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

18 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
49. x = r cos 8 = ( 4 + 2 sec 8) cos 8 = 4 cos 8 + 2. Now, r -+ oo =>
(4 + 2sec8)-+ 00 => 8-+ (~) - or 8 -+ e2"' )+ [since we need only
consider 0 $ 8 < 271' ], so lim x =
. r-oo 8 -+-rr/2-
lim ( 4 cos 8 + 2) = 2. Also,
T -+ -oo => ( 4 + 2 sec B) -+ -oo => 8 -+ ( ~) + or 8 -+ ( 3 ;)-, so
lim x = lim ( 4 cos 8 + 2) = 2. Therefore, lim x = 2 => x = 2 is a vertical asymptote.
r--oo 8--+rr/2+ T"--+±oo
(6,0)
51 . To show that x = 1 is an asymptote we must prove lim x = 1.
r - ±oo
x = (r) cosO= (sin 8 tan 8) cos 8 = sin 2 ().Now; r -+ oo => sinO tan8-+ oo =>
8 -+ (f)- ,so lim x= lim sin 2 8 = l.Aiso,r-+-oo => sinBtanB-->-oo =>
r --+oo 0-+'Tr/2- .
x=l
e-+ (~) + ,so lim X = lim sin 2 {;I,;, 1. Therefore, lim X= 1 . => X= 1 is
. r ...... - oo 0--+1fj 2+ ,·-±oo
a vertical asymptote. Also notice that X = sin 2 e ~ 0 for all e, and X = sin 2 {;I $ 1 for all e. And Xi= 1, since the curve is not
defined at odd multiples of ~. Therefore, the curve lies entirely within the vertical strip 0 $ x < 1.
53. (a) We see that the curve r = 1 + csin 8 crosses itself at the origi n, where r = 0 (in fact the inner loop corresponds to
negative r-values,) so we solve the equation of the lima~on for r = 0 c sin 8 = - 1 sin 8 = - 1/ c. Now if
lei < 1, then this equation has no solution and hence there is no inner loop. But if c < - 1, then on the interval (0, 27T)
the equation has the two solutions 8 = sin- 1 ( -1/c) and 8 = 1r- sin- 1 ( - 1/c), and if c > 1, the solutions are
(} = 1r + sin - 1 (1/c) and ()= 21r- sin- 1 (1(c). Tn each case, r < 0 for (} between the two solutions, indicating a loop.
(b) For 0 < c < 1, the dimple (if it exists) is characterized by the fact that y has a local maximum at 8 = 3 ;
. So we
determ.ine for what c-valu~s
d 2 ~ is negative at(} = 3 2~, since by the Second Derivative Test tllis indicates a maximum:
~ . .
y = rsin8 =sinO + c sin 2 (} =>
dy
dB = cos8 + 2csinfJ cos8 = cosfJ + csin 28 =>.
d 2 y
- ., = - sin8 + 2ccos 28.
dB~
At 8 = 3 2 ,.., this is equal to - (- 1) + 2c( -1) = 1 - 2c, which is negative only for c > %. A similar argument shows that
for - 1 < c < 0, y only has a local minimum at e = ~ (indicating a dimple) for c < -~ . .
55. r = 2 sin (} => x = r cos(} = 2 sin fJ cos 8 = sin 28, y = 1· sin 8 = 2 sin 2 B =>
dy = dyjd(J = 2 · 2 sin fJ cosO= sin 28 = tan 2
B
dx dxjdB cos2fJ·2 cos28
When 8 =
7T dy ( 7r) 7T rr; ·
6 , d.'t =tan 2 · 6 =tan = v 3. [Another method: Use Equation 3.]
3
57. r= l / 8 => x=rcos8 = (cosfJ) j B, y=rsin8=(sinfJ)/ 8 =>
.dy - 0 + 7r(- 1) - 'Tr
When 8 = 1r, dx = -(-1) -1r(O) = l
dy dyjd8 sin.O( - 1/ 8 2 ) + (1/ 8) cos 8 8 2 - sin (}+(} cosO
dx = dxj dO = cos8(- 1/ fJ 2 ) - (1/ 8) sinB · (} 2 = - cosO - Osin B
= -1r.
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