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

SECTION 12.6 CYLINDERS AND QUADRIC SURFACES 0 139
1
To restrict the z-range as in the second graph, we can use the option view= - 4 . . 4 in Maple's p l"ot3d command, or
Plo.t Range - > {- 4 , 4} in Mathematica's Pl ot 3D command.
39. Solving the equation for z we get z = ±y' 4x 2 + y 2 , so we plot separately z = y' 4x 2 + y2 and z = - y' 4x2 + y2.
41.
z z=2 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 .
The parabola
y=x2
X
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 y'(x + 1) 2 + y 2 + z 2 and the distance from P to the plane x = 1 is lx- 11 /Vf'i = lx - 11
(by Equation 12.5.9). So lx- 11 = y'(x + 1)2 + y 2 + z 2 ¢::> (x- 1) 2 = (x + 1) 2 + v. 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.
2 2 2
47. (a) An equation for an ellipsoid centered at the origin with intercepts x = ±a, y = ± b, and z = ±cis x 2
+ Y 2
+ .=.._ = 1.
a b c 2
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, so an equation is
x2 y2 z2 .
(6378.137) 2 + (6378.137) 2 + (6356.523)2 = 1
• • ~ ~ I ~
(b) Traces to z = k are the Circles (6378.137)2 + (6378.137)2 = 1 - (6356.523)2
2 2 = (6378.137)2- (6378.137)2
k2
x + y ~6.~ .
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