Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the rate of 2 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer.

Answer 1

The volume of the water in the tank at time \(t\) is
\[V(t)=\frac{\pi}{3}r(t)h(t)\]where \(r\) is the radius of the cone formed by the water and \(h\) is the depth.

Using similar triangles, you know that the right triangle formed by the cone of water has legs proportional to those of the tank itself:
\[\frac{16}{4}=\frac{h(t)}{r(t)}\implies r(t)=\frac{1}{4}h(t)\]which means the volume is
\[V(t)=\frac{\pi}{12}h(t)^2\]Differentiating with respect to \(t\), you have
\[\frac{\mathrm dV}{\mathrm dt}=\frac{\pi}{6}h(t)\frac{\mathrm dh}{\mathrm dt}\]You're given the flow rate \(\dfrac{\mathrm dV}{\mathrm dt}=-2\), and you're asked to find the rate of change of the depth \(\dfrac{\mathrm dh}{\mathrm dt}\) when \(h=8\). Plug in everything you know and solve for \(\dfrac{\mathrm dh}{\mathrm dt}\).