Related Rates

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June 17th 2006, 08:23 AM
ron007

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

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

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

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

Therefore: . $\frac{dy}{dt}\:=\:\frac{800}{625}\:=\:1.28$ ft/min.
June 17th 2006, 09:37 AM
ron007

I have that idea but I was confussed. But THANKS A LOT!!!