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A water tank has the shape of an inverted circular cone with the base radius $2$m and height $4$m. If water is being pumped into the tank at a rate of $2 m^3/min$, find the rate at which the water level is rising when the water is 3m deep.

I believe you are supposed to create a similar triangle. I've solved the problem my way, but the answer seems iffy.

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  • $\begingroup$ Why not show us what you've done, and we'll see whether it's iffy or not? $\endgroup$ – Gerry Myerson Mar 17 '14 at 8:06
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    $\begingroup$ One of the most "popular" related rates problems: see, for instance, math.stackexchange.com/questions/424139/… $\endgroup$ – colormegone Mar 17 '14 at 8:06
  • $\begingroup$ @GerryMyerson, I would, but I joined today and I still have no idea how to format this solution so it doesn't look messy. $\endgroup$ – user135905 Mar 17 '14 at 8:09
  • $\begingroup$ @RecklessReckoner Thank you! That helped a lot. I think my answer is correct, I did essentially the same thing with my different numbers. $\endgroup$ – user135905 Mar 17 '14 at 8:11
  • $\begingroup$ You can follow links on the "help" menu to information on formatting. Or you can just write it as you wish, and count on someone to come along and edit it into shape. $\endgroup$ – Gerry Myerson Mar 17 '14 at 8:11
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Hint:

  • What is the volume $V$ in terms of the height $h$ if the radius is half the height?
  • Take the derivative with respect to time to find $\dfrac{dV}{dt}$ in terms of $\dfrac{dh}{dt}$ and $h$
  • Apply the known rate of change in volume and the known height to find the rate of change in height
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