Thomas Calculus 13th [Solutions]

Section 14.6 Tangent Planes and Differentials 1019
e
0.1 cos(0.1) 1.11, the max of | f yy ( x , y ) |
0.1
on R is e cos(0.1) 1.11, and the max of | fxy
( x , y ) | on R is
0.1 2 e sin(0.1) 0.12 M 1.11; thus 1
2
| E( x, y)| (1.11) | x| | y| (0.555)(0.1 0.1) 0.0222
2
38. f (1, 1) 0, f ( , ) 1 (1, 1) 1, ( , ) 1
x x y x y y (1,1) 1 ( , ) 0 1( 1) 1( 1)
x
f f x y y
f L x y x y
x y 2; f 1 1
xx ( x, y) , f ( , ) , ( , ) 0;
2 yy x y f
2 xy x y | x 1| 0.2 0.98 x 1.2 so the max of
x
y
| fxx
( x, y )| on R is 1
2
(0.98)
1
(0.98)
2
1.04 M 1.04; thus
1.04; | y 1| 0.2 0.98 y 1.2 so the max of | f yy ( x, y )| on R is
1
2 2
E( x, y)| (1.04) | x 1| + | y 1| (0.52)(0.2 0.2) 0.0832
2
39. (a) f (1, 1, 1) 3, fx (1, 1,1) y z| (1,1,1) 2, f y (1, 1, 1) x z| (1,1,1) 2, fz
(1, 1, 1) y x| (1,1,1) 2
L( x, y, z) 3 2( x 1) 2( y 1) 2( z 1) 2x 2y 2z
3
(b) f (1, 0, 0) 0, fx (1, 0, 0) 0, f y (1, 0, 0) 1, fz
(1, 0, 0) 1
L( x, y, z) 0 0( x 1) ( y 0) ( z 0) y z
(c) f (0, 0, 0) 0, fx (0, 0, 0) 0, f y (0, 0, 0) 0, fz
(0, 0, 0) 0 L( x, y, z) 0
40. (a) f (1, 1, 1) 3, fx
(1, 1,1) 2 x | (1,1,1) 2 f y (1, 1,1) 2 y | (1,1,1) 2, fz (1, 1, 1) 2 z| (1,1,1) 2
L( x, y, z) 3 2( x 1) 2( y 1) 2( z 1) 2x 2y 2z
3
(b) f (0,1, 0) 1, fx
(0, 1, 0) 0, f y (0,1, 0) 2, f z (0, 1, 0) 0
L( x, y, z) 1 0( x 0) 2( y 1) 0( z 0) 2y
1
(c) f (1, 0, 0) 1, fx
(1, 0, 0) 2, f y (1, 0, 0) 0, f z (1, 0, 0) 0
L( x, y, z) 1 2( x 1) 0( y 0) 0( z 0) 2x
1
41. (a)
f (1, 0, 0) 1, f (1, 0, 0) x
x
1,
2 2 2
x y z (1, 0, 0)
y
f y (1, 0, 0) 0,
2 2 2
x y z (1, 0, 0)
f (1, 0, 0) z
z
0 L( x, y, z) 1 1( x 1) 0( y 0) 0( z 0) x
2 2 2
x y z (1, 0, 0)
(b) f (1, 1, 0) 2, f (1, 1, 0) 1 , (1,1, 0) 1
x
f ,
2
y
f (1,1, 0) 0
2 z
L( x, y, z) 2 1 ( x 1) 1 ( y 1) 0( z 0) 1 x 1 y
2 2 2 2
(c) f (1, 2, 2) 3, f (1, 2, 2) 1 , (1, 2, 2) 2
x
f ,
3 y
f 2
3 z (1, 2, 2)
3
L( x, y, z) 3 1 ( x 1) 2 ( y 2) 2 ( z 2) 1 x 2 y 2 z
3 3 3 3 3 3
42. (a)
y cos xy
f ,1, 1 1, f , 1,1 0,
2 x 2
z
,1,1
2 2
z ,1,1
2
2
2
xcos
xy
f y , 1,1 0,
2
z
,1,1
sin xy
fz
, 1, 1 1 L( x, y, z) 1 0 x 0( y 1) 1( z 1) 2 z
(b) f (2, 0, 1) 0, f x (2, 0, 1) 0, f y (2, 0,1) 2, fz
(2, 0, 1) 0
L( x, y, z) 0 0( x 2) 2( y 0) 0( z 1) 2y
2
Copyright
2014 Pearson Education, Inc.