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## Homework Statement

Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area [tex]400 \pi[/tex] square feet. The depth,

*h*, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

A) Write an expression for the volume of the water in the conincal tank as a function of

*h*.

B) At what rate is the volume of the water in the conical tank changing when h=3?

C) Let

*y*be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?

## Homework Equations

[tex]V = \frac{1}{3} \pi r^2 h[/tex]

## The Attempt at a Solution

First I took the derivative of the volume function using the product rule:

[tex]\frac{dv}{dt} = \frac{2}{3} \ast \pi \ast r \ast \frac{dr}{dt} \ast h + \frac{1}{3} \ast \pi \ast r^2 \ast \frac{dh}{dt}[/tex]

I know that dh/dt = (h-12), but I'm not sure how to find dr/dt and I'm also not sure if I use 4 as the radius (b/c the radius of the water in the cone is constantly changing.) I'm guessing there is another relevant equation I'm missing, or something obvious. I know that the ratio of height of the cone to radius is constant, but I'm not sure if that helps or not.

If anyone could help me figure these out, I'm pretty sure I can do B) and C) on my own. Thanks in advance!

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