Question

Please can someone help me with this: A swimming pool is 50m long and 20 m wide. Its depth decreases linearly along the length from 3m to 1m. It is initially empty and is filled at a rate of 1m^3/min. How fast is the water level rising 250 min after the filling begins? How long will it take to fill the pool?

Answer 1

Well I know the first question can be answered simply by finding the volume of the pool seeing as it fills 1 cubic meter per minute.
The second one, you need to find out the slope at 250 minutes of the formula that tells how many cubic meters are going into the pool per minute.
I hope this helps.

Answer 2

You can calculate the volume by this way, I think: volume of a pyramid with the basis 2 m x 20 m and height = 50 m plus the volume of a cube with height = 1 m, width = 20 m and 50 m long.

Answer 3

Creeksider is late in finding this question and perhaps this answer is too late to be of help, but the question is interesting so here's an outline of how to solve it.

This is a problem in related rates. In these problems you need to find the rate at which a changes with respect to c, using the rate a changes with respect to b and the rate b changes with respect to c:\[\frac{ da }{ dc }=\frac{ da }{ db }\frac{ db }{ dc }\]In this case we need dh/dt, the change in the height of the water with respect to time. We're given dv/dt, the change in volume with respect to time, so our task is to find dh/dv and multiply by dv/dt. As it happens, dv/dt is 1, so we have our answer when we find dh/dv.

You always want to begin a problem like this by drawing a diagram. In this case the diagram you need is a cross section of the pool. It's 3m deep at one end and 1m deep at the other, with a bottom that slants upward 2m over a distance of 50m, for a slope of 1/25. The resulting figure is a trapezoid (called a trapezium outside the U.S.), but for our purposes it's an inverted triangle with a rectangle on top.

We need a formula that relates v to h. Volume will be the area of this cross section multiplied by 20, the width of the pool. Until h reaches 2, we are using the area formula for a triangle, which is 1/2 times h times b, the base of the triangle. We're looking at this triangle upside down, so the base is at the top. We need to avoid introducing a separate variable for the base, but that's easy because we know the slope of the line representing the bottom of the pool. For any given h in the range from 0 to 2, b will be 25h. So the formula for v is 20 times one-half times h times 25h, or 250h^2.

From here you can proceed in two directions. The most obvious is to solve for h and then differentiate, applying the resulting formula to the existing volume of 250. You have to confirm that the existing volume produces a height in the range from 0 to 2, where the shape of the pool is triangular. The other approach is to use implicit differentiation directly on the equation v = 250h^2. This will give you dv/dh in terms of h, so you need to figure out what h is when v = 250, but that's pretty easy. Either way, you'll want to check to see that your answer makes sense: think about how large the surface area of the pool is and how big a difference in height you're likely to get by adding one cubic meter of water.

The last part is simple and does not require calculus. Find the area of the triangle, add the area of the rectangle on top of it, and multiply by the width of the pool.