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Application of derivatives to geometry

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Of all cylinders inscribed in sphere of radius [itex]R[/itex] largest area of side([itex]M[/itex]) has cylinder which hight is [itex]R\sqrt{2}[/itex]. Prove.

    3. The attempt at a solution
    I understand how to prove this i only have problem with derivative:

    [itex]M=2*r*Pi*H[/itex] and [itex]r=\frac{\sqrt{(2R)^2 - H^2}}{2}[/itex]

    [itex]M=Pi*H*\sqrt{(2R)^2 - H^2}[/itex]

    [itex]M^{'}=Pi*(\sqrt{(2R)^2 - H^2}) + \frac{Pi*H}{2*\sqrt{(2R)^2 - H^2}}[/itex]

    Then
    [itex]M^{'}=Pi*\frac{8R^2 -2H^2 + H}{2\sqrt{(2R)^2 - H^2}}[/itex] so how can i from this get that [itex]R*\sqrt{2} = H[/itex]

    My textbook says that [itex]M^{'}=Pi*\frac{4R^2 -2H^2}{\sqrt{(4R)^2 - H^2}}[/itex] therefore [itex]R*\sqrt{2} = H[/itex]
     
    Last edited: Jan 13, 2013
  2. jcsd
  3. Jan 13, 2013 #2

    Dick

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    Your differentiation is wrong. You need to use the chain rule when you find the derivative of the second factor in your product rule.
     
    Last edited: Jan 13, 2013
  4. Jan 13, 2013 #3
    So [itex]M^{'}=Pi*(\sqrt{(2R)^2 - H^2}) + \frac{Pi*H}{2*\sqrt{(2R)^2 - H^2}}*(8R -2H)[/itex]

    [itex]M^{'}=Pi*(\sqrt{(2R)^2 - H^2}) + \frac{Pi*H*(4R - H)}{\sqrt{(2R)^2 - H^2}}[/itex]

    [itex]M^{'}=Pi*((\sqrt{(2R)^2 - H^2}) + \frac{H*(4R - H)}{\sqrt{(2R)^2 - H^2}})[/itex]

    [itex]M^{'}=Pi*((\sqrt{(2R)^2 - H^2}) + \frac{H*(4R - H)}{\sqrt{(2R)^2 - H^2}})[/itex]

    [itex]M^{'}=Pi*(\frac{(4R^2 + 4RH - 2H^2)}{\sqrt{(2R)^2 - H^2}})[/itex]

    This time 4RH is making me a problem i feel like there is a better way to factor out this expression
     
  5. Jan 13, 2013 #4

    Dick

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    The derivative of (2R)^2-H^2 is just -2H. R is a constant!
     
  6. Jan 13, 2013 #5
    :cry::cry: I can't believe it. I spent almost 2 hours on this silly problem because of silly mistakes. Thank you.:smile:
     
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