Question:
Water is dripping from a filter in the shape of an inverted right circular cone at a rate of $\rm5\ cm^3/s$. The altitude of the filter is $\rm10\, cm$ and its base radius is $\rm 5\ cm$.At the same time, water is pouring into the filter. Given that the water level is rising at a rate of $\rm \frac{1}{2\pi}\ cm/s$ when the depth of water is $\rm8\ cm$, what is the pouring rate of water into the filter at this moment?
I have tried using similar triangles to find the radius of the volume of water at $\rm 8\ cm$, and I think I am supposed to find the rate of pouring water into the filter, which is the change of volume minus the loss of water. How can I organize all this into one equation?
Using similar triangles, if $h$ is the height of water in the conical filter, and $r$ is the radius of the water cone base, then
$ \dfrac{h}{r} = \dfrac{10}{5} = 2 $
The volume of the water cone is
$V_w = \dfrac{1}{3} \pi r^2 h $
And from the first equation we deduce that $r = \dfrac{1}{2} h $
Hence,
$V_w = \dfrac{1}{12} \pi h^3 $
Differentiating with respect to time
$\dfrac{d V_w}{dt} = \dfrac{1}{4}\pi h^2 \dfrac{dh}{dt} $
The left hand side is $F - 5$ where $F$ is the volumetric flow rate of the poured water into the filter. And $\dfrac{dh}{dt} = \dfrac{1}{2\pi} $, and $h = 8$
Hence,
$F - 5 = \dfrac{1}{4} \pi (8)^2 \dfrac{1}{2\pi} = 8 $
Hence, $F = 13 \text{ cm}^3 / \text{sec} $