Page 209 - Schaum's Outline of Theory and Problems of Advanced Calculus

200 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8

APPLICATIONS TO ERRORS
ffiffiffiffiffiffiffi
p
8.28. The period T of a simple pendulum of length l is given by T ¼ 2 l=g. Find the (a) error and
2
(b) percent error made in computing T by using l ¼ 2m and g ¼ 9:75 m=sec ,if the true values
2
are l ¼ 19:5m and g ¼ 9:81 m=sec .
g
(a) T ¼ 2 l 1=2 1=2 . Then
s ffiffiffiffiffi
l
1 3=2
ð l
dT ¼ð2 g 1=2 1 1=2 dlÞþð2 l 1=2 Þð g ffiffiffiffi dl dg
2 2 dgÞ¼ p lg g 3 ð1Þ
Error in g ¼ g ¼ dg ¼þ0:06; error in l ¼ l ¼ dl ¼ 0:5
The error in T is actually T, which is in this case approximately equal to dT. Thus, we have
from (1),
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 0:05Þ ðþ0:06Þ¼ 0:0444 sec (approx.)
3
Error in T ¼ dT ¼ p
ð2Þð9:75Þ ð9:75Þ
r ffiffiffiffiffiffiffiffiffi
2
The value of T for l ¼ 2; g ¼ 9:75 is T ¼ 2 ¼ 2:846 sec (approx.)
9:75
dT 0:0444
¼ 1:56%:
T 2:846
ðbÞ Percent error (or relative error) in T ¼ ¼
1
1
Another method: Since ln T ¼ ln 2 þ ln l ln g,
2
2

dT 1 dl 1 dg 1 0:05 1 þ0:06

¼ 1:56%
T ¼ 2 l 2 g ¼ 2 2 2 9:75 ð2Þ
as before. Note that (2)can be written
1 1 Percent error in g
Percent error in T ¼ 2 Percent error in l 2
MISCELLANEOUS PROBLEMS
ð 1 x 1
8.29. Evaluate dx.
0 ln x
In order to evaluate this integral, we resort to the following device. Define
x 1
ð 1
dx > 0
ð Þ¼
0 ln x
Then by Leibnitz’s rule
1 @ x 1 1 x ln x 1 1
ð ð ð
0
0 @ ln x 0 ln x 0 þ 1
ð Þ¼ dx ¼ dx ¼ x dx ¼
Integrating with respect to , ð Þ¼ lnð þ 1Þþ c. But since ð0Þ¼ 0; c ¼ 0; and so ð Þ¼ lnð þ 1Þ.
Then the value of the required integral is ð1Þ¼ ln 2.

The applicability of Leibnitz’s rule can be justified here, since if we define Fðx; Þ¼ ðx 1Þ= ln x,
0 < x < 1, Fð0; Þ¼ 0; Fð1; Þ¼ ,then Fðx; Þ is continuous in both x and for 0 @ x @ 1 and all finite
> 0.
8.30. Find constants a and b for which
ð
2
2
fsin x ðax þ bxÞg dx
Fða; bÞ¼
0
is a minimum.