# Related Rates Equation

I feel like I understand what's going on, I would just like a walk-through of this problem to make sure I have grasped the concept and that I am completing the problem correctly.

A $50$-ft. by $60$-ft. swimming pool is being drained at the rate of $30$ ft$^3$/min. At what rate is the water level in the pool decreasing?

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Why don't you figure out how long it would take to drain away 1' of water? That is, compute the volume of water in the pool if it was just 1' deep, divide by the rate and figure out how many minutes that takes. Then express that in feet per minute. – copper.hat Nov 16 '12 at 6:26
My teacher wants us to use the related rates equations of the derivative with respect to time. Otherwise I would do it that way. :) – Courtney Nov 16 '12 at 6:29
Why don't you write the formula for the volume of water in the pool in terms of the area and depth? The area is constant, so the rate of change of volume and depth are related by this equation. You are given $\frac{d V}{dt}$, figure out $\frac{dh}{dt}$. – copper.hat Nov 16 '12 at 6:31

If the volume of water in the pool is given by:

$V=A\times h$ [ft$^{3}$]

Where $A$ is the area of the pool (constant, $A=50\times 60$ sq.ft) and $h$ is the depth of the water [ft].

You want to work out $\frac{dh}{dt}$, given that $\frac{dV}{dt}=-30$ ft$^{3}$/min.

Hint:

$\frac{dV}{dt}=A\frac{dh}{dt}$

(so $\frac{dh}{dt}=\frac{1}{A}\frac{dV}{dt} = \frac{-30}{300}=\frac{-1}{10}$ ft/min)

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