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A tank is in the form of a cone with the point downward, and the height and diameter are each 10 feet. How fast is the water pouring in at the moment when it is 5 feet deep and the surface is rising at the rate of 4 feet per minute?

The volume of a cone is $$V_c = \frac{1}{12} \pi D^2 H.$$

Given that the diameter of the cone is equal to its height, the volume of the cone may be written $$V_c = \frac{1}{12} \pi H^3.$$

So, $$\frac{dV}{dt} = \frac{1}{4} \pi H^2 \frac{dH}{dt} = \frac{1}{4} \pi (5)^2\times 4 = 25 \pi \textrm{ ft^3/min}$$

So, my answer: $25 \pi \textrm{ ft^3/min}.$

The book's answer: $\frac{25 \pi}{12} \textrm{ ft^3/min}.$

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    $\begingroup$ Your calculation seems right to me. $\endgroup$
    – saulspatz
    Aug 26 '19 at 2:02
  • $\begingroup$ Thanks again Saul :) If I have trouble on the other problem, would you mind if I respond with another comment? $\endgroup$ Aug 26 '19 at 2:06
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    $\begingroup$ Not at all, but it's getting late here, so I might not see it until tomorrow. If you do have trouble, the best thing to do is to edit your question, explaining what you tried and where you got hung up. Then someone (me or somebody else) will be able to give you the most appropriate answer. $\endgroup$
    – saulspatz
    Aug 26 '19 at 2:09
  • $\begingroup$ Alright. Sounds good Saul. I'll do that. Thanks for your help and have a good night! $\endgroup$ Aug 26 '19 at 2:09
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It might be a typo in the book. Your calculations look legit.

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  • $\begingroup$ Thank you Joe!. $\endgroup$ Aug 26 '19 at 2:07

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