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Steven O.
Guest
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 Posted:
Sun Feb 06, 2005 8:40 am Post subject:
A little help with some basic calculus |
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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.
www.SpyingOnTheCollegeOfYourChoice.com
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Ari
Guest
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| Posted:
Sun Feb 06, 2005 8:40 am Post subject:
Re: A little help with some basic calculus |
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From
ln R1 = a T1 + Ts and ln R2 = a T2 + Ts
exponentiate both sides giving
R1 = exp(a T1 + Ts) and R2 = exp(a T2 + Ts).
Now use your "small eXponent approximation" e^X = 1 + X
giving
R1 = 1 + a T1 + Ts and R2 = 1 + a T2 + Ts.
Now form the ratio R2/R1, then isolate R2.
Aristotle Polonium
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Badger
Guest
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| Posted:
Sun Feb 06, 2005 8:40 am Post subject:
Re: A little help with some basic calculus |
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On Sun, 06 Feb 2005 04:56:37 GMT, Steven O. <null@null.com> wrote:
[snip]
| Quote: | 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
|
I'll make a guess here that your "constant of integration"
could be ( - a Ts ), so that
ln R = a T - a Ts = a (T - Ts)
exp[ ln R ] = exp[ a (T - Ts) ]
exp[ a (T - Ts) ] = 1 + a (T - Ts) approx., so that
R = 1 + a (T - Ts)
If the Prof will buy into that, then
| Quote: | Assume R1 corresponds to T1, and R2 to T2, then....
ln R1 = a T1 + Ts, and ln R2 = a T2 + Ts
|
R1 = a (T1 - Ts) and R2 = a (T2 - Ts)
R1/R2 = [ a (T1 - Ts) ] / [ a (T2 - Ts) ]
R1 = { [ a (T1 - Ts) ] / [ a (T2 - Ts) ] } R2
As I said, just a guess on my part, FWIW.
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Ari
Guest
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| Posted:
Sun Feb 06, 2005 8:40 am Post subject:
Re: A little help with some basic calculus |
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"Steven O." <null@null.com> wrote in message news:b68b011nureekgu9ghd8hfgbrbcmvqkt9q@4ax.com...
| Quote: | 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.
www.SpyingOnTheCollegeOfYourChoice.com
|
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Is the equation
R2 = {[ 1 + a(T1 - Ts) ] / [1 + a(T2 - Ts)]} R2
correct? Should R2 appear on both sides of the equation?
Aristotle Polonium
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + |
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Steven O.
Guest
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| Posted:
Mon Feb 07, 2005 2:10 am Post subject:
Re: A little help with some basic calculus |
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Thanks to everyone who helped out on this one.
Steve O.
On Sun, 06 Feb 2005 04:56:37 GMT, Steven O. <null@null.com> wrote:
| Quote: | 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.
www.SpyingOnTheCollegeOfYourChoice.com
|
"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.
www.SpyingOnTheCollegeOfYourChoice.com |
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