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Question

A trough is 10ft long and its ends have the shape of isosceles triangles that are 3ft across the top and have a height of 1ft. If the trough is being filled with water at a rate of 12ft2/min, how fast is the water level rising when the water is 6 inches deep?

Answer 1

Let
h = height of the water in feet
V = volume water
t = time
b = base (at surface of water)
dV/dt = 12 ft³/min

Find dh/dt when h = 1/2 foot.

The ratio of base to height of the water is:
b/h = 3/1 = 3
b = 3h

V = (1/2)12bh = 6bh = 6(3h)h = 18h²

Take the derivative.

dV/dh = 2*18h = 36h

dh/dt = (dV/dt)/(dV/dh) = 12 / (36h) = 1/(3h)
dh/dt = 1/[3(1/2)] = 2/3 ft/min = 8 inches/min