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Frustrated by simple algebra problem

  1. Feb 22, 2014 #1
    1. The problem statement, all variables and given/known data
    For small changes in temperature, the formula for the expansion of a metal rod under a change of temperature is:
    L-Lo = aLo(t - to)
    L= length at temp. t
    Lo= initial length at temp. to
    a= constant that depends on metal
    A) express L as a linear function of t. Find the slope and y intercept(hint: treat the other quantities as constants.)


    2. Relevant equations
    Y=mx+b
    Y-Yo=m(x-xo)



    3. The attempt at a solution
    Solution 1?: the slope is aLo as stated by the equation, and the y intercept is Lo...
    Distribute
    L= aLot-aLoto+Lo
    L=Lo(at-ato+1)
    Doesnt work, still has variable to, lo, etc.

    Solution 2?: assume a=1
    L=Lo(t-to+1)
    Slope= Lo? Y intercept also lo?

    Solution 3: Lo, a and to=0
    L=0... doesnt work

    Or, pretend all constants=1
    L-1=1(t-1)
    L=t
    Seems too specific.

    I don't really know from which other angle to tackle this problem, I would appreciate some ideas, although I can imagine the answer should be obvious to me. I don't know how to turn this fully into y=mx+b. Thanks in advance.
     
  2. jcsd
  3. Feb 22, 2014 #2

    SteamKing

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    Are to, lo variables? Or are they constants which look like variables?
     
  4. Feb 22, 2014 #3
    They are constants i suppose, because they are the initial condition which doesnt change. But i dont understand what "treat them as constants" means... does it mean i could just ignore them?

    Tried another way..
    From L=aLot-aLoto+lo
    Y intercept= aLoto+lo
    Slope=aLo
     
    Last edited: Feb 22, 2014
  5. Feb 22, 2014 #4

    SammyS

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    That y intercept is -aL0t0 + L0 .

    Either way might be correct.

    The initial result is correct if the y-axis is at t = t0 .

    The result is the post quoted here is correct if the y-axis is at t = 0 .

    The slope is the same in both cases.

    The result you have in your Original Post makes more sense to me.
     
  6. Feb 22, 2014 #5

    Ray Vickson

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    Your final answer here is not quite correct. If we use 'intercept' to have its normal meaning, then you need Intercept = L_0 - a L_0 t_0 (not L_0 + a L_0 t_0 as you wrote).

    "Treating them as constants" means that they are not the variables in this problem, but certainly you cannot ignore them. The only "variables" here are L and t. In a specific problem the parameters L_0 and t_0 would be given some numerical values, but that would still allow L and t to vary.
     
    Last edited: Feb 22, 2014
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