# Rate Problem Involving Changing Height..

• Aug 26th 2012, 10:14 AM
kspkido
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.
Attachment 24604I 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.
• Aug 26th 2012, 11:47 AM
skeeter
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 $\frac{dh}{dt}$
• Aug 26th 2012, 05:32 PM
kspkido
Re: Rate Problem Involving Changing Height..
hmmm may I ask how did you derive that equation? thank you
• Aug 26th 2012, 06:38 PM
HallsofIvy
Re: Rate Problem Involving Changing Height..
Do you know how to find the area of a trapezoid?
• Aug 26th 2012, 07:16 PM
kspkido
Re: Rate Problem Involving Changing Height..
A=(b1+b2)h/2 yep....
• Aug 27th 2012, 10:19 AM
skeeter
Re: Rate Problem Involving Changing Height..
Quote:

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

note the sketch ...