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

D
PROBLEMS PLUS
1. Since three-dimensional situations are often difficult to visualize and work with, let
us first try to find an analogous problem in two dimensions. The analogue of a cube
is a square and the analogue of a sphere is a circle. Thus a similar problem in two
dimensions is the following: if five circles with the same radius rare contained in a
square of side 1 m so that the circles touch each other and four of the circles touch
two sides ofthe square, find r.
The diagonal of the square is .J2. The diagonal is a lso 4r + 2x . But x is the diagonal of a smaller square of side r . Therefore
x = .J2r ~ .J2=4r+2x =4r+2.J2r= (4+2.J2)r ~ r = 4
/{,/2.
Let's use these ideas to solve the original three-dimensional problem. The diagonal of the cube is \/12 + 12 + 12 = v'a.
The diagonal of the cube is also 4r + 2x where x is the diagonal of a smaller cube with edger. Therefore
r;; r;; v'3 2 v'a - 3
x = Jr 2 + 1·~ + r 2 = v'3 r ~ v'a = 4r+2x .=4r+2v3r = (4+2v3)r.Thus r= r;;=
4 + 2v3 2
The radius of each ball is ( v'3 - ~) m.
3. (a) We find the line of intersection L as in Example 12.5. 7(b). Observe that the point ( - 1, c, c) lies on both planes. Now since
L lies in both planes, it is perpendicular to both of the normal vectors n1 and n 2, and thus parallel to their cross product
j
k
n 1 x n 2 = c 1 1 = (2c, -c 2 + 1, -c 2 - 1). So symmetric equations of L can be written as
1 - c c
-- x +1 = y - c z - c
~
1 =
- 2c (..- - ~ c + 1
, provided that c I 0, ±1.
If c = 0, then the two planes are given by y + z = 0 and x = - 1, so symmetric· equations of L are x = - 1, y = - z. If
c = - 1, then the two planes are given by -x + y + z = - 1 and x + y + z = -1, and they intersect in the line x = 0,
y = - z - 1. If c = 1, then the two planes are given by x + y + z = 1 and x - y + z = 1, and they intersect in the line
y = 0, X = 1- Z.
(b) If we set z = t in the symmet~i c equations and solve for x andy separately, we get x + 1 = (t - c)( - 2 c)
c 2 + 1
(t- c)(c 2 - 1)
y - c = c2 +1
- 2ct + (c 2 - 1) (c 2 - 1)t + 2c El. . . fr th .
~ x = c 2 + 1
, Y = & + 1
. unmatmg c om ese equattons, we
have x 2 + y 2 = t 2 + 1. So the curve traced out by L in the plane z = tis a circle with center~~ (0, 0, t) and
radius J fl + 1.
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