Hello, lysserloo!

We can solve this without using Trig.

A coffee filter in the shape of a right circular cone with a base radius of 6cm and depth 10cm

contains water, which drips out through a hole at the bottom at a constant rate of 2 cm³/sec.

How fast is the water level falling when the depth is 8 cm? Code:

: - 6 - : - 6 - :
- *-------+-------*
: \ | /
: \ | r /
: \- - + - -/
10 \ | /
: \ h| /
: \ | /
: \|/
- *

The volume of the water is: .$\displaystyle V \:=\:\tfrac{\pi}{3}r^2h$ .[1]

From the similar right triangles: .$\displaystyle \frac{r}{h} \:=\:\frac{6}{10} \quad\Rightarrow\quad r \:=\:\tfrac{3}{5}h $

Substitute into [1]: .$\displaystyle V \:=\:\tfrac{\pi}{3}\left(\tfrac{3}{5}h\right)^2h \quad\Rightarrow\quad V \:=\:\frac{3\pi}{25}h^3$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \:=\:\frac{9\pi}{25}h^2\left(\frac{dh}{dt}\right) $ .[2]

We are given: .$\displaystyle h = 8,\;\;\frac{dV}{dt} = \text{-}2$

Substitute into [2]: .$\displaystyle -2 \;=\;\frac{9\pi}{25}(8^2)\cdot\frac{dh}{dt} $

Therefore: .$\displaystyle \frac{dh}{dt} \:=\:-\frac{25}{288\pi}$ cm/sec.