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?

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

\[V= 7bh\]

\[V= 7h^2\]

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

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

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

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