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Math Help - Calculus 3 - Laplace Equation

  1. #1
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    Calculus 3 - Laplace Equation

    Hi all
    Ran into a problem on a quiz that for the life of me I thought I was solving correctly but the proof wasnt working.

    Anyhow the question was
    Prove that
    Fxx + Fyy = 0

    F(x,y) = LN(SQRT(X^2+Y^2))

    My solution

    Property of logs converted my problem to
    1/2 LN(X^2+Y^2)
    Derived with respect to X = Fx (x,y) = x/(x^2+y^2)

    Derived with respect to Y = Fy (x,y) = y/(x^2+y^2)

    Derived Fx with respect to X = Using the power rule for the function (x(x^2+y^2)^-1)
    = (1)(x^2+y^2)^-1) + (x)(2x)(-1)(x^2+y^2)^-2)
    =1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

    So Fxx = 1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

    Derived Fy with respect to y = Using the power rule for the function (y(x^2+y^2)^-1)
    = (1)(x^2+y^2)^-1) + (y)(2y)(-1)(x^2+y^2)^-2)
    =1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)

    So Fyy = =1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)

    If I do Fxx + Fyy I dont think I get 0.

    Any suggestions on where I went wrong please!

    Thanks
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  2. #2
    A Plied Mathematician
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    Re: Calculus 3 - Laplace Equation

    I don't think your second derivatives are correct. Check this out. I usually prefer to use the quotient rule, because, while it may be a little more calculus up front, it can often be a lot less algebra later. In particular, you don't have to find a common denominator when you use the quotient rule.
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  3. #3
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    Re: Calculus 3 - Laplace Equation

    Quote Originally Posted by Ackbeet View Post
    I don't think your second derivatives are correct. Check this out. I usually prefer to use the quotient rule, because, while it may be a little more calculus up front, it can often be a lot less algebra later. In particular, you don't have to find a common denominator when you use the quotient rule.


    That helps alot, but I still dont see where I went wrong with the product rule. I keep doing it and keep getting the same wrong answer.
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  4. #4
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    Re: Calculus 3 - Laplace Equation

    Quote Originally Posted by Spoolx View Post
    Hi all
    Ran into a problem on a quiz that for the life of me I thought I was solving correctly but the proof wasnt working.

    Anyhow the question was
    Prove that
    Fxx + Fyy = 0

    F(x,y) = LN(SQRT(X^2+Y^2))

    My solution

    Property of logs converted my problem to
    1/2 LN(X^2+Y^2)
    Derived with respect to X = Fx (x,y) = x/(x^2+y^2)

    Derived with respect to Y = Fy (x,y) = y/(x^2+y^2)

    Derived Fx with respect to X = Using the power rule for the function (x(x^2+y^2)^-1)
    = (1)(x^2+y^2)^-1) + (x)(2x)(-1)(x^2+y^2)^-2)
    =1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)
    Should be 1/(x^2+y^2) - 2x^2/(x^2+y^2)^2).

    Get a common denominator and simplify.

    So Fxx = 1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

    Derived Fy with respect to y = Using the power rule for the function (y(x^2+y^2)^-1)
    = (1)(x^2+y^2)^-1) + (y)(2y)(-1)(x^2+y^2)^-2)
    =1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)
    Same mistake here.

    So Fyy = =1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)

    If I do Fxx + Fyy I dont think I get 0.

    Any suggestions on where I went wrong please!

    Thanks
    Follow Math Help Forum on Facebook and Google+

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