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

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164 D CHAPTER 13 VECTOR FUNCTIONS<br />

for x > - ~ ln. 2, the maximum curvature is attain<strong>ed</strong> at the point (-~ ln 2, e(- In 2 ) 1 2 ) = (- ~ ln 2, ~).<br />

Since lim e"'(1 + e 2 x)- 3 1 2 = 0, ~~:(x) approaches 0 as x---+ oo.<br />

:.~: - oo<br />

33. (a) C appears to be changing direction more quickly at P than Q, so we would expect the curvature to be great~r<br />

,<br />

at P.<br />

(b) First we sketch approximate osculating circles at P and Q. Using the<br />

axes scale as a guide, we measure the radius of the osculating circle<br />

at P to be approximately 0.8 units, thus p = .!. =><br />

K,<br />

~ 1.3. Similar!y,'we estimate the radius of the<br />

p .8<br />

"'= .!. ~ - 0<br />

1<br />

c<br />

osculating circle at Q to be 1.4 units, son, = .!. ~ ....!...<br />

4 ~ p 1.<br />

0.7.<br />

X<br />

j6x-' 1 1 6<br />

[1 + (-2x-3)z]3/2 = :z;4 (1 + 4x- 6)3/ 2'<br />

The appearance of the two humps in this graph is perhaps a little surprising, but it is<br />

explain<strong>ed</strong> by the fact that y = x - 2 increases asymptotically at the origin from both<br />

-1<br />

directions, and so its graph has very little bend there. [Note th~t r;,(O) is undefin<strong>ed</strong>.]<br />

37. r(t) = (tet, e- 1 ,../2t) => .r '(t) = ((t+l)et,-e-'·,../2), r"(t) = ((t+2)et,e-t, O). Then<br />

r' (t) x r" (t) = ( -.J2e-t, ../2(t + 2)e 1 , 2t + 3), · Jr ' (t) x r " (t)l = y'2e- 21 + 2(t + 2) 2 e 2 t + (2t + 3)2,<br />

l r'(t)J = . l(t + 1 )2e2t + e zt + 2 , ( ) lr'(t) x r "(t)l y'2e- 2 t +'2(t + 2)2e 2 t + (2t + 3)2<br />

v and "' t = lr'(t)13 = ((t + 1)2e2t + e-2t + 2]3/2<br />

We plot the space curve and its curvature function for - 5 :::; t :::; 5 below.<br />

K(l)<br />

0.6<br />

y<br />

- 5<br />

5 I<br />

From the graph of r;,(t) we see that curvature is maximiz<strong>ed</strong> fort = 0, so the curve bends most sharply at the point (0, 1, 0).<br />

The curve bends more gradually as we move away from this point, becoming almost linear. This is reflect<strong>ed</strong> in the curvature<br />

graph, where n,(t) becomes nearly 0 as /tl increases.<br />

39. Notice that the curve b has two inflection points at which the graph appears alrnos,t straight. We would expect the curvature to<br />

be 0 or nearly 0 at these values, bu't the curve a isn't near 0 there. Th~ s, a must be the graph of y = f ( x) rather than the graph<br />

of curvature, and b is the graph .of y = K-( x).<br />

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