Calculus Problem [Calculus I] The sand is falling on a conical pile of sand at a rate of $12\,\text{m}^2 / \text{min},$ so that the diameter of the base of the pile is always $3/2$ of the height. Find the speed at which the height increases when the pile is $2$ meters. 
So, I don't even know where to start, I don't know what to do; the problem seems more of a physics problem but it is from a Final Calculus Exam. A hint would be nice.
I don't know if it is important or not but this is the second part of the exercise, while the first part was to find the volume of a cone using an integral. 
 A: Note: Your post says that sand is falling on a conical pile at $12\text{m}^2/\text{min}$, which doesn't make much sense–how can an area fall onto a pile? Perhaps the problem asked for $12\text{m}^3/\text{min}$. If that's not the case and the area of the base is increasing by $12\text{m}^2/\text{min}$, try and relate area and height as I do in this answer.
This is a related rates problem. First, write down everything you know:
$$d = \frac 32 h \quad V = \frac 13 \pi r^2h \quad \frac{\mathrm dV}{\mathrm dt} = 12\text{m}^3/\text{min}$$
The goal of the exercise is to find $\frac{\mathrm dh}{\mathrm dt}$. We can do this by relating $V$ and $h$. Realize that $12\text{m}^2/\text{min}$ represents the change in volume over time. Since we know the radius is one-half the diameter, and the diameter is $3/2$ the height, we can relate them as such, rewriting purely in terms of $h$:
$$d = \frac 32 h \implies r = \frac 34 h$$
$$V = \frac 13 \pi r^2h \implies V = \frac 13 \pi\left(\frac 34h\right)^2h \implies V = \frac 3{16} \pi h^3$$
Now differentiate and equate:
$${\mathrm dV\over \mathrm dt} = {\mathrm d\over \mathrm dt}\left(\frac 3{16} \pi h^3\right)\implies 12 = \frac 9{16}\pi h^2{\mathrm dh\over \mathrm dt}$$
Now isolate $\frac{\mathrm dh}{\mathrm dt}$ to solve. Since you are trying to find $\frac{\mathrm dh}{\mathrm dt}$ when $h = 2\text{m}$, plug in $2$ for $h$.
A: First, write down an equation that describes the relationship between variables in your problem. In this case, the variables can be reduced to volume and height of the cone.
$$V=\frac{1}{3}Ah=\frac{1}{3}\pi r^2h=(\frac{1}{3}\pi\frac{9}{16})h^3$$
Take derivative of the equation above with respect to $t$. Since you know $\frac{dV}{dt}=12$, you can plug that in to find $\frac{dh}{dt}$.
