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Deriving an Equation from a graph

  1. Apr 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Hello everyone,

    I'm in need of some assistance in regards to deriving an equation from a graph. I have been tasked to graph information which I've collected through experimentation and then write an equation to represent the graph.

    I'm stuck at steps 3 and 4 and I am not really sure on how to approach the questions. I know that I have a power graph and if I square the frequencies that I obtained I come to a linear graph.

    Here is a direct link to the graphs that I've made:

    http://s7.photobucket.com/albums/y266/murdoc8888888/?action=view&current=hmwrk2.jpg




    2. Relevant equations


    [itex]Fc=(mv^2)/r[/itex]

    [itex]f= 1 / T [/itex]


    3. The attempt at a solution


    I'm not exactly sure on how to derive an equation from curved graph at all, the linear graph though is simple.

    (Y2 - Y1) / (X2 - X1)

    is the equation to find slope and from there I would use 'y' intercept form and derive that equation.

    I don't know how to do this with a power graph though.
     
  2. jcsd
  3. Apr 17, 2007 #2

    Hootenanny

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    Firstly, you only need to derive an equation from linear graph (the one with the straight line), not the curved one. So looking at you graph, you have a (roughly :s) straight line when you plot Fc vs. f2 and therefore you can say that that Fc[itex]\propto[/itex]f2. Now, you should be able to write an expression with a constant of proportionality, and what is the general expression for a straight line...?
     
  4. Apr 17, 2007 #3
    Well the general expression for a straight line is Ax + By + C = 0, or y = mx + b.


    I know it's a very rough linear graph but my teacher only wants the use of whole numbers such as 2 or 3 rather than decimals as exponents. So from here how would I write the proportionality statement of Fc[itex]\propto[/itex]f^2 using the equation of the linear graph?

    Fc[itex]\propto[/itex]f^2 if I remember correctly, becomes an equation such as [itex]Fc = Kf^2[/itex] where K is a constant. Is this true?

    P.S. I'm new to the latex system so please bare with me ...
     
    Last edited: Apr 17, 2007
  5. Apr 17, 2007 #4
    Well, on your straight line graph, what is represented by "y" in y=mx+b?
    What is represented by "x" in y=mx+b?
     
  6. Apr 17, 2007 #5
    y is the same as f (x) where f is centripetal force, and x represents frequency. So could I then say that Fc(frequency) = mx + b ? or am I missing something here?
     
  7. Apr 17, 2007 #6
    I was just wondering if my logic is correct.
     
  8. Apr 18, 2007 #7

    Hootenanny

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    Your very close but not quite there. What did you plot on your linear graph? Did you really plot Fc vs. f, or did you plot Fc vs. f2....?
     
  9. Apr 18, 2007 #8
    Alright,

    [itex]Fc v f^2[/itex] becomes an equation when you add a constant. I'll make it 'k', would 'k' then represent the slope of the linear line?

    [itex]Fc = kf^2 [/itex] ????
     
    Last edited: Apr 18, 2007
  10. Apr 18, 2007 #9

    Hootenanny

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    Looks good to me :approve:. Depending on your data you may have to add in a constant (which would be the y axis intercept + C), but this constant would be due to the standard error of your data. And yes k would represent the gradient of the line, if you have a proper line fitting program I would use that, but failing that excel does a reasonable job of plotting lines of bet fit.

    P.S. What is your academic level?
     
    Last edited: Apr 18, 2007
  11. Apr 18, 2007 #10
    I'm not sure what you mean academic level. I'm currently a senior student in high school.
     
  12. Apr 18, 2007 #11

    Hootenanny

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    Then my guess would be the line of best fit drawn by excel would be sufficient for High School level, but it might be worth checking with your teacher. :smile:
     
  13. Apr 18, 2007 #12
    I see I see,

    Nope my teacher won't accept excel, it needs to be hand drawn.

    Thanks for the help!
     
  14. Apr 18, 2007 #13

    Hootenanny

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    Ahh, the fun of hand plotting graphs :grumpy: It was a pleasure :smile:
     
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