# A water tank has the shape of an upside-down cone with radius 2 m and height 5 m. The water is running out of the tank through a small hole at the bottom. What is the rate of change of the water level when flow-out reads 3 m^3/min at height, h= 4 m?

May 29, 2016

rate of change is $6.2 \times {10}^{-} 3 m {s}^{-} 1 = 0.37 \frac{m}{\min}$

#### Explanation:

We are given the rate of change of volume
$\frac{\mathrm{dV}}{\mathrm{dt}} = 3 {m}^{3} / \min = 0.05 {m}^{3} / s$
and are asked to find the rate of change of water level , $\frac{\mathrm{dh}}{\mathrm{dt}}$.

In terms of a differential equation we can express this as
$\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{\mathrm{dV}}{\mathrm{dh}} \cdot \frac{\mathrm{dh}}{\mathrm{dt}} = 0.05$ which we will call equation1

we want to find $\frac{\mathrm{dh}}{\mathrm{dt}}$ so we will need to find $\frac{\mathrm{dV}}{\mathrm{dh}}$.

The volume of a cone is given by
$V = \pi {r}^{2} \frac{h}{3}$

And in this case $\frac{r}{h} = \frac{2}{5}$ so $r = \frac{2}{5} h$

Substituting this into the Volume equation gives:

$V = \pi {r}^{2} \frac{h}{3} = \pi {\left(\frac{2}{5} h\right)}^{2} \left(\frac{h}{3}\right) = \pi \frac{4}{75} {h}^{3}$

We can now differentiate:

$\frac{\mathrm{dV}}{\mathrm{dh}} = \pi \frac{12}{75} {h}^{2}$

When h = 4, then $\frac{\mathrm{dV}}{\mathrm{dh}} = \pi \frac{12}{75} {h}^{2} = 2.56 \pi$

So if we substitute this back into equation 1:

$\left(2.56 \pi\right) \cdot \left(\frac{\mathrm{dh}}{\mathrm{dt}}\right) = 0.05$

$\frac{\mathrm{dh}}{\mathrm{dt}} = \frac{0.05}{2.56 \pi} = 6.2 \times {10}^{-} 3 m {s}^{-} 1 = 0.37 \frac{m}{\min}$