How fast the height of the cone is changing? Suppose that sand is collecting in the shape of a cone in such a way that the base radius of the cone is always one-third of its height. If $3\,\mathrm{cm^3/min}$ is the rate at which the sand is being added to the cone, how fast is the height of the cone changing when it is $7\,\mathrm{cm}$ tall? 
I got $Dv/dt=3cm^3 $ and $r=1/3h$. So I'm looking for $dh/dt$ when $h=7$. I'm not sure if I'm doing the problem right.
 A: As Wolfgang Kais states in his answer, the formula for the volume of a cone is
$$V = \pi r^2 \cfrac{h}{3} \tag{1}\label{eq1}$$
As the question, and OP, states, the radius is always one-third the height, so $r = \frac{h}{3}$. Thus, \eqref{eq1} becomes
$$V = \pi \cfrac{h^3}{27} \tag{2}\label{eq2}$$
Differentiating with respect to $t$ gives
$$\cfrac{dV}{dt} = \pi \cfrac{h^2}{9} \cfrac{dh}{dt} \tag{3}\label{eq3}$$
This is what the OP has already determined, and states in a comment. Now, it's given that $\frac{dV}{dt} = 3\text{ cm}^3$/min and it asks for $\frac{dh}{dt}$ when $h = 7$ cm. Plugging these values into \eqref{eq3} gives
$$3 = \pi \cfrac{49}{9} \cfrac{dh}{dt} \Rightarrow \cfrac{dh}{dt} = \cfrac{27}{49\pi}  \tag{4}\label{eq4}$$
Thus, the height at $7$ cm is increasing at a rate of $\frac{27}{49\pi}$ cm/min.
A: If you were a physicist, you would express $h$ as a function of $V$, expand the fraction $\frac{dh}{dt}$ by multiplying it with $1=\frac{dV}{dV}$, find that $\frac{dh}{dt}=\frac{dh}{dt}\frac{dV}{dV}=\frac{dh}{dV}\frac{dV}{dt}$ (the "chain rule") and only need the deviation $\frac{dh}{dV}$ (because you already have $\frac{dV}{dt}$).
Hint: $V=\pi r^2\cdot\frac13h=\pi\left(\frac13h\right)^3$, so $h = 3\left(\frac{V}{\pi}\right)^{1/3}$, and if $h=7$ then $V=\pi\left(\frac73\right)^3$.
Added:
As seen in the comments, it seems to be unclear how to continue with these hints, so here's how to move on from there:
We have: $h=3\left(\frac{V}{\pi}\right)^{1/3}$ and thus $\frac{dh}{dV}=\frac{1}{\pi}\left(\frac{V}{\pi}\right)^{-2/3}$
Also, we know from the prerequisites: $\frac{dV}{dt} = 3\,\mathrm{cm^3/min}$.
As deduced, before, $\frac{dh}{dt}=\frac{dh}{dV}\frac{dV}{dt}$ of which we now know both factors and can calculate:
$$
\begin{align}
\frac{dh}{dt} & =\frac{dh}{dV}\cdot\frac{dV}{dt}=\frac{1}{\pi}\left(\frac{V}{\pi}\right)^{-2/3}\cdot\frac{dV}{dt}=\frac{1}{\pi}\left(\left(\frac{h}{3}\right)^3\right)^{-2/3}\cdot\frac{dV}{dt}=\frac{9}{\pi h^2}\cdot3\,\mathrm{\frac{cm^3}{min}}\\
& =\frac{27}{\pi h^2}\,\mathrm{\frac{cm^3}{min}}
\end{align}
$$
Now, we can evaluate this for the given height $h=7\,\mathrm{cm}$ and finally get an increase rate of
$$
\frac{27}{\pi h^2}\,\mathrm{\frac{cm^3}{min}}=\frac{27}{\pi\cdot 49\,\mathrm{cm^2}}\,\mathrm{\frac{cm^3}{min}}=\frac{27}{49\pi}\,\mathrm{\frac{cm}{min}}
$$
