29.06.2016 Views

Thomas Calculus 13th [Solutions]

  • No tags were found...

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

1064 Chapter 14 Partial Derivatives<br />

(b) The direction toward (4, 6) is determined by v<br />

3 4<br />

3 (4 1) i (6 2) j 3i 4j u i j<br />

5 5<br />

f u<br />

14<br />

5 .<br />

44. (a) True (b) False (c) True (d) True<br />

45. f 2xi j 2zk<br />

f j 2 k,<br />

(0, 1, 1)<br />

f j,<br />

(0,0,0)<br />

f j 2k<br />

(0, 1,1)<br />

46. f 2yj<br />

2zk<br />

f<br />

(2,2,0)<br />

4 j,<br />

f<br />

(2, 2,0)<br />

4 j,<br />

f<br />

(2,0,2)<br />

4 k ,<br />

f<br />

(2,0, 2)<br />

4k<br />

47.<br />

48.<br />

f 2xi j 5k f 4i j 5k Tangent Plane: 4( x 2) ( y 1) 5( z 1) 0<br />

(2, 1,1)<br />

4x y 5z 4; Normal Line: x 2 4 t, y 1 t, z 1 5t<br />

f 2xi 2yj k f 2i 2j k Tangent Plane: 2( x 1) 2( y 1) ( z 2) 0<br />

(1,1,2)<br />

2x 2y z 6 0; Normal Line: x 1 2 t, y 1 2 t, z 2 t, z 2 t<br />

49. z 2x z<br />

x 2 2<br />

x y x (0,1,0)<br />

2y<br />

z 2 0<br />

0<br />

and z 2 y<br />

y 2 2<br />

x y<br />

z<br />

y<br />

(0,1,0)<br />

2;<br />

thus the tangent plane is 2( y 1) ( z 0) 0 or<br />

2<br />

50. z 2 2<br />

2x x y z<br />

1 and<br />

x<br />

x 1,1,<br />

1 2<br />

1 ( x 1) 1 ( y 1) z 1 0 or x y 2z<br />

3 0<br />

2 2 2<br />

2<br />

z<br />

y<br />

2 2 2<br />

2<br />

y x y z<br />

1<br />

y<br />

1,1,<br />

2<br />

; thus the tangent plane is<br />

1<br />

2<br />

Copyright<br />

2014 Pearson Education, Inc.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!