Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 10 .~ POLAR COORDINATES 0 19
59. r = cos 2() => x = r cos() = cos 2() cos(}, y = r sin() = cos 2(} !lin(} =>
dx = dx/ d() = cos 2() (- sin(})+ cos() ( -2 sin 2())
dy dy/ df) cos 2(} cos(}+ sin(} (-2 sin 2())
1f dy
When() =
4 , dx -,0,!-( J2---:2 /:---27) +_('-:-v-2 -:::2/,.--!2)~(-_2;,_) - _- J2_2 - 1
O(-J2/ 2) + (J2/ 2}(-2) - -J2 - .
61. r = 3cos0 => x = rcosO = 3cos0 cosO,_ y = rsinO = 3cos0 sinO => ,
~ = -3 sin 2 9 +3cos 2 ,() = 3cos20 = 0 => 2() = ~ or 3 ; 0 = %or 3 ;.
So the tangent is horizontal at ( ~, %) and ( - -32, 3 ;) [same as ( -32, -:a:)] .
~~ = - 6 sin() cos~= -3 sin 20 = 0 => 29 = 0 or 1r 0 = 0 or~· So the tangent is vertical at (3, 0) and (o, ~).
63. r = 1 + cos 0 · => x = r cos 9 = cos() (1 +cos 9), y = r sin 0 = sin 0 (1 + cos 9) =>
·.~ = (1 +cosO) cosO- sin 2 0 = 2cos 2 0 + cos9 - 1 = (2cos9 - 1)(cos(J + 1) = 0 => cos()=~ or - 1 =>
() = 'i· 1r, or 5 ; => horizontal tangent at(~ , 'i), (0, tr), and(~ ,
~~ = - ( 1 + cos 8) sin (} - cos ()sin() = - sin (} ( 1 + 2 cos 9) = 0 => sin 8 = 0 or cos 9 = - ~ =>
5 ; ) .
() = 0, 1r, 2 ;, or 4 ; => vertical tangent ~t (2, 0), (!, 2 ;), and (!, -t;).
Note that the ~gent is horizontal, not vertical when 8 = 7r, since e : j :: = 0.
65. r = a sin() + b cos 9 => r 2 = ar sin() + br cos 8 '=> x 2 + y 2 = ay + bx =>
x 2 - bx + (~b? + y 2 - ay + ( ~a) 2 = (~b) 2 + (ta? => (x - ~b) 2 + (y - ~ a? = Ha 2 + b 2 ), and this is a circle
with center (!b, ~a) and radius. ~.Ja 2 + b 2 •
67. r = 1 + 2 sin(8 / 2): The parameter intei-val is [0, 47r]. 69. r = e•inB - 2cos(40).
The parameter interval is [0, 21l'j.
71. r = 1 + cos 999 8. The parameter interval is [0, 2tr).
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