FIlling a pool – Related rates


A swimming pool is

25 meters long
10 meters wide
1 meter deep in one end
6 meters deep in the other end

The bottom of the pool is linearly slanted.

The pool is filled with water at a rate of 2000 liters/minute.

How fast does the water level rise at the point in time when the depth of the water is 3 meters?

My work

I have drawn a figure, as I often do.

Since we’re dealing with a volume of water, I thought I might start with expressing the volume of the water as a function of the distances of the body of water.

Since the water will be the area of the right triangle, times the width of the pool, I have V=12ghwV = \frac12 g h w where gg is the horizontal edge of the triangle, hh is the current depth of the water, and ww is the width of the pool.

Knowing that h=3h = 3 and w=10w = 10 gives me V=15gV = 15g.

At this point, I’m stuck, and would appreciate any help!



2 Answers


There are 10001000 liters in a cubic meter, so the fill rate is 22 m3^3/min. The slope of the bottom of the pool is 0.20.2 (or −0.2-0.2, depending on your point of view). So when the water is 33 m deep at the deep end, the horizontal water surface is 1515 m long. Since the pool is 1010 m wide, the surface area at that point is 150150 m.

Instantaneously, then, since you are only interested in first-order changes in water level, you can just pretend the pool has a surface area of 150150 m. How fast does the water level increase if you are pouring water in at 22 m3^3/min?

Find an expression for the Volume in terms of the height. Using similar triangles, you can show that the area of the triangle when the height reaches hh, the length is 5h5h so that the volume is V(h)=50h2V(h) = 50 h^2. It should now be easy for you to calculate dVdh\frac{dV}{dh} and find the value you desire.