Related Rates — Question 10

Question 10 - Solution

Let:

$V$: volume of water

$r$: radius of water surface

$h$: depth of water

Given:

$\dfrac{dV}{dt} = 3 \, \text{m}^3/\text{min}$

Tank dimensions: height = 12 m, top radius = 4 m

Volume of cone: $V = \dfrac{1}{3} \pi r^2 h$

Use similar triangles: $$\frac{r}{h} = \frac{4}{12} = \frac{1}{3} \Rightarrow r = \frac{h}{3}$$

Substitute into volume formula: $$V = \frac{1}{3} \pi \left( \frac{h}{3} \right)^2 h = \frac{\pi}{27} h^3$$

Differentiate: $$\frac{dV}{dt} = \frac{\pi}{9} h^2 \cdot \frac{dh}{dt}$$

Substitute known values: $$3 = \frac{\pi}{9} (6)^2 \cdot \frac{dh}{dt} = \frac{\pi}{9} \cdot 36 \cdot \frac{dh}{dt} \Rightarrow 3 = 4\pi \cdot \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{3}{4\pi}$$

Final Answer: $$\boxed{\frac{dh}{dt} = \frac{3}{4\pi} \, \text{m/min} \approx 0.239 \, \text{m/min}}$$