Calculus
posted by Abrich .
A water trough is 6 m long, and its crosssection 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?

Draw a figure, noting that the trapezoid has 30cm wide triangles at each end.
When the water is y cm high, its surface is 20+2*30*(y/40) cm across the crosssection. 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