Inverted Cone related rate problem HELP!

Jun 2010
1
0
Here the the problem that I can't crack on my final:

A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cc/min. When the depth of the water in the cone is 8 cm the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to its radius?

Thanks in advance!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
The volume of a cone is given by \(\displaystyle V= \frac{1}{3}\pi r^2 h\). If the "ratio of height to radius" is k, then h/r= k so that h= kr or r= h/k. Put that into the formula for volume to get \(\displaystyle V= \frac{1}{3}\pi (h/k)^2h= \frac{1}{3}\pi h^2/k^2[/itex]

Differentiate that with respect to t to find [itex]\frac{dV}{dt}= f(h,k)\frac{dh}{dt}\). Now, you are given that dV/dt= -10 cc/min, dh/dt= -2 cc/min and h= 8 cm. Solve for k.