Question

A conical tank has height $\text{3 m}$ and radius $\text{2 m}$ at the top.

Water flows in at a rate of ${\text{2 m}}^3{\text{min}}^{-1}$.

How fast is the water level rising when it is $\text{2 m}$?

Answer 1

Find the how fast is the water level rising when it is $\text{2 m}$:

Given data

Height of the conical tank is $\text{3 m}$

Radius of the conical tank is $\text{2 m}$

Rate of flow of water is ${\text{2 m}}^3{\text{min}}^{-1}$

Step-1: The formula of the conical tank is expressed using the formula:

$V=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Get the rate at which the water is flowing in the tank as,

$$\begin{gathered} \frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}+\frac{dV}{dh}\frac{dh}{dt} \\ \frac{dV}{dt}=\frac{d\left({\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{2}h\right)}{dr}\frac{dr}{dt}+\frac{d\left(\frac{1}{3}\mathrm{\pi }{r}^{2}h\right)}{dh}\frac{dh}{dt} \\ \frac{dV}{dt}=\frac{2}{3}\mathrm{\pi }rh\frac{dr}{dt}+\frac{1}{3}\mathrm{\pi }{r}^2\frac{dh}{dt}......\left(1\right) \end{gathered}$$

From the given data we can write it as

$$\begin{gathered} \Rightarrow \text{r=2 m} \\ \Rightarrow \text{h=3 m} \\ \Rightarrow \frac{dV}{dt}=2{\text{m}}^3{\text{min}}^{-1} \\ \text{⇒}\frac{dr}{dt}=0 \end{gathered}$$

Step-2 : Substituting the above values in the Equation (1)

$$\begin{gathered} \frac{dV}{dt}=\frac{2}{3}\mathrm{\pi }rh\frac{dr}{dt}+\frac{1}{3}\mathrm{\pi }{r}^2\frac{dh}{dt} \\ \Rightarrow 2=0+\frac{1}{3}\mathrm{\pi }{\left(2\right)}^2\frac{\mathrm{dh}}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{3}{2\mathrm{\pi }}\mathrm{m}.{\min }^{-1} \end{gathered}$$

$\Rightarrow \frac{dh}{dt}=0.48m.mi{n}^{-1}$

Hence, the rate at which the water level rises when the level is $\text{2 m}$ is $0.48{\text{m.min}}^{-1}$