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Homework Help: Quick question - Basic differential techniques used or not used

  1. Dec 20, 2013 #1
    Okay, so for context, I am trying to find the linear expansion coeffient - d(L)/[dT.L] , where L is the length and T the temperature. I am given two data points, and asked to manipulate :

    (d(L)/dT).sinθ = -((dθ/dT).cos))

    In order to deduce the linear expansion coefficnet.
    The two data points given are: T1=293K, θ1=48.75°, T2=1200K, θ2=45.66

    My Problem Lies in the integration tecniques , not used, and how we deduce the changes in both variables:

    I agree with my books solution up to the point of attaining:


    Then it is just a matter of how you evaluate the RHS to deduce the quantity desired - LHS.

    Now my book then does, RHS:
    [cot 48.75°.(48.75-45.66).∏/180]/1200-293

    Which I am quite confused by:
    1) When we evaluate cot θ, how have we simply picked out one angle and neglected the other value?
    2) Why do we not integrate - when I do this and attain cot θ dθ = In sin θ, plug in the limits, and then divide by ΔT, in the same way as the method above - still not linking the variables, I get the incorrect answer !
    3) When I link the data points together, I.e attain a value for dθ/dT from the gradient of these two data points, and then staying in line with the method above - so multiplying by cot 48.75°, I again get an incorrect answer.

    Many,many thanks to anyone who can shed some light on this, greatly appreciated.

    Sorry if this is more of a physics problem
  2. jcsd
  3. Dec 20, 2013 #2
    I could help more if I better understood your notation. What is T.L? If L is the length, is it a function of T? Or are L and T independent variables? Finally, the introduction of the sin and cos suggest to me you have a 2 dimensional problem here - so what "length" do you mean? Do you mean area? Do you have a circle and mean the diameter? Do you mean the length of the line through (cosθ ,sinθ)?

    In your expression ((dθ/dT)cos) I presume you mean cosθ ?

    Sorry for all this -- I am not a physicist.

    Re your first question, I see the dθ/dT at the point θ = 48.75 is being approximated by the divided difference. Did you lose a minus sign in that expression?

    You could have used the other data point. Either of them gives you an approximation because you don't really know dθ/dT. In each case you are estimating dL/dT at the particular θ you are using. I would presume they would not be the same, but since the θ 's are close the approximation to dL/dT should be good either way.

    Re your integration, I don't understand exactly what you did. If you are treating it as a separation of variables problem, you do want all your θ on the right side.
  4. Dec 21, 2013 #3
    Thanks for your reply.
    Oh I understand now, thanks - the minus sign has been taken into consideration in the denominator - I have reversed the order of subtraction of data points 1 and 2 compared to the numerator.

    The integration... thanks I think I see now. So it wouldnt be possible to solve this via integration , as you would have to take the dT to the other side, which does not give the desired quantity (linear expansion coefficent) on the LHS?

    It was meant to be more of an approximation , so integrate over the θ, as we have an extra θ term, but for change in T, just taking the difference as a valid approximation, as in the above method.
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