SOLUTION 16: Consider the given side view of the conical tank of height $10$ meters and base radius $8$ meters.

Consider the cone of water of depth $h$ meters and the right triangle with legs $r$ and $h$ in the side view of the conical tank. Assume that lengths $r$ and $h$ are both functions of time $t$.

RECALL: The volume of a right circular cone of height $h$ and base radius $r$ is

$$V = \displaystyle{ 1 \over 3 } \pi r^2 h$$

GIVEN: $\ \ \ \displaystyle{ dV\over dt } = \pi \ m^3/min.$

FIND: $\ \ \ \displaystyle{ dh \over dt }$ when $\ \$ a.) $\ h=1 \ m.$ $\ \$ b.) $\ h=9 \ m.$

First consider the two given similar right triangles. Use them to write $r$ in terms of $h$.

Using similar triangles we can conclude that

$$\displaystyle{ r \over h } = { 8 \over 10 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ r \over h } = { 4 \over 5 } \ \ \ \ \longrightarrow$$ $$r = \displaystyle{ 4 \over 5 }h$$

The volume equation can now be rewritten as

$$V = \displaystyle{ 1 \over 3 } \pi \Big\{ { 4 \over 5 }h \Big\}^2 h \ \ \ \ \longrightarrow$$ $$V = \displaystyle{ 16 \over 75 } \pi h^3$$

Differentiating the volume equation, we get

$$D \{ V \} = D\{ \displaystyle{ 16 \over 75 } \pi h^3 \} \ \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = \displaystyle{ 16 \over 75 } \pi \cdot 3h^2 { dh \over dt } \ \ \ \ \longrightarrow$$ $$(**) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle{ dV \over dt } = \displaystyle{ 16 \over 25 } \pi h^2 { dh \over dt } \ \ \ \ \longrightarrow$$

$\Big($ a.) Now let $\displaystyle{ dV \over dt } = \pi$ and $\ h=1 . \Big)$

$$\pi = \displaystyle{ 16 \over 25 } \pi (1)^2 { dh \over dt } \ \ \ \ \longrightarrow$$

$\Big($Divide both sides of the equation by $\pi. \Big)$

$$1 = \displaystyle{ 16 \over 25 } { dh \over dt } \ \ \ \ \longrightarrow$$ $$\displaystyle{ dh \over dt } = { 25 \over 16 } =1.5625 \ meters/min.$$

$\Big($ b.) Now let $\displaystyle{ dV \over dt } = \pi$ and $\ h=9$ in equation (**). $\Big)$

$$\pi = \displaystyle{ 16 \over 25 } \pi (9)^2 { dh \over dt } \ \ \ \ \longrightarrow$$ $$1 = \displaystyle{ 16 \over 25 } (81) { dh \over dt } \ \ \ \ \longrightarrow$$ $$1 = \displaystyle{ 1296 \over 25 } { dh \over dt } \ \ \ \ \longrightarrow$$ $$\displaystyle{ dh \over dt } = { 25 \over 1296 } \approx 0.019 \ meters/min.$$