# Thread: Error In Problem Creation?

1. ## 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-- $\displaystyle v = (\pi/3)(h^3/9)$ I took the derivative with respect to time and got $\displaystyle \frac{dh}{dt} = \frac{dV}{dt}\frac{9}{\pi h^2}$

For a) I got 9/pi, but the answer is 4/pi. Did I set the equations up wrong, how about my derivative?

2. ## Re: Error In Problem Creation?

$\displaystyle r = 6$ , $\displaystyle h = 12$

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

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

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

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

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

try again