Finding the rate of rising water. Water is pouring into a conical tank at a rate of 8 cubic feet per minute.  If the height of the tank is 12ft, and the radius of its circular opening is 6ft, how fast is the water level rising when the water is 4ft deep? 
This is an in-class example that my professor used. The work below is how he solved the problem. 

Volume of water is $V=V(t)$
Depth of water is $h=h(t)$
The relationship between V and h is: 
$$V=\frac 13 \pi r^2h$$
$r=\frac rh$, and $\frac rh=\frac 6{12}=\frac 12$
So, $V=\frac13\pi(\frac h2)^2h=\pi h^3$   

How did he get the $\frac h2$?

$\frac {dv}{dt}=\frac {\pi}{12}3h^2(\frac {dh}{dt})= \frac {\pi}4h^2(\frac {dh}{dt})$
When $h=4$, $\frac {dv}{dt}=8$.
$8=\frac {\pi}4(4)^2(\frac {dh}{dt})$
$\frac {dh}{dt}=\frac {\pi}2$


Let's see if I understand this. I've tried a similar problem with different values.
Water is pouring into a conical tank at a rate of 8 cubic feet per minute.  If the height of the tank is 10ft, and the radius of its circular opening is 5ft, how fast is the water level rising when the water is 4ft deep?
Volume of water is $V=V(t)$
Depth of water is $h=h(t)$
The relationship between $V$ and $h$ is$$V=\frac 13\pi r^2 h$$
$h_0=$ the height of the cylinder $=10ft$, $r_0=$ the radius of the opening $=5ft$
$$\frac rh=\frac {r_0}{h_0}=\frac 5{10}=\frac 12$$
$$r=h(\frac 12)=\frac h2$$
so, $$V=\frac 13 \pi (\frac h2)^2h=\frac {\pi}{12}h^3$$
When $h=4ft$ and $\frac {dV}{dt}=8ft^3/min$,
$$\frac {dV}{dt}=\frac {\pi}{12}h^3(\frac {dh}{dt})$$
=$$8ft^3/min=\frac {\pi}{12}(4)^3(\frac {dh}{dt})$$
= $$8=\frac {\pi}{12}(64)(\frac {dh}{dt})$$
= $$\frac {dh}{dt}=\frac 8{\frac {64\pi}{12}}=\frac {96}{64\pi}$$
=$$\frac {dh}{dt}=\frac 3{2\pi}ft^3/min$$
 A: Hint:
You can calculate the volume of the water it takes to fill the conical tank up to a height of $h$. You will get some equation that tells you the function $V(h)$.
The function $V(h)$ will be invertible, so you will be able to calculate its inverse. The inverse tells you what the height if the water is when the tank is filled up to a volume of $V$, i.e. it will be a function $h(V)$.
Now, since you know that $V$ is a function of time, you know that $h$ is a function of time, i.e. $h = h(V(t))$. Using the chain rule, you can calculate the terivative of $V$ which is what you need.

I advise you to perform each of the three steps I listed above. If you are unsure of the solution, you can post it here (by editing your original question) and I will comment on it (tell you if it is correct or not).

I see that your professor wrote down that $V=\frac{1}{3}\pi\left(\frac h2\right)^2 h$ and that that is the part that is confusing you. The justification (that $r=\frac rh$) is completely wrong and due either to some mistake in writing (either your professors or yours). Here is how I would justify it:
If $r_0$ is the radius of the opening and $h_0$ is the height of the conic hole, let $h$ be the height of the water and $r$ be the radius of the water surface (which is a hole). Then the water volume is $$V=\frac13 \pi r^2 h.$$
Now, the ratio of $h$ to $r$ is equal to the ratio of $h_0$ to $r_0$ (similar triangles, if this is unclear, draw a picture). This means that $$\frac rh = \frac{r_0}{h_0} = \frac{6}{12} = \frac12$$
or, with some reorganization:
$$r = h\frac{1}{2} =  \frac{h}{2}$$
Which means that $$V=\frac{1}{3}\pi r^2 h = \frac13 \pi\left(\frac h2\right)^2 h$$
A: here is a much shorter way to figure out the answer. you know that the depth of the  water in the tank is twice the radius of the surface. there when the depth is $4 \, ft,$  radius is $2\, ft.$  the surface area at that instant is 
$ \pi 2^2  = 4\pi \, ft^2.$  the water is pouring in at the rate $8 \, ft^3 /min$ so in $dt \, min$ time it would have added $8\, dt \, ft^3$ this water must be spread over $4 \,ft^2$ to a depth $dh \,ft$  equating the two volumes we get $$4\pi dh =  8 dt \implies \frac{dh}{dt} = 2 \, ft / min.$$
