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A spherical container of r meters is being filled with a liquid at a rate of \rho \,{{\rm {m}}}^{3}/{{\rm {min}}}. At what rate is the height of the liquid in the container changing with respect to time?

Let V be the volume already in the container, h be the height of the liquid in the container and t be the time since some initial starting point. We are given that {\frac  {dV}{dt}}=\rho \,{{\rm {m}}}^{3}/{{\rm {min}}} and we are asked for the related rate {\frac  {dh}{dt}}. The other rate needed for this problem is the rate at which the height is changing with respect to the volume, namely {\frac  {dh}{dV}}. In order to calculate this, we need a relation between the height and the volume of the liquid at any particular point in time.

We will use a trick from integral calculus to get the volume in terms of the height. Let a silhouette of our container be described by the implicit equation x^{2}+(y-r)^{2}=r^{2}, a circle of radius r whose bottom is on the x-axis. Using the slicing method for finding a volume of revolution, we get that V=\int _{0}^{h}\pi x^{2}\,dy=\int _{0}^{h}\pi (r^{2}-(y-r)^{2})\,dy=\left.\pi \left(r^{2}y-{\frac  {1}{3}}(y-r)^{3}\right)\right|_{{y=0}}^{h}. This simplifies into the formula V=\pi \left(r^{2}h-{\frac  {1}{3}}(h-r)^{3}-{\frac  {1}{3}}r^{3}\right). Then {\frac  {dV}{dh}}=\pi \left(r^{2}-(h-r)^{2}\right)=\pi h(2r-h), so {\frac  {dh}{dV}}={\frac  {1}{\pi h(2r-h)}}.

Then we get the general forumla for the desired rate: {\frac  {dh}{dt}}={\frac  {dh}{dV}}\cdot {\frac  {dV}{dt}}={\frac  {1}{\pi h(2r-h)}}\cdot \rho ={\frac  {\rho }{\pi h(2r-h)}}\,{{\rm {m/min}}}. This is a general formula which will give us the rate at which the height is changing at any given height (assumed to be between 0 and 2r).