calculus
posted by Gabe .
A trough is 15 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft^3/min, how fast is the water level rising when the water is 8 inches deep?
Does 1.4 ft/min sound right? The computer is saying it's wrong.

The water surface area multiplied by the rate of water level rising equals the volume flow rate. Let's write this as
A*dh/dt = dV/dt = 14 ft^/min.
All you need is the area of water surface, A. At a depth of 8 inches, water is 2/3 of the way to the top and the width of the water channel is (2/3)*4 = 8/3 feet. The total surface area is
(8/3) ft x 15 ft = 40 ft^2
Therefore
dy/dt = 14 ft^3/min/40 ft^2 = 0.35 ft/min 
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