Expert: Sherry Wallin - 8/30/2010

Question
Hi, I need a really fast answer (by tomorrow) for this problem:

A water pipe of fixed length L has a carrying capacity that depends on the inner diameter of the pipe. The pipe initially has inner diameter D, but over many years as mineral deposits accumulate inside the pipe, its carrying capacity is reduced.

L=length of pipe, D= diameter, t = mineral deposits

Using equation f(t) = pi L (D/2 -t)^2, answer the following:

Show that f(1/4 D) = 1/4 * f(0). Explain in your own words what your diagram says about the carrying capacity of the pipe.

Answer
Maria~
   I am not sure what diagram you are referring to but here is how you would show f(1/4 D) = 1/4 * f(0):

Show each one individually:
f((1/4)D)= f(D/4)= pi*L(D/2 - D/4)^2 = pi*L(D/4)^2 = (pi*LD^2)/16

(1/4)f(0)= 1/4)(pi*L(D/2-0)^2) = (1/4)(pi*L(D/2)^2) = [(1/4)pi*LD]/4
= pi*LD/16

Note the end results are the sams so they are equal.

Math Prof