505
PART 1." Solutions to Odd-Numbered Exercises and Practice T~sts
Section 10.6 Polar Coordinates
[] In polar coordinates you do not have unique representation of points. The point (r, 0) can be represented by
(r, 0 -+ 2n~) or by (-r, 0 + (2n + 1)qr) where n is any integer. The pole is represented by (0, 0) where 0
is any angle.
[] To convert from polar coo~.inates to rectapgular coordinates, use the following relationships.
x = r cos 0
y = r sin 0
[] To convert from rectangular coordinates to polar coordinates, use the following relationships.
r = +_ x,/~ y2
tan 0 = y/x
If 0 is in the same quadrant as the point (x, y), then r is positive. If 0 is in the opposite quadrant as the point
(x, y), then r is negative.
[] You should be able to convert rectangular equations to polar form and vice versa.
Solutions to Odd-Numbered Exercises
1. Polar coordinates: , ~
X~ 4COS
=0
y = 4 sin
=4
Rectaogular coordinates: (0, 4)
3. Polar coordinates: (- 1, ~)
x = -1 cos
-2
2
Rectangular coordinates: (~, ~)
(0, 4)
2
Three additional representations
(-- 4, -~), (4, -?), (- 4, -~)
Three additional representations:
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
506
11.
Three additional representations:
Three additional representations:
7,n"
¢r
I1~"
13.
15.
x = -1 cos -3~r
~
y = -1 sin -3~r V’~
Rectangularcoordinates:
2
17. Polar coordinates: (0, -~) (origin!)
x=0 cos(- 7_~)
=0
x = 32 cos
= 0, y = 32 sin
Rectangular coordinates: (0, 32)
,--0,~n(-~)--0
Rectangular coordinates: (0, 0)
= 32
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PART l: Solutions to Odd-Numbered Exercises and Practice Tests
21. Polar coordinates: (~/~, 2.36)
x = ~ cos(2.36) ~ - 1.004
y = ~/~ sin(2.36) ~ 0.996
2
Rectangular coordinates: (- 1.004, 0.996)
23. (r, 0)=(2,~"~) =:’ (x,y) =(-l.414,1.414)=(--J~,-f~)
25. (r,O)= (-4.5,1.3) ~ (x,y)= (S1.204,-4.336)
27. Rectangular coordinates: (-7, 0)
r=7, tan 0= 0, 0=0
29. Rectangular coordinates: (1, 1)
r=.~/~, tan0= 1, 0
Polar coordinates: (7, ~r), (-7,0)
31. Rectangular coordinates: (-3, 4)
r= ,/~+ 16=5, tan0= -~, 0~ 2.214
Polar coordinates: (5, 2.214), (-5, 5.356)
"rr
4
33. Rectangular coordinates: (r= ~/~+ 3 = ~f~, tan0= 1, 0
4
Polar coordinates: (./’~,~), (_./.~,~.rr)
I23
508 PART I: Solutions to Odd-Numbered Exercises and Practice Tests
35. Rectangular coordinates: (4, 6)
37. (x,y) = (3,-2) ==~ r=
3
r = ~ = 2x/~, tan 0 = ~, 0 ~ 0.983
-0:588
0=
(r, O) ~" (v/~, --0.588)
Polar coordinates: (2x/~, 0.983), (-2x/~, 4.124)
39. (x, y) = (-4’~, 2) ==~ r = ~ + 22 -~ v/ff
O= arctan(-~) = 0.857
4/3
O= aretan(5~) ~ 0.490
(r. 0) = (x/if, 0.857)
43. (x,y)= (0,-5) ::~ (r, 0)= (5,-1.57!) = (5,
45. (a) x2+y2=49
47. (a) x2+y~-2ax=O
49. (a)
r~- 2acos O= 0
r=7
(b) x2 + y2 = a2 (a > O)
x = 12
r cos 0 = 12
r(r - 2a cos O) = 0
r = 2acos 0
(b) xz + yz - 2ay = O
r = 12 see 0
x=a
(b)
r co8 0 = a
r2- 2asinO= 0
r(r - 2 a sin’O) -~ 0
r = 2asin 0
~3. (a) y2 = x3
51. (a) xy=4
(r cos O)(r sin 0) = 4
r2 cos Osin 0 = 4
ta(2 cos O sin 0) = 8
r~ sin 20 = 8
r2 = 8csc20
2(r cos O)(r sin O) = 1
rz(2 cos 0 sin O) = 1
r~sin20 = 1
r~ = csc 20
(r sin 0)2 = (r cos 0)3
sin2 0 = rcos3 0
sin2 O
r----------3tan20see0
cos 0
(b) x2 = y3
(r cos 0)2 = (r sin 0)3
cos2 0 = r sin~ 0
cos~ 0
~ = 0cot~ Ocse a
r = sin3
509
$$.
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
r = 4 sin 0
ta = 4r sin 0
x~ + yZ = 4y
x~ + y~-4y=O
57.
o=
tan 0 -
r=4
&= 16
2
x + y2 = 16
6’
3
x3
,/~x - 3y = 0
61. r = -3cscO
ta = cos 0
r3 = r cos 0
63.
r sin 0 = -3
(X2 + y2)3/2 = X
y= -3
x2 + y2 =
(x~ + y2)3 = x~
65.
69.
r = 2 sin30
r = 2(3 sin 0 - 4 sins O)
r~=6tasinO- 8tasin~0
(~ +Y~)~ = 6(~z +YgY -- 8Y~
(x~ + y~)~ = 6x2y - 2y3
6
2- 3sinO
r(2 - 3 sin O) = 6
2r = 6 + 3rsinO
2(±-~ = 6 + 3y
4(x~ + yZ) = (6 + 3y)z
4x~ + 4y~ = 36 + 36y + 9y~
4xz -- 5yz -- 36y -- 36 -- 0
67.
71,
r=3
ra = 9
xZ+y2=9
r= 3sec0
r cos 0 --- 3
tan 0 = tan ~Y
x-y=O
77. True, the distances from the origin are the same.
1
1 - cos 0
r- rcos O= 1
4/~+y:z - X = 1
x2+y~= 1 +2x+x2
y~= 2x + 1
75.
73.
r
6
3
2
x-3-0
510
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
79. (a) (r1, 01) = (xl, yl) where xI = r~ cos 01 and yl = r~ sin 0r
(r~, 0z) = (x~, y~) where xz = r~ cos 0~ and y~ = r~ sin 02.
Then xlz + yl2 = r~~ COS2 01 + r~~ sin~ 0~ = rl2 and x~~ + y:Z = r~. Thus,
~ = ~(x~ - x~)~ + ~ - y~)~
= ~x~~ - ~x~ + xz~ + y~: - 2y~y~ + y~
= 4(x~~ + Y~) + (x22 + Y22) - 2(x~x2 +
= ~rx2 + r~~ - 2(r~r~ cos O~ cos O~ + r~r~ sin 01 sin 0:)
= ~r~a + r~2 - 2r~r2 cos(O~ - 0:).
(b) If O~ = 02, the points are on the same line through the origin. In this case,
~ = ~? + r? - a,,,~ ~o~(0) = ~ ~)~ = Ir, - ’~l(c) If O~ - O~ = 90~, d = ~ r~2, the Pythagorean Theorem.
81. D = _
Dx= _
= 5-21 = -16
83. D=
=-11-21=-32
1
-3
0 -2
-
=35
1
Dx=
D =,
08 1-3
- =40
_53 -_113 =_15_33=_48
3
O~, -32 2
x= D = -IV=
Dy_-48=3
Y= D --16
8
Oy =
0
0
88
~
3 -2
= 56
Dz=
~ -31
_D~, 40 8
X D 35 7
z=D_~.=56__= 8
D 35 5
85. (x + 5)8. ax3 = 175,000x3. a = 175,000
87. (2x - y)~2. axTy5 _-- -- 101,376xTy5. a = - 101,376