Find the ratio between the radius and the height 
A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cm$^3$/min. When the depth of water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to the radius?

So,
${dV \over dt} = -10$
$h=8$
${dh \over dt}= -2$
Now,
$${V}= {1 \over 3} \pi r^2h$$
However, it seems like I am not given enough information.
What do I do?

I tried:
$${dV \over dh} = {2\over 3} \pi r {dr \over dh}$$
However, it feels like I am not given enough information since I don't have ${dr \over dh}$
 A: You made an error in your calculation of a derivative:
$\frac23 \pi r \frac{dr}{dh} \neq \frac{dV}{dh}.$
In fact, 
$$\frac23 \pi r \frac{dr}{dh} = \frac{d}{dh}\left(\frac13 \pi r^2\right).$$
So you got the derivative of $\frac13 \pi r^2$ instead of $\frac13 \pi r^2 h.$
But I think there's an easier way to work on this problem without computing that derivative in this way at all.

Think about how you would find the volume of the cone using the disk method.
You slice the cone into thin horizontal disks.
If the disk at height $y$ above the bottom point of the inverted cone has area $A(y),$ the volume of a cone of height $h$ is
$$  V(h) = \int_0^h A(y) \,dy. $$
By the second fundamental theorem of calculus, therefore,
$$ \frac{dV}{dh} = A(h). $$
Note that all of this is true for any three-dimensional shape that you can fill with water.
If a three-dimensional shape grows or shrinks by adding or removing material on a flat top surface (like the water in a vessel),
the rate at  which the volume changes (with respect to height) is exactly equal to the area of that top surface.
As others have shown, you have enough information in the equations you already wrote to compute the value of $\frac{dV}{dh}$ at the instant when $h = 8.$
So you know $A(8),$ the area of the circular "base" (actually top) of the inverted cone when its height is $8.$ Knowing the area, you can find the radius. Knowing the radius and the height, you can find the ratio of one to the other.
A: The following is a basic diagram of a vertical cross-section of an inverted cone:

The volume is in the lower part of the cone, with it being
$$V = \frac{1}{3}\pi r^2 h \tag{1}\label{eq1}$$
Also, you are asked to find the ratio of the cone's total height to it's radius, i.e.,
$$\frac{h_0}{r_0} = c \tag{2}\label{eq2}$$
where $c$ is a constant, and the value to be found.
Next, note that $r$ and $h$ are not independent of each other. By similar triangles, and using the reciprocal of \eqref{eq2}, we have that
$$\frac{r}{h} = \frac{r_0}{h_0} = \frac{1}{c} \; \Rightarrow r = \frac{h}{c} \tag{3}\label{eq3}$$
Thus, substituting this into \eqref{eq1} gives
$$V = \frac{1}{3}\pi\left(\frac{h}{c}\right)^2h = \frac{\pi}{3c^2}h^3 \tag{4}\label{eq4}$$
Thus, differentiating wrt to time gives
$$\frac{dV}{dt} = \frac{\pi}{c^2}h^2 \frac{dh}{dt} \tag{5}\label{eq5}$$
Using the known values of $\frac{dV}{dt} = -10$ always, and $\frac{dh}{dt} = -2$ at $h = 8$, in \eqref{eq5} will allow you to solve for $c$. I take it you can finish the rest yourself.
