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.
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 $$ [\text{rate of motion of boundary}] = \frac{3\text{ cm}^3/\text{sec}}{\pi^3 \text{ cm}^2} = \frac 3 {\pi^3}\, \frac{\text{cm}}{\text{sec}}. $$