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Interpreting derivatives at a point

  1. Dec 16, 2015 #1
    1. The problem statement, all variables and given/known data
    How should I interpret the derivative of the following function at C=6%
    ##N = \frac {I} {C} - D##


    2. Relevant equations
    ## \frac {dN} {dC} = \frac {-I} {C^2} ##

    At I =12 and C=6%, I am getting
    ## \frac {dN} {dC} = 3,333 ##

    I am not sure what to make of this large number. I want to be able to say for every 1/100th increase in C, N changes by X dollars, or something like that. The following document, for example, shows one such example.
    http://pblpathways.com/calc/C11_3_4.pdf

    How can I interpret the 3,333 figure?
     
  2. jcsd
  3. Dec 16, 2015 #2

    Ray Vickson

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    You can think of ##dN/dC = -3,333.333...## (not +3,333) as the rate of change in ##N## per unit change in ##C##. In other words, if you were to increase ##C## from ## 0.06 = 6 \%## to ##.06+1 = 1.06 = 106 \%##, ##N## would be expected to undergo a large decrease.
     
  4. Dec 16, 2015 #3
    Thanks Ray.
    I am confused. Let's say, originally I =12, C=6%, and D=75
    As per the first equation N =125
    Now If I changed C from 6% to 106%, N would be -64. Not sure how that change is related to -3,333.
     
  5. Dec 16, 2015 #4

    Ray Vickson

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    I said the change would be expected to be large; I did not say it would be exactly -3333.3333. The ##r = -3333.333## figure is a rate, which means that for small ##\Delta C## the change in ##N## will be approximately ##r \Delta C##. However, the rate varies nonlinearly with ##C##, so you cannot expect to be able to apply that rate over an interval that is not "small". You can still expect the result to be large, however.

    To clarify: if you change ##C## from 6% to 6.1%, you are changing ##C## from 0.6 to .061, so ##\Delta C = 0.01##. You would estimate the change in ##N## as approximately ##-3,333.3333 \times 0.001 \doteq -3.333##. This is not too far from the actual change of ## -3.279##.

    The point (which you seem to have missed) is that the large value of the derivative is "misleading" in a way, and has to do with the units you are choosing (percent instead of the actual, numerial value).
     
  6. Dec 16, 2015 #5
    Thanks Ray. Now I get the most of it.
    I still however not clear on the 'misleading' part. I tried to manually calculate the change required in C to get -3,333 times decline in N, but couldn't find it.

    If a 0.001 increase in C results in a 3.3 decline in N, what change in C would result in a 3,333 decline in N?

    I also looked at this problem using different numbers, for example I=10,000, C=200.
    n this case ## \frac {dN} {dC} = -2.5 ##

    So we can interpret that if C changes from 200 to 201, N would decline by 2.5, which makes sense. I am not sure how to interpret the answer for the original problem.
     
    Last edited: Dec 16, 2015
  7. Dec 16, 2015 #6

    WWGD

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    Problem is that the rate of change is a local characteristic, although the rate may be a good approximation in a small neighborhood of the point (unless you have a linear function, in which case the rate of the change is constant). Still, for this approximation you can solve for DC/DN = -3,333.
     
  8. Dec 16, 2015 #7

    Ray Vickson

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    No increase in ##C## can lead to a 3,333 decrease in ##N##. Plot the graph of ##N = 12/C## for ##C > 0## to see why.
     
  9. Dec 16, 2015 #8
    Did you mean C < 0 ?
    When I use less than 0 values, I get -ve N values.
     
  10. Dec 16, 2015 #9
    Thanks WWGD
    I am not sure what you mean here. Can you please explain a bit more?
    Thanks.
     
  11. Dec 16, 2015 #10

    Mark44

    Staff: Mentor

    If C > 0 but close to zero, a very small change in C leads to very large decrease in N.

    I'm pretty sure Ray did not mean C < 0.
     
  12. Dec 16, 2015 #11

    Ray Vickson

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    No, I did not mean C < 0. We start from C = 0.06, and then look at increases (as I clearly stated). You will never get C < 0 that way.

    Anyway, if we look at decreases (down from C = 0.06) the derivative predicts a large rate of increase in N. This will keep happening until the system goes crazy by passing through ##C = 0## and encountering singularities.
     
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