Subject: Differentiation
Name: Jon
Who are you: Student

A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deeps at the deep end. Water is being pumped into the pool at 1/4 cubic meters per minute, an there is 1 meter of water at the deep end.

a) what percent of the pool is filled?

b) at what rate is the water level rising?

Hi Jon.

I'll show you how this is solved but I'll change it a bit to avoid giving you the answer outright.

Let's say the swimming pool is square, 6 m on a side with a sloped bottom from 3 meters at the deep end to 1 meters at the shallow end. Water is being pumped in at 1/2 cubic meters per minute. What rate is the water level rising when the water level is 1 meter from the bottom of the deep end?

The first part of the solution is to find an expression for the volume of water based on the depth. We begin by recognizing that the sloped part is the only part that matters when the water level (which we'll label D and measure from the bottom of the deep end upwards positive) is in the 1 meter deep neighbourhood. So we can simplify our
visualization. The cross-section of the volume is a right triangle D meters high by 3D meters wide. This is because the depth of the slope is 3-1=2 meters and the width of the slope is 6 meters, so the ratio between height and width is 2:6 or 1:3, thus the triangle is D by 3D. The area of the triangle is 3D²/2. Multiply this by the length of the pool (6 meters) and you get 9D² as the volume of the water based on the depth (at least this is true at the water level, the value of D, we are interested in - 1 m).

The change in volume over time is the derivative of this expression with respect to time. Let's write this out:

V = 9D²
(d/dt)V = (d/dt)(9D²)

By using this (Leibnitz) notation, it is clear that you are taking the derivative of 9D², but with respect to t, not D. But the depth D does depend on time t of course, so this is exactly the way to think of the chain rule.

So we have the situation you've seen in textbooks where

df/dc = (df/db)(dg/dc) where a = f(b) and b = g(c)

In this case then,

dV/dt = d/dD(9D²)(dD/dt)

The derivative of 9D² (with respect to D now) is just 18D, so
dV/dt = 18D (dD/dt)

We're almost done. dD/dt is the rate of change of depth with respect to time (exactly what we're asked to find) and the value dV/dt is the rate of change of volume with respect to time, and we're told in the question what that is: 1/2 cubic meter per minute, so we can solve for dD/dt:

1/2 = 18D (dD/dt)
1/(36D) = dD/dt

The last step is to plug in the value of D that we are measuring for:
1 meter deep. So dD/dt = 1/36 meters/minute. That's the answer.

Now you try following this method for your own answer.
Stephen La Rocque.