Question

A water trough is 30 ft. long and cross-section has the shape of an isosceles trapezoid with 10 in. Wide at the bottom, 25 in. wide at the top, and has a height 15 in. How fast, in in/min, is the water level raising when the trough is being filled with water at the rate of 1.9 ft^3 / min and the water is 9 in. deep?

Answer 1

0.04 ft/min

Explanation:

You want the rate of rise of water in a 30 ft trough with a trapezoidal cross section 25 inches wide at the top and 10 inches wide at the bottom. The water is 9 in deep and the trough is being filled at 1.9 ft³/min.

Area

The width of the water surface is 9/15 = 3/5 of the way between 10 inches and 25 inches. It is ...

(3/5)(25) +(2/5)(10) = 15 +4 = 19 . . . . . inches

Then the surface area of the water in the trough is ...

(19 in)/(12 in/ft) · (30 ft) = 47.5 ft²

Rising

The rate of rise is the rate of change of volume, divided by the area of the surface:

(1.9 ft³/min)/(47.5 ft²) = (19/475) ft/min = 0.04 ft/min

The water is rising at the rate of 0.04 feet per minute.

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Additional comment

We could write an equation for the volume as a function of depth and differentiate that. The result would be the same. Attention must be given to units (inches vs. feet).

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