# Related Rates Word Problem

• Nov 27th 2011, 04:51 PM
TheDarkNight93
Related Rates Word Problem
"Coffee is draining from a conical filter(height 6" diameter 6") into a cylindrical coffeepot (diameter 8" height not given) at a rate of 10in^3/min. How fast is the level in the pot rising and how fast is the level in the cone falling when the coffee is 5"deep?"

For the first part, I got as far as h'=(v'-2pirh)/(pir^2) or h'= (10-8pih)/(16pi) but I'm not sure how to find h without aditional conditions. Am I even headed in the right direction?

I havent the slightest idea how to go about the second part. Is the 5" deep is referring to the coffee in the pot?

Any and all help appreciated.
• Nov 27th 2011, 05:19 PM
skeeter
Re: Related Rates Word Problem
$\frac{dV}{dt} = -10$ cubic in per min for the cone

$\frac{dV}{dt} = 10$ cubic in per min for the cylindrical pot

coffee volume in the cone ...

$V = \frac{\pi}{3} r^2 h$

since $\frac{r}{h} = \frac{1}{2}$ ...

$V = \frac{\pi}{12} h^3$ , where $h$ = depth of coffee in the cone

coffee volume in the pot

$V = 16\pi y$ , where $y$ = depth of coffee in the pot

as far as the 5" depth info ... I'd say it's depth of coffee in the cone since the question wants the rate of change of depth in the cone at that time.

try it again ...
• Nov 27th 2011, 05:43 PM
TheDarkNight93
Re: Related Rates Word Problem
That definately helped. I got both answers, but on the volume of the cone I'm not quite sure how you went from v= (pi/3)r^2h to v=(pi/12)h^3
• Nov 27th 2011, 06:08 PM
skeeter
Re: Related Rates Word Problem
Quote:

Originally Posted by TheDarkNight93
That definately helped. I got both answers, but on the volume of the cone I'm not quite sure how you went from v= (pi/3)r^2h to v=(pi/12)h^3

$V = \frac{\pi}{3} r^2 h$

note that for the cone ...

$\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$

therefore, $r = \frac{h}{2}$

substitute $\frac{h}{2}$ for $r$ in the volume formula above to get V strictly in terms of h