Water Depth Rate in a Draining Conical Bucket

We model the water in the cone using similar triangles.

1) Geometry of the bucket

The bucket is a cone with total height $60$ cm and top radius $20$ cm. If the water depth is $h$, then the water surface radius $r$ satisfies

$$\frac{r}{h} = \frac{20}{60} = \frac{1}{3} \quad\Rightarrow\quad r = \frac{h}{3}.$$

2) Volume as a function of depth

The volume of a cone is

$$V = \frac{1}{3}\pi r^2 h.$$

Substitute $r = h/3$:

$$V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{9} \cdot h = \frac{\pi}{27}h^3.$$

Differentiate with respect to time:

$$\frac{dV}{dt} = \frac{\pi}{9}h^2\frac{dh}{dt}.$$

3) Net rate of volume change

Water flows in at

$$120\pi \text{ cm}^3/\text{s}$$

and drains out at

$$10\pi\sqrt{h} \text{ cm}^3/\text{s}.$$

So the net volume rate is

$$\frac{dV}{dt} = 120\pi - 10\pi\sqrt{h}.$$

4) Solve for $dh/dt$

Set the two expressions for $dV/dt$ equal:

$$\frac{\pi}{9}h^2\frac{dh}{dt} = 120\pi - 10\pi\sqrt{h}.$$

Cancel $\pi$:

$$\frac{1}{9}h^2\frac{dh}{dt} = 120 - 10\sqrt{h}.$$

Thus,

$$\frac{dh}{dt} = \frac{9(120 - 10\sqrt{h})}{h^2}.$$

5) Evaluate at $h=36$

Since $\sqrt{36} = 6$,

$$\frac{dh}{dt} = \frac{9(120 - 10\cdot 6)}{36^2} = \frac{9(120 - 60)}{1296} = \frac{9\cdot 60}{1296} = \frac{540}{1296} = \frac{5}{12}.$$

Final Answer

$$\boxed{\frac{dh}{dt} = \frac{5}{12}\ \text{cm/s}}$$

So the water depth is increasing at an exact rate of $\frac{5}{12}$ cm/s when the water is 36 cm deep.