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

SECTION 10.3 POLAR COORDINATES 0 13
67. Iff' is continuous and j' (t) i= 0 for a$ t $ b,. then either!' (t) > 0 for all tin (a, b) or j' (t) < 0 for all t in [a, bj. Thus, f
. . '
is monotonic (in fact, strictly increasing or stri 1 ( fl± - xfl ) xy - xif ' . . ·
dt di = dt ;; = xz ~ dt = 1 + (iJ/x) 2 x2 = xz +iF. Usmg the Cham Rule, and tne
"a·cttllat s- rot (d:r;)2+(!!11..)2dt d .• /(d:r;)2+(!!11..)2 ('2+ •2)1/2 b h
dt dt ~ 'Jt - y dt dt = x y , we ave t at
! i J
0
d¢> _ d¢>/dt _ (xii- xfl ) 1 _ xii - xiJ . _- I d¢> 1-1 xy- xy 1- l±ii - xiJI
ds - ds/dt - :i;2 + y2 (±2 + y2)1/2 - (~2 + 1;2)3/2 · So"'- - ds - (±2 + y2)3/2 - (±2 + y2)3/2 ·
d !( )
. 1 .. 0 d . dy .. d2y
(b) x = x_an y = x ~ x = 'x = an y = dx' y = dx2'
- II· (d~y/dx 2 ) - 0 . (dyjcb:) i - ld 2 y/ dx 2 1
So"' - (1 + (dyjdx) 2 )31 2 - (1 + (dyjdx)2] 31 2 •
71. X = 8 - sin8 ~ :i; = 1----' cosB -~ X = sin8, and y = 1 - cos8 ~ iJ =sine ~ y = cos8. Therefore,
icosB - cos 2 e- sin 2 el icosB - (cos 2 8 + sin 2 8)1 JcosB - II .
"' = = . 2 = . The top of the arch 1s
[(1 - cos B) 2 + sin 2 B)S/ 2 (1 - 2 cos e + cos 2 e + sm 8)31 2 (2 - 2 cos 8)3/2
characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when B = (2n -
l )1r,
d b · B · ti . fi Jcos 1r -11 1-1- 11 1
sotaken=1 an su stttute =1rmto 1e ex~ress ton or ~~;: ~t= ( 2 _ 2 cos1r) 3 / 2 = 12
_ 2
(- 1
))3/ 2
=4·
73. The coordinates of T are .(r cosB, r sin B). Since TP was unwound from
arc T A, T P has length rB. Also LPTQ = LPT R - LQT R = t 1r - 8,
soP has coordinates x = r cos B + r8 cos(~1r ~B) = r(cos B + B sin 8),
y = rsin B - rBsin (~1r - 8) = r(sin8- 8cos9).
X
10.3 Polar Coordinates
1. (a) (2, i) By adding 21r to i, we obtain the point ( 2, 7 ;). The direction
opposite i is 4 3
.,., so (- 2, ~ ) is a point that satisfies the r < 0
· requirement.
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