Related Rates Coffee Filter Problem

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SUMMARY

The discussion centers on a related rates problem involving a conical coffee filter and a cylindrical coffee pot. The filter has a diameter and height of 15 cm, and coffee drains at a rate of 2 cm3/sec. The user correctly identifies a typo in the drainage rate and calculates the rate of change of the coffee depth in the filter as 8/(169 * π) cm/s. The user seeks clarification on the calculation for the rate of change of the coffee depth in the pot.

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manik
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Coffee drains from a conical filter, with diameter of 15cm and a height of 15 cm into a cylindrical coffee pot, with a diameter of 15 cm at 2 cm^2/ sec. At a certain time, the coffee in the filter is 13 cm deep.

a) How fast is the depth of coffee in the filter changing at that time?
b) How fast is the depth of coffee in the pot changing at that time?

Apparently this is an assignment question. I think that 2cm^2 / sec is a typo and should be replaced by 2cm^3 / sec.

http://img242.imageshack.us/img242/3917/attemptau7.jpg

<< that is my attempt for problem a), is this right? Can someone make any suggestions.. thanks in advance
 
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I'm getting 8/(169 * Pi) cm/s as my answer. I think one place where you did something wrong was you did not square h/2 in place of r in the original volume equation. It should have looked something like this.

v = 1/3 * PI * r^2 * h --> r = h/2 --> v = 1/3*Pi*(h/2)^2*h which gives

v = 1/12*Pi*h^3 By differentiation we obtain:

dv = 1/4*Pi*h^2*dh solving for dh we get (4*dv)/(Pi*h^2) or 8/(169*Pi) cm/s.
 
thanks i got a, but can someone tell me for b i don't get it
 

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