# word problem invovling rate change

• Mar 11th 2010, 07:59 PM
ascendancy523
word problem invovling rate change
Here's my problem

Sand falls out of the end of a slurry at the rate of 60 cc/sec. The pile forms a circular cone, the ratio of whose base diameter to height is 3. When the pile is of height 10 cm., at what rate is the height of the pile increasing?

So far I have $V=\frac{1}{3}(\pi)r^2h$ where $r=\frac{2h}{3}$ but then I can't seem to figure out the rest of the problem. Any help is greatly appreciated.
• Mar 11th 2010, 08:08 PM
sa-ri-ga-ma
It is given that the ratio of whose base diameter to height is 3
So d/h = 3
2r/h = 3
r = 3h/2.
Volume of the cone is
V = 1/3*π*r^2*h. Substitute r = 3h/2 and find dV/dt to find dh/dt.
• Mar 12th 2010, 08:02 AM
ascendancy523
Quote:

Originally Posted by sa-ri-ga-ma
It is given that the ratio of whose base diameter to height is 3
So d/h = 3
2r/h = 3
r = 3h/2.
Volume of the cone is
V = 1/3*π*r^2*h. Substitute r = 3h/2 and find dV/dt to find dh/dt.

So I'm getting the answer wrong still, maybe you can check my work?

$V=\frac{1}{3}(\pi)(\frac{3h}{2})^2*h
$

$\\
\\
=\frac{1}{3}(\pi)\frac{9h^2}{4}*h$

$
\\
\\
=\frac{9\pi}{12}(h)^3$

$
\frac{dV}{dt}=\frac{9\pi}{4}h^2$

$
\\
60=\frac{9\pi}{4}(10^2)\frac{dh}{dt}$

I'm getting a negative answering which I know isn't right, Did I setup my calculations right or did I do some mathematical error?
• Mar 12th 2010, 08:07 AM
sa-ri-ga-ma
Quote:

Originally Posted by ascendancy523
So I'm getting the answer wrong still, maybe you can check my work?

$V=\frac{1}{3}(\pi)(\frac{3h}{2})^2*h
$

$\\
\\
=\frac{1}{3}(\pi)\frac{9h^2}{4}*h$

$
\\
\\
=\frac{9\pi}{12}(h)^3$

$
\frac{dV}{dt}=\frac{9\pi}{4}h^2$

$
\\
60=\frac{9\pi}{4}(10^2)\frac{dh}{dt}$

I'm getting a negative answering which I know isn't right, Did I setup my calculations right or did I do some mathematical error?