Related Rate of Cylindrical Cone (Filling + Leaking) So, we are only told of how to solve related rates with one underlying problem. Either a cone is leaking or Being filled up at some point $x$ but I never encountered both working at the same time.
Here's an example of the worksheet problem:
"Water is poured at the rate of $8 {ft^3\over min}$ into a conical-shaped tank, $20 ft$ deep and
$10 ft$ in diameter at the top. If the tank has a leak in the bottom and the water level is rising at
the rate of $1 {in\over min}$, when the water is $16 ft$ deep, how fast is the water leaking?"
What I definitely know how to do is to fin the radius of the current volume of liquid by proportion oof similar triangles. and so the radius is $4$.
I also know that $$ V_{cone}={1 \over 3}\pi r^2h  $$
I am given with ${dv\over dt}$ (change in volume by filling) which is $8 {ft^3\over min}$ and ${dh\over dt}$ (change in height by filling w/ respect to $t$) which is $1 {in\over min}$ and so I am left with one missing item: ${dr\over dt}$. Is that the proper interpretation of the problem? What I feel is that there'es something missing because I think I haven't implemented the leaking part of the equation.
 A: So we know the following:
Rate at which water is entering cone, lets call that $\dfrac{df}{dt}$.
Rate at which the height of the water in the cone is rising, $\dfrac{dh}{dt}$.
We need to find the leak rate, call it $\dfrac{dk}{dt}$.
My hint was that the change in volume of water in the tank, $\dfrac{dv}{dt}$, satisfies
$\dfrac{dv}{dt}=\dfrac{df}{dt}-\dfrac{dk}{dt}$
We have only one of the things we need, but we can find $\dfrac{dv}{dt}$ using our other given (this is the trickiest part). Note that the cone has fixed proportions (the relationship between $r$ and $h$ is a constant fixed ratio). Hence we can rewrite $v=\dfrac{1}{3}\pi r^2h$ entirely in terms of $h$. This uses the dimensions of the tank given in the problem. Hence $r=\dfrac{1}{4}h$. Substituting for $r$, we obtain
$v=\dfrac{1}{3}\pi \left( \dfrac{1}{4}h \right)^2h=\dfrac{1}{48}h^3$
Differentiate implicitly with respect to time
$\dfrac{dv}{dt}=\dfrac{3}{48}h^2\dfrac{dh}{dt}$
But we know what $h$ and $\dfrac{dh}{dt}$ are from the givens in the problem (you found $h$ yourself).
Thus you can find $\dfrac{dv}{dt}$ and use that and then given for $\dfrac{df}{dt}$ to solve for $\dfrac{dk}{dt}$.
