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First derivative test help

  1. Jun 14, 2010 #1
    The curve y = x^2√(1 − x^2) for x ≥ 0 and its maximum point M. Find the coordinates of M (x coordinates) ^ --> means power

    dy/dx is zero at stationary point

    so i make use of the chain rule

    and after simplification i get my answer as

    x^3 = 2 x

    so x^2 = 2 and x = square root of 2

    but unfortunately the answer in the marking scheme
    is square root 2/3

    How is it so, please help me, where is the error...!!!!!!
     
  2. jcsd
  3. Jun 14, 2010 #2

    rl.bhat

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    Re: Differenciation

    [tex]y = x^2sqrt{(1-x^2)}[/tex]

    [tex]\frac{dy}{dx} = x^2\frac{-2x}{2\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0[/tex]

    [tex]\frac{-x^3}{\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0[/tex]

    Now simplify.
     
  4. Jun 14, 2010 #3

    HallsofIvy

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    Re: Differenciation

    We can't tell you where your error is until you tell us how you got this! Your error is somewhere in that calculation.

     
  5. Jun 15, 2010 #4
    Re: Differenciation


    There is a mistake cause I get

    [tex]y = x^2\sqrt{(1-x^2)}[/tex]

    [tex]\frac{dy}{dx} = \frac{-3\cdot x^3-2 \cdot x}{\sqrt{(1-x)} \cdot \sqrt{(1+x)}}[/tex]

    and then use the "First derivative test" shown here

    http://www.math.hmc.edu/calculus/tutorials/extrema/

    to find the maximum point of the function
     
    Last edited: Jun 15, 2010
  6. Jun 15, 2010 #5

    Mentallic

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    Re: Differenciation

    What you have is wrong. Show us what you did and we'll steer you in the right direction. The key here is to notice that you need to use the product rule.
     
    Last edited: Jun 15, 2010
  7. Jun 15, 2010 #6
    Re: Differenciation

    Dude according to Wolfram integrator my result is correct..
     
  8. Jun 15, 2010 #7

    Mentallic

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    Re: Differenciation

    So you didn't actually calculate the derivative yourself? Because with my understanding of derivatives, what rl.bhat got is correct and what you have is not what he has... calculator or not...
     
  9. Jun 15, 2010 #8
    Re: Differenciation

    I did calculate it and get the result above !
     
  10. Jun 15, 2010 #9

    Mentallic

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    Re: Differenciation

    Then post what you did so I can point at where you've gone wrong.
     
  11. Jun 15, 2010 #10
    Re: Differenciation

    you got me pappa :redface:

    after retrying the calculation I get

    [tex]f'(x) = 2x\cdot \sqrt{(1-x^2)} + x^2 \cdot \frac{-2x}{2\cdot \sqrt{1-x^2}}[/tex]

    but still the first derivative test is the way to go to find the maximum of f..
     
  12. Jun 15, 2010 #11

    Mentallic

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    Re: Differenciation

    There you go :smile:
     
  13. Jun 16, 2010 #12
    Re: Differenciation


    hey its so strange, i got the answer that is square root 2/3..

    I make use of the dy/dx = v du - u dv / v^2

    so i put x^2 / (1-x^2)^-1/2 and i solve...woOo so puzzling.. anyway thanks to all who helped me..!!!!
     
  14. Jun 16, 2010 #13

    Mentallic

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    Re: Differenciation

    It's not puzzling at all that you apply a different rule of differentiation to solve the same function! If it were different, there are obvious flaws in our "rules".

    It works for anything you like, say we wanted to differentiate x2, well we use the obvious rule to turn it into 2x, but say we instead used the quotient rule on [tex]\frac{1}{x^{-2}}[/tex] then you would find the same answer, even though more work is required :smile:

    But be wary, while it may not seem like the same answer at first sight, they are still equivalent so you would need to simplify/re-arrange in order to show they're equal.
     
  15. Jun 16, 2010 #14

    HallsofIvy

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    Re: Differenciation

    You could also have done that as
    [tex]y= x^2(1- x^2)^{1/2}[/tex]
    with the chain rule and product rule.
     
  16. Jun 16, 2010 #15
    Re: Differenciation

    Another way that works is

    [tex] y^2=x^4-x^6 [/tex]

    [tex] 2yy' = 4x^3-6x^5 [/tex]

    [tex] y' = \frac{4x^3-6x^5}{2\sqrt{x^4-x^6}} [/tex]

    of course y'=0 implies

    [tex] 4x^3-6x^5 = 0 [/tex]

    [tex]x = \sqrt{2}/3 [/tex] works
     
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