Rate at which water is draining from the tank?

Quote:

Originally Posted by A5HLEY

Hi all! I'm having trouble with some Calc homework, and I'm hoping you can help me. Here goes:

V= 4600 * [1-(t/38)]^2 , 0 [< or equal to] t [< or equal to] 38

Find the rate at which the water is draining from the tank after:

a). 4 minutes
b). 9 minutes
c). 17 minutes

I think you just take a derivative and the plug in for t, but I'm having a hard time with finding the derivative.

$V(t)=4600(1-\frac{t}{38})^2$

Taking the derivative w.r.t time

$V'(t)=9200(1-\frac{t}{38})\frac{-1}{38}$

Plug and chug.. the rate of change is negative hence the volume is decreasing