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Dif. equ.

  1. Apr 13, 2005 #1
    if [tex]x^2+y^2=25[/tex], what is the value of [tex]\frac{d^2y}{dx^2}[/tex] at the point (4,3)?

    [tex]x^2+y^2=25[/tex]
    [tex]2x+2y\frac{dy}{dx}=0[/tex]
    [tex]\frac{dy}{dx}=-\frac{2x}{2y}[/tex]
    [tex]\frac{dy}{dx}=-\frac{3}{4}[/tex]
    [tex]\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(-x)}{x^2}[/tex]
    [tex]\frac{d^2y}{dx^2}=\frac{-3+-\frac{3}{4}(-4)}{4^2}[/tex]
    [tex]\frac{d^2y}{dx^2}=0[/tex]

    where did I go wrong?
     
  2. jcsd
  3. Apr 13, 2005 #2

    jamesrc

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    When you computed the 2nd derivative, I assume you used the quotient rule, but you have a typo in your denominator and you missed a sign in your second term.
     
  4. Apr 13, 2005 #3
    here's what I have [tex]\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(x)}{y^2}[/tex]

    is that correct?
     
  5. Apr 13, 2005 #4
    [tex]x^2+y^2=25[/tex]
    [tex]2x+2y\frac{dy}{dx}=0[/tex]
    [tex]\frac{dy}{dx}=-\frac{x}{y}[/tex]
    [tex]\frac{dy}{dx}=-\frac{4}{3}[/tex]
    [tex]\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(x)}{y^2}[/tex]
    [tex]\frac{d^2y}{dx^2}=\frac{-3+-\frac{4}{3}(4)}{3^2}[/tex]
    [tex]\frac{d^2y}{dx^2}=-\frac{25}{27}[/tex]

    is that correct?
     
    Last edited: Apr 13, 2005
  6. Apr 13, 2005 #5

    dextercioby

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    Nope,the differentiation is okay,but the arithmetics is terrible...

    Daniel.
     
  7. Apr 13, 2005 #6
    whoops forgot to change my previous numbers that I just copied, it's edited
     
  8. Apr 13, 2005 #7

    dextercioby

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    It looks okay,now.

    Daniel.
     
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