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Derivative Transformation with Law of Cosines

  1. Nov 17, 2012 #1
    Hi there,

    The Law of Cosines can be stated as

    [itex]a^2 = b^2 + c^2 - 2bccos(A)[/itex]

    where [itex]a[/itex],[itex]b[/itex], and [itex]c[/itex] are the sides of a triangle, and [itex]A[/itex] is the angle opposite the side [itex]a[/itex]. I have a function, [itex]f(b,c,A)[/itex], with an associated set of partial derivatives [itex](\frac{∂f}{∂c})_{b,A}[/itex] etc. What I want to do is to use a coordinate transformation to get the related derivatives [itex](\frac{∂f}{∂c})_{b,a}[/itex] etc. This looks like a multivariable partial derivative problem with a constraint. Using the chain rule, it seems to me that

    [itex](\frac{∂f}{∂c})_{b,A}=\frac{∂f}{∂c}+\frac{∂f}{∂a} \frac{∂a}{∂c}[/itex]

    [itex](\frac{∂f}{∂c})_{b,a}=\frac{∂f}{∂c}+\frac{∂f}{∂A} \frac{∂A}{∂c}[/itex]

    where I think [itex]\frac{∂f}{∂c}[/itex] is the same in both expressions. I can calculate the [itex]\frac{∂a}{∂c}[/itex] and [itex]\frac{∂A}{∂c}[/itex] parts using the Law of Cosines, but then I don't know what to do with [itex]\frac{∂f}{∂a}[/itex] and [itex]\frac{∂f}{∂A}[/itex], i.e. if I use the transformation [itex]\frac{∂f}{∂a}=\frac{∂f}{∂A}\frac{∂A}{∂a}[/itex] then it looks like [itex](\frac{∂f}{∂c})_{b,A}=(\frac{∂f}{∂c})_{b,a}[/itex] but numerical results and intuition tell me otherwise.

    In essence, I don't know how to find the partial derivative [itex](\frac{∂f}{∂c})_{b,a}[/itex] given [itex](\frac{∂f}{∂c})_{b,A}[/itex] and that the variables are connected via the Law of Cosines.

    Any assistance with this problem would be greatly appreciated :)
     
  2. jcsd
  3. Nov 17, 2012 #2
    You are getting confused because you use the same symbol, f, to denote different functions. Do not do it. Let f be a function of (b, c, A). Then let F be a function of (b, c, a). Obviously, f(b, c, A) = F(b, c, a(b, c, A)), where a(b, c, A) is the cosine law. You need to obtain the derivatives of F, use the chain rule.
     
  4. Nov 17, 2012 #3
    Hi Voko,

    Thanks for the response, but I'm still a little stuck here. This is how I continued:

    [itex]dF = (∂F/∂c)dc + (∂F/∂b)db + (∂F/∂a)da[/itex]

    Now [itex](∂F/∂c)_{b,a} = (∂F/∂c)[/itex] (is this not just [itex](∂f/∂c)[/itex]?)

    And [itex](∂F/∂c)_{b,A} = (∂F/∂c) + (∂F/∂A)(∂A/∂c)[/itex]

    But [itex](∂A/∂c) = (c-bcosA)/a[/itex]

    Hence, [itex](∂F/∂c)_{b,A} - (∂F/∂c)_{b,a} = (∂A/∂c)(c-bcosA)/a[/itex]

    However, this doesn't conform to what I have when I substitute in numerical values. I'm clearly not getting something here, could you give me a more forceful push?
     
  5. Nov 17, 2012 #4
    Just to make sure we are on the same page. What you really have is ## \partial f/\partial b, \ \partial f/\partial c, \ \partial f/\partial A ##, correct? And you need to find ## \partial F/\partial b, \ \partial F/\partial c, \ \partial F/\partial a ##.

    Then [tex]

    \frac {\partial f} {\partial b} = \frac {\partial} {\partial b} F (b, c, a(b, c, A)) = \frac {\partial F} {\partial b} + \frac {\partial F} {\partial a} \frac {\partial a } {\partial b}

    \\

    \frac {\partial f} {\partial c} = \frac {\partial} {\partial c} F (b, c, a(b, c, A)) = \frac {\partial F} {\partial c} + \frac {\partial F} {\partial a} \frac {\partial a } {\partial c}

    \\

    \frac {\partial f} {\partial A} = \frac {\partial} {\partial b} F (b, c, a(b, c, A)) = \frac {\partial F} {\partial a} \frac {\partial a } {\partial A}

    [/tex]

    This is a linear system; solve for ## \partial F/\partial b, \ \partial F/\partial c, \ \partial F/\partial a ##.
     
  6. Nov 17, 2012 #5
    Thanks a heap Voko, that did the trick! :)
     
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