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Solve partial derivatives from a table

  1. Sep 2, 2011 #1
    Let a represent the area, p the perimeter, d the diagonal, b the breadth, and L the length of a rectangle. One can easily write down from analytical geometry all the various relationships between the above variables, and from these obtain directly a variety of partial differential quantities.

    Thus, a = bL

    hence [itex]({\frac{\partial a}{\partial b}})[/itex]L = L, [itex]({\frac{\partial a}{\partial L}})[/itex]b = b, and so on, in this way a complete family of partial differentials could be evaluated. Alternatively, however, one partial differential can be obtained from another through the use of the various rules given above.

    The following table of partial differentials is given.

    [itex]({\frac{\partial a}{\partial L}})[/itex]b = b
    [itex]({\frac{\partial L}{\partial b}})[/itex]d = -(b/L)
    [itex]({\frac{\partial p}{\partial b}})[/itex]L = 2
    [itex]({\frac{\partial a}{\partial b}})[/itex]L = L
    [itex]({\frac{\partial d}{\partial b}})[/itex]L = (b/d)
    [itex]({\frac{\partial L}{\partial b}})[/itex]p = -1

    In the following problems you are asked to find the stated partial derivative first by the direct method, and then by using only the partial derivatives given in the table above, plus the various rules for interconverting derivatives.

    1. Find [itex]({\frac{\partial a}{\partial L}})[/itex]d


    I'm not going to type all my work out, but basically what I did was form an equation where the 3 variables are used and what I ended up with is

    a = (d2-L2)1/2L

    and taking the partial derivative of that, holding d as a constant and using the product rule, I end up with

    [itex]\frac{d^2 - 2L^2}{\sqrt{d^2 - L^2}}[/itex]


    Can someone confirm if what I did was correct? What I need help on is how to use the table to solve the problem, so if someone can give me a few hints on how to begin, it would be much appreciated.
     
  2. jcsd
  3. Sep 2, 2011 #2

    SammyS

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    What you have found is indeed, [itex]\displaystyle \left({\frac{\partial a}{\partial L}}\right)_{\!d}\,,[/itex] although I didn't see that in the table.
     
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