Asked 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.
Does 1.4 ft/min sound right? The computer is saying it's wrong.
Answers
Answered by
drwls
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
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
Answered by
Zhao
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