Page 205 - Schaum's Outline of Theory and Problems of Advanced Calculus

196 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8

!
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
ð b cos x p b 1
8.19. Prove that ln dx ¼ ln b þ p ffiffiffiffiffiffiffiffiffiffiffiffiffi if a; b > 1.
2
0 a cos x a þ a 1
ð dx
From Problem 5.58, Chapter 5, ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi ; > 1:
0 cos x 1
2
Integrating the left side with respect to from a to b yields

ð ð b d ð b ð b cos x
ln dx
lnð cos xÞ dx ¼
dx ¼
0 a cos x 0 a 0 a cos x
Integrating the right side with respect to from a to b yields
!
ffiffiffiffiffiffiffiffiffiffiffiffiffi
p
2
ð d p b b 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2 b þ
¼ ln
p ffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lnð þ 1 Þ p ffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
0 1 a a þ a 1
and the required result follows.
MAXIMA AND MINIMA
8.20. Prove that a necessary condition for f ðx; yÞ to have a relative extremum (maximum or minimum)
at ðx 0 ; y 0 Þ is that f x ðx 0 ; y 0 Þ¼ 0, f y ðx 0 ; y 0 Þ¼ 0.
If f ðx 0 ; y 0 Þ is to be an extreme value for f ðx; yÞ,then it must be an extreme value for both f ðx; y 0 Þ and
f ðx 0 ; yÞ. But a necessary condition that these have extreme values at x x ¼ 0 and y ¼ y 0 , respectively, is
f x ðx 0 ; y 0 Þ¼ 0, f y ðx 0 ; y 0 Þ¼ 0(using results for functions of one variable).
8.21. Let f be continuous and have continuous partial derivatives of order two, at least, in a region R
with the critical point P 0 ðx 0 ; y 0 Þ an interior point. Determine the sufficient conditions for relative
extrema at P 0 .
In the case of one variable, sufficient conditions for a relative extrema were formulated through the
second derivative [if positive then a relative minimum, if negative then a relative maximum, if zero a possible
point of inflection but more investigation is necessary]. In the case of z ¼ f ðx; yÞ that is before us we can
expect the second partial derivatives to supply information. (See Fig. 8-6.)

Fig. 8-6

First observe that solutions of the quadratic equation

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p 2
4B 4AC
2 2B
At þ 2Bt þ C ¼ 0are t ¼
2A
2
Further observe that the nature of these solutions is determined by B AC.Ifthe quantity is positive
the solutions are real and distinct; if negative, they are complex conjugate; and if zero, the two solutions are
coincident.