SOLUTION: A water pipe of fixed length L has a carrying capacity that depends on the inner diameter of the pipe. The pipe initally has inner diameter D, but over many years, as mineral depos

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: A water pipe of fixed length L has a carrying capacity that depends on the inner diameter of the pipe. The pipe initally has inner diameter D, but over many years, as mineral depos      Log On


   



Question 172696: A water pipe of fixed length L has a carrying capacity that depends on the inner diameter of the pipe. The pipe initally has inner diameter D, but over many years, as mineral deposits accumulate inside the pipe, its carrying capacity is reduced.
a. give a quadratic function f(t) that models the carrying capacity of the pipe as a function of the thickness t of mineral deposits. Your function will be in terms of the constants D and L and the variable t. (HINT: the volume of a cylinder is pi*r^2*h)
b. show that f(1/4d)= 1/4*f(0), and my book says to make a diagram illustrating this fact. If you could make the diagram that would be great, but I would be satisfied with an explanation of what to put in the diagram and how it should look.

I would really appreciate any help on this problem!
Please and Thank you in advance.
Please explain step by step.

Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
a) Consider the pipe as a hollow right circular cylinder, the variable 't' being the decrease in inner radius due to the mineral deposits.
So, carrying capacity of the pipe now i.e. present internal volume is f%28t%29+=+pi+%2A+%28%28D+-+2t%29%2F2%29%5E2+%2A+L


The blue inner circle is the final inner surface of the pipe and the red circle is the previous inner surface.
b) Put t = 0 in f%28t%29+=+pi+%2A+%28%28D+-+2t%29%2F2%29%5E2+%2A+L
f%280%29+=+pi+%2A+%28%28D+-+2x0%29%2F2%29%5E2+%2A+L+=+pi%2AD%5E2%2AL%2F4

Put t = D/4 in f%28t%29+=+pi+%2A+%28%28D+-+2t%29%2F2%29%5E2+%2A+L
f%28D%2F4%29+=+pi+%2A+%28%28D+-+2%2A%28D%2F4%29%29%2F2%29%5E2+%2A+L+=+pi%2AD%5E2%2AL%2F16

So you can clearly see that f%28D%2F4%29+=+1%2F4+%2A+f%280%29.