Thomas Calculus 13th [Solutions]

Chapter 14 Practice Exercises 1065
51. F ( cos x)
i j f ( ,1) i j the tangent
line is ( x ) ( y 1) 0 x y 1; the normal
line is y 1 1( x ) y x 1
52. f xi yj f (1,2) i 2j the tangent line
is ( x 1) 2( y 2) 0 y 1 x 3 ; the normal
2 2
line is y 2 2( x 1) y 2x
4
53. Let
2
f ( x, y, z) x 2y 2z 4 and g( x, y, z) y 1. Then
f 2 xi 2 j 2 k | 2 i 2 j 2 k and
1,1, 1
2
54. Let
i j k
g j f g 2 2 2 2i 2k the line is x 1 2 t, y 1, z 1 2t
2
0 1 0
2
f ( x, y, z) x y z 2 and g( x, y, z) y 1. Then f i 2 yj k | 1,1,
1 i 2j k and
i j k
g j f g 1 2 1 i k the line is x 1 t, y 1, z 1 t
2 2
0 1 0
2 2
55. f , 1 , f 1 1
4 4 2 x , cos x cos y
4 4 ( /4, /4) , f , sin sin |
2 y
x y
4 4 ( /4, /4) 2
L( x, y) 1 1 x 1 y 1 1 x 1 y ; f ( , ) sin cos , ( , ) sin cos ,
2 2 4 2 4 2 2 2 xx x y x y f yy x y x y and
fxy ( x, y) cos x sin y . Thus an upper bound for E depends on the bound M used for | fxx
|, | f xy |, and | f yy | .
2
2
2 2 2
With M we have | E( x, y)| 1 x y
(0.2) 0.0142;
2
2 2 4 4 4
2
2
with M 1, | E( x, y)| 1 (1) x y 1 (0.2) 0.02.
2 4 4 2
56. f (1, 1) 0, fx
(1,1) y | (1,1) 1, f y (1,1) x 6 y | (1,1) 5 L( x, y) ( x 1) 5( y 1) x 5y
4;
fxx
( x, y) 0, f yy ( x, y ) 6, and fxy
( x, y ) 1 maximum of | fxx
|, | f yy |, and | f xy | is 6 M 6
1 2 1
2
| E( x, y)| (6) | x 1| | y 1| (6)(0.1 0.2) 0.27
2 2
57. f (1, 0, 0) 0, fx (1, 0, 0) y 3 z | (1,0,0) 0, f y (1, 0, 0) x 2 z | (1,0,0) 1, fz
(1, 0, 0) 2y 3 x | (1,0,0) 3
L( x, y, z) 0( x 1) ( y 0) 3( z 0) y 3 z ; f (1,1, 0) 1, fx (1, 1, 0) 1, f y (1, 1, 0) 1, fz
(1,1, 0) 1
L( x, y, z) 1 ( x 1) ( y 1) 1( z 0) x y z 1
58.
f 0, 0, 4 1, fx
0, 0, 4 2 sin x sin ( y z ) | 0,0, 0, f y 0, 0, 4
2 cos x cos ( y z ) | 0,0,
1,
4 4
fz
0, 0, 4 2 cos x cos ( y z ) 0, 0,
1 L ( x , y , z ) 1 1( y 0) 1 z
4 1 y z
4
;
4
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