a) $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$
using similar triangles:
$\vec { V } = \frac { 1 } { 3 } \pi \left( \frac { 1 } { 3 } h \right) ^ { 2 } h = \frac { 1 } { 27 } \pi h ^ { 3 }$
$$\frac { d h } { d t } = h - 12 , V = \frac { 1 } { 3 } \pi r ^ { 2 } h$$
b) we want dv when $h = 3$
$$\begin{aligned}
V & = \frac { 1 } { 27 } \pi h ^ { 3 } d t \\
d y & = \frac { 1 } { a } \pi h ^ { 2 } d h \\
d t & = \frac { 1 } { a t } \pi h ^ { 2 } ( h - 12 ) \\
& = \frac { 1 } { 9 } \pi \left( 3 ^ { 2 } \right) ( - 9 ) \\
& = - 9 \pi
\end{aligned}$$
$- 9 \pi \mathrm { ft } ^ { 3 } / \min$
c) we want $\frac { d y } { d t }$ when $h = 3$
$$\begin{aligned}
& \pi R ^ { 2 } = 400 \pi \\
& R ^ { 2 } = 400 \\
& R = 20
\end{aligned}$$
volume of culinder $= \pi R ^ { 2 } 4$
$\frac { d V } { d t } = 2 \pi R \varphi \frac { d R } { d t } + \pi R ^ { 2 } \frac { d y } { d t }$
$d V = \pi R ^ { 2 } d y \quad d R = 0$ since $R$ is a constant
$d t d t d t$
$9 \pi = 400 \pi \frac { d \| } { d t }$
$d u = q \quad t +$ Imin
d1 400
or:
$$\begin{aligned}
& y = 400 \pi y \\
& \frac { d y } { d t } = 400 \pi d y \\
& \frac { d y } { d t } = \frac { 9 \pi } { 400 \pi } = \frac { 9 } { 400 }
\end{aligned} \text { Himin }$$