Thread: word problem invovling rate change

1. 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.

2. 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.

3. 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?

4. 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?
Where is the negative answer?
dh/dt = 60*4/9π*10^2.