Answer
(a) $-\frac{1}{24\pi}\approx-1.33cm/min$
(b) $r=\sqrt {26y-y^2}$ m
(c) $-\frac{5}{288\pi}\approx-0.55cm/min$.
Work Step by Step
Identify the given quantities: $\frac{dV}{dt}=-6m^3/min$, $R=13m$
(a) With $V=\frac{\pi}{3}y^2(3R-y)=\pi y^2R-\frac{\pi}{3}y^3$ and $y_0=8m$, $\frac{dV}{dt}=2\pi yR\frac{dy}{dt}-\pi y^2\frac{dy}{dt}=(2(8)(13\pi)-(8)^2\pi)\frac{dy}{dt}=144\pi\frac{dy}{dt}$. Since $\frac{dV}{dt}=-6m^3/min$, we have $\frac{dy}{dt}=-\frac{6}{144\pi}=-\frac{1}{24\pi}\approx-0.0133m/min=-1.33cm/min$
(b) Using the figure given in the Exercise, we have $13^2=r^2+(13-y)^2$ or $r^2=13^2-(13-y)^2=26y-y^2$ and $r=\sqrt {26y-y^2}$ m
(c) At $y=8m$, $r=\sqrt {26(8)-8^2}=12$ m. With $r^2=26y-y^2$, we have $2r\frac{dr}{dt}=(26-2y)\frac{dy}{dt}=(26-2(8))\frac{dy}{dt}=10(-\frac{1}{24\pi})=-\frac{5}{12\pi}$. Thus $\frac{dr}{dt}=-\frac{5}{12(24)\pi}=-\frac{5}{288\pi}\approx-0.0055m/min=-0.55cm/min$.
