Calc.RR2

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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 $V=\int_0^h\pi x^2\,dy=\int_0^h\pi(r^2-(y-r)^2)\,dy=\left.\pi\left(r^2y-\frac{1}{3}(y-r)^3\right)\right|_{y=0}^h$ . This simplifies into the formula $V=\pi\left(r^2h-\frac{1}{3}(h-r)^3-\frac{1}{3}r^3\right)$ . 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).

Calc.RR2

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