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 A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. 12f 3 ft 31 (a) if water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.2 feet deep? X ft/min (b) if the water is rising at a rate of 3/8 inch per minute when h - 2.3, determine the rate at which water is being pumped into the trough fo/min

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Therefore ,the water level is rising  by 0.1388 ft/min when h is 1.2 feet deep and  the water is being pumped at rate of 0.8625 cubic feet per min when water is rising at 3/8 inch/min

What is volume?

A three-dimensional space's occupied volume is measured. It is typically quantified using a variety of imperial or US-standard units as well as SI-derived units. The concept of length and volume are related.

Here,

Given: A trough has a 12 foot length and a 3 foot top width. Three-foot-high isosceles triangles make up its ends. 12f 3 ft 31

So the first case:

using this we find the rate of Volume,

So,

=> dV/dt = length * height *dh/dt

=> dV/dt = 12 * 1.2 *dh/dt

=> 2 = 12*1.2* dh/dt

=> 2 = 14.4 * dh/dt

=>dh/dt = 2/14.4

=>dh/dt = 0.1388 ft/min

Therefore ,the water level is rising  by 0.1388 ft/min when h is 1.2 feet deep .

Second case:

Since water level is rising by 3/8 inch per min =0.03125 feet/min

=> dV/dt = length * height *dh/dt

=> dV/dt = 12 * 2.3 *0.03125

=>dV/dt = 0.8625 cubic feet per min

Therefore , the water is being pumped at rate of 0.8625 cubic feet per min.

To know more about volume , visit

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