A trough (a place where horses and sheep drink water) has the shape of a half-cylinder with length 4 m and radius 20 cm. Water is poured into the trough at a constant rate of 300 cm3/s. At what rate (in cm/s) is the level of the water in the trough rising at the moment the level of the water is 10 cm below the upper surface of the trough?

Question

Steve · Answer

when the water is y down from the top, the surface subtends an angle θ, so the cross-section has area

1/2 r^2(θ - sinθ)

where cosθ = y/r

So, plugging in our numbers,

v = (400)(1/2)(20^2)(arccos(y/20)-√(20^2-y^2))
v = 800(arccos(y/20)-√(400-y^2))
so,

dv/dt = 800(y-1)/√(400-y^2) dy/dt

Now, when y=10, we have

300 = 7200/√300 dy/dt
dy/dt = √300/7200 = √3/720 = 0.0024 cm/s

seems kinda small, right? Let's see whether is is. When the water is 10 cm deep, the surface has an area of 400*10√3 = 6928 cm^2

So, since the volume is changing at 300cm^3/s, we'd expect the height to be changing at 300/6928 = 0.043 cm/s.

I'll think on it a bit. In the meantime, maybe you can see where my mistake is.

emy · Answer

Thank you for the solution! I think the mistake is at the calculations of finding dy/dt, I think it is 0.72. Maybe I am wrong but one more time I thank you in advance.