Posted by **Abrich** on Saturday, November 24, 2012 at 12:02am.

A water trough is 6 m long, and its cross-section has the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 50 cm wide at the top and 40 cm high. If the trough is being filled with water at the rate of 0.2 cubic meters/min, how fast is the water level rising when the water is 25 cm deep?

- Calculus -
**Steve**, Saturday, November 24, 2012 at 12:32am
Draw a figure, noting that the trapezoid has 30-cm wide triangles at each end.

When the water is y cm high, its surface is 20+2*30*(y/40) cm across the cross-section. That's 20 + 3y/2.

So, at depth y, the volume

v = (20 + 20+3y/2)/2 * y * 600

= 450y^2 + 12000y

dv/dt = 900y dy/dt + 12000 dy/dt

since .2 m^3 = 200000 cm^3

200000 = 900*25 dy/dt + 12000 dy/dt

dy/dt = 5.80 cm/min

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