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Exponential Graph

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Is a function that has a square root an exponential graph since you can rewrite a square root as x^1/2?


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2014 #2

    Ray Vickson

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    I cannot figure out what you are asking, since your question is not written as a proper sentence.

    However, you seem to be confusing "exponential" with "exponent"; exponent = power n in the formula x^n. This is a "power" function, not an "exponential" function.
     
  4. Jan 30, 2014 #3
    ... So is a square root function linear?
     
  5. Jan 30, 2014 #4

    BvU

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    Nope. "function(a times x)" is not equal to "a times function (x)"
     
  6. Jan 30, 2014 #5
    How would you classify this equation?: v=A√h
     
  7. Jan 30, 2014 #6

    Ray Vickson

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    Its a power law of the form ##v = A h^n## with ##n = 1/2##.
     
  8. Jan 30, 2014 #7
    Hm... I'm trying to write a lab report and the question asks if the graph is linear or exponential. When I plugged in my values for v (the y axis) and √h (the x axis)to excel the graph came up as looking linear, but with a very slight curve to it. I know I have my axes right because my lab sheet told us which values to put on each axis.
     
  9. Jan 30, 2014 #8

    Ray Vickson

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    You could do an excel "least-squares fit" to the data log(v) vs. log(h). If a power law ##v = A h^m## applies then you get a linear equation ##\log(v) = \log(A) + m \log(h)## between the new variables ##y = \log(v)## and ##x = \log(h)##. In the straight-line fit, the intercept would be log(A) and the slope would be the exponent, m.

    On the other hand, if the relationship is "exponential", that means that ##v = A b^h##, and you would get a linear relationship between ##y = \log(v)## and ## x = h## itself; that is, ##\log(v) = \log(A) + \log(b)\: h##.

    In EXCEL you can just find the best straight line fit to both data sets, and thus determine the best functional form for your data (together with parameter estimates for A, b, m, etc).
     
    Last edited: Jan 30, 2014
  10. Jan 30, 2014 #9

    BvU

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    If you measured v and varied h, you could try a plot of v2 as a function of h ?
    Error handling will be a bit more difficult then, though.
     
  11. Jan 30, 2014 #10
    I'll give it a shot, thanks for your help everybody!
     
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