Question

A cylindrical tank with radius 5 m is being filled with water at a rate of \(3\,{{\rm{m}}^{\rm{3}}}{\rm{/min}}\). How fast is the height of the water increasing?

Short answer

The height of the water is increasing at \(\frac{3}{{25\pi }}\,{\rm{m/min}}\)

Step-by-step solution

The volume of the cylinder

It is known that the volume of the cylinder whose height is \(h\) and radius is \(r\) equals to \(V = \pi {r^2}h\).

The rate of change of volume

The rate of change of volume with respect to time is given by \(\frac{{dV}}{{dt}}\).

Here, \(V = \pi {r^2}h\). Here, radius of the cylinder is fixed. Differentiate \(V\) with respect to \(t\):

\[\begin{array}{l}\frac{{dV}}{{dt}} = \frac{d}{{dt}}\left( {\pi {r^2}h} \right)\\\frac{{dV}}{{dt}} = \pi {r^2} \cdot \frac{{dh}}{{dt}}\end{array}\]

Hence, \(\frac{{dV}}{{dt}} = \pi {r^2} \cdot \frac{{dh}}{{dt}}\).

Increase in area

Here, given that the volume of the water is increasing at the rate of \(3\,\,{{\rm{m}}^3}{\rm{/min}}\) when the radius of the cylinder is \(5\,{\rm{m}}\), means \(\frac{{dV}}{{dt}} = 3\)and \(r = 5\).

Substitute in \(\frac{{dV}}{{dt}} = \pi {r^2} \cdot \frac{{dh}}{{dt}}\) :

\(\begin{aligned}3 &= \pi {\left( 5 \right)^2} \cdot \frac{{dh}}{{dt}}\\3 &= 25\pi \frac{{dh}}{{dt}}\\\frac{{dh}}{{dt}} &= \frac{3}{{25\pi }}\end{aligned}\)

Hence, the height of the water is increasing at \(\frac{3}{{25\pi }}\,{\rm{m/min}}\).