Q:

Water is being poured into an inverted right conical vessel

Water is being poured into an inverted right conical vessel whose apex angle at $90^\circ$ at a constant rate of $3 \text{ cm}^3/\text{s}$. At what rate is the water level rising when the depth is $π$ cm. 
Hi,
Can anyone help me with this question.  I don't know where should i start. 
Thanks. 

A:

Strangely, many mathematicians seem to do this problem the hard way: Find the volume as a function of time, and hence the depth as a function of time, and then differentiate the depth with respect to time. 
Here's the easy way:
\begin{align}
3\,\frac{\text{cm}^3}{\text{sec}} & = [\text{rate of change of volume}] \\[10pt]
& = [\text{size of moving boundary}] \times [\text{rate of motion of boundary}] \\[10pt]
& = \pi r^2\times [\text{rate of motion of boundary}] \\[10pt]
& = \pi\cdot(\pi^2 \text{ cm}^2) \times [\text{rate of motion of boundary}] \\
& \qquad\qquad (\text{The radius is $\pi$ cm because the vertex angle is $90^\circ.$}) \\[10pt]
& = \pi^3 \text{ cm}^2 \times [\text{rate of motion of boundary}]. 
\end{align}
Therefore
$$
[\text{rate of motion of boundary}] = \frac{3\text{ cm}^3/\text{sec}}{\pi^3 \text{ cm}^2} = \frac 3 {\pi^3}\, \frac{\text{cm}}{\text{sec}}. 
$$

