Question

Water pours out a rate of $Q$ from a tap, into a cylindrical vessel of radius $r$. Find the rate at which the height of water level rises when the height is $h$.

Answer 1

Hint:
We will first write the expression of the volume of water when $r$ is the radius and $h$ is the height till where the water rises. Then, differentiate the expression of volume, $V = \pi {r^2}h$ by differentiating it with respect to time, $t$. Also, substitute the value of \[\dfrac{{dV}}{{dt}}\] as \[Q\] and hence, find the value of \[\dfrac{{dh}}{{dt}}\]

Complete step by step solution:
We are given that the water is poured into a cylindrical vessel with radius $r$ and the level of water rises to height $h$.
Then, the volume of the water in the cylindrical tank is given by $V = \pi {r^2}h$
We want to find the rate at which the water is poured in the vessel.
Rate is given by the amount of water pours in unit time.
We will differentiate the expression of volume to find the rate of water pouring in the vessel.
If the volume is given by $V = \pi {r^2}h$, where radius is fixed, then differentiate it with respect to time, $t$
That is,
\[\dfrac{{dV}}{{dt}} = \pi {r^2}\dfrac{{dh}}{{dt}}\]
Now, the term \[\dfrac{{dV}}{{dt}}\] represents the rate of flow.
But, we are given that the rate of flow is given by \[Q\]
Then,
\[Q = \pi {r^2}\dfrac{{dh}}{{dt}}\]
We need to find out the rate at which the height of water level rises when the height is $h$
\[\dfrac{{dh}}{{dt}} = \dfrac{Q}{{\pi {r^2}}}\]

Note:
When we differentiate a constant term which is multiplied to the variable, then constant term remains the same. But, if only a constant term is differentiated then the result is zero. Students must know that the rate of flow is obtained by dividing Volume by time.