A cup has a radius of 2" at the bottom and 6" on the top. It is 10" high. 4 Minutes ago, water started pouring at 10 cubic " per minute. How fast was the water level rising 4 minutes ago? How fast is the water level rising now? What will the rate be when the glass is full?

Fair warning, I've been told my methods are unconventional.

Okay first we have to find the volume of your cup any any given level. (not a point in time, we're not there yet.)
The radius is R =(2/5) y + 2 . (just simple algebra on the dimensions of the cup)
So the volume at any given height h is given by V(h)= \pi \int_0^h (\frac{2}{5}y + 2)^2 dy (By the washer method.)
Now the easy thing to do would be to expand the polynomial and take the integral from there, but what we want is the change in height at any given time. So simply solving the integral will give us more work later on, and like I tell my students, I'm lazy. What we're going to do is solve by substitution. I hope the reasons will become obvious. So we have:

V(h)= \pi \int_0^h (\frac{2}{5}y + 2)^2 dy
Let u= \frac{2}{5}y+2 and du=\frac{2}{5}
V(h)= \frac{5}{2}\pi \int_2^{\frac{2}{5}h+2} u^2 du
V(h)= \frac{5}{6}\pi u^3|_2^{\frac{2}{5}h+2}
V(h)= \frac{5 \pi }{6} ((\frac{2}{5}h+2)^3-8) .

Now we have the volume for our height. We also know that the volume for a time t [\math] is given by [math] 10 t [\math]. So we can say:
[math] \frac{5 \pi }{6} ((\frac{2}{5}h+2)^3-8) = 10 t for 0 \leq h \leq 10 [\math] and [math] t \geq 0 [\math].

Now there are some important things to keep in mind. First of all I'm defining [math] t=0 [\math] as when we start filling the cup. So "now" is defined as [math] t=4 [\math]. Secondly, weird things happen to this when we get [math] h <0 [\math], so solving this for all real numbers for [math] h [\math] would be hellacious, but we don't care because all we care about is when we are actually filling the cup. So now we solve for [math] h [\math], and we get:

[math] \frac{5 \pi }{6} ((\frac{2}{5}h+2)^3-8) = 10 t
(\frac{2}{5}h+2)^3 = \frac{12}{\pi} t+8
\frac{2}{5}h+2 = \sqrt[3]{\frac{12}{\pi} t+8}
h = \frac{5}{2}(\sqrt[3]{\frac{12}{\pi} t+8}-2)

So now we have the height as a function of time. But we're still not done, the question was what is the change in height, and that's a derivative. So we still need to find the derivative of our function, so:

h(t) = \frac{5}{2}(\frac{12}{\pi} t+8)^{1/3}-2)
h^{\prime}(t) = \frac{5}{2} \frac{1}{3}(\frac{12}{\pi} t+8)^{-2/3}(\frac{12}{\pi})
h^{\prime}(t) = \frac{10}{\pi} (\frac{12}{\pi} t+8)^{-2/3}

So that will give you the change in volume for any time t \geq 0 [\math]. All you need to do is take the time [math] t [\math] that you want and "plug and chug". To get the height when the cup is full take our formula
[math] \frac{5 \pi }{6} ((\frac{2}{5}h+2)^3-8) = 10 t from way back when and just set [math] h=10 [\math] and solve for [math] t [\math]. I'll leave the actual computations up to you, like I said, I'm lazy.

I hope this helps :) I hope all my latex comes out okay.

Damn it, I mixed up some of the forward slashes and back slashes. Sorry. It would be real nice if a mod could change it for me. It would be real nice. If any mod is listening, I sure hope he/she could find it in there heart to change my previous post and change all the "\" in [\math] to a "/". Any moderator at all. ;)