# If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes.

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

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

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:

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

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