Question

Water is poured into a reservoir in the shape of a cone 6 feet
tall with a radius of 4 feet. If the water level is rising at the
constant rate of 0.5 feet per second, how fast is the water
being poured in at the instant the depth is 2 feet?

( I don't understand what to do besides knowing I have to find the derivative of cone volume with respect to dv/dt and dh/dt)

Answer 1

\[V=\frac{1}{3} \pi h r^2 \]Take the total derivative of the above\[dV=\frac{1}{3} \pi r^2 dh+\frac{2}{3} \pi h r dr \]and plug in the numbers.\[\left\{r\to 4,h\to 6,dh\to \frac{1}{2},h\to 2\right\} \]\[dV=\frac{8 \pi }{3}\left.\text{ft}^3\right/\sec \]