# Thread: Rate Problem Involving Changing Height..

1. ## Rate Problem Involving Changing Height..

Hello So first this is the problem:

A Swimming pool is 20 ft. wide, 40 ft. long, 3 ft. deep at the shallow end and 9 ft. deep at it's deepest point. A cross section is shown in the figure. If the pool is being filled at a rate of 0.8 cu ft. per min., how fast is the water level rising when the depth at the deepest end is 5ft.
I understand that dV/dT= .8cu ft/min but what I don't understand is how to set the problem up.. V=hlw where length and width are constant but the height is changing.. 9 is also the critical point since it's the highest.. but how do we set this problem up. with the height changing? dh/dt?
Thank you Very Much for your help... This is my first post.. I love browsing in this forum. cheers.

2. ## Re: Rate Problem Involving Changing Height..

when $h < 6 \, ft$ ...

$V = 20(h^2 + 12h)$

you are given $\frac{dV}{dt}$ and $h$ ... take the time derivative of the volume equation and solve for $\frac{dh}{dt}$

3. ## Re: Rate Problem Involving Changing Height..

hmmm may I ask how did you derive that equation? thank you

4. ## Re: Rate Problem Involving Changing Height..

Do you know how to find the area of a trapezoid?

5. ## Re: Rate Problem Involving Changing Height..

A=(b1+b2)h/2 yep....

6. ## Re: Rate Problem Involving Changing Height..

Originally Posted by kspkido
hmmm may I ask how did you derive that equation? thank you
note the sketch ...