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

152 D CHAPTER 13 VECTOR FUNCTIONS
11. The corresponding parametric equations are x = 1, y = cos t, z = 2 sin t.
Eliminating the parameter in y and z gives ~ 2 + (z/ 2) 2 = cos 2 t + sin 2 t = 1
or y 2 + z 2 / 4 = 1. Since x = 1, the curve is an ellipse centered at (1, 0, 0) in
the plane x = 1.
X
13. The parametric equations are X = e. y = t 4 • z = t 6 . These are positive
for t =I 0 and 0 when t = 0. So the curve lies entirely in the first octant.
The projection of the graph onto the xy-plane is y = x 2 , y > 0, a half parabola.
Onto the xz-plane z = x 3 , z > 0, a half cubic, and the yz-plane, y 3 = z 2 •
I
I
I
I
I
I
I
I
I
, I
,'
,,,- y=xl
,,'
X
I
15. Tbe projection of the curve onto the xy-plane is given by r (t) = (t , sin t, 0} [we use 0 for the z-component] whose graph
is the curve y =sin x, z = 0. Similarly, the projection onto the xz-plane is r (t) = (t, 0, 2 cost), whose graph is the cosine
wave z = 2 cos x, y = 0, and the projection onto the yz-plane is r (t) = (0, sin t, 2 cost) whose graph is the ellipse
2
- I
y
-2
xy-plane
xz-plane
yz-plane
From the projection onto the yz-plane we see that the curve lies on an elliptical
cylinder with axis the x-axis. The other two projections show that the curve
oscillates both vertically and horizontally as we move in the x-direction,
suggesting that the curve is an elliptical helix that spirals along the cylinder.
17. Taking ro = (2, 0, 0) and r1 = (6, 2, -2), we have from Equation 12.5.4
r (t) = {1 - t)ro +tr1 = {1- t) {2,0, 0) +t{6, 2,-2), 0 ~ t ~ 1 or r {t) = (2 + 4t,2t, - 2t);O ~ t ~ 1.
Parametric equations are x = 2 + 4t, y = 2t, z = -2t, 0 ~ t ~ 1.
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