# Thread: conical rate of change question

1. ## conical rate of change question

My teacher is giving us AP practice tests to get in some extra practice before the AP test in May, but I still haven't gtten used to rate of change functions.

"Water is draining at the rate of 48pi ft^3/second from the vertex at the bottom of a conical tank whose diameter at its base is 40 feet and whose height is 60 feet."

a) Find an expression for the volume of water in the tank in terms of its radius at the surface of the water.

I've gotten from V=1/3*pi*r^2*h to dV/dt=pi/3*(r^2*dh/dt+2rh*dr/dt) by taking the drivative, but I get stuck here. what do I need to do to finish finding the expression?

b) At what rate is the radius of the water in the tank shrinking when the radius is 16 feet?
c) How fast is the height of the water in the tank dropping at the instant that the radius is 16 feet?

If someone could give me a formula for these 2 that would be great.

Any help offered is greatly appreciated. Thanks in advance.

2. Originally Posted by Jziffra
My teacher is giving us AP practice tests to get in some extra practice before the AP test in May, but I still haven't gtten used to rate of change functions.

"Water is draining at the rate of 48pi ft^3/second from the vertex at the bottom of a conical tank whose diameter at its base is 40 feet and whose height is 60 feet."

a) Find an expression for the volume of water in the tank in terms of its radius at the surface of the water.
$\displaystyle \frac{r}{h} = \frac{\textcolor{red}{20}}{\textcolor{blue}{60}} = \frac{1}{3}$

$\displaystyle 3r = h$

$\displaystyle V = \frac{\pi}{3} r^2 h = \frac{\pi}{3} r^2 (3r) = \pi r^3$