A swimming pool is 40 feet long, 20 feet wide, 8 feet deep at the deep end, and 3 feet deep at the shallow end; the bottom is rectangular. If the pool is filled by pumping water into it at the rate of 40 cubic feet per minute, how fast is the water level rising when it is 3 feet deep at the deep end?

The error in Steve's answer is that v = 1/2 * y * 8y * 20. The height of the line of the pool bottom is determined by h = 1/8 *l, so the length of the cross section is 8h, not h/8. The correct answer is 1/12 ft^3/min.

1/12 ft^3/min is the selected answer in Varberg Early Transcendentals for this problem. Please check answers carefully.

Luckily, all we are working with is a triangular section, until the water is 5' deep; then we have to work with a trapezoid.

Anyway, looking at the side of the pool, when the water is y' deep, we have a triangular prism with a cross-section that is y by y/8 (since the pool bottom is a line with slope 5/40 = 1/8

So, the volume is
v = 1/2 * y*y/8 * 20 = 5y^2/4
dv/dt = 5y/2 dy/dt
40 = 5*3/2 dy/dt
dy/dt = 16/3 ft/min

To find how fast the water level is rising, we need to find the rate of change of the volume of the water with respect to time.

Let's first find the equation that relates the volume of the water to the depth of the water at any given point.

Since the bottom of the pool is rectangular, we can consider two separate regions: the shallow end region and the deep end region.

The volume, V, of the shallow end region can be calculated using the formula for the volume of a rectangular prism: V = length x width x height. In this case,
Vshallow = (20 feet x 8 feet x h), where h is the depth of the water at the shallow end.

The volume, V, of the deep end region can also be calculated using the formula for the volume of a rectangular prism: V = length x width x height. In this case,
Vdeep = [(40 feet - 20 feet)x (8 feet - 3 feet) x h], where h is the depth of the water at the deep end.
We subtract the shallow end dimensions from the overall pool dimensions to get the dimensions of the deep end region.

To find the total volume of the water in the pool, we need to sum up the volumes of the shallow and deep end regions:
Vtotal = Vshallow + Vdeep

Now, let's differentiate the total volume equation with respect to time (t):
dVtotal/dt = d(Vshallow)/dt + d(Vdeep)/dt

Since the water is being pumped into the pool at a constant rate of 40 cubic feet per minute, we have:
dVtotal/dt = 40

We are asked to find how fast the water level is rising (dh/dt) when the depth at the deep end (h) is 3 feet.

To find dh/dt, we need to differentiate both Vshallow and Vdeep with respect to t.

Differentiating Vshallow with respect to time, we get:
d(Vshallow)/dt = (20 feet x 8 feet x (dh/dt)) = 160(dh/dt)

Differentiating Vdeep with respect to time, we get:
d(Vdeep)/dt = [(40 feet - 20 feet) x (8 feet - 3 feet) x (dh/dt)] = 20 x 5(dh/dt) = 100(dh/dt)

Now, let's substitute these values into our earlier equation:
40 = 160(dh/dt) + 100(dh/dt)

Combining like terms, we have:
40 = 260(dh/dt)

Finally, solving for dh/dt, we get:
dh/dt = 40/260 = 2/13 feet per minute

Therefore, the water level is rising at a rate of 2/13 feet per minute when the depth at the deep end is 3 feet.

IDK any math ahahahahah but I like cheese!

The error in Emory's answer is that the units should be ft/min since it is asking how fast the depth of the water is increasing. But otherwise the 1/12 is correct!