roots of Complex Numbers
An nth root of a complex number z is any complex number „ such that „ n z. De Moivre’s
Theorem gives us a method for calculating the nth roots of any complex number.
n th ROOTS OF COMPLEX NUMBERS
If z r 1cos u i sin u2 and n is a positive integer, then z has the n distinct nth
roots
„k r 1/n c cos a
u 2kp
u 2kp
b i sin a
bd
n
n
for k 0, 1, 2, . . . , n 1.
PROOF
To find the nth roots of z, we need to find a complex number „ such that
„n z
Let’s write z in polar form:
z r 1cos u i sin u 2
One nth root of z is
„ r 1/n a cos
u
u
i sin b
n
n
since by De Moivre’s Theorem, „ n z. But the argument u of z can be replaced by
u 2kp for any integer k. Since this expression gives a different value of „ for k 0, 1,
2, . . . , n 1, we have proved the formula in the theorem.
■
The following observations help us use the preceding formula.
FINDING THE nth ROOTS OF z r11 cos u i sin u22
1. The modulus of each nth root is r1/n.
2. The argument of the first root is u/n.
3. We repeatedly add 2p/n to get the argument of each successive root.
These observations show that, when graphed, the nth roots of z are spaced equally on
the circle of radius r1/n.
EXAMPLE 8
Finding Roots of a Complex Number
Find the six sixth roots of z 64, and graph these roots in the complex plane.
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SECTION 8.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 561
S O L U T I O N In polar form, z 641cos p i sin p 2 . Applying the formula for nth
roots with n 6, we get
„k 641/6 c cos a
p 2kp
p 2kp
b i sin a
bd
6
6
for k 0, 1, 2, 3, 4, 5. Using 641/6 2, we find that the six sixth roots of 64 are
We add 2p/6 p/3 to each argument
to get the argument of the next root.
Im
2i „⁄
„¤
„‚
0
_2
„‹
2 Re
„fi
_2i „›
F I G U R E 9 The six sixth roots of
z 64
„0 2 a cos
p
p
i sin b 13 i
6
6
„1 2 a cos
p
p
i sin b 2i
2
2
„2 2 a cos
5p
5p
i sin
b 13 i
6
6
„3 2 a cos
7p
7p
i sin
b 13 i
6
6
„4 2 a cos
3p
3p
i sin
b 2i
2
2
„5 2 a cos
11p
11p
i sin
b 13 i
6
6
All these points lie on a circle of radius 2, as shown in Figure 9.
■
NOW TRY EXERCISE 85
When finding roots of complex numbers, we sometimes write the argument u of the
complex number in degrees. In this case the nth roots are obtained from the formula
„k r 1/n c cos a
u 360°k
u 360°k
b i sin a
bd
n
n
for k 0, 1, 2, . . . , n 1.
EXAMPLE 9
Finding Cube Roots of a Complex Number
Find the three cube roots of z 2 2i, and graph these roots in the complex plane.
S O L U T I O N First we write z in polar form using degrees. We have
r 222 22 212 and u 45. Thus
We add 360/3 120 to each
argument to get the argument of the
next root.
Im
z 212 1cos 45° i sin 45°2
Applying the formula for nth roots (in degrees) with n 3, we find that the cube roots of
z are of the form
„k A212B 1/3 c cos a
Ϸ2 i
„⁄
45° 360°k
45° 360°k
b i sin a
bd
3
3
where k 0, 1, 2. Thus the three cube roots are
„‚
0
_Ϸ2
„¤
Ϸ2 Re
„1 12 1cos 135° i sin 135°2 1 i
12122 1/3 123/2 2 1/3 21/2 12
„2 12 1cos 255° i sin 255°2 0.366 1.366i
_Ϸ2 i
F I G U R E 1 0 The three cube roots of
z 2 2i
„0 12 1cos 15° i sin 15°2 1.366 0.366i
The three cube roots of z are graphed in Figure 10. These roots are spaced equally on a
circle of radius 12.
NOW TRY EXERCISE 81
■
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562
| Polar Coordinates and Parametric Equations
CHAPTER 8
Solving an Equation Using the nth Roots Formula
EXAMPLE 10
Solve the equation z 6 64 0.
S O L U T I O N This equation can be written as z 6 64. Thus the solutions are the sixth
roots of 64, which we found in Example 8.
■
NOW TRY EXERCISE 91
8.3 EXERCISES
SKILLS
CONCEPTS
1. A complex number z a bi has two parts: a is the
part, and b is the
we graph the ordered pair 1
2. Let z a bi.
,
5–14
part. To graph a bi,
2 in the complex plane.
(a) The modulus of z is r .
(b) We can express z in polar form as z ,
where r is the modulus of z and u is the argument of z.
p
p
i sin b in rectangular form is
6
6
7. 2
8. 6
9. 5 2i
10. 7 3i
11. 13 i
12. 1 .
Im
16. z 1 i13
17. z 8 2i
19–20
plane.
z
i
3 4i
5
17–18 ■ Sketch the complex number z and its complex conjugate
z on the same complex plane.
(b) The complex number graphed below can be expressed in
or in polar form as
13
i
3
12 i 12
14.
2
15. z 1 i
.
rectangular form as
6. 3i
15–16 ■ Sketch the complex number z, and also sketch 2z, z,
and 12 z on the same complex plane.
. The complex number
z
5. 4i
13.
3. (a) The complex number z 1 i in polar form is
z 2 a cos
Graph the complex number and find its modulus.
, and an argument of
z is an angle u satisfying tan u z
■
■
18. z 5 6i
Sketch z1, z2, z1 z2, and z1z2 on the same complex
19. z1 2 i, z2 2 i
20. z1 1 i, z2 2 3i
21–28
0
1
. The number 16 has
These roots are
,
fourth roots.
,
, and
. In the complex plane these roots all lie on a circle
of radius
. Graph the roots on the following graph.
22. 5z a bi 0 a 1, b 16
23. 5z @ 0 z 0 36
25. 5z @ 0 z 0 26
27. 5z a bi 0 a b 26
24. 5z @ 0 z 0 16
26. 5z @ 2 0 z 0 56
28. 5z a bi 0 a b6
29–52 ■ Write the complex number in polar form with
argument u between 0 and 2p.
Im
i
0
Sketch the set in the complex plane.
21. 5z a bi 0 a 0, b 06
Re
4. How many different nth roots does a nonzero complex number
have?
■
4 Re
29. 1 i
30. 1 13 i
31. 12 12 i
32. 1 i
33. 213 2i
34. 1 i
35. 3i
36. 3 3 13 i
37. 5 5i
38. 4
39. 413 4i
40. 8i
41. 20
42. 13 i
43. 3 4i
44. i12 2i2
45. 3i11 i 2
46. 211 i 2
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SECTION 8.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 563
47. 41 13 i2
48. 3 3i
49. 2 i
83. The fourth roots of 81i
50. 3 13 i
51. 12 12 i
52. pi
84. The fifth roots of 32
53–60 ■ Find the product z1z2 and the quotient z1/z2. Express your
answer in polar form.
p
p
53. z1 cos p i sin p, z2 cos i sin
3
3
p
p
3p
3p
i sin
54. z1 cos i sin , z2 cos
4
4
4
4
p
p
4p
4p
i sin
b
55. z1 3 a cos i sin b , z2 5 a cos
6
6
3
3
9p
p
9p
p
i sin
b , z2 2 a cos i sin b
56. z1 7 a cos
8
8
8
8
57. z1 41cos 120° i sin 120°2 ,
z2 21cos 30° i sin 30°2
58. z1 121cos 75° i sin 75°2 ,
z2 3 121cos 60° i sin 60°2
60. z1 45 1cos 25° i sin 25° 2 ,
z2 15 1cos 155° i sin 155° 2
61. z1 13 i, z2 1 13 i
63. z1 2 13 2i, z2 1 i
64. z1 12 i, z2 3 313 i
65. z1 5 5i, z2 4
66. z1 413 4i, z2 8i
67. z1 20, z2 13 i
68. z1 3 4i, z2 2 2i
Find the indicated power using De Moivre’s Theorem.
71. 12 13 2i2
73. a
70. 11 13 i2
5
12
12 12
ib
2
2
75. 12 2i 2 8
77. 11 i 2 7
79. 12 13 2i2 5
5
72. 11 i2 8
74. 1 13 i2 10
1
13 15
ib
76. a 2
2
78. 13 13 i2 4
80. 11 i2 8
81–90 ■ Find the indicated roots, and graph the roots in the complex plane.
81. The square roots of 4 13 4i
82. The cube roots of 4 13 4i
89. The fourth roots of 1
90. The fifth roots of 16 1613i
91–96
■
Solve the equation.
91. z 4 1 0
92. z 8 i 0
93. z 3 4 13 4i 0
94. z 6 1 0
95. z 3 1 i
96. z 3 1 0
DISCOVERY
■
DISCUSSION
■
WRITING
98. Sums of Roots of Unity Find the exact values of all
three cube roots of 1 (see Exercise 97) and then add them. Do
the same for the fourth, fifth, sixth, and eighth roots of 1.
What do you think is the sum of the nth roots of 1 for any n?
62. z1 12 12 i, z2 1 i
69. 11 i 2
88. The fifth roots of i
s, s„, s„ 2, s„ 3, . . . , s„ n1
61–68 ■ Write z1 and z2 in polar form, and then find the product
z1z2 and the quotients z1/z2 and 1/z1.
20
87. The cube roots of i
2p
2p
i sin
where n is a positive
n
n
2
integer. Show that 1, „, „ , „ 3, . . . , „ n1 are the n
distinct nth roots of 1.
(b) If z 0 is any complex number and s n z, show that
the n distinct nth roots of z are
z2 251cos 150° i sin 150°2
■
86. The cube roots of 1 i
97. (a) Let „ cos
59. z1 41cos 200° i sin 200°2 ,
69–80
85. The eighth roots of 1
99. Products of Roots of Unity Find the product of the
three cube roots of 1 (see Exercise 97). Do the same for the
fourth, fifth, sixth, and eighth roots of 1. What do you think
is the product of the nth roots of 1 for any n?
100. Complex Coefficients and the Quadratic Formula
The quadratic formula works whether the coefficients of the
equation are real or complex. Solve these equations using the
quadratic formula and, if necessary, De Moivre’s Theorem.
(a) z 2 11 i 2z i 0
(b) z 2 iz 1 0
(c) z 2 12 i 2z 14 i 0
❍ DISCOVERY
PROJECT
Fractals
In this project we use graphs of complex numbers to create
fractal images. You can find the project at the book companion
website: www.stewartmath.com
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564
CHAPTER 8
| Polar Coordinates and Parametric Equations
8.4 P LANE C URVES AND PARAMETRIC E QUATIONS
Plane Curves and Parametric Equations Eliminating the Parameter Finding Parametric Equations for a Curve Using Graphing Devices to Graph
Parametric Curves
So far, we have described a curve by giving an equation (in rectangular or polar coordinates) that the coordinates of all the points on the curve must satisfy. But not all curves in
the plane can be described in this way. In this section we study parametric equations,
which are a general method for describing any curve.
▼ Plane Curves and Parametric Equations
We can think of a curve as the path of a point moving in the plane; the x- and
y-coordinates of the point are then functions of time. This idea leads to the following
definition.
PLANE CURVES AND PARAMETRIC EQUATIONS
If f and g are functions defined on an interval I, then the set of points 1f 1t2, g1t22
is a plane curve. The equations
y g1t 2
x f1t2
where t I, are parametric equations for the curve, with parameter t.
EXAMPLE 1
Sketching a Plane Curve
Sketch the curve defined by the parametric equations
x t 2 3t
yt1
S O L U T I O N For every value of t, we get a point on the curve. For example, if t 0,
then x 0 and y 1, so the corresponding point is 10, 12 . In Figure 1 we plot the
points 1x, y2 determined by the values of t shown in the following table.
t
x
y
2
1
0
1
2
3
4
5
10
4
0
2
2
0
4
10
3
2
1
0
1
2
3
4
t=5
y
t=4
t=3
t=2
1
5
t=1
10 x
t=0
t=_1
t=_2
FIGURE 1
As t increases, a particle whose position is given by the parametric equations moves
along the curve in the direction of the arrows.
NOW TRY EXERCISE 3
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 8.4
| Plane Curves and Parametric Equations 565
If we replace t by t in Example 1, we obtain the parametric equations
x t 2 3t
y t 1
© Bettmann /CORBIS
The graph of these parametric equations (see Figure 2) is the same as the curve in Figure 1,
but traced out in the opposite direction. On the other hand, if we replace t by 2t in Example
1, we obtain the parametric equations
MARIA GAETANA AGNESI (1718–
1799) is famous for having written Instituzioni Analitiche, one of the first calculus textbooks.
Maria was born into a wealthy family
in Milan, Italy, the oldest of 21 children.
She was a child prodigy, mastering many
languages at an early age, including
Latin, Greek, and Hebrew. At the age of
20 she published a series of essays on
philosophy and natural science.After
Maria’s mother died, she took on the task
of educating her brothers.In 1748 Agnesi published her famous textbook,
which she originally wrote as a text for
tutoring her brothers.The book compiled and explained the mathematical
knowledge of the day.It contains many
carefully chosen examples, one of which
is the curve now known as the“witch of
Agnesi”(see Exercise 64, page 571). One
review calls her book an“exposition by
examples rather than by theory.” The
book gained Agnesi immediate recognition.Pope Benedict XIV appointed her to
a position at the University of Bologna,
writing, “we have had the idea that you
should be awarded the well-known chair
of mathematics, by which it comes of itself that you should not thank us but we
you.” This appointment was an extremely high honor for a woman, since
very few women then were even allowed to attend university.Just two
years later, Agnesi’s father died, and she
left mathematics completely.She became a nun and devoted the rest of her
life and her wealth to caring for sick and
dying women, herself dying in poverty at
a poorhouse of which she had once
been director.
x 4t 2 6t
y 2t 1
The graph of these parametric equations (see Figure 3) is again the same, but is traced out
“twice as fast.” Thus, a parametrization contains more information than just the shape of
the curve; it also indicates how the curve is being traced out.
t=_5
y
t=_4
y
t=2
t=_3
t=1
t=_2
1
1
5
t=_1
10 x
5
t=0
10
x
t=0
t=1
t=_1
t=2
F I G U R E 2 x t 2 3t, y t 1
F I G U R E 3 x 4t 2 6t, y 2t 1
▼ Eliminating the Parameter
Often a curve given by parametric equations can also be represented by a single rectangular equation in x and y. The process of finding this equation is called eliminating the parameter. One way to do this is to solve for t in one equation, then substitute into the other.
EXAMPLE 2
Eliminating the Parameter
Eliminate the parameter in the parametric equations of Example 1.
S O L U T I O N First we solve for t in the simpler equation, then we substitute into the
other equation. From the equation y t 1, we get t y 1. Substituting into the equation for x, we get
x t 2 3t 1y 12 2 31y 12 y 2 y 2
Thus the curve in Example 1 has the rectangular equation x y 2 y 2, so it is a
parabola.
■
NOW TRY EXERCISE 5
Eliminating the parameter often helps us identify the shape of a curve, as we see in the
next two examples.
EXAMPLE 3
Modeling Circular Motion
The following parametric equations model the position of a moving object at time t (in
seconds):
x cos t
y sin t
t
0
Describe and graph the path of the object.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
566
| Polar Coordinates and Parametric Equations
CHAPTER 8
y
S O L U T I O N To identify the curve, we eliminate the parameter. Since cos2 t sin2 t 1
and since x cos t and y sin t for every point 1x, y 2 on the curve, we have
π
t= 2
(ç t, ß t)
t
t=π
x 2 y 2 1cos t 2 2 1sin t2 2 1
t=0
0
This means that all points on the curve satisfy the equation x 2 y 2 1, so the graph is a
circle of radius 1 centered at the origin. As t increases from 0 to 2p, the point given by the
parametric equations starts at 11, 02 and moves counterclockwise once around the circle, as
shown in Figure 4. So the object completes one revolution around the circle in
2p seconds. Notice that the parameter t can be interpreted as the angle shown in the figure.
x
(1, 0)
t=2π
t= 3π
2
■
NOW TRY EXERCISE 25
FIGURE 4
EXAMPLE 4
Sketching a Parametric Curve
Eliminate the parameter, and sketch the graph of the parametric equations
y 2 cos2 t
x sin t
y
(_1, 2)
S O L U T I O N To eliminate the parameter, we first use the trigonometric identity
cos2 t 1 sin2 t to change the second equation:
(1, 2)
y 2 cos2 t 2 11 sin2t2 1 sin2 t
Now we can substitute sin t x from the first equation to get
y 1 x2
x
0
FIGURE 5
so the point 1x, y2 moves along the parabola y 1 x 2. However, since 1 sin t 1,
we have 1 x 1, so the parametric equations represent only the part of the parabola
between x 1 and x 1. Since sin t is periodic, the point 1x, y2 1sin t, 2 cos2 t2
moves back and forth infinitely often along the parabola between the points 11, 22 and
11, 22 , as shown in Figure 5.
■
NOW TRY EXERCISE 17
▼ Finding Parametric Equations for a Curve
It is often possible to find parametric equations for a curve by using some geometric properties that define the curve, as in the next two examples.
EXAMPLE 5
Finding Parametric Equations for a Graph
Find parametric equations for the line of slope 3 that passes through the point 12, 62 .
y
S O L U T I O N Let’s start at the point 12, 62 and move up and to the right along this line.
Because the line has slope 3, for every 1 unit we move to the right, we must move up
3 units. In other words, if we increase the x-coordinate by t units, we must correspondingly increase the y-coordinate by 3t units. This leads to the parametric equations
3t
6
t
x2t
To confirm that these equations give the desired line, we eliminate the parameter. We
solve for t in the first equation and substitute into the second to get
0
2
FIGURE 6
y 6 3t
y 6 31x 22 3x
x
Thus the slope-intercept form of the equation of this line is y 3x, which is a line of slope
3 that does pass through 12, 62 as required. The graph is shown in Figure 6.
NOW TRY EXERCISE 29
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 8.4
| Plane Curves and Parametric Equations 567
Parametric Equations for the Cycloid
EXAMPLE 6
As a circle rolls along a straight line, the curve traced out by a fixed point P on the circumference of the circle is called a cycloid (see Figure 7). If the circle has radius a and
rolls along the x-axis, with one position of the point P being at the origin, find parametric equations for the cycloid.
P
P
P
FIGURE 7
S O L U T I O N Figure 8 shows the circle and the point P after the circle has rolled through
an angle u (in radians). The distance d1O, T2 that the circle has rolled must be the same
as the length of the arc PT, which, by the arc length formula, is au (see Section 6.1). This
means that the center of the circle is C1au, a2 .
Let the coordinates of P be 1x, y2 . Then from Figure 8 (which illustrates the case
0 u p/2), we see that
y
a
P
C (a¨, a)
¨
x d1O, T2 d1P, Q2 au a sin u a1u sin u 2
Q
y d1T, C2 d1Q, C2 a a cos u a11 cos u 2
y
x
x
T
0
so parametric equations for the cycloid are
x a1u sin u 2
a¨
y a11 cos u 2
■
NOW TRY EXERCISE 57
FIGURE 8
The cycloid has a number of interesting physical properties. It is the “curve of quickest
descent” in the following sense. Let’s choose two points P and Q that are not directly above
each other, and join them with a wire. Suppose we allow a bead to slide down the wire under the influence of gravity (ignoring friction). Of all possible shapes into which the wire
can be bent, the bead will slide from P to Q the fastest when the shape is half of an arch of
an inverted cycloid (see Figure 9). The cycloid is also the “curve of equal descent” in the
sense that no matter where we place a bead B on a cycloid-shaped wire, it takes the same
time to slide to the bottom (see Figure 10). These rather surprising properties of the cycloid
were proved (using calculus) in the 17th century by several mathematicians and physicists,
including Johann Bernoulli, Blaise Pascal, and Christiaan Huygens.
P
B
Cycloid
FIGURE 9
B
FIGURE 10
▼ Using Graphing Devices to Graph Parametric Curves
8
_6.5
B
Q
Most graphing calculators and computer graphing programs can be used to graph parametric equations. Such devices are particularly useful in sketching complicated curves
like the one shown in Figure 11.
6.5
EXAMPLE 7
_8
F I G U R E 1 1 x t 2 sin 2t,
y t 2 cos 5t
Graphing Parametric Curves
Use a graphing device to draw the following parametric curves. Discuss their similarities
and differences.
(a) x sin 2t
(b) x sin 3t
y 2 cos t
y 2 cos t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
568
| Polar Coordinates and Parametric Equations
CHAPTER 8
S O L U T I O N In both parts (a) and (b) the graph will lie inside the rectangle given by
1 x 1, 2 y 2, since both the sine and the cosine of any number will be between 1 and 1. Thus, we may use the viewing rectangle 31.5, 1.54 by 32.5, 2.54 .
2.5
_1.5
1.5
_2.5
(a) x=ß 2t, y=2 ç t
(a) Since 2 cos t is periodic with period 2p (see Section 5.3) and since sin 2t has period p, letting t vary over the interval 0 t 2p gives us the complete graph,
which is shown in Figure 12(a).
(b) Again, letting t take on values between 0 and 2p gives the complete graph shown
in Figure 12(b).
Both graphs are closed curves, which means they form loops with the same starting
and ending point; also, both graphs cross over themselves. However, the graph in Figure
12(a) has two loops, like a figure eight, whereas the graph in Figure 12(b) has three loops.
2.5
_1.5
1.5
_2.5
(b) x=ß 3t, y=2 ç t
FIGURE 12
■
NOW TRY EXERCISE 43
The curves graphed in Example 7 are called Lissajous figures. A Lissajous figure is
the graph of a pair of parametric equations of the form
x A sin v1t
y B cos v2t
where A, B, v1, and v 2 are real constants. Since sin v1t and cos v 2t are both between
1 and 1, a Lissajous figure will lie inside the rectangle determined by A x A,
B y B. This fact can be used to choose a viewing rectangle when graphing a Lissajous figure, as in Example 7.
Recall from Section 8.1 that rectangular coordinates 1x, y2 and polar coordinates 1r, u2
are related by the equations x r cos u, y r sin u. Thus we can graph the polar equation r f1u2 by changing it to parametric form as follows:
Since r f 1u 2
x r cos u f1u2 cos u
y r sin u f1u2 sin u
Replacing u by the standard parametric variable t, we have the following result.
POLAR EQUATIONS IN PARAMETRIC FORM
The graph of the polar equation r f 1u2 is the same as the graph of the
parametric equations
x f1t2 cos t
EXAMPLE 8
32
_32
y f1t2 sin t
Parametric Form of a Polar Equation
Consider the polar equation r u, 1 u 10p.
(a) Express the equation in parametric form.
(b) Draw a graph of the parametric equations from part (a).
32
SOLUTION
(a) The given polar equation is equivalent to the parametric equations
x t cos t
_32
F I G U R E 1 3 x t cos t, y t sin t
y t sin t
(b) Since 10p 31.42, we use the viewing rectangle 332, 324 by 332, 324 , and we
let t vary from 1 to 10p. The resulting graph shown in Figure 13 is a spiral.
NOW TRY EXERCISE 51
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 8.4
| Plane Curves and Parametric Equations 569
8.4 EXERCISES
CONCEPTS
1. (a) The parametric equations x f 1t 2 and y g 1t 2 give the
coordinates of a point 1x, y 2 1f 1t 2 , g 1t 2 2 for appropriate
values of t. The variable t is called a
.
(b) Suppose that the parametric equations x t, y t 2,
t 0, model the position of a moving object at time t.
When t 0, the object is at 1 ,
2, and when t 1,
the object is at 1 ,
2.
(c) If we eliminate the parameter in part (b), we get the equation y . We see from this equation that the
path of the moving object is a
13. x 2 sin t, y 2 cos t, 0 t p
14. x 2 cos t, y 3 sin t, 0 t 2p
15. x sin2 t, y sin4 t
16. x sin2 t, y cos t
17. x cos t, y cos 2t
18. x cos 2t, y sin 2t
19. x sec t, y tan t, 0 t p/2
20. x cot t, y csc t, 0 t p
21. x tan t, y cot t, 0 t p/2
22. x sec t, y tan2 t, 0 t p/2
23. x cos2 t, y sin2 t
24. x cos3 t, y sin3 t, 0 t 2p
.
2. (a) True or false? The same curve can be described by parametric equations in many different ways.
(b) The parametric equations x 2t, y 12t2 2 model the position of a moving object at time t. When t 0, the object
is at 1 ,
2, and when t 1, the object is at 1 ,
2.
(c) If we eliminate the parameter, we get the equation
, which is the same equation as in Exercise
y
1(b). So the objects in Exercises 1(b) and 2(b) move along
the same
but traverse the path differently. Indicate the position of each object when t 0 and when
t 1 on the following graph.
25–28 ■ The position of an object in circular motion is modeled
by the given parametric equations. Describe the path of the object
by stating the radius of the circle, the position at time t 0, the
orientation of the motion (clockwise or counterclockwise), and the
time t that it takes to complete one revolution around the circle.
25. x 3 cos t,
y 3 sin t
27. x sin 2t, y cos 2t
26. x 2 sin t, y 2 cos t
28. x 4 cos 3t, y 4 sin 3t
29–34 ■ Find parametric equations for the line with the given
properties.
29. Slope 12 , passing through 14, 12
30. Slope 2, passing through 110, 20 2
31. Passing through 16, 72 and 17, 82
y
32. Passing through 112, 72 and the origin
33. Find parametric equations for the circle x 2 y 2 a 2.
34. Find parametric equations for the ellipse
y2
x2
21
2
a
b
1
0
35. Show by eliminating the parameter u that the following parametric equations represent a hyperbola:
x
1
x a tan u
36. Show that the following parametric equations represent a part
of the hyperbola of Exercise 35:
SKILLS
x a1t
3–24 ■ A pair of parametric equations is given. (a) Sketch the curve
represented by the parametric equations. (b) Find a rectangularcoordinate equation for the curve by eliminating the parameter.
3. x 2t, y t 6
y At 12 B 2
Sketch the curve given by the parametric equations.
37. x t cos t, y t sin t, t 0
3t
,
1 t3
y
3t 2
1 t3
40. x cot t, y 2 sin2 t, 0 t p
7. x 1t, y 1 t
41. If a projectile is fired with an initial speed of √0 ft/s at an angle
a above the horizontal, then its position after t seconds is
given by the parametric equations
8. x t 2, y t 4 1
11. x 4t 2, y 8t 3
■
39. x 5. x t 2, y t 2, 2 t 4
1
9. x , y t 1
t
37–40
y b2t 1
38. x sin t, y sin 2t
4. x 6t 4, y 3t, t 0
6. x 2t 1,
y b sec u
10. x t 1,
y
t
t1
12. x 0 t 0 , y 0 1 0 t 0 0
x 1√0 cos a 2 t
y 1√0 sin a 2t 16t 2
(where x and y are measured in feet). Show that the path of the
projectile is a parabola by eliminating the parameter t.
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570
CHAPTER 8
| Polar Coordinates and Parametric Equations
42. Referring to Exercise 41, suppose a gun fires a bullet into the
air with an initial speed of 2048 ft/s at an angle of 30 to the
horizontal.
(a) After how many seconds will the bullet hit the ground?
(b) How far from the gun will the bullet hit the ground?
(c) What is the maximum height attained by the bullet?
58. (a) In Exercise 57 if the point P lies outside the circle at a distance b from the center (with b a), then the curve traced
out by P is called a prolate cycloid. Show that parametric
equations for the prolate cycloid are the same as the equations for the curtate cycloid.
(b) Sketch the graph for the case in which a 1 and b 2.
43–48 ■ Use a graphing device to draw the curve represented by
the parametric equations.
59. A circle C of radius b rolls on the inside of a larger circle of radius a centered at the origin. Let P be a fixed point on the
smaller circle, with initial position at the point 1a, 02 as shown
in the figure. The curve traced out by P is called a hypocycloid.
43. x sin t, y 2 cos 3t
44. x 2 sin t, y cos 4t
45. x 3 sin 5t, y 5 cos 3t
y
46. x sin 4t, y cos 3t
47. x sin1cos t 2, y cos1t 3/2 2 ,
0 t 2p
C
48. x 2 cos t cos 2t, y 2 sin t sin 2t
¨
49–52 ■ A polar equation is given. (a) Express the polar equation
in parametric form. (b) Use a graphing device to graph the parametric equations you found in part (a).
49. r 2u/12, 0 u 4p
4
51. r 2 cos u
x
52. r 2sin u
53. x t 3 2t, y t 2 t
54. x sin 3t, y sin 4t
55. x t sin 2t, y t sin 3t
56. x sin1t sin t 2, y cos1t cos t 2
y
II
(a) Show that parametric equations for the hypocycloid are
x 1a b 2 cos u b cos a
ab
ub
b
y 1a b 2 sin u b sin a
ab
ub
b
(b) If a 4b, the hypocycloid is called an astroid. Show that
in this case the parametric equations can be
reduced to
y
x a cos3 u
0
(a, 0)
0
50. r sin u 2 cos u
53–56 ■ Match the parametric equations with the graphs
labeled I–IV. Give reasons for your answers.
I
b P
0
x
x
y a sin3 u
Sketch the curve. Eliminate the parameter to obtain an
equation for the astroid in rectangular coordinates.
60. If the circle C of Exercise 59 rolls on the outside of the larger
circle, the curve traced out by P is called an epicycloid. Find
parametric equations for the epicycloid.
III
y
IV
0
0
61. In the figure, the circle of radius a is stationary, and for every
u, the point P is the midpoint of the segment QR. The curve
traced out by P for 0 u p is called the longbow curve.
Find parametric equations for this curve.
y
x
y
2a
x
R
y=2a
57. (a) In Example 6 suppose the point P that traces out the curve
lies not on the edge of the circle, but rather at a fixed point
inside the rim, at a distance b from the center (with b a).
The curve traced out by P is called a curtate cycloid (or
trochoid). Show that parametric equations for the curtate
cycloid are
x au b sin u
y a b cos u
(b) Sketch the graph using a 3 and b 2.
P
a
Q
¨
0
x
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SECTION 8.4
62. Two circles of radius a and b are centered at the origin, as
shown in the figure. As the angle u increases, the point P
traces out a curve that lies between the circles.
(a) Find parametric equations for the curve, using u as the
parameter.
(b) Graph the curve using a graphing device, with a 3 and
b 2.
(c) Eliminate the parameter, and identify the curve.
y
a
b
P
¨
0
x
| Plane Curves and Parametric Equations 571
65. Eliminate the parameter u in the parametric equations for the
cycloid (Example 6) to obtain a rectangular coordinate equation for the section of the curve given by 0 u p.
A P P L I C AT I O N S
66. The Rotary Engine The Mazda RX-8 uses an unconventional engine (invented by Felix Wankel in 1954) in which the
pistons are replaced by a triangular rotor that turns in a special
housing as shown in the figure. The vertices of the rotor maintain contact with the housing at all times, while the center of
the triangle traces out a circle of radius r, turning the drive
shaft. The shape of the housing is given by the parametric
equations below (where R is the distance between the vertices
and center of the rotor):
x r cos 3u R cos u
63. Two circles of radius a and b are centered at the origin, as
shown in the figure.
(a) Find parametric equations for the curve traced out by the
point P, using the angle u as the parameter. (Note that the
line segment AB is always tangent to the larger circle.)
(b) Graph the curve using a graphing device, with a 3 and
b 2.
y
y r sin 3u R sin u
(a) Suppose that the drive shaft has radius r 1. Graph
the curve given by the parametric equations for the
following values of R: 0.5, 1, 3, 5.
(b) Which of the four values of R given in part (a) seems to
best model the engine housing illustrated in the figure?
A
a
P
b
¨
0
B
67. Spiral Path of a Dog A dog is tied to a circular tree
trunk of radius 1 ft by a long leash. He has managed to wrap
the entire leash around the tree while playing in the yard, and
he finds himself at the point 11, 02 in the figure. Seeing a
squirrel, he runs around the tree counterclockwise, keeping the
leash taut while chasing the intruder.
(a) Show that parametric equations for the dog’s path (called
an involute of a circle) are
x
64. A curve, called a witch of Agnesi, consists of all points P determined as shown in the figure.
(a) Show that parametric equations for this curve can be written as
x 2a cot u
y 2a sin2 u
(b) Graph the curve using a graphing device, with a 3.
y=2a
y
x cos u u sin u
y sin u u cos u
[Hint: Note that the leash is always tangent to the tree, so
OT is perpendicular to TD.]
(b) Graph the path of the dog for 0 u 4p.
y
C
T
1
A
P
1
a
D
¨
O
1
x
¨
0
x
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572
CHAPTER 8
DISCOVERY
| Polar Coordinates and Parametric Equations
■
DISCUSSION
■
WRITING
68. More Information in Parametric Equations In this
section we stated that parametric equations contain more
information than just the shape of a curve. Write a short
paragraph explaining this statement. Use the following
example and your answers to parts (a) and (b) below in your
explanation.
The position of a particle is given by the parametric equations
x sin t
y cos t
where t represents time. We know that the shape of the path of
the particle is a circle.
(a) How long does it take the particle to go once around the
circle? Find parametric equations if the particle moves
twice as fast around the circle.
(b) Does the particle travel clockwise or counterclockwise
around the circle? Find parametric equations if the
particle moves in the opposite direction around the circle.
69. Different Ways of Tracing Out a Curve The curves
C, D, E, and F are defined parametrically as follows, where
the parameter t takes on all real values unless otherwise stated:
C: x t, y t 2
D: x 1t,
y t, t 0
E: x sin t, y sin2 t
F: x 3t,
y 32t
(a) Show that the points on all four of these curves satisfy the
same rectangular coordinate equation.
(b) Draw the graph of each curve and explain how the curves
differ from one another.
CHAPTER 8 | REVIEW
■ CONCEPT CHECK
1. Describe how polar coordinates represent the position of a
point in the plane.
2. (a) What equations do you use to change from polar to rectangular coordinates?
(b) What equations do you use to change from rectangular to
polar coordinates?
3. How do you sketch the graph of a polar equation r f1u2 ?
4. What type of curve has a polar equation of the given form?
(a) r a cos u or r a sin u
(b) r a11 cos u 2 or r a11 sin u2
(c) r a b cos u or r a b sin u
(d) r a cos nu or r a sin nu
5. How do you graph a complex number z? What is the polar
form of a complex number z? What is the modulus of z? What
is the argument of z?
6. (a) How do you multiply two complex numbers if they are
given in polar form?
(b) How do you divide two such numbers?
7. (a) State De Moivre’s Theorem.
(b) How do you find the nth roots of a complex
number?
8. A curve is given by the parametric equations
x f 1t2 , y g1t2 .
(a) How do you sketch the curve?
(b) How do you eliminate the parameter?
■ EXERCISES
1–6 ■ A point P1r, u 2 is given in polar coordinates.
(a) Plot the point P. (b) Find rectangular coordinates for P.
1. A12, p6 B
2. A8, 3p
4 B
5. A4 13, 5p
3 B
6. A6 12, p4 B
3. A3, 7p
4 B
4. A 13, 2p
3 B
7–12 ■ A point P1x, y 2 is given in rectangular coordinates.
(a) Plot the point P. (b) Find polar coordinates for P with r 0.
(c) Find polar coordinates for P with r 0.
7. 18, 82
9. 1612, 6122
11. 13, 132
8. 1 12, 162
10. 13 13, 3 2
12. 14, 4 2
13–16 ■ (a) Convert the equation to polar coordinates and
simplify. (b) Graph the equation. [Hint: Use the form of the
equation that you find easier to graph.]
13. x y 4
15. x y 4x 4y
2
2
14. xy 1
16. 1x 2 y 2 2 2 2xy
17–24 ■ (a) Sketch the graph of the polar equation.
(b) Express the equation in rectangular coordinates.
17. r 3 3 cos u
18. r 3 sin u
19. r 2 sin 2u
20. r 4 cos 3u
21. r sec 2u
22. r 2 4 sin 2u
23. r sin u cos u
24. r 2
4
2 cos u
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CHAPTER 8
25–28 ■ Use a graphing device to graph the polar equation.
Choose the domain of u to make sure you produce the entire
graph.
25. r cos1u/3 2
47–48
27. r 1 4 cos1u/32
29–34 ■ A complex number is given. (a) Graph the complex
number in the complex plane. (b) Find the modulus and argument.
(c) Write the number in polar form.
29. 4 4i
30. 10i
31. 5 3i
32. 1 13 i
33. 1 i
34. 20
35–38
■
4
37. 1 13 i2 4
■
0 t p/2
0t2
Use a graphing device to draw the parametric curve.
y sin 3t
48. x sin1t cos 2t 2 , y cos1t sin 3t 2
49. In the figure, the point P is the midpoint of the segment QR
and 0 u p/2. Using u as the parameter, find a parametric
representation for the curve traced out by P.
y
R
Use De Moivre’s Theorem to find the indicated power.
35. 11 13 i 2
39–42
■
y 1 sin t,
1
2
2, y 2 ,
t
t
47. x cos 2t,
6p u 6p
y t2 1
45. x 1 cos t,
46. x 26. r sin19u/4 2
28. r u sin u,
44. x t 2 1,
36. 11 i 2
38. a
8
P
1
1
13 20
ib
2
2
Find the indicated roots.
39. The square roots of 16i
| Review 573
Q
¨
0
1 x
40. The cube roots of 4 4 13 i
41. The sixth roots of 1
42. The eighth roots of i
43–46 ■ A pair of parametric equations is given. (a) Sketch
the curve represented by the parametric equations. (b) Find a
rectangular-coordinate equation for the curve by eliminating the
parameter.
43. x 1 t 2, y 1 t
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CHAPTER 8
TEST
1. (a) Convert the point whose polar coordinates are 18, 5p/4 2 to rectangular coordinates.
(b) Find two polar coordinate representations for the rectangular coordinate point
16, 2132 , one with r 0 and one with r 0 and both with 0 u 2p.
2. (a) Graph the polar equation r 8 cos u. What type of curve is this?
(b) Convert the equation to rectangular coordinates.
3. Graph the polar equation r 3 6 sin u. What type of curve is this?
4. Let z 1 13 i.
(a) Graph z in the complex plane.
(b) Write z in polar form.
(c) Find the complex number z 9.
5. Let z1 4 a cos
Find z1z2 and
7p
7p
i sin
b
12
12
and z2 2 a cos
5p
5p
i sin
b.
12
12
z1
.
z2
6. Find the cube roots of 27i, and sketch these roots in the complex plane.
7. (a) Sketch the graph of the parametric curve
x 3 sin t 3
y 2 cos t
10 t p2
(b) Eliminate the parameter t in part (a) to obtain an equation for this curve in
rectangular coordinates.
8. Find parametric equations for the line of slope 2 that passes through the point 13, 5 2 .
574
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FOCUS ON MODELING
The Path of a Projectile
Modeling motion is one of the most important ideas in both classical and modern physics.
Much of Isaac Newton’s work dealt with creating a mathematical model for how objects
move and interact—this was the main reason for his invention of calculus. Albert Einstein
developed his Special Theory of Relativity in the early 1900s to refine Newton’s laws of
motion.
In this section we use coordinate geometry to model the motion of a projectile, such as
a ball thrown upward into the air, a bullet fired from a gun, or any other sort of missile. A
similar model was created by Galileo, but we have the advantage of using our modern
mathematical notation to make describing the model much easier than it was for Galileo!
▼ Parametric Equations for the Path of a Projectile
y
√‚t
√‚t ß ¨
¨
0
√‚t ç ¨
x
Suppose that we fire a projectile into the air from ground level, with an initial speed √0
and at an angle u upward from the ground. If there were no gravity (and no air resistance), the projectile would just keep moving indefinitely at the same speed and in the
same direction. Since distance speed time, the projectile would travel a distance
√0t, so its position at time t would therefore be given by the following parametric equations (assuming that the origin of our coordinate system is placed at the initial location
of the projectile; see Figure 1):
x 1√0 cos u2t
FIGURE 1
y 1√0 sin u 2t
No gravity
But, of course, we know that gravity will pull the projectile back to ground level. By
using calculus, it can be shown that the effect of gravity can be accounted for by
subtracting 12 gt 2 from the vertical position of the projectile. In this expression, g is
the gravitational acceleration: g 32 ft/s2 9.8 m/s2. Thus we have the following parametric equations for the path of the projectile:
x 1√0 cos u 2t
EXAMPLE
y 1√0 sin u2t 12 gt 2
With gravity
The Path of a Cannonball
Find parametric equations that model the path of a cannonball fired into the air with an
initial speed of 150.0 m/s at a 30 angle of elevation. Sketch the path of the cannonball.
S O L U T I O N Substituting the given initial speed and angle into the general parametric
equations of the path of a projectile, we get
x 1150.0 cos 30° 2t
y 1150.0 sin 30° 2t 12 19.82t 2
Substitute
√0 150.0, u 30
x 129.9t
y 75.0t 4.9t 2
Simplify
This path is graphed in Figure 2.
y
500
F I G U R E 2 Path of a cannonball
x
(meters)
■
575
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576
Focus on Modeling
▼ Range of a Projectile
How can we tell where and when the cannonball of the above example hits the ground?
Since ground level corresponds to y 0, we substitute this value for y and solve for t:
0 75.0t 4.9t 2
0 t175.0 4.9t 2
t0
or
t
75.0
15.3
4.9
Set y 0
Factor
Solve for t
The first solution, t 0, is the time when the cannon was fired; the second solution means
that the cannonball hits the ground after 15.3 s of flight. To see where this happens, we
substitute this value into the equation for x, the horizontal location of the cannonball.
x 129.9115.32 1987.5 m
The cannonball travels almost 2 km before hitting the ground.
Figure 3 shows the paths of several projectiles, all fired with the same initial speed but
at different angles. From the graphs we see that if the firing angle is too high or too low,
the projectile doesn’t travel very far.
y
¨=85*
¨=75*
¨=60*
¨=45*
¨=30*
¨=15*
¨=5*
0
x
F I G U R E 3 Paths of projectiles
Let’s try to find the optimal firing angle—the angle that shoots the projectile as far as
possible. We’ll go through the same steps as we did in the preceding example, but we’ll
use the general parametric equations instead. First, we solve for the time when the projectile hits the ground by substituting y 0:
0 1√0 sin u 2t 12 gt 2
0 t1√0 sin u 12 gt 2
0 √0 sin u The Granger Collection, New York
t
2√0 sin u
g
GALILEO GALILEI (1564–1642) was
born in Pisa, Italy. He studied medicine
but later abandoned this in favor of science and mathematics. At the age of 25,
by dropping cannonballs of various
sizes from the Leaning Tower of Pisa, he
demonstrated that light objects fall at
the same rate as heavier ones. This contradicted the then-accepted view of
Aristotle that heavier objects fall more
quickly. He also showed that the dis-
1
2 gt
Substitute y 0
Factor
Set second factor equal to 0
Solve for t
tance an object falls is proportional to the square of the time it has
been falling, and from this he was able to prove that the path of a projectile is a parabola.
Galileo constructed the first telescope and, using it, discovered the
moons of Jupiter. His advocacy of the Copernican view that the earth
revolves around the sun (rather than being stationary) led to his being
called before the Inquisition. By then an old man, he was forced to recant his views, but he is said to have muttered under his breath,“Nevertheless, it does move.” Galileo revolutionized science by expressing scientific principles in the language of mathematics. He said,“The great
book of nature is written in mathematical symbols.”
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The Path of a Projectile
577
Now we substitute this into the equation for x to see how far the projectile has traveled
horizontally when it hits the ground:
x 1√0 cos u 2t
1√0 cos u 2 a
Parametric equation for x
2√0 sin u
b
g
Substitute t 12√0 sin u 2 /g
2√ 20 sin u cos u
g
Simplify
√ 20 sin 2u
g
Use identity sin 2u 2 sin u cos u
We want to choose u so that x is as large as possible. The largest value that the sine of any
angle can have is 1, the sine of 90. Thus we want 2u 90, or u 45. So to send the
projectile as far as possible, it should be shot up at an angle of 45. From the last equation in the preceding display, we can see that it will then travel a distance x √ 20 / g.
PROBLEMS
1. Trajectories Are Parabolas From the graphs in Figure 3 the paths of projectiles
appear to be parabolas that open downward. Eliminate the parameter t from the general parametric equations to verify that these are indeed parabolas.
2. Path of a Baseball Suppose a baseball is thrown at 30 ft/s at a 60 angle to the
horizontal from a height of 4 ft above the ground.
(a) Find parametric equations for the path of the baseball, and sketch its graph.
(b) How far does the baseball travel, and when does it hit the ground?
3. Path of a Rocket Suppose that a rocket is fired at an angle of 5 from the vertical with
an initial speed of 1000 ft/s.
(a) Find the length of time the rocket is in the air.
(b) Find the greatest height it reaches.
(c) Find the horizontal distance it has traveled when it hits the ground.
(d) Graph the rocket’s path.
4. Firing a Missile
The initial speed of a missile is 330 m/s.
(a) At what angle should the missile be fired so that it hits a target 10 km away? (You should
find that there are two possible angles.) Graph the missile paths for both angles.
(b) For which angle is the target hit sooner?
5. Maximum Height Show that the maximum height reached by a projectile as a function
of its initial speed √0 and its firing angle u is
y
√ 02 sin2 u
2g
6. Shooting Into the Wind Suppose that a projectile is fired into a headwind that pushes
it back so as to reduce its horizontal speed by a constant amount „. Find parametric equations for the path of the projectile.
7. Shooting Into the Wind Using the parametric equations you derived in Problem 6,
draw graphs of the path of a projectile with initial speed √0 32 ft/s, fired into a headwind
of „ 24 ft/s, for the angles u 5, 15, 30, 40, 45, 55, 60, and 75. Is it still true that
the greatest range is attained when firing at 45? Draw some more graphs for different angles,
and use these graphs to estimate the optimal firing angle.
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578
Focus on Modeling
8. Simulating the Path of a Projectile The path of a projectile can be simulated on a
graphing calculator. On the TI-83, use the “Path” graph style to graph the general parametric
equations for the path of a projectile, and watch as the circular cursor moves, simulating the
motion of the projectile. Selecting the size of the Tstep determines the speed of the
“projectile.”
(a) Simulate the path of a projectile. Experiment with various values of u. Use √0 10 ft/s
and Tstep 0.02. Part (a) of the figure below shows one such path.
(b) Simulate the path of two projectiles, fired simultaneously, one at u 30 and the other at
u 60. This can be done on the TI-83 using Simul mode (“simultaneous” mode). Use
√0 10 ft/s and Tstep 0.02. See part (b) of the figure. Where do the projectiles
land? Which lands first?
(c) Simulate the path of a ball thrown straight up 1u 90°2 . Experiment with values
of √0 between 5 and 20 ft/s. Use the “Animate” graph style and Tstep 0.02.
Simulate the path of two balls thrown simultaneously at different speeds. To better distinguish the two balls, place them at different x-coordinates (for example,
x 1 and x 2). See part (c) of the figure. How does doubling √0 change the
maximum height the ball reaches?
2
2
2
3
0
(a)
3
0
(b)
3
0
(c)
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