Solving for Rate of Change of Depth in Swimming Pool

[stripes]

Homework Statement

A swimming pool is 5 m wide, 10 m long, 1 m deep at the shallow end, and 3 m deep at its deepest point. A cross-section is shown in the figure. If the poole is being filled at a rate of 0.1 m^3 per minute, how fast is the water level rising when the depth at the deepest point is 1 m?

Homework Equations

None really...but attached is a picture. It is not drawn to scale and all values are given in meters. The blue water given is not given to scale, and is merely used as an aid.

The Attempt at a Solution

Okay, so we know dV/dt = .1 m^3/min, we want dx/dt (let x = water level in the pool (height)), and we need to find some sort of relationship between x and V, and subsequently a relationship between dV/dt and dx/dt.

After much thought, I figured that the volume of the pool will be equal to 50x - 8.5, and i figured that out from:

v = height x length x width minus the small blocks of unused space in the pool

however, that's wrong because when the water level is below 2 m, we're not subtracting some of those blocks...

so i figured that the blocks form triangles, and i could find the rate of change of those triangles' lengths with respect to the height in the pool, x, but i determined that these blocks or triangles aren't remaining as triangles when the pool water level is below 2 meters...they form quadrilaterals...

so I'm stuck, i can't seem to find a relationship between the height of the pool and the volume of the pool...once i find that, it's easy, differentiate implicitly with respect to time, plug in your known values, and then solve for dV/dt...but again, my problem here is finding a relationship between the two.

Any help is greatly appreciated, thanks so much in advance.

 

[lippyka]

The volume of the pool is given by V = 50x - 8.5(2-x)^2, where x is the height of the pool. To find the rate of change of the depth, we need to differentiate with respect to time. So, dV/dt = 50dx/dt - (8.5)(2-x)d(2-x)/dtIf we assume that the water is filling the pool at a constant rate of 0.1 m^3 per minute, then dV/dt = 0.1. We can substitute this into our equation and solve for dx/dt:0.1 = 50dx/dt - (17)(2-x)dx/dtdx/dt = 0.1/(50 - 17(2-x))When the deepest point of the pool is 1 m, we have x = 2. Substituting this into our equation gives:dx/dt = 0.1/(50 - 17) = 0.0024 m/minTherefore, when the deepest point of the pool is 1 m, the water level is rising at a rate of 0.0024 m/min.