Volume in Conical Tank at t = 5 min
We find the height of water in a conical tank such that:
$$ V = \frac{1}{3} \pi x^3 $$
given that water is flowing in at a rate $dV/dt = 1 m^3 / min$.
Parker Glynn-Adey is an Assistant Professor of Mathematics, Teaching Stream, at UTSC.