3 Polar Form of Complex Numbers; De Moivre's Theorem
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594
CHAPTER 9
| 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 P.2). 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 ϩ bi and the origin in the complex plane is 2a 2 ϩ 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 ϭ a ϩ bi is
0 z 0 ϭ 2a 2 ϩ b 2
FIGURE 4
Calculating the Modulus
EXAMPLE 3
Find the moduli of the complex numbers 3 ϩ 4i and 8 Ϫ 5i.
The plural of modulus is moduli.
SOLUTION
0 3 ϩ 4i 0 ϭ 232 ϩ 42 ϭ 125 ϭ 5
0 8 Ϫ 5i 0 ϭ 282 ϩ 1Ϫ52 2 ϭ 189
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NOW TRY EXERCISE 9
Absolute Value of Complex Numbers
EXAMPLE 4
Graph each set of complex numbers.
(a) C ϭ 5z @ 0 z 0 ϭ 16
(b) D ϭ 5z @ 0 z 0 Յ 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
ă
0
FIGURE 7
a
Re
Let z a ϩ bi be a complex number, and in the complex plane let’s draw the line segment joining the origin to the point a ϩ bi (see Figure 7). The length of this line segment is r ϭ 0 z 0 ϭ 2a 2 ϩ b 2. If u is an angle in standard position whose terminal side
SECTION 9.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 595
coincides with this line segment, then by the definitions of sine and cosine (see Section 6.2)
a ϭ r cos u
b ϭ r sin u
and
so z ϭ r cos u ϩ ir sin u ϭ r(cos u ϩ i sin u). We have shown the following.
POLAR FORM OF COMPLEX NUMBERS
A complex number z ϭ a ϩ bi has the polar form (or trigonometric form)
z ϭ r 1cos u ϩ i sin u 2
where r ϭ 0 z 0 ϭ 2a ϩ b and tan u ϭ 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 ϩ i
(b) Ϫ1 ϩ 13i
(c) Ϫ413 Ϫ 4i
These complex numbers are graphed in Figure 8, which helps us find their
SOLUTION
arguments.
Im
Im
_1+œ∑3 i
(d) 3 ϩ 4i
Im
Im
4i
œ∑3 i
3+4i
1+i
i
ă
ă
ă
ă
0
1
Re
_1
0
Re
0
_4
3
_4
3-4i
(a)
(b)
Re
0
3
Re
_4i
(c)
(d)
FIGURE 8
tan u 11 1
u
tan u ϭ
p
4
13
ϭ Ϫ 13
Ϫ1
u ϭ 2p
3
tan u ϭ
Ϫ4
1
ϭ
Ϫ4 13
13
u ϭ 7p
6
1 ϩ i ϭ 12 a cos
p
p
ϩ i sin b
4
4
(b) An argument is u ϭ 2p/3 and r ϭ 11 ϩ 3 ϭ 2. Thus
Ϫ1 ϩ 13 i ϭ 2 a cos
tanϪ1 43
2p
2p
ϩ i sin
b
3
3
(c) An argument is u ϭ 7p/6 (or we could use u ϭ Ϫ5p/6), and r ϭ 148 ϩ 16 ϭ 8.
Thus
Ϫ4 13 Ϫ 4i ϭ 8 a cos
tan u ϭ 43
uϭ
(a) An argument is u ϭ p/4 and r ϭ 11 ϩ 1 ϭ 12. Thus
7p
7p
ϩ i sin
b
6
6
(d) An argument is u ϭ tanϪ1 43 and r ϭ 232 ϩ 42 ϭ 5. So
3 ϩ 4i ϭ 53cosAtanϪ1 43 B ϩ i sinAtanϪ1 43 B 4
NOW TRY EXERCISES 29, 31, AND 33
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596
CHAPTER 9
| Polar Coordinates and Parametric Equations
The Addition Formulas for Sine and Cosine that we discussed in Section 8.2 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 ϭ r1 1cos u1 ϩ i sin u1 2
z2 ϭ r2 1cos u2 ϩ i sin u2 2
and
then
z1z2 ϭ r1r2 3cos1u1 ϩ u2 2 ϩ i sin1u1 ϩ u2 2 4
z1
r1
ϭ 3cos1u1 Ϫ u2 2 ϩ i sin1u1 Ϫ 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 ϭ r1r2 1cos u1 ϩ i sin u1 2 1cos u2 ϩ i sin u2 2
ϭ r1r2 3cos u1 cos u2 Ϫ sin u1 sin u2 ϩ i1sin u1 cos u2 ϩ cos u1 sin u2 2 4
ϭ r1r2 3cos1u1 ϩ u2 2 ϩ i sin1u1 ϩ 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 ϭ 2 a cos
p
p
ϩ i sin b
4
4
and
z2 ϭ 5 a cos
p
p
ϩ i sin b
3
3
Find (a) z1z2 and (b) z1/z2.
SOLUTION
(a) By the Multiplication Formula
z1z2 ϭ 122 152 c cos a
ϭ 10 a cos
p
p
p
p
ϩ b ϩ i sin a ϩ b d
4
3
4
3
7p
7p
ϩ i sin
b
12
12
To approximate the answer, we use a calculator in radian mode and get
z1z2 Ϸ 101Ϫ0.2588 ϩ 0.9659i2
ϭ Ϫ2.588 ϩ 9.659i
■
SECTION 9.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 597
(b) By the Division Formula
z1
2
p
p
p
p
ϭ c cos a Ϫ b ϩ i sin a Ϫ b d
z2
5
4
3
4
3
ϭ
2
p
p
c cos aϪ b ϩ i sin aϪ b d
5
12
12
2
p
p
ϭ a cos
Ϫ i sin b
5
12
12
Using a calculator in radian mode, we get the approximate answer:
z1 2
Ϸ 10.9659 Ϫ 0.2588i 2 ϭ 0.3864 Ϫ 0.1035i
z2 5
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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 ϭ r 1cos u ϩ i sin u2 , then for any integer n
z n ϭ r n 1cos nu ϩ 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 ϭ zz ϭ r 2 3cos1u ϩ u2 ϩ i sin1u ϩ u2 4
ϭ r 2 1cos 2u ϩ i sin 2u 2
Now we multiply z 2 by z to get
z 3 ϭ z 2z ϭ r 3 3cos12u ϩ u2 ϩ i sin12u ϩ u 2 4
ϭ r 3 1cos 3u ϩ i sin 3u 2
Repeating this argument, we see that for any positive integer n
z n ϭ r n 1cos nu ϩ i sin nu 2
A similar argument using the Division Formula shows that this also holds for negative
integers.
■
EXAMPLE 7
Find
A 12
ϩ
Finding a Power Using De Moivre’s Theorem
1 10
2 iB .
SOLUTION
Since 12 ϩ 12 i ϭ 12 11 ϩ i2 , it follows from Example 5(a) that
12
p
1
p
1
a cos ϩ i sin b
ϩ iϭ
2
2
2
4
4
598
CHAPTER 9
| Polar Coordinates and Parametric Equations
So by De Moivre’s Theorem
a
1
1 10
12 10
10p
10p
ϩ ib ϭ a
b a cos
ϩ i sin
b
2
2
2
4
4
ϭ
25
5p
5p
1
a cos
ϩ i sin
b ϭ 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 ϭ 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 ϭ r 1cos u ϩ i sin u2 and n is a positive integer, then z has the n distinct nth
roots
„k ϭ r 1/n c cos a
u ϩ 2kp
u ϩ 2kp
b ϩ i sin a
bd
n
n
for k ϭ 0, 1, 2, . . . , n Ϫ 1.
PROOF
To find the nth roots of z, we need to find a complex number „ such that
„n ϭ z
Let’s write z in polar form:
z ϭ r 1cos u ϩ i sin u 2
One nth root of z is
„ ϭ r 1/n a cos
u
u
ϩ i sin b
n
n
since by De Moivre’s Theorem, „ n ϭ z. But the argument u of z can be replaced by
u ϩ 2kp for any integer k. Since this expression gives a different value of „ for k ϭ 0, 1,
2, . . . , n Ϫ 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 ϭ Ϫ64, and graph these roots in the complex plane.
SECTION 9.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 599
S O L U T I O N In polar form, z ϭ 641cos p ϩ i sin p 2 . Applying the formula for nth
roots with n ϭ 6, we get
„k ϭ 641/6 c cos a
p ϩ 2kp
p ϩ 2kp
b ϩ i sin a
bd
6
6
for k ϭ 0, 1, 2, 3, 4, 5. Using 641/6 ϭ 2, we find that the six sixth roots of Ϫ64 are
We add 2p/6 ϭ p/3 to each argument
to get the argument of the next root.
Im
2i
Ô
0
_2
2 Re
„ﬁ
_2i „›
F I G U R E 9 The six sixth roots of
z ϭ Ϫ64
„0 ϭ 2 a cos
p
p
ϩ i sin b ϭ 13 ϩ i
6
6
„1 ϭ 2 a cos
p
p
ϩ i sin b ϭ 2i
2
2
„2 ϭ 2 a cos
5p
5p
ϩ i sin
b ϭ Ϫ 13 ϩ i
6
6
„3 ϭ 2 a cos
7p
7p
ϩ i sin
b ϭ Ϫ 13 Ϫ i
6
6
„4 ϭ 2 a cos
3p
3p
ϩ i sin
b ϭ Ϫ2i
2
2
„5 ϭ 2 a cos
11p
11p
ϩ i sin
b ϭ 13 Ϫ i
6
6
All these points lie on a circle of radius 2, as shown in Figure 9.
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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 ϭ r 1/n c cos a
u ϩ 360°k
u ϩ 360°k
b ϩ i sin a
bd
n
n
for k ϭ 0, 1, 2, . . . , n Ϫ 1.
EXAMPLE 9
Finding Cube Roots of a Complex Number
Find the three cube roots of z ϭ 2 ϩ 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 ϭ 222 ϩ 22 ϭ 212 and u ϭ 45Њ. Thus
We add 360Њ/3 ϭ 120Њ to each
argument to get the argument of the
next root.
Im
z ϭ 212 1cos 45° ϩ i sin 45°2
Applying the formula for nth roots (in degrees) with n ϭ 3, we find that the cube roots of
z are of the form
„k ϭ A212B 1/3 c cos a
œ∑2 i
„⁄
45° ϩ 360°k
45° ϩ 360°k
b ϩ i sin a
bd
3
3
where k ϭ 0, 1, 2. Thus the three cube roots are
0
_2
Ô
2 Re
1 12 1cos 135° ϩ i sin 135°2 ϭ Ϫ1 ϩ i
12122 1/3 ϭ 123/2 2 1/3 ϭ 21/2 ϭ 12
„2 ϭ 12 1cos 255° ϩ i sin 255°2 Ϸ Ϫ0.366 Ϫ 1.366i
_œ∑2 i
F I G U R E 1 0 The three cube roots of
z ϭ 2 ϩ 2i
„0 ϭ 12 1cos 15° ϩ i sin 15°2 Ϸ 1.366 ϩ 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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600
| Polar Coordinates and Parametric Equations
CHAPTER 9
Solving an Equation Using the nth Roots Formula
EXAMPLE 10
Solve the equation z 6 ϩ 64 ϭ 0.
S O L U T I O N This equation can be written as z 6 ϭ Ϫ64. Thus the solutions are the sixth
roots of Ϫ64, which we found in Example 8.
■
NOW TRY EXERCISE 91
9.3 EXERCISES
SKILLS
CONCEPTS
1. A complex number z ϭ a ϩ bi has two parts: a is the
part, and b is the
we graph the ordered pair 1
2. Let z ϭ a ϩ bi.
,
5–14
part. To graph a ϩ bi,
2 in the complex plane.
(a) The modulus of z is r ϭ
.
(b) We can express z in polar form as z ϭ
,
where r is the modulus of z and u is the argument of z.
p
p
ϩ i sin b in rectangular form is
6
6
7. Ϫ2
8. 6
9. 5 ϩ 2i
10. 7 Ϫ 3i
11. 13 ϩ i
12. Ϫ1 Ϫ
.
Im
16. z ϭ Ϫ1 ϩ i13
17. z ϭ 8 ϩ 2i
19–20
plane.
z
i
3 ϩ 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
Ϫ12 ϩ i 12
14.
2
15. z ϭ 1 ϩ i
.
rectangular form as
6. Ϫ3i
15–16 ■ Sketch the complex number z, and also sketch 2z, Ϫz,
and 12 z on the same complex plane.
. The complex number
zϭ
5. 4i
13.
3. (a) The complex number z ϭ Ϫ1 ϩ i in polar form is
z ϭ 2 a cos
Graph the complex number and find its modulus.
, and an argument of
z is an angle u satisfying tan u ϭ
zϭ
■
■
18. z ϭ Ϫ5 ϩ 6i
Sketch z1, z2, z1 ϩ z2, and z1z2 on the same complex
19. z1 ϭ 2 Ϫ i, z2 ϭ 2 ϩ i
20. z1 ϭ Ϫ1 ϩ i, z2 ϭ 2 Ϫ 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 ϭ a ϩ bi 0 a Ͼ 1, b Ͼ 16
23. 5z @ 0 z 0 ϭ 36
25. 5z @ 0 z 0 Ͻ 26
27. 5z ϭ a ϩ bi 0 a ϩ b Ͻ 26
24. 5z @ 0 z 0 Ն 16
26. 5z @ 2 Յ 0 z 0 Յ 56
28. 5z ϭ a ϩ bi 0 a Ն 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 ϭ a ϩ bi 0 a Յ 0, b Ն 06
Re
4. How many different nth roots does a nonzero complex number
have?
■
4 Re
29. 1 ϩ i
30. 1 ϩ 13 i
31. 12 Ϫ 12 i
32. 1 Ϫ i
33. 213 Ϫ 2i
34. Ϫ1 ϩ i
35. Ϫ3i
36. Ϫ3 Ϫ 3 13 i
37. 5 ϩ 5i
38. 4
39. 413 Ϫ 4i
40. 8i
41. Ϫ20
42. 13 ϩ i
43. 3 ϩ 4i
44. i12 Ϫ 2i2
45. 3i11 ϩ i2
46. 211 Ϫ i 2
SECTION 9.3
| Polar Form of Complex Numbers; De Moivre’s Theorem 601
47. 41 13 ϩ i2
48. Ϫ3 Ϫ 3i
49. 2 ϩ i
83. The fourth roots of Ϫ81i
50. 3 ϩ 13 i
51. 12 ϩ 12 i
52. Ϫpi
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 ϭ cos p ϩ i sin p, z2 ϭ cos ϩ i sin
3
3
p
p
3p
3p
ϩ i sin
54. z1 ϭ cos ϩ i sin , z2 ϭ cos
4
4
4
4
p
p
4p
4p
ϩ i sin
b
55. z1 ϭ 3 a cos ϩ i sin b , z2 ϭ 5 a cos
6
6
3
3
9p
p
9p
p
ϩ i sin
b , z2 ϭ 2 a cos ϩ i sin b
56. z1 ϭ 7 a cos
8
8
8
8
57. z1 ϭ 41cos 120° ϩ i sin 120°2 ,
z2 ϭ 21cos 30° ϩ i sin 30°2
58. z1 ϭ 121cos 75° ϩ i sin 75°2 ,
z2 ϭ 3 121cos 60° ϩ i sin 60°2
60. z1 ϭ 45 1cos 25° ϩ i sin 25° 2 ,
z2 ϭ 15 1cos 155° ϩ i sin 155° 2
61. z1 ϭ 13 ϩ i, z2 ϭ 1 ϩ 13 i
63. z1 ϭ 2 13 Ϫ 2i, z2 ϭ Ϫ1 ϩ i
64. z1 ϭ Ϫ 12 i, z2 ϭ Ϫ3 Ϫ 313 i
65. z1 ϭ 5 ϩ 5i, z2 ϭ 4
66. z1 ϭ 413 Ϫ 4i, z2 ϭ 8i
67. z1 ϭ Ϫ20, z2 ϭ 13 ϩ i
68. z1 ϭ 3 ϩ 4i, z2 ϭ 2 Ϫ 2i
Find the indicated power using De Moivre’s Theorem.
71. 12 13 ϩ 2i2
73. a
70. 11 Ϫ 13 i2
5
12
12 12
ib
ϩ
2
2
75. 12 Ϫ 2i 2 8
77. 1Ϫ1 Ϫ i 2 7
79. 12 13 ϩ 2i2 Ϫ5
5
72. 11 Ϫ i2 8
74. 1 13 Ϫ i2 Ϫ10
1
13 15
ib
76. a Ϫ Ϫ
2
2
78. 13 ϩ 13 i2 4
80. 11 Ϫ i2 Ϫ8
81–90 ■ Find the indicated roots, and graph the roots in the complex plane.
81. The square roots of 4 13 ϩ 4i
82. The cube roots of 4 13 ϩ 4i
89. The fourth roots of Ϫ1
90. The fifth roots of Ϫ16 Ϫ 1613i
91–96
■
Solve the equation.
91. z 4 ϩ 1 ϭ 0
92. z 8 Ϫ i ϭ 0
93. z 3 Ϫ 4 13 Ϫ 4i ϭ 0
94. z 6 Ϫ 1 ϭ 0
95. z 3 ϩ 1 ϭ Ϫi
96. z 3 Ϫ 1 ϭ 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 ϭ 12 Ϫ 12 i, z2 ϭ 1 Ϫ i
69. 11 ϩ i 2
88. The fifth roots of i
s, s„, s„ 2, s„ 3, . . . , s„ nϪ1
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
ϩ i sin
where n is a positive
n
n
2
integer. Show that 1, „, „ , „ 3, . . . , „ nϪ1 are the n
distinct nth roots of 1.
(b) If z 0 is any complex number and s n ϭ z, show that
the n distinct nth roots of z are
z2 ϭ 251cos 150° ϩ i sin 150°2
■
86. The cube roots of 1 ϩ i
97. (a) Let „ ϭ cos
59. z1 ϭ 41cos 200° ϩ 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 ϩ 11 ϩ i 2z ϩ i ϭ 0
(b) z 2 Ϫ iz ϩ 1 ϭ 0
(c) z 2 Ϫ 12 Ϫ i 2z Ϫ 14 i ϭ 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
602
CHAPTER 9
| Polar Coordinates and Parametric Equations
9.4 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 ϭ g1t 2
x ϭ f1t2
where t ʦ 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 ϭ t 2 Ϫ 3t
yϭtϪ1
S O L U T I O N For every value of t, we get a point on the curve. For example, if t ϭ 0,
then x ϭ 0 and y ϭ Ϫ1, so the corresponding point is 10, Ϫ12 . In Figure 1 we plot the
points 1x, y2 determined by the values of t shown in the following table.
t
x
y
Ϫ2
Ϫ1
0
1
2
3
4
5
10
4
0
Ϫ2
Ϫ2
0
4
10
Ϫ3
Ϫ2
Ϫ1
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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SECTION 9.4
| Plane Curves and Parametric Equations 603
If we replace t by Ϫt in Example 1, we obtain the parametric equations
x ϭ t 2 ϩ 3t
y ϭ Ϫt Ϫ 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 609). 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 ϭ 4t 2 Ϫ 6t
y ϭ 2t Ϫ 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 ϭ t 2 ϩ 3t, y ϭ Ϫt Ϫ 1
F I G U R E 3 x ϭ 4t 2 Ϫ 6t, y ϭ 2t Ϫ 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 ϭ t Ϫ 1, we get t ϭ y ϩ 1. Substituting into the equation for x, we get
x ϭ t 2 Ϫ 3t ϭ 1y ϩ 12 2 Ϫ 31y ϩ 12 ϭ y 2 Ϫ y Ϫ 2
Thus the curve in Example 1 has the rectangular equation x ϭ y 2 Ϫ y Ϫ 2, so it is a
parabola.
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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 ϭ cos t
y ϭ sin t
Describe and graph the path of the object.
tՆ0
604
| Polar Coordinates and Parametric Equations
CHAPTER 9
y
S O L U T I O N To identify the curve, we eliminate the parameter. Since cos2 t ϩ sin2 t ϭ 1
and since x ϭ cos t and y ϭ sin t for every point 1x, y 2 on the curve, we have
t= 2
(ỗ t, ò t)
t
t=
x 2 ϩ y 2 ϭ 1cos t 2 2 ϩ 1sin t2 2 ϭ 1
t=0
0
This means that all points on the curve satisfy the equation x 2 ϩ y 2 ϭ 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
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NOW TRY EXERCISE 25
FIGURE 4
EXAMPLE 4
Sketching a Parametric Curve
Eliminate the parameter, and sketch the graph of the parametric equations
y ϭ 2 Ϫ cos2 t
x ϭ 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 ϭ 1 Ϫ sin2 t to change the second equation:
(1, 2)
y ϭ 2 Ϫ cos2 t ϭ 2 Ϫ 11 Ϫ sin2t2 ϭ 1 ϩ sin2 t
Now we can substitute sin t ϭ x from the first equation to get
y ϭ 1 ϩ x2
x
0
FIGURE 5
so the point 1x, y2 moves along the parabola y ϭ 1 ϩ x 2. However, since Ϫ1 Յ sin t Յ 1,
we have Ϫ1 Յ x Յ 1, so the parametric equations represent only the part of the parabola
between x ϭ Ϫ1 and x ϭ 1. Since sin t is periodic, the point 1x, y 2 ϭ 1sin t, 2 Ϫ cos2 t2
moves back and forth infinitely often along the parabola between the points 1Ϫ1, 22 and
11, 22 , as shown in Figure 5.
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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
xϭ2ϩt
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 ϭ 6 ϩ 3t
y ϭ 6 ϩ 31x Ϫ 22 ϭ 3x
x
Thus the slope-intercept form of the equation of this line is y ϭ 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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