Page 159 - How To Solve Word Problems In Calculus

5.
A

a
θ
B
θ

C
b

Let θ be the angle between the pipe and the vertical as shown in the
diagram.

|AB| |BC|
= sec θ = csc θ
a b
|AB|= a sec θ |BC|= b csc θ

If L represents the length of the pipe,

L =|AB|+|BC|

L (θ) = a sec θ + b csc θ

We differentiate with respect to θ and find the critical value of this
function.

L (θ) = a sec θ tan θ − b csc θ cot θ

sin θ
tan θ =
cos θ
0 = a sec θ tan θ − b csc θ cot θ
cos θ
cot θ =
a sec θ tan θ = b csc θ cot θ sin θ
1
sec θ =
sin θ cos θ cos θ
a = b 1
2
2
cos θ sin θ csc θ =
sin θ
3
3
a sin θ = b cos θ
146