# Calc.RR2

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).