After **time** t, the following is the relation that represents the **volume** V of **water** that **remains in the tank** as per **Torricelli’s Law**.\[{5000\left(1-\frac{t}{40}\right)}^2=V,\ \ where\ 0\le t\le 40\]

As the water is draining from the tank, calculate its **rate** after (a)5min and (b)10min.

Also, find the **time** at which the **rate of water draining** from the tank is **fastest** and **slowest**.

The aim of this article is to find the **rate of water draining** from the tank at a certain instance of **time** and find the time of **fastest** and **slowest drain rate**.

The basic concept behind this article is the use of **Torricelli’s Equation** to calculate the **rate of flow**.

The **Rate of Flow of a given volume** $V$ is calculated by taking the **first derivative** of **Torricelli’s Equation** with respect to **time** $t$.

\[Rate\ of\ Flow=\frac{d}{dt}(Torricelli\prime s\ Equation\ for\ Volume)=\frac{d}{dt}(V)\]

## Expert Answer

Given that:

**Torricelli’s Equation** for the **Volume of Water** remaining in the Tank is:

\[{5000\left(1-\frac{t}{40}\right)}^2=V,\ \ where\ 0\le t\le 40\]

To calculate the **rate** at which **water is draining** at different instances of **time** $t$, we will be taking the** first derivative** of **Torricelli’s Equation** with respect to time $t$.

\[\frac{d}{dt}\left(V\right)=\frac{d}{dt}V(t)\]

\[\frac{d}{dt}V(t)=\frac{d}{dt}\left[{5000\left(1-\frac{t}{40}\right)}^2\right]\]

\[V^\prime(t)=5000\times2\left(1-\frac{t}{40}\right)\times\left(-\frac{1}{40}\right)\]

\[V^\prime(t)=-250\left(1-\frac{t}{40}\right)\]

The **negative sign** indicates that the **rate** at which the water is draining is **decreasing** with **time**.

To calculate the **rate at which water is draining** from the tank after $5min$, substitute $t=5$ in the above equation:

\[V^\prime(5)=-250\left(1-\frac{5}{40}\right)\]

\[V^\prime(5)=-218.75\frac{Gallons}{Min}\]

To calculate the **rate at which water is draining** from the tank after $10min$, substitute $t=10$ in the above equation:

\[V^\prime(10)=-250\left(1-\frac{10}{40}\right)\]

\[V^\prime(10)=-187.5\frac{Gallons}{Min}\]

To calculate the **time** at which **rate of water draining** from the tank is **fastest** or **slowest**, take the following assumptions from the given **minimum** and **maximum range** of $t$

\[1st\ Assumption\ t=0\ min\]

\[2nd\ Assumption\ t=40\ min\]

For **1st assumption** of $t=0$

\[V^\prime(0)=-250\left(1-\frac{0}{40}\right)\]

\[V^\prime(0)=-250\frac{Gallons}{Min}\]

For **2nd assumption** of $t=40$

\[V^\prime(40)=-250\left(1-\frac{40}{40}\right)\]

\[V^\prime(40)=0\frac{Gallons}{Min}\]

Hence, it proves that the **rate at which the water is draining** is **fastest** when $V^\prime(t)$ is **maximum** and **slowest** when $V^\prime(t)$ is **minimum**. Thus, the **fastest rate** at which the water is draining is at the **start** when $t=0min$ and the **slowest** at the **end** of the drain when $t=40min$. As the time passes, the **rate of drain** becomes **slower** until it becomes $0$ at $t=40min$

## Numerical Result

The **rate** at which **water is draining** from the tank after $5min$ is:

\[V^\prime(5)=-218.75\frac{Gallons}{Min}\]

The **rate** at which **water is draining** from the tank after $10min$ is:

\[V^\prime(10)=-187.5\frac{Gallons}{Min}\]

The **fastest rate of the drain** is at the **start** when $t=0min$ and the **slowest** at the **end** when $t=40min$.

## Example

Water is draining from a tank containing $6000$ **gallons of water**. After **time** $t$, the following is the relation that represents the **volume** $V$ of water that remains in the tank as per **Torricelli’s Law**.

\[{6000\left(1-\frac{t}{50}\right)}^2=V,\ \ where\ 0\le t\le 50\]

Calculate its **rate of drain** after $25min$.

**Solution**

\[\frac{d}{dt}V(t)=\frac{d}{dt}\ \left[{\ 6000\left(1-\frac{t}{50}\right)}^2\ \right]\]

\[V^\prime(t)=-240\left(1-\frac{t}{50}\right)\]

To calculate the **rate** at which **water is draining from the tank** after $25min$, substitute $t=5$ in the above equation:

\[V^\prime(t)=-240\left(1-\frac{25}{50}\right)\]

\[V^\prime(t)=-120\frac{Gallons}{Min}\]