Answer
a) $dh=\frac{10}{9\pi}\frac{in}{min}$
b) $dh=-\frac{8}{5\pi}\frac{in}{min}$
Work Step by Step
Given: $dV=10\frac{in^3}{min}$
as the pot is cylindrical, we have: $V_{pot}=\pi r^2h$
when r=3 then $V_{pot}=9\pi h$
$V_{cone}=\frac{1}{3}\pi r^2h$ and $h=2r,r=\frac{h}{2},V_{cone}=\frac{1}{12}\pi h^3$
a) $V_{pot}=9\pi h$
$dV=9\pi dh$
$10=9\pi dh,dh=\frac{10}{9\pi}\frac{in}{min}$
b) $V_{cone}=\frac{1}{12}\pi h^3$
$dV=\frac{1}{4}\pi h^2dh$
so $-10=\frac{\pi (5)^2}{4}dh$ (the cone is having a decreasing rate, hence the negative sign)
$dh=-\frac{8}{5\pi}\frac{in}{min}$
Thus, the final answer is:
a)$dh=\frac{10}{9\pi}\frac{in}{min}$
b)$dh=-\frac{8}{5\pi}\frac{in}{min}$
