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A trough is 12 feet long and 3 feet across the top. Water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when the depth h is 1 foot? The water is rising at a rate of 3/8 inch per minute when h = 2 feet. Determine the rate at which water is being pumped into the trough.

This question aims to find the rate at which water flows and the speed of water in a trough. 

The question depends on the concepts of the volume of a body and the velocity of water flowing. Determining the volume equation with respect to time will give us the rate of change in water flowing. The equation of the volume for prism is given as:

\[ Volume\ V = \dfrac{ 1 }{ 2 } b \times h \times l \]

Expert Answer

The formula of volume having depth instead of length is written as:

\[ V = \dfrac{ 1 }{ 2 } b \times h \times d \]

Here, d is the depth.

If the base and height are 3 feet, it is an isosceles triangle and the depth is 12 feet. By putting values in the formula:

\[ V = \dfrac{ 1 }{ 2 } b \times h \times 12 \]

\[ V = 6bh \]

\[V = 6h^2 \]

Taking derivative on both sides:

\[ \dfrac{ dV }{ dt } = 12h \dfrac{ dh }{ dt } ….. Eq.1 \]

\[ \dfrac { dh } { dt } = \dfrac { 1 } { 12 h }  \dfrac { dV } { dt } \]

To find the speed at which the water level rises when the depth of the trough is 1 foot. Here, h = 1 and $ \frac { dV } { dt } = 2 $. By putting values in the above equation:

\[ \frac{ dh }{ dt } = \frac{ 1 }{ 12(1) } (2) \]

\[ \frac{ dh }{ dt } = \frac{ 1 }{ 6 } ft\min\]

To find the rate at which the water is being pumped into the trough water level at a rate of 3/8 inch per minute when h=2 feet.

\[ \frac{ dh }{ dt } = \frac{ 3 }{ 8 } in/min = \frac{ 1 }{ 32 } ft/min\]

By putting values in the equation:

\[ V =  6h^2\]

\[ \dfrac{dV}{dt} = 12h \dfrac{dh}{dt} \]

\[ \dfrac{dV}{dt} = 12(2) ( \dfrac{ 1 }{ 32 }) \]

\[ \dfrac{dV}{dt} = \dfrac{ 3 }{ 4 } ft^3/min\]

Numerical Results

The speed of rising water level in the trough is $\frac{1}{6} ft\min$. The rate at which the water is being pumped into the trough is calculated to be:

\[ \dfrac{dV}{dt} = \dfrac{3}{4} {ft}^3/min \]

Example

A trough is 14 feet long and 4 feet across the top. The ends of the trough are isosceles triangles having an altitude of 3 feet. The water is pumped into the trough at 6 cubic feet per minute. Determine how fast the water level is rising when the depth h is 2 feet?

\[V= \frac{1}{2} b\times h \times 14 \]

\[V= 7bh\]

\[V= 7h^2\]

\[\frac{dh}{dt} = \frac{1}{14h} \frac{dV}{dt}\]

\[ \frac{ dh }{ dt } = \frac{ 1 }{ 14 (2) } (6)\]

\[ \frac{ dh }{ dt } = \frac { 3 }{14} ft/min \]

\[ \dfrac{ dh }{ dt } = 0.214 ft/min \]

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