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Math Help - quickly differentiate this please

  1. #1
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    quickly differentiate this please

     v= {10 \pi r^2(6-r) /3

    the whole is meant to be over 3
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  2. #2
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    You have

    v = \frac{10\pi r^2(6-r)}{3}.

    Start by expanding the brackets. You should be able to find the derivative term by term.
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  3. #3
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    is it
     v' = \frac{ 60 \pi r^2 - 10 \pi r^3 }{3} =0

      v' = \pi r^2(20 - \frac{ 10}{3}r) =0

      r= 6 , 0 zero is excluded as 0 for a length (in the original question cannot make sense)
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  4. #4
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    No, your expansion gives

    v = \frac{60\pi r^2 - 10\pi r^3}{3}

     = 20\pi r^2 - \frac{10\pi}{3} r^3

    but you haven't differentiated it yet...
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  5. #5
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    ok i really gotta practice with pi in these eqn's.

     v = 20\pi r^2 - \frac{10\pi}{3} r^3
     v' = 40 \pi r - 10 \pi r^2 for stat points v' = 0
    10 \pi r (4 - r) = 0
    therefore
     r= 0 or 4

    0 doesnt make sense so 4 only?
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  6. #6
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    Your calculations are correct.

    But it should be possible to say that if you don't have a radius you don't have a volume, so r = 0 is definitely a minimum...
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