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Is this differentiation correct?

  1. Sep 9, 2013 #1
    Hi,

    Would you be able to tell me if my differentiation in the attached file correct?
    Note that N, U and D are constants.

    I am trying to understand how I changes with a change in S.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Sep 9, 2013 #2

    HallsofIvy

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    No, it isn't.
     
  4. Sep 9, 2013 #3

    SteamKing

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    What happens to the exponent when we differentiate?
     
  5. Sep 9, 2013 #4
    Are you sure?
     
  6. Sep 9, 2013 #5
    Oh! How about this?

    dI/dS = (- N U/ D2 ) / ( S U/ D + 1)2
     
  7. Sep 9, 2013 #6

    SteamKing

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    That looks better.
     
  8. Sep 10, 2013 #7
    Thanks SteamKing. I am still struggling with one question:
    The reason behind this calculation was to understand the sensitivity of I to S. I wanted to know how much I changes for a 1% change in in S. It seems the differentiation doesn't answer that.

    What should I do to understand the sensitivity?

    Thanks.
     
  9. Sep 11, 2013 #8

    Mark44

    Staff: Mentor

    It's much simpler to write the equation in this form:
    I = N(US + D)-1

    Differentiating with respect to S yields
    dI/dS = -N * (US + D)-2 * U = ##\frac{-NU}{(US + D)^2}##

    Also, to find the sensitivity of I to small changes in S, write the equation above using differentials.


    dI = ##\frac{-NU}{(US + D)^2} dS ##

    For small changes in S, you can approximate dS by ΔS and dI by ΔI.
     
  10. Sep 17, 2013 #9
    How do I use this equation to find the change in I for a 1% change in S?
     
  11. Sep 17, 2013 #10
    What Mark is saying (I believe) is that if you replace dI with change in I (the answer you want to get) and dS with change in S (which you have; 1% = 0.01), then you're set. The concept is this: dI/dS is the change of dI over dS; the change of output compared to input. So, in dI/dS, set dS to what it is; the change in input, or 0.01. Now multiply both sides by 0.01 to get dI = (rest of equation). There, you've got a ratio! ^_^
     
  12. Sep 23, 2013 #11
    I am still a bit confused. Please see the attached file.
    N=100, U= 110, S = 9 and D=120

    With these values I = 9.01%
    Also these values give me dI/dS = -0.009

    Now, for dS = 1%, I get dI= -0.01%. However, this result doesn't make any sense.
    For example, if I actually increase S by 1% from 9.00 to 9.09, I changes from 9.01% to 8.93%, a decline of -0.08%, much higher than the 0.01% indicated above.

    What am I missing?
     

    Attached Files:

  13. Sep 23, 2013 #12

    Mark44

    Staff: Mentor

    You didn't show what you did to get your value of dI, so it's hard to say what you're missing. Using the formula I wrote in post #8, I get dI ≈ -8.93 X 10-5.

    I used the values you show in the PDF:
    N = 100
    U =110
    S = 9
    D = 120
    ΔS = .01
    So dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .01##

    Edit: Corrections to the above
    When S increases by 1%, that means that ΔS = .09, not .01.
    This results in a value for dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .09 ≈ -8.035 X 10^{-4}##
     
    Last edited: Sep 23, 2013
  14. Sep 23, 2013 #13
    Sorry about that. You are right, that is how I also calculated it dI= -0.0000893
    See the attached Excel file.
     

    Attached Files:

    Last edited: Sep 23, 2013
  15. Sep 23, 2013 #14

    Mark44

    Staff: Mentor

    Why are you converting to percentages? I think they are throwing you off.

    If S increases by 1%, ΔS = .09, so the corrected value for ΔI is about -.0008035.

    Calculating I directly with S = 9 results in .09009 (approx.)
    Calculating I directly with S = 9.09 results in .08929

    So I has decreased (making ΔI negative), and by direct calculation we see that ΔI = -.00080. This agrees reasonably well with the value obtained using differentials.

    Using the values I showed in post 12, I get dI ≈ -0.0000893

    Calculating I directly with S = 9, I get I ≈ .09009
    Increasing S by 1% (to 9.09), I get I ≈ .089293
     
  16. Sep 23, 2013 #15
    Got it. Thank you Mark44. With that correction, it is clear to me now.
    :thumbs:
     
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