SOLUTION: A trough is 8 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a ra

Question 1127030: A trough is 8 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 13 ft3/min, how fast is the water level rising when the water is 8 inches deep?
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

The triangular cross section of the trough has a base that is 5 times the depth (height). So at any time while the trough is being filled, the cross section of the "solid" formed by the trough and the surface of the water is an isosceles triangle with its base 5 times the depth.

We want to find the rate of change of the depth at a particular time; that means we need an expression for the volume of water as a function of the depth only.

The water forms a triangular prism whose volume is the length of the trough times the area of the cross section.

The length of the trough is a constant 8 feet.

When the depth of the water is x feet, the base of the triangular cross section is 5x; the area of the triangular cross section (one-half base times height) is .

So the volume of water, in terms of its depth, is .

The rate of change of the depth is then the derivative with respect to time of the volume function.

We are given dV/dt as 13 ft^3/min, so

The rate of change of the depth when x is 8 inches = 2/3 feet is

The water level is rising at a rate of 39/80 feet per minute when the depth is 8 inches.
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