1 Attachment(s)

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.

Re: Rate Problem Involving Changing Height..

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

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

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

Re: Rate Problem Involving Changing Height..

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

Re: Rate Problem Involving Changing Height..

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

Re: Rate Problem Involving Changing Height..

1 Attachment(s)

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 ...