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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.

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  • $\begingroup$ The rate at which the water is rising is the difference between rate the water is coming in and the rate at which it's leaking out. Can you see how to go about solving this now? (The process at least.) $\endgroup$ – Justin Benfield Mar 12 '16 at 18:47
  • $\begingroup$ I think your concept is right but I don't know how will I implement it onto an equation $\endgroup$ – user312617 Mar 12 '16 at 18:50
  • $\begingroup$ See Related Rates: How fast is the water leaking from a conical-shaped tank?. $\endgroup$ – N. F. Taussig Mar 13 '16 at 12:40
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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 $\dfrac{dk}{dt}$.

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  • $\begingroup$ Note: mind the units on $\frac{dh}{dt}$. $\endgroup$ – Justin Benfield Mar 12 '16 at 19:05
  • $\begingroup$ I think I get it now. Thanks heaps :) $\endgroup$ – user312617 Mar 13 '16 at 6:20

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