Question

Water is leaking out of an inverted conical tank at a rate of $10,000{\text{cm}}^{\text{3}}\text{/min}$ at the same time that water is being pumped into the tank at a constant rate.

The tank has a height of $6\text{m}$, and the diameter at the top is $4\text{m}$.

If the water level is rising at a rate of $20\text{cm/min}$ when the height of the water is $2\text{m}$, find the rate at which water is being pumped into the tank.

(Note: $\frac{dV}{dt}=\left(\text{rate of pumped in water}\right)-\left(\text{rate of leak}\right)$ ).

Answer 1

Step-1: Determine the expression of the Volume of water in the tank:

It is given that the shape of the tank is that of an inverted cone of height $6\text{m}$ and diameter of circular top $4\text{m}$.

Thus, the dimensions of the conical vessel can be written as, height $H=6\text{m}$ and radius of base $R=\frac{4\text{m}}{2}=2\text{m}$.

Since the vessel is conical, the water filling it will also take shape of an inverted cone that is a part of the whole conical vessel.

Let, the height of the conical volume occupied by the water be $h$ and the radius of the circular top of that volume be $r$.

Thus, the volume of water in the vessel is given by $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$.

Also, since this cone is similar in shape to the whole vessel, their dimensions will be proportional, namely:

$\begin{array}{rcl}\frac{r}{R}& =& \frac{h}{H}\\ & \Rightarrow & \frac{r}{2}=\frac{h}{6}\\ & \Rightarrow & r=\frac{h\times 2}{6}\\ & \Rightarrow & r=\frac{h}{3}\end{array}$

Thus, $r=\frac{h}{3}$ and hence the volume of water can be written entirely in terms of $h$ by substituting $r=\frac{h}{3}$ in $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$.

$\begin{array}{rcl}V& =& \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\\ & =& \frac{1}{3}\mathrm{\pi }{\left(\frac{\mathrm{h}}{3}\right)}^2\mathrm{h}\\ & =& \frac{\mathrm{\pi }}{27}{\mathrm{h}}^3\end{array}$

Thus, the volume of water is $\begin{array}{rcl}V& =& \frac{\mathrm{\pi }}{27}{\mathrm{h}}^3\end{array}$.

Step-2: Find an expression for the rate of increase in the volume of water:

Differentiate the equation $\begin{array}{rcl}V& =& \frac{\mathrm{\pi }}{27}{\mathrm{h}}^3\end{array}$ to get the rate of increase in the volume of water with respect to time.

$\begin{array}{rcl}V& =& \frac{\mathrm{\pi }}{27}{\mathrm{h}}^3\\ & \Rightarrow & \frac{dV}{dt}=\frac{d}{d\mathrm{t}}\left(\frac{\mathrm{\pi }}{27}{\mathrm{h}}^3\right)\\ & \Rightarrow & \frac{d\mathrm{V}}{d\mathrm{t}}=\frac{\mathrm{\pi }}{27}\frac{d}{d\mathrm{t}}\left({\mathrm{h}}^3\right)\\ & \Rightarrow & \frac{d\mathrm{V}}{d\mathrm{t}}=\frac{\mathrm{\pi }}{27}\times 3{\mathrm{h}}^2\times \frac{d\mathrm{h}}{d\mathrm{t}}\\ & \Rightarrow & \frac{d\mathrm{V}}{d\mathrm{t}}=\frac{\mathrm{\pi }}{9}{\mathrm{h}}^2\times \frac{d\mathrm{h}}{d\mathrm{t}}\end{array}$

Thus, the rate of increase in the volume of water with respect to time is $\begin{array}{rcl}\frac{d\mathrm{V}}{d\mathrm{t}}& =& \frac{\mathrm{\pi }}{9}{\mathrm{h}}^2\times \frac{d\mathrm{h}}{d\mathrm{t}}\end{array}$.

Step-3: Determine an expression for the rate of pumping water in.

It is also given that the rate of leak is $10,000{\text{cm}}^{\text{3}}\text{/min}$ and the rate of pumped in water is unknown so assume it to be $x{\text{cm}}^3\text{/min}$.

The rate of increase of volume of water $\begin{array}{rcl}\frac{d\mathrm{V}}{d\mathrm{t}}& =& \frac{\mathrm{\pi }}{9}{\mathrm{h}}^2\times \frac{d\mathrm{h}}{d\mathrm{t}}\end{array}$is the rate of pumped in water $x{\text{cm}}^3\text{/min}$ less the rate of leak $10,000{\text{cm}}^{\text{3}}\text{/min}$, that is, $\frac{dV}{dt}=\frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}=x-10000$.

Isolate $x$ from the equation $\frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}=x-10000$:

$\begin{array}{rcl}\frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}& =& x-10000\\ & \Rightarrow & \frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}+10000=x\end{array}$

Thus, the rate of pumped in water is $x=\frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}+10000$.

Step-4: Determine the rate of pumping water in:

Finally, it is given that the rate of rise of the water level is $20\text{cm/min}$ when the height of water is $2\text{m}$, mathematically ${\overline{)\frac{dh}{dt}}}_{h=2\text{m}}=20\text{cm/min}$.

Substitute $\begin{array}{rcl}h& =& 2\text{m}=200\text{cm}\end{array}$, and ${\overline{)\frac{dh}{dt}}}_{h=2\text{m}}=20\text{cm/min}$ in the equation $x=\frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}+10000$.

Then solve for $x$:

$\begin{array}{rcl}x& =& \frac{\mathrm{\pi }}{9}{h}^2\times \frac{dh}{dt}+10000\\ & \Rightarrow & x=\frac{\mathrm{\pi }}{9}{\left(200\right)}^2\times {\overline{)\frac{dh}{dt}}}_{x=2\text{m}}+10000\\ & \Rightarrow & x=\frac{40000\mathrm{\pi }}{9}\times 20+10000\\ & \Rightarrow & x=10000\left(\frac{80\mathrm{\pi }}{9}+1\right)\\ & \Rightarrow & x=10000\left(\frac{80\mathrm{\pi }+9}{9}\right)\end{array}$

Hence, the rate of pumped water is $\mathbf{10000}\left(\frac{80\pi +9}{9}\right)\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{3}}\mathbf{\text{/min}}$.

This is approximately$\mathbf{289252}\mathbf{.}\mathbf{68}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{3}}\mathbf{\text{/min}}$.