Thomas Calculus 13th [Solutions]

96. Let
Chapter 14 Practice Exercises 1075
2
f ( x, y, z) xy yz zx x z 0. If the tangent plane is to be parallel to the xy -plane, then f is
perpendicular to the
xy -plane f i 0 and f j 0. Now f ( y z 1) i ( x z) j ( y x 2 z)
k
so that f i y z 1 0 y z 1 y 1 z,
and f j x z 0 x z.
Then
2 2 1
2
z(1 z) (1 z) z z( z) ( z) z 0 z 2z 0 z or z 0. Now z 1 x 1
and
1 1
,
1
,
1
2 2 2 2
2 2
y is one desired point; z 0 x 0 and y 1 (0,1, 0) is a second desired point.
97.
f
1 2
f x y z x f x y z x g y z
x
2
( i j k ) ( , , ) ( , ) for some function
1
2
2
f
z
g
z
1
2
2
C 1 2 1 2 1 2 1 2 1 2
g( y, z) y z C f ( x, y, z)
x y z
2 2 2 2 2
C
f g
g y
y y
g( y, z) y h( z ) for some function h z h ( z ) h ( z ) z C for some arbitrary
constant
1 2
f (0, 0, a)
a C and
2
claimed.
1
2
2
f (0, 0, a) ( a) C f (0, 0, a) f (0, 0, a ) for any constant a, as
df
ds
f (0 su ,0 su ,0 su ) f (0, 0, 0)
lim , s 0
s 0
s
98.
1 2 3
u,(0, 0, 0)
however,
2 2 2 2 2 2
1 2 3
s u s u s u 0
lim , s 0
s 0
s
2 2 2
1 2 3
s u u u
lim lim | u| 1;
s 0
s
s 0
y
f x i j z k fails to exist at the origin (0, 0, 0)
2 2 2 2 2 2 2 2 2
x y z x y z x y z
99. Let f ( x, y, z) xy z 2 f yi xj k . At (1,1, 1), we have f i j k the normal line is
x 1 t, y 1 t, z 1 t , so at t 1 x 0, y 0, z 0 and the normal line passes through the origin.
2 2 2
100. (b) f ( x, y, z) x y z 4 f 2xi 2yj 2zk
at (2, 3, 3) the gradient is f 4i 6j 6k
which is normal to the surface
(c) Tangent plane: 4x 6y 6z 8 or
2x 3y 3z
4
Normal line: x 2 4 t, y 3 6 t, z 3 6t
2 yz
2 2 y
101. (a) y, z are independent with w x e and z x y w w x w w z
y x y y y z y
yz 2 2
2 x yz yz
2 2
y
xe zx e (1) yx e (0); z x y 0 2x x 2 y x ; therefore,
y
y y x
w yz y 2 yz 2 yz
2xe zx e 2y zx e
y
z
x
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