# Help with Related Rates problem

I'm slowly working my way through a ton of these problems but have come across one that has me stumped.

Here's the problem in full:

Water is leaking out of an inverted conical tank at a rate of $0.0081$${\frac {m^3}{min}}. At the same time water is being pumped into the tank at a constant rate. The tank has height 8 meters and the diameter at the top is 3 meters. If the water level is rising at a rate of 0.23$${\frac {m}{min}}$ when the height of the water is 2 meters, find the rate at which water is being pumped into the tank.

So what I've done so far:

Volume of the water $V=\pi r^2\frac {H}{3}$

Height of the water $H = 2$

Radius of the tank at the $r = 0.375$ (via similar triangles when H = 2)

$C$ is the rate at which water is being pumped in and is the rate I am trying to find.

$0.0081$ is the rate at which it is leaking out.

$\frac {dV}{dt} = C - 0.0081$

$C = \frac {dV}{dt} + 0.0081$

So what I need to do is find is $\frac {dV}{dt}$ when H = 2 and then use that to find C.

I set up the problem appropriately, and used implicit differentiating to get:

(using the product an chain rules)

$\frac {dV}{dt} = (2\pi r\frac {dr}{dt})*(\frac {H}{3})+\frac {\pi r^2}{3}\frac {dH}{dt}$

This process is the same as all the other RR problems I've done, and at this point I'd simply sub in the known values and solve for the rate I'm looking for. The issue here is I don't know what $\frac {dr}{dt}$ is. $H=2$, $r=0.375$, and $\frac {dH}{dt} = 0.23$.

Please let me know what I'm missing/what I need to do. Also if I made any mistakes along the way. Thanks in advanced.

We can use similar triangles to get rid of a variable. Since we're given $\frac{dH}{dt}$, we want to eliminate $r$ by expressing it as a function of $H$. To this end, notice that at any point in time, we have that: $$\frac{r}{H} = \frac{3/2}{8} \iff r = \frac{3}{16}H$$ Substituting before differentiating, we obtain: \begin{align*} V &= \frac{\pi}{3}\left(\frac{3}{16}H\right)^2H =\frac{3\pi}{256}H^3 \\ \frac{dV}{dt} &= \frac{9\pi}{256}H^2 \cdot \frac{dH}{dt} \\ C &= 0.0081 + \frac{9\pi}{256}H^2 \cdot \frac{dH}{dt} \\ \end{align*} Subsituting the known values at the given snapshot in time, we obtain: $$C = 0.0081 + \frac{9\pi}{256}(2)^2 \cdot (0.23) = 0.109711\ldots \frac{\text m^3}{\text {min}}$$