# Quick question - Basic differential techniques used or not used

1. Dec 20, 2013

### binbagsss

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:

(d(L)/(dT.L))=(-dθ/dT).cotθ

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. Dec 20, 2013

### brmath

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 (cosÎ¸ ,sinÎ¸)?

In your expression ((dÎ¸/dT)cos) I presume you mean cosÎ¸ ?

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

Re your first question, I see the dÎ¸/dT at the point Î¸ = 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 dÎ¸/dT. In each case you are estimating dL/dT at the particular Î¸ you are using. I would presume they would not be the same, but since the Î¸ '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 Î¸ on the right side.

3. Dec 21, 2013