Page 94 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 4] DERIVATIVES 85

MISCELLANEOUS PROBLEMS
2
2
4.34. If x ¼ gðtÞ and y ¼ f ðtÞ are twice differentiable, find (a) dy=dx; ðbÞ d y=dx .
(a)Letting primes denote derivatives with respect to t,we have
dy dy=dt 0
if g ðtÞ 6¼ 0
f ðtÞ
0
dx ¼ dx=dt ¼ g ðtÞ
0

d f ðtÞ d f ðtÞ
0
0
2
d y d dy d f ðtÞ dt g ðtÞ dt g ðtÞ
0
0
0
ðbÞ 2 ¼ ¼ ¼ ¼
dx dx dx dx g ðtÞ dx=dt g ðtÞ
0
0
1 g ðtÞf ðtÞ f ðtÞg ðtÞ g ðtÞf ðtÞ f ðtÞg ðtÞ
00
0
0
0
0
00
00
00
if g ðtÞ 6¼ 0
0
2 3
¼ ¼
0
0 0
g ðtÞ ½g ðtފ ½g ðtފ
1=x 2
e ; x 6¼ 0 (a) f ð0Þ¼ 0; 00
0
4.35. Let f ðxÞ¼ .Prove that ðbÞ f ð0Þ¼ 0.
0; x ¼ 0
e 1=h 2 0 e 1=h 2
f þ ð0Þ¼ lim f ðhÞ f ð0Þ ¼ lim ¼ lim
0
h h h
ðaÞ
h!0þ h!0þ h!0þ
If h ¼ 1=u,using L’Hospital’s rule this limit equals
2 u 2 u 2
u
lim ue ¼ lim u=e ¼ lim 1=2ue ¼ 0
u!1 u!1 u!1
Similarly, replacing h ! 0þ by h ! 0 and u !1 by u ! 1,we find f ð0Þ¼ 0. Thus
0
f þ ð0Þ¼ f ð0Þ¼ 0, and so f ð0Þ¼ 0.
0
0
0
2 2
0 0 e 1=h 2h 3 0 2e 1=h 2u 4
f þ ð0Þ¼ lim f ðhÞ f ð0Þ ¼ lim ¼ lim ¼ lim 2 ¼ 0
00
ðbÞ 4
h h h u!1 e u
h!0þ h!0þ h!0þ
by successive applications of L’Hospital’s rule.
Similarly, f ð0Þ¼ 0 and so f ð0Þ¼ 0.
00
00
In general, f ðnÞ ð0Þ¼ 0for n ¼ 1; 2; 3; .. .
4.36. Find the length of the longest ladder which can be carried around the corner of a corridor, whose
dimensions are indicated in the figure below, if it is assumed that the ladder is carried parallel to
the floor.
A O
The length of the longest ladder is the same as the shortest
straight line segment AB [Fig. 4-10], which touches both outer
a
walls and the corner formed by the inner walls. a sec

As seen from Fig. 4-10, the length of the ladder AB is
L ¼ a sec þ b csc
L is a minimum when b csc
dL=d ¼ a sec tan b csc cot ¼ 0
B
3
3
i.e.; a sin ¼ b cos or tan ¼ p ffiffiffiffiffiffiffiffi b
3
b=a
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a 2=3 þ b 2=3 a 2=3 þ b 2=3
Then sec ¼ ; csc ¼ Fig. 4-10
a 1=3 b 1=3
so that L ¼ a sec þ b csc ¼ða 2=3 þ b 2=3 3=2
Þ
Although it is geometrically evident that this gives the minimum length, we can prove this analytically
2
2
by showing that d L=d for ¼ tan 1 p ffiffiffiffiffiffiffiffi
3
b=a is positive (see Problem 4.78).