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Homework Help: Exponential modeling of G-force

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Derive an equation of the form y=ax^b to model the given data: (35, 0.01) (28, 0.03) (20, 0.1) (15, 0.3) (11, 1) (9,3) (6, 10) (4.5, 30)

    2. Relevant equations

    Well, I know the answer is y = (7790 +/- 1246)x^(-3.698+/-0.1036) because that's what LoggerPro spits out, but I don't know how to derive it correctly without a program.

    3. The attempt at a solution

    Among the many, many attempts:

    y = Ax^b
    b = log(y)-log(A)
    ****** log(x)

    Then I inserted numbers from two different data points then set the equations equal to each other, resulting in:

    log(0.1)-log(A) = log(1)-log(A)
    *** log(20) ****** *** log(11)

    Which, after a step or two, became:

    log(20) = log(11)log(0.1) +log(11)
    ************* log(A)

    So, log(A) = log(11)log(0.1)
    *********** log (20) - log(11)

    Yielding an answer of A ~ -4.01

    However, that isn't right, so I didn't even try to solve for b.

    Could anyone please help? This is a big assignment, so sorry if I bump this thread a bit until I get help. And please take me through the steps, because I don't want to copy, I want to understand. I just need some help getting there. Thanks.

    P.S. Ignore the asterisks. They're there only to get the denominators in the right place.
  2. jcsd
  3. Apr 25, 2010 #2
    I would use the method of least squares.
  4. Apr 25, 2010 #3


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    so starting with
    [tex] y = ax^b [/tex]

    taking logs
    [tex] log(y) = log(ax^b) = log(x^b) + log(a)= b.log(x) + log(a)[/tex]

    so plotting up log(y) vs log(x) should be a graph of a straight line... if you can work out the line of best fit for that straight line, you should be able to cacaluate a & b from them...
  5. Apr 25, 2010 #4


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    if its really experimental data, the data is not perfect, so you can't use a single pair of data points - you should use the whole data set to find the line of best fit...

    as zachzach points of least squares is a good idea
  6. Apr 25, 2010 #5
    @ zach: Thanks, but no idea how to do that. And I posted this in the wrong subforum, because I'm in advanced pre-calc. So no calculus to help me please.

    @lane: it's not experimental. That's an interesting idea though. I'll try it. Thanks a lot.
  7. Apr 25, 2010 #6


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    then just plot the points on a log-log graph & draw a straight line of best fit on graph paer & work out the gradient & intercept
  8. Apr 25, 2010 #7
  9. Apr 25, 2010 #8
    Thanks again, but that looks like gibberish to me. I've never used sigma outside of physics, for example. Maybe it wouldn't be that hard to learn, but I did google least squares and none of it really made sense to me.
  10. Apr 25, 2010 #9


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    least squares is just a statistical method to give you the line of best fit through a set of data points

    its derived by calculus, but doesn't require any to use it...
  11. Apr 25, 2010 #10
    Good point.
  12. Apr 25, 2010 #11


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