Related Rate Problem Involving Conical tank Formula. I do not understand why this is incorrect.

Question

A right conical tank with the point oriented down, a height of 15 feet, and a radius of 4 feet has sprung a leak. How fast is the volume of water in the tank changing when the water is 2 feet high and the water level changes at a rate of 14 inches per minute?

My result was

896π/225

This is apparently incorrect.

The following equation I used was

$$ \frac{dV}{dt}=\frac{1}{3}pi(\frac{4h}{15})^2(\frac{dH}{dt})$$

Answer 1

From

$$\begin{align*} \frac rh &= \frac{4}{15}\\ V &= \frac13 \pi r^2h\\ &= \frac13\pi \left(\frac{4h}{15}\right)^2h \end{align*}$$

Differentiating $V$ with respect to $t$ gives

$$\begin{align*} \frac{dV}{dt} &= \frac13\pi \left[\left(\frac{4h}{15}\right)^2\cdot\frac{dh}{dt} + h\cdot 2\left(\frac{4h}{15}\right)\cdot\frac{4}{15}\frac{dh}{dt}\right]\\ &= \frac13 \pi\cdot 3\left(\frac{4h}{15}\right)^2\cdot\frac{dh}{dt}\\ &= \pi\left(\frac{4h}{15}\right)^2\frac{dh}{dt}\\ \end{align*}$$

(Using product rule and chain rule; same result if you first simplify $V$ as a multiple of $h^3$.)

Substituting $h=2\text{ ft}$ and $\frac{dh}{dt} = -14 \text{ in min$^{-1}$}$ gives:

$$\begin{align*} \frac{dV}{dt} &= \pi \left(\frac{4\cdot2}{15}\text{ ft}\right)^2 \left(-14 \text{ in min$^{-1}$}\right)\\ &= -\frac{896\pi}{225}\text{ ft$^{2}$ in min$^{-1}$} \end{align*}$$

So I had to point out, what unit did you intend to use? Some may consider the absolute rate of change as $\frac{896\pi}{225}$ correct, if you had included its corresponding unit.

Answer 2

HINTS

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

Differentiate with respect to $t$

$$ \frac{r}{h}=\frac{\dot r}{\dot h }=\frac{4}{15}$$

$$ \dot V= \frac{\pi}{3}(2 r \dot r h + r^2\dot h) $$

Eliminate $r$, plug in $\dot r$ .