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Related Rates Question

  1. Oct 14, 2007 #1
    Here is the solution to one of my problems:

    [​IMG]

    When inserting dV/dt to differentiate the equation as a function of time, why doesn't the book use the power rule on r^2 and multiply the entire equation by 2? I thought when dr/dt was put into the equation you had to differentiate?
     
  2. jcsd
  3. Oct 14, 2007 #2
    i don't see why there is a need to multiply by 2? if anything, dividing by 4 would clean things up.

    and there is no need to apply the power rule on r^2 b/c it was already differentiated.

    sorry i can't give you a more detailed answer.
     
  4. Oct 14, 2007 #3

    rock.freak667

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    Homework Helper

    Why would they multiply by 2?

    they differentiated it correctly.Unless you are differentiating the wrong statement.

    [tex]\frac{d}{dt}(\frac{4}{3}\pi r^3)= (3(\frac{4}{3})\pi r^2))\frac{dr}{dt}[/tex]

    which is [tex]=4\pi r^2 \frac{dr}{dt}[/tex]
     
  5. Oct 14, 2007 #4
    ahh yeah I was looking at the formula as being 4pi*r^2 instead of cubed, even though it was right in front of me in it's correct form!

    Another question given this same example, why do you only differentiate r^2 -- why not 4 and pi as well?
     
  6. Oct 14, 2007 #5
    r^3***

    and we don't differentiate 4pi bc it's a constant

    you could do the product rule on them but you would yield the same results

    y=2x
    y'=2
     
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