Originally Posted by

**Lambin** The first problem:

**Making coffee. **Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of $\displaystyle 10\frac{in^3}{min}$.

**a. **How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?

**b. **How fast is the level in the cone falling then?

Some facts given from the figure:

(1) The conical filter has a diameter of 6 inches at the top, and a depth of 6 inches also. It appears to be in the shape of a right circular cone.

(2) The coffeepot has a diameter 6 inches and in the shape of a right cylinder.

**Attempt:**

To address question (a), the level of in the pot rising would refer to $\displaystyle \frac{dh}{dt}$. I began with the equation that relates the variables V, h, and r and treat them as differentiable functions of t with the exception of r; the radius is constant for a right cylinder.

Therefore,

$\displaystyle V={\pi}r^2h$, the volume of a right cylinder. (1)

$\displaystyle \frac{dV}{dt}={\pi}r^2\frac{dh}{dt}$ (2)

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{{\pi}r^2}$, (3)

Now, I make an assertion here by recognizing that volume is conserved between the conical filter and the coffeepot; for, any volume loss from the filter should be transferred to the coffeepot and no where else. This is neglecting any effects of absorption of the coffee onto the filter paper, which I believe was a reasonable assumption in this problem. However, the problem poses a condition "when the coffee in the cone is 5" deep."

From here, I think that $\displaystyle \frac{dV}{dt}$ is dependent on the depth of the coffee in the cone which is sensible considering Torricelli's law, but I'm not sure where to go from here. I arrived at the equation,

$\displaystyle \frac{dV}{dt}=\frac{\pi}{3}[r^2\frac{dh}{dt}+2rh\frac{dr}{dt}] $ which I think I must solve for $\displaystyle \frac{dV}{dt}$ and substitute the value into equation (3). I must have gone awry somewhere.

Help?