Thomas Calculus 13th [Solutions]

Section 14.6 Tangent Planes and Differentials 1017
23. (a) The unit tangent vector at
1 3
, in the direction of motion is
2 2
u 3 1
2 i 2 j;
1 3 1 3
T (sin 2 y) i (2x cos 2 y) j T , sin 3 i cos 3 j D T , T
2 2 u
u
2 2
3
sin 3 1 cos 3 0.935° C / ft
2 2
(b) r( t) (sin 2 t) i (cos 2 t) j v( t) (2cos 2 t) i (2sin 2 t)
j and v 2;
dT T dx T dy
dt x dt y dt
T v T v ( DuT
) ,
v
v v where u v
v
; at
1 3
, we have
2 2
3
u i 1 j
2 2
from part (a)
dT
dt
3
sin 3 1 cos 3 2 3 sin 3 cos 3 1.87° C/sec
2 2
2 2
24. (a) T (4 x yz) i xzj xyk T (8, 6, 4) 56i 32j 48 k;
r( t) 2t i 3tj t k the particle is
at the point P (8, 6, 4) when t 2; v( t) 4ti 3j 2 tk v(2) 8i 3j 4k u v
v
8 i 3 j 4 k D T (8, 6, 4) T 1 56 8 32 3 48 ( 4) 736 ° C/m
89 89 89
u
u
89 89
(b) dT T dx T dy
T v ( T u)
v at t 2, dT D 736
dt x dt y dt
t 2 (2) 89 736° C/sec
dt u T v
89
25. (a) f (0, 0) 1, fx
( x, y) 2 x f x (0,0) 0, f y ( x, y) 2 y f y (0,0) 0
L( x, y) 1 0( x 0) 0( y 0) 1
(b) f (1,1) 3, f x (1, 1) 2, f y (1, 1) 2 L( x, y) 3 2( x 1) 2( y 1) 2x 2y
1
26. (a) f (0, 0) 4, fx
( x, y) 2( x y 2) f x (0, 0) 4, f y ( x, y) 2( x y 2) f y (0, 0) 4
L( x, y) 4 4( x 0) 4( y 0) 4x 4y
4
(b) f (1, 2) 25, f x (1, 2) 10, f y (1, 2) 10 L( x, y) 25 10( x 1) 10( y 2) 10x 10y
5
27. (a) f (0, 0) 5, fx
( x, y ) 3 for all ( x, y ), f y ( x, y ) 4 for all ( x, y) L( x, y) 5 3( x 0) 4( y 0)
3x
4y
5
(b) f (1,1) 4, f x (1, 1) 3, f y (1, 1) 4 L( x, y) 4 3( x 1) 4( y 1) 3x 4y
5
2 4 3 3
28. (a) f (1, 1) 1, fx ( x, y) 3 x y fx (1, 1) 3, f y ( x, y) 4 x y f y (1,1) 4
L( x, y) 1 3( x 1) 4( y 1) 3x 4y
6
(b) f (0, 0) 0, fx
(0, 0) 0, f y (0, 0) 0 L( x, y) 0
x
x
29. (a) f (0, 0) 1, fx
( x, y) e cos y f x (0, 0) 1, f y ( x, y) e sin y f y (0, 0) 0
L( x, y) 1 1( x 0) 0( y 0) x 1
(b)
f 0, 0, f 0, 0,
2 x 2
f y 0, 1 L( x, y) 0 0( x 0) 1 y y
2 2 2
30. (a)
(b)
2 y x
f 0, 0 1, fx
x, y e fx
0, 0 1
L( x, y) 1 1( x 0) 2 y 0 x 2y
1
2 y x
f y ( x, y) 2 e f y (0, 0) 2
3 3 3 3 3 3 3 3 3
f (1, 2) e , fx
(1, 2) e , f y (1, 2) 2 e L( x, y) e e ( x 1) 2 e ( y 2) e x 2e y 2e
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