A little help with some basic calculus

Feb 06, 2005 2 Replies
S

Okay, this is a homework problem, but I'm middle-aged and already have my B.A., and just returning to school to pick up a little more knowledge. Besides, you can clearly see (below) that I'm giving this my best shot. So, someone please help me out here. For a basic electronics class, we are given that the temperature coefficient for the resistance of a material is given by:



a (for alpha) = (1/R)(dR/dT), and we are asked to show that:



R2 = {[ 1 + a(T1 - Ts) ] / [1 + a(T2 - Ts)]} R2



where Ts is the "reference temperature".



But here is how the math works out for me:



a = (1/R)(dR/dT)



dR/R = a dT Take indefinite integral of both sides....



ln R = a T + Ts, where Ts is said reference temperature



Assume R1 corresponds to T1, and R2 to T2, then....



ln R1 = a T1 + Ts, and ln R2 = a T2 + Ts



ln R1 - ln R2 = a (T1 - T2), so Ts is already gone.



ln (R1/R2) = a (T1 - T2), exponentiate both sides...



R1/R2 = exp (a [T1 - T2])



exp x is approximately 1 + x, so we have,



R1 = R2 { (1 + a [T1 - T2]) }, which is not the professor's desired result. Did the professor screw up, or have I forgotten some basic calculus or algebra in twenty years since college? Thanks in advance for all replies....



Steve O.


"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.

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A

This is wrong. The constant of integration isn't Ts.

Your maths looks fine to me, apart from that constant of integration - which cancels-out anyway

Maybe your teacher did it this way :

[ ln R ] Limits Rs -> R = [ aT ] Limits Ts -> T

ln (R1/Rs) = a[T1-Ts] ln (R2/Rs) = a[T2-Ts]

R1/Rs = exp( a[T1-Ts] ) R2/Rs = exp( a[T1-Ts] )

If we apply the approximation now ...

R1/Rs ~ 1 + a[T1-Ts] eqn 1 R2/Rs ~ 1 + a[T1-Ts] eqn 2

Divide eqn 1 / eqn 2

R1/R2 = { 1 + a[T1-Ts] } / { 1 + a[T2-Ts] }

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J

On Sun, 06 Feb 2005 04:57:47 GMT, Steven O. wrote: If I follow the intent, I get:

dR/R = a dT ln R + C1 = a*T + C2 ln R = a*T + C2 - C1 ln R = a*T + C

solve C for initial conditions, which is Rref and Tref

C = ln Rref - a*Tref

Plugging back in, we get:

ln R = a*T + ln Rref - a*Tref ln R = a*T - a*Tref + ln Rref ln R = a*(T - Tref) + ln Rref ln R - ln Rref = a*(T-Tref) ln (R/Rref) = a*(T-Tref) R/Rref = e^(a*(T-Tref)) R/Rref = e^(a*T)/e^(a*Tref)

or,

R/Rref = A*e^(a*T), where A = e^(-a*Tref)

I think that puts me in rough agreement with you, to this point. Substituting

1+x for e^x:

R/Rref = (1+a*(T-Tref)) R = RRef * (1+a*(T-Tref))

I suppose if there were two different R's, say R1 and R2, then their ratio would be:

R1/R2 = [RRef * (1+a*(T1-Tref))] / [RRef * (1+a*(T2-Tref))] R1/R2 = [(1+a*(T1-Tref))] / [(1+a*(T2-Tref))]

That would look about like your teacher's ratio, wouldn't it?

Jon

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