Page 212 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 212
CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 203
2
2
8.46. The temperature at any point ðx; yÞ in the xy plane is given by T ¼ 100xy=ðx þ y Þ. (a)Find the direc-
tional derivative at the point ð2; 1Þ in a direction making an angle of 608 with the positive x-axis. (b)In
what direction from ð2; 1Þ would the derivative be a maximum? (c) What is the value of this maximum?
p ffiffiffi 1
Ans. (a)12 3 6; (b)ina direction making an angle of tan 2with the positive x-axis, or in the
p ffiffiffi
direction i þ 2j;(c)12 5
8.47. Prove that if Fð ; ; zÞ is continuously differentiable, the maximum directional derivative of F at any point is
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2
@F 1 @F @F
2
given by þ þ .
@ 2 @ @z
DIFFERENTIATION UNDER THE INTEGRAL SIGN
ð 1= d ð 1= 1 1 1
2
2
2
8.48. If ð Þ¼ cos x dx, find . Ans. x sin x dx cos p cos 2
ffiffiffi
ffiffi d ffiffi 2 2
p p
ð 2
1 x dF
8.49. (a)If Fð Þ¼ tan dx, find by Leibnitz’s rule. (b) Check the result in (a)bydirect integration.
0 d
Ans. ðaÞ 2 tan 1 1 2
2
lnð þ 1Þ
m
ð 1 1 ð 1 ð 1Þ m!
m
p
p
8.50. Given x dx ¼ ; p > 1. Prove that x ðln xÞ dx ¼ mþ1 ; m ¼ 1; 2; 3; ... .
0 p þ 1 0 ðp þ 1Þ
!
ffiffiffiffiffiffiffiffiffiffiffiffiffi
p
ð 1 2
8.51. Prove that lnð1 þ cos xÞ dx ¼ ln 1 þ ; j j < 1.
0 2
ð 2
2
8.52. Prove that lnð1 2 cos x þ Þ dx ¼ ln ; j j < 1 . Discuss the case j j¼ 1.
0 0; j j > 1
ð dx 59
8.53. Show that :
3 ¼ 2048
0 ð5 3cos xÞ
INTEGRATION UNDER THE INTEGRAL SIGN
1 ð 2 2 ð 1
ð ð
2
2
2
2
8.54. Verify that ð x Þ dx d ¼ ð x Þ d dx.
0 1 1 0
ð 2
8.55. Starting with the result ð sin xÞ dx ¼ 2 ,prove that for all constants a and b,
0
ð 2
2 2 2 2
fðb sin xÞ ða sin xÞ g dx ¼ 2 ðb a Þ
0
ð 2 dx 2
8.56. Use the result ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi ; > 1to prove that
0 þ sin x 1
2
9
ð 2 5 þ 3 sin x
ln dx ¼ 2 ln
0 5 þ 4 sin x 8
ð =2 dx cos 1
8.57. (a)Use the result ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi ; 0 @ < 1toshow that for 0 @ a < 1; 0 @ b < 1
0 1 þ cos x 1 2
=2 1 þ b cos x
ð
2
2
1
sec x ln dx ¼ fðcos 1 aÞ ðcos 1 bÞ g
0 1 þ a cos x 2
ð =2 5 2
(b) Show that 1 .
2 72
sec x lnð1 þ cos xÞ dx ¼
0
