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Math Help - partial differentiation

  1. #1
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    partial differentiation

    r^2 = (x - a)^2 + (y - b)^2 + (z -c)^2

    Find \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}

    Got the following for \frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}

    2rdr = 2(x - a) + 2(y - b) + 2(z - c)

    dr = \frac{(x - a) + (y - b) + (z - c)}{r}

    Here, I go astray. The answer provided to me for d^2r is \frac{2}{r}.

    I get \frac{\partial^2 r}{\partial x^2} = \frac{(1)r - ((x - a) + (y - b) + (z - c))(1)}{r^2}

    Likewise for the other partials leads to :
     d^2r = \frac{3(r - ((x - a) + (y - b) + (z - c))}{r^2}

    Obviously not doing something right.

    -Scott
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ScottO View Post
    r^2 = (x - a)^2 + (y - b)^2 + (z -c)^2

    Find \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}

    Got the following for \frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}

    2rdr = 2(x - a) + 2(y - b) + 2(z - c)

    dr = \frac{(x - a) + (y - b) + (z - c)}{r}

    Here, I go astray. The answer provided to me for d^2r is \frac{2}{r}.

    I get \frac{\partial^2 r}{\partial x^2} = \frac{(1)r - ((x - a) + (y - b) + (z - c))(1)}{r^2}

    Likewise for the other partials leads to :
     d^2r = \frac{3(r - ((x - a) + (y - b) + (z - c))}{r^2}

    Obviously not doing something right.

    -Scott
    yes, i don't know what you are doing.

    when we find the partial derivative with respect to, say x, we treat all other variables as constants.

    let me do \frac {\partial^2 r}{\partial x^2} for you. you do the rest

    r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2

    \Rightarrow 2r~\frac {\partial r}{\partial x} = 2(x - a)

    Now \frac {\partial^2 r}{\partial x^2} means we find the second derivative of r with respect to x

    \Rightarrow 2 ~\frac {\partial^2 r}{\partial x^2} = 2


    based on what you said the answer was, you may be looking for it in a different form.

    you may want to write: r = \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}

    and then use the chain rule

    \Rightarrow \frac {\partial^2 r}{\partial x^2} = 1
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    yes, i don't know what you are doing.

    when we find the partial derivative with respect to, say x, we treat all other variables as constants.

    let me do \frac {\partial^2 r}{\partial x^2} for you. you do the rest

    r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2

    \Rightarrow 2r~\frac {\partial r}{\partial x} = 2(x - a)

    Now \frac {\partial^2 r}{\partial x^2} means we find the second derivative of r with respect to x

    \Rightarrow 2 ~\frac {\partial^2 r}{\partial x^2} = 2

    \Rightarrow \frac {\partial^2 r}{\partial x^2} = 1
    OK. What I did was find the first total partial ...

    dr = \frac {\partial r}{\partial x} + \frac {\partial r}{\partial z} + \frac {\partial r}{\partial z} = \frac{2(x -a)}{2r} + \frac{2(y -a)}{2r} + \frac{2(z -a)}{2r}  = \frac{(x - a) + (y - b) + (z - c)}{r}

    Then I started from there on the second total partial, using the quotient rule since I had r in the denominator. What you're showing is I should find the individual partials, \frac {\partial r}{\partial x}, then \frac {\partial^2 r}{\partial x^2}, for each term, and then add them.

    Which gives an answer of d^2r = \frac {\partial^2 r}{\partial x^2} + \frac {\partial^2 r}{\partial y^2} + \frac {\partial^2 r}{\partial z^2} = 3.

    -Scott
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ScottO View Post
    OK. What I did was find the first total partial ...

    dr = \frac {\partial r}{\partial x} + \frac {\partial r}{\partial z} + \frac {\partial r}{\partial z} = \frac{2(x -a)}{2r} + \frac{2(y -a)}{2r} + \frac{2(z -a)}{2r}  = \frac{(x - a) + (y - b) + (z - c)}{r}

    Then I started from there on the second total partial, using the quotient rule since I had r in the denominator. What you're showing is I should find the individual partials, \frac {\partial r}{\partial x}, then \frac {\partial^2 r}{\partial x^2}, for each term, and then add them.

    Which gives an answer of d^2r = \frac {\partial^2 r}{\partial x^2} + \frac {\partial^2 r}{\partial y^2} + \frac {\partial^2 r}{\partial z^2} = 3.

    -Scott
    yes. what you typed confused me at first. i don't think it was wise to do it like that. find each piece first and then add them, because the statement dr = ... which you had was not correct. for various notational and conceptual violations
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