Question

If a cylindrical tank with radius $5$ meters is being filled with water at a rate of $3$ cubic meters per minute, how fast is the height of the water increasing?

Answer 1

Finding the increasing rate of height of water:

Step-1: Finding the volume of the tank

Formula to be used: We know that the volume of a cylindrical tank of radius $r$ unit and height $h$ unit is $V=\pi r^{2}h$cubic unit.

Here, $r=5m$. Suppose the height of the tank be $hm$. Hence, the volume of the tank is

$$\begin{gathered} V=\pi r^{2}h \\ =\pi \times 5^2\times h \\ =25h\pi m^3 \end{gathered}$$

Step-2: Finding the rate of increasing of the height of water

Here, the rate is time $\left(t\right)$ related. Thus the increasing rate of the height of water with respect to time will be $\frac{dh}{dt}$.

Now, differentiating both sides of $V=25h\pi$ with respect to $t$, we get:

$$\begin{gathered} \frac{dV}{dt}=25\pi \left(\frac{dh}{dt}\right) \\ \Rightarrow \frac{dh}{dt}=\frac{{\displaystyle \frac{dV}{dt}}}{25\pi } \end{gathered}$$

Given that $\frac{dV}{dt}=3m^3/min$. So, $\frac{dh}{dt}=\frac{3}{25\pi }{m}^3/min$.

Therefore, the height of water increases $\frac{3}{25\pi }{m}^3/min$.