Question

Water is poured into a conical paper cup at the rate of 3/2 in3/sec. If the cup is 6 inches tall and the top has a radius of 6 inches, how fast is the water level rising when the water is 4 inches deep?

Answer 1

So.. hmm notice the picture below

r = h, due to similar triangles ratio, whatever "h" may be

now, we're given how much water is being poured into the cup per sec
namely dv/dt   the volume's rate

so.. hmmm $\bf \textit{volume of a cone}=V=\cfrac{\pi r^2 h}{3}\quad however\implies r=h \\\\\\ thus\implies V=\cfrac{\pi h^2 h}{3}\implies V=\cfrac{\pi h^3}{3}\\\\ ------------------------\\\\ now \\\\\\ \cfrac{dv}{dt}=\cfrac{\pi }{3}\cdot 3h^2\cdot \cfrac{dh}{dt}\implies \cfrac{dv}{dt}=\pi h^2\cdot \cfrac{dh}{dt} \\\\\\ \cfrac{\frac{dv}{dt}}{\pi h^2}=\cfrac{dh}{dt}\quad \begin{cases} \frac{dv}{dt}=\frac{3}{2}\\\\ \left. \frac{dh}{dt} \right|_{h=4} \end{cases}\implies \cfrac{\frac{3}{2}}{\pi 4^2}=\cfrac{dh}{dt}$