# Error In Problem Creation?

• May 13th 2012, 12:11 PM
Bashyboy
Related Rates Problem
The problem is as follows:

A conical tank with its vertex down is 12 ft high and 12 ft in diameter at the top. Water is being pumped in at the rae of 4 cubic feet per minute. Find the rate at which the water level is rising: a) when the water is 2 ft deep; b) and 8 ft deep.

Since they did not give me the rate at which the radius changes, I used the relationship 3r = h, giving me h/3 = r; then I squared each side to and substituted into the equation--which is given-- $v = (\pi/3)(h^3/9)$ I took the derivative with respect to time and got $\frac{dh}{dt} = \frac{dV}{dt}\frac{9}{\pi h^2}$

For a) I got 9/pi, but the answer is 4/pi. Did I set the equations up wrong, how about my derivative?
• May 13th 2012, 05:11 PM
skeeter
Re: Error In Problem Creation?
$r = 6$ , $h = 12$

$\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$

so ... $h = 2r \implies r = \frac{h}{2}$

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

$V = \frac{\pi}{3} \left(\frac{h}{2}\right)^2 \cdot h$

$V = \frac{\pi}{12} h^3$

try again