# Related Rates

• Jun 17th 2006, 08:23 AM
ron007
Related Rates
A swimming pool is 25 ft wide, 60 ft long, 3 ft deep at the shallow end, and 15
ft at its deepest point. If the pool is being filled at a rate of 800 ft^3/min,
at what rate is the water level rising when the depth at the deepest point is
5ft.

I'm stuck in this exercise. I just need a tip for the exercise. Thanks for you
help.

I did this:
Let x feet be the wide, and y the long, t for time measured in minutes.

dx/dt= 25 ft and I need to find dy/dt when y = 60ft.

I know that I have to write an equation that relates the various quantities of
the problem. But I don't know what equation I should put. :confused:
• Jun 17th 2006, 09:34 AM
Soroban
Hello, Ron!

Quote:

A swimming pool is 25 ft wide, 60 ft long, 3 ft deep at the shallow end, and 15 ft at its deepest point.
If the pool is being filled at a rate of 800 ft³/min, at what rate is the water level rising
when the depth at the deepest point is 5ft?
Side view of the pool . . . Ignore the top 3 feet.
Code:

        : - - - -  60 - - - - - :   -    *-----------------------*   :    |      x          *   :  - +---------------*   12  : |:::::::::::*   :  y |:::::::*   :  : |:::*   -  - *
The area of the water is: .$\displaystyle \,A\,=\,\frac{1}{2}xy$
From similar right triangles, we have: .$\displaystyle \,\frac{x}{y}\,=\,\frac{60}{12}\;\;\Rightarrow\;\; x\,=\,5y$
The area of the water is: .$\displaystyle \,A\:=\;\frac{1}{2}(5y)(y)\:=\:\frac{5}{2}y^2$
The volume of the water is: . $\displaystyle V\:=\:\frac{5}{2}y^2 \times 25\:=\:\frac{125}{2}y^2$

Differentiate with respect to time: . $\displaystyle \frac{dV}{dt}\:=\:125y\cdot\frac{dy}{dt}$

We are given: $\displaystyle \frac{dV}{dt} = 800$ ft³/min and $\displaystyle y = 5$

So we have: .$\displaystyle 800 \:=\:125(5)\cdot\frac{dy}{dt}$

Therefore: .$\displaystyle \frac{dy}{dt}\:=\:\frac{800}{625}\:=\:1.28$ ft/min.
• Jun 17th 2006, 09:37 AM
ron007
I have that idea but I was confussed. But THANKS A LOT!!!