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Small change (derivative)

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Given that [tex]y=2+\frac{8}{x}[/tex]. Find the approximate change in x which will cause y to increase from 6 to 6.01


    2. Relevant equations
    [tex]\frac{\delta y}{\delta x}=\frac{dy}{dx}[/tex]


    3. The attempt at a solution
    I did it with two ways :

    1.
    when y = 6 --> x = 2
    when y = 6.01 --> [tex]x=\frac{800}{401}[/tex]

    approximate change in x = [tex]\frac{800}{401}-2 = -\frac{2}{401}[/tex]


    2.
    [tex]\delta y = 0.01[/tex]

    [tex]\frac{\delta y}{\delta x}=\frac{dy}{dx}[/tex]

    [tex]\frac{\delta y}{\delta x}=\frac{-8}{x^2}~ \text{where x = 2}[/tex]

    [tex]\delta x = -\frac{1}{200}[/tex]


    My question is : Can I use the first method? I don't think I can but I don't know the reason why I may not use it.
     
  2. jcsd
  3. Oct 5, 2009 #2

    rock.freak667

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    Homework Helper

    I think your first method assumes 'y' changed linearly with 'x'. But it's been a long time since I've some small changes. So it is always best to use the second method.


    Also it is δy/δx ≈ dy/dx, it is only equal when you take the limit δx→0
     
  4. Oct 5, 2009 #3

    Dick

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    Your first method gives you the EXACT value of the change. The second method gives you the APPROXIMATE value of the change. 1/200=0.005. 2/401 is about 0.00499. They are nearly equal, which is what you expect.
     
  5. Oct 5, 2009 #4
    Hi rock.freak667 and Dick

    Thanks a lot for you both ! :smile:
     
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