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Math Help - Rate of change problem

  1. #1
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    Rate of change problem

    the radius of a cone is increasing at a rate of 3 in/sec, and the height of the cone is 3 times the radius. Find the rate of change for the volume of that cone when the radius is 7 inches.

    The textbook is no help and the prof doesnt teach.

    any help will be appreciated
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  2. #2
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    Re: Rate of change problem

    You should know that the volume of a cone is V_{cone} = \dfrac{1}{3} \pi r^3h and you are given \dfrac{dr}{dt} = 3 \text{  and  } h = 3r

    The question is asking you to find \dfrac{dV}{dt}.


    From the chain rule we know that \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}


    Do you know how to proceed from there?

    Spoiler:
    Substitute 3r for h in the equation for the volume of a cone - this will now become V = \dfrac{1}{3}\pi r^3 \cdot 3r = \pi r^4
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  3. #3
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    Re: Rate of change problem

    I was stuck on H= Pi r^3. Basically, i am stuck at where you stopped. Pi R^4 works also.
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    Re: Rate of change problem

    If V = \pi r^4 then what is \dfrac{dV}{dr}?
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    Re: Rate of change problem

    Pi 4R^3 x 3.
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    Re: Rate of change problem

    That is the answer to the final question: \dfrac{dV}{dt} = 12\pi r^3 .

    To find the rate of change at 7 inch radius sub in r=7 into the above equation
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  7. #7
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    Re: Rate of change problem

    I did that before but it is different from the answer i have. The answer for the question is 441pi = 1385.4 from the final review sheet. I tried every way possible that I know of but couldnt get 441pi = 1385.4.

    edit - what you did seem to be the correct way but the outcome is different from the answer on my final review sheet. I was doing your way before i came here, but my answer never match the solution on the review sheet.
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  8. #8
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    Re: Rate of change problem

    the correct solution is \frac{dV}{dt} = 4116\pi \, in^3/sec

    ... looks like a digit was left off the end of the given "answer".
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  9. #9
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    Re: Rate of change problem

    4116 was what i got numerous time. I guess the Math department made a mistake. Thank you for all of your help I really appreciate it.
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