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Experimental methodology

  1. Feb 18, 2012 #1
    I've been given a set of data from an experiment:

    x= 0, 1, 2, 5, 7, 10
    Y=(3), (5), (8,7), (25,4) (40), (66,2)

    My mission is to come up with a function that i can use, based on these values I've been given. I've tride to LN the values so I get a straight line (Y=kx +m), but I don't know what the next step is. Im stuck. Really need your help!

    Thanks,
     
  2. jcsd
  3. Feb 18, 2012 #2
    LN?
    Natural Log?

    What do you mean when you say y=(8,7)?
    Is there two different values for y?
    If there is then you're not going to get a function that gives you those values back..
     
  4. Feb 18, 2012 #3

    When x is 2 then y= 8,7
    when x is 5 then y= 25,4

    Yes, natural log. You need to come up with a function that works with the values I got. I think it's a power function.
     
  5. Feb 18, 2012 #4
    If you suspect that it is a power function then the starting point is to write a power equation relating y and x. What would you write?
     
  6. Feb 18, 2012 #5
    I think it shoul look like this: Y = C +ax^n
    The problem is to get the values a and n.
    C is given since y = 3 when x =0 --> C=3.

    I don't know what to do next, Ive tried a few things but now I'm just lost :S
     
  7. Feb 18, 2012 #6
    If you have y=3 when x=0 I would rewrite the equation as (y-3) = ax^n
    Then you can take Ln(y-3) and Ln(x) to plot a graph.
    If you get a straight line then the gradient =n and the intercept = Ln(a)
    You cannot take Ln(y) and Ln (C + ax^n)
    Hope this helps.
    I will have a go with your numbers and see what I get !!
     
  8. Feb 18, 2012 #7
    I got a pretty good straight line and my calculated gradient an intercept reproduced the table of values to better than 10%
    I will be interested to see what you get.
     
  9. Feb 18, 2012 #8
    I don't really understand what you will do from here:Then you can take Ln(y-3) and Ln(x) to plot a graph.

    What function did you come up with?
     
  10. Feb 18, 2012 #9
    Are you OK with the step
    Y = 3 + ax^n ? You cannot take Ln(3 + ax^n) so I rearranged the expression to be
    Y-3 = ax^n and this is the power law equatio that can be analyzed
    Now take logs......Ln(y-3) = Ln(a) + nLn(x)
    This equation is in the form of a straight line graph ( y = mx + C)
    So plot Ln( y-3) against Ln(x) and you should get a straight line.
    The slope of the line is n and where the line cuts the Ln( y-3) axis is Ln(a)
    You need to make another table to get the values ( y-3), Ln(y-3) and Ln(x)
    Hope this helps!
     
  11. Feb 19, 2012 #10
    I tried it for x =7, y=40.

    Ln (40-3) = ln (a) + n ln (7)
    3,6109 = ln (a) + n * 1,94

    The problem now is that I have two variables? I don't know how to figure them out. You told me to plot the result but It's difficult with two variables. Mabye Im just stupid?
     
  12. Feb 19, 2012 #11
    No one who can help me? :cry:
     
  13. Feb 19, 2012 #12
    I will write my working out for you and send it....give me about 10mins...dont despair
     
  14. Feb 19, 2012 #13
    Thanks a lot! :smile:
     
  15. Feb 19, 2012 #14
    here you go, not brilliant copy but I hope you can see what I did
     

    Attached Files:

  16. Feb 19, 2012 #15
    Thank you. That's similar to what I did the first time I tried but I guess I got a bit confused :) I managed to get a function that is very precise now :D Thank you for your valuable help! It feels great when you finally understand :D
     
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