# Related rates fill rate problem

1. Nov 24, 2011

### kentarch

1. The problem statement, all variables and given/known data

A swimming pool is 50m long and 20m 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 90 minutes after the filling began?

3. The attempt at a solution

I'm really lost on this problem, so my attempt at a solution isn't much of one. I assume that I will be using two separate volume formulas: one for the area with the sloped bottom, and one for the rectangle above that. Beyond that I'm lost.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 25, 2011

### HallsofIvy

Staff Emeritus
Well, it is not so much two different formulas for different areas as it is for different times. Look at the pool from the side. It is a trapezoid with bases, at the ends, of length 1 and 3 and height, at the top of length 50. Its volume will be the area of that trapezoid times the width, 20.

But until the water has covered the bottom, its figure, on the side of the pool will be a right triangle with height x, the height of the water, and base determined by where the line giving the bottom crosses that height. Taking the origin of a coordinate system at the lowest point in the pool, positive x-axis up, The bottom of the pool is the line from (0, 0) to (2, 50). That has equation y= 25x. So the area of the side covered by water will be $(1/2)(25x)(x)= (25/2)x^2$. The volume will be $(25/2)x^2(20)= 250x^2$. That will be true for x, the height of the water, from 0 to 2. After that, it will be the total volume of the water, $250(4)= 1000$, plus the volume of the rectangular solid above it, (20)(50)(x- 2).