Solução_Calculo_Stewart_6e

F.
128 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ETTX.10
CHAPTER 12
39. Solving the equation for z we get z = ± 4x 2 + y 2 , so we plot separately z = 4x 2 + y 2 and z = − 4x 2 + y 2 .
41.
43. The surface is a paraboloid of revolution (circular paraboloid) with vertex at the origin, axis the y-axis and opens to the right.
Thus the trace in the yz-plane is also a parabola: y = z 2 , x =0. The equation is y = x 2 + z 2 .
45. Let P =(x, y, z) be an arbitrary point equidistant from(−1, 0, 0) and the plane x =1. Then the distance from P to
(−1, 0, 0) is (x +1) 2 + y 2 + z 2 and the distance from P to the plane x =1is |x − 1| / √ 1 2 = |x − 1|
(by Equation 13.5.9 [ ET 12.5.9]). So |x − 1| = (x +1) 2 + y 2 + z 2 ⇔ (x − 1) 2 =(x +1) 2 + y 2 + z 2 ⇔
x 2 − 2x +1=x 2 +2x +1+y 2 + z 2 ⇔ −4x = y 2 + z 2 . Thus the collection of all such points P is a circular
paraboloid with vertex at the origin, axis the x-axis, which opens in the negative direction.
47. (a) An equation for an ellipsoid centered at the origin with intercepts x = ±a, y = ±b,andz = ±c is x2
a 2 + y2
b 2 + z2
c 2 =1.
Here the poles of the model intersect the z-axis at z = ±6356.523 and the equator intersects the x-andy-axes at
x = ±6378.137, y = ±6378.137,soanequationis
(b) Traces in z = k are the circles
x 2 + y 2 =(6378.137) 2 −
x 2
(6378.137) + y 2
2 (6378.137) + z 2
2 (6356.523) =1 2
x 2
(6378.137) + y 2
2 (6378.137) =1− k 2
2
2 6378.137
k 2 .
6356.523
(6356.523) 2 ⇔