Thomas Calculus 13th [Solutions]

Section 14.6 Tangent Planes and Differentials 1023
63. z f ( x, y) g( x, y, z) f ( x, y) z 0 gx ( x, y, z) fx ( x, y), g y ( x, y, z) f y ( x, y ) and
gz ( x, y, z) 1 gx ( x0 , y0, f ( x0 , y0)) fx ( x0 , y0 ), g y ( x0, y0, f ( x0, y0 )) f y ( x0 , y 0)
and
gz
( x0 , y0 , f ( x0 , y 0 )) 1 the tangent plane at the point P 0 is
fx
( x0 , y0)( x x0 ) f y ( x0 , y0 )( y y0 ) [ z f ( x0 , y 0 )] 0 or
z fx
( x0 , y0)( x x0 ) f y ( x0, y0 )( y y0 ) f ( x0 , y0)
( t cos t) i ( t sin t)
j
64. f 2xi 2yj 2(cos t t sin t) i 2(sin t t cos t)
j and v ( t cos t) i ( t sin t)
j u v
| v|
2 2
( t cos t) ( t sin t)
(cos t) i (sin t)
j since t 0 ( Du
f ) P f u 2(cos t t sin t)(cos t) 2(sin t t cos t)(sin t) 2
65. f 2xi 2yj 2zk 2(cos t) i 2(sin t) j 2tk and
0
v ( sin t) i (cos t)
j k u
sin t i cos t j 1 k ( D f ) f (2cos ) sin t (2sin ) cos t (2 ) 1 2
u P u t t t
t
2 2 2 0
2 2 2 2
( Du f ) , ( D f )(0) 0
4 2 2
u and
( Du
f )
4 2 2
v
| v|
( sin t) i (cos t)
j k
2 2 2
( sin t) (cos t) 1
66.
67.
1 1 1/2 1 1/2
r ti t j ( t 3) k v t i t j
1
k;
t 1 x 1, y 1, z 1 P
4 2 2 4
0 (1, 1, 1) and
1 1 1
2 2
v(1) i j k;
f ( x, y, z) x y z 3 0 f 2xi 2 yj k f (1,1, 1) 2i 2 j k;
2 2 4
therefore v
1
4 ( f ) the curve is normal to the surface
1 1/2 1 1/2
r ti t j (2t 1) k v t i t j 2 k;
t 1 x 1, y 1, z 1 P
2 2
0 (1, 1, 1) and
1 1
2 2
v(1) i j 2 k;
f ( x, y, z) x y z 1 0 f 2xi 2 yj k f (1, 1,1) 2i 2 j k;
2 2
now v (1) f (1, 1, 1) 0, thus the curve is tangent to the surface when t 1
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