Answer
a. $\frac{10}{9\pi} in/min$.
b. $-\frac{8}{5\pi}in/min$
Work Step by Step
Given $\frac{dV_1}{dt}=-10in^3/min$ from the top cone:
a. The rate of volume increase of the bottom cylinder is $\frac{dV_2}{dt}=10in^3/min$.
Using $V_2=\pi r_2^2h_2=\pi 3^2h_2=9\pi h_2$, we have $\frac{dV_2}{dt}=9\pi\frac{dh_2}{dt}=10$; thus $\frac{dh_2}{dt}=\frac{10}{9\pi} in/min$.
b. For the cone, $V_1=\frac{1}{3}\pi r^2 h$ and $\frac{r}{h}=\frac{3}{6}=\frac{1}{2}$ or $r=\frac{1}{2}h$.
Thus $V_1=\frac{1}{12}\pi h^3$ and $\frac{dV_1}{dt}=\frac{3}{12}\pi h^2\frac{dh}{dt}=-10$.
At $h=5in$, we get $\frac{1}{4}\pi 5^2\frac{dh}{dt}=-10$, which gives $\frac{dh}{dt}=-\frac{8}{5\pi}in/min$
