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Understanding sensitivity of a variable using derivatives

  1. Oct 29, 2015 #1
    Hi,

    Please see the attached file.

    1. The problem statement, all variables and given/known data

    I am trying to figure out K’s sensitivity to x and y. I wish to know which variable K is more sensitive to. Both x and y are positive decimals less than 1.

    I want to be able to say a 1% increase in y results in a p% change in K. As the attached file shows, the derivatives are not helping me much here.

    How should I go about this problem?

    2. Relevant equations
    See the attached file.

    3. The attempt at a solution
    See the attached file.
     

    Attached Files:

  2. jcsd
  3. Oct 29, 2015 #2

    Ray Vickson

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    Science Advisor
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    If you type up your work I will be happy to look at it. Otherwise, not (along with many other helpers in this forum).
     
  4. Oct 29, 2015 #3

    Mark44

    Staff: Mentor

    Calculate the differential of K, which is ##dK = \frac{\partial K}{\partial x}dx + \frac{\partial K}{\partial y}dy##
    To get a value for the differential, you will need values for x, y, dx, and dy.

    BTW, everything you did in longhand and posted as a PDF can be done using LaTeX, and posted directly into this pane. We prefer to see the work here, rather than having to open a PDF or open a window on another site. Posting work as an image makes it difficult for us to insert comments at the precise point where an error appears. We have a LaTeX tutorial here -- https://www.physicsforums.com/help/latexhelp/
     
  5. Oct 29, 2015 #4
    Please see the problem in Latex form.


    ## K = \frac {1-y} {1-y-x} ## ---------------------- 1

    differentiating K by y

    ## \frac {dK} {dy} = \frac {(-1). (1-y-x) - (1-y) . (-1)} { (1-y-x)^2 } ##

    ## \frac {dK} {dy} = \frac {x} { 1 + y^2 +x^2 + 2xy -2x-2y } ## ---------------------- 2


    differentiating eqn 1 by x

    ## \frac {dK} {dx} = \frac {0 - (1-y) . (-1)} { (1-y-x)^2 } ##

    ## \frac {dK} {dx} = \frac {1-y} { 1 + y^2 +x^2 + 2xy -2x-2y } ## ---------------------- 3
     
  6. Oct 29, 2015 #5

    Mark44

    Staff: Mentor

    This is really a partial derivative, which in LaTeX looks like this: ## \frac{\partial K} {\partial y}##, and renders like this ##\frac {\partial K} {\partial y}##.
    Also, there's really no point in expanding ##(1 - y - x)^2##. Same comments for the work below.

    Otherwise, both partials look fine.
    Now plug things into the equation for the differential of K (dK) that I gave in post #3. Since you are holding x fixed, then dx = 0, but you'll need some values for x and y. If all you know is that x < 1 and y < 1, then you might be able to get an upper bound on the size of dK relative to a dy.
     
  7. Oct 29, 2015 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    I don't think you can make any convincing general statement like "a ##1\%## change in ##y## produces a ##p\%## change in ##K##". The effect of a ##1\%## change in ##y## depends very much on the base values of ##x## and ##y##, and things become especially tricky when the denominator is small (close to vanishing, but not quite), because in that case the derivatives may present a very misleading picture of the actual changes for given ##\Delta x## and/or ##\Delta y##.
     
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