Linear fit for data (resistance of a wire)

[Cade]

Homework Statement

The resistance for a wire with cross-sectional radius r is given by R = k/(r^2). Convert this equation to a straight line for which k can be determined from the slope.

Homework Equations

Equation of a line is:
y = mx + c

The Attempt at a Solution

R = k/(r^2)
ln(R) = ln(k/(r^2))
ln(R) = ln(k) - ln(r^2)
ln(R) = ln(k) - 2*ln(r)

I'm not sure how to linearize this equation, this is where I get stuck. ln(R) would represent y on the graph. k ought to equal the slope, but I can't see how to do that.

 

[technician]

What you have done is a perfectly good analysis but k is given by the intercept (lnk) on the graph.
The slope of the ln-ln graph is -2. This is the value of ln-ln graphs...enables you to determine the power law of a set of data.
Since R =k/r^2 a graph of R against 1/(r^2) will be a straight line with gradient k

 

[Cade]

The problem is that if I plot given values of R on the y-axis = R and given values of r on the x-axis = (1/(r^2)), my graph looks like this:

Edit: This is my dataset
R r
0.81 0.0001
0.73 0.00015
0.60 0.0002
0.38 0.00025
0.1 0.0003

 

[sophiecentaur]

Why not plot a graph of R against 1/r²?
That should give a straight line y=kx which should pass through the origin as when r = 0, so should R. The c would be zero.

It is handy to get used to the idea of choosing what to plot against what in order to get a straight line graph. I remember needing quite a few examples before it became 'obvious' what to do.

 

[Cade]

That graph I showed is R against 1/(r^2). My problem is that it is very curved, so I don't have a best-fit line.

 

[Cade]

If I use some known value of k to produce my own table of r and R from R = k/(r^2), I do get a linear fit, so I guess I'm missing something involving the dataset.

I have a similar small question, how would I convert this to a straight line such that a and b can be found? My known values are x and y.
y = a*(1 + b*(x^2))

y = a + a*b*(x^2)
ln(y) = ln(a) + ln(a*b*(x^2))
ln(y) = ln(a) + ln(a) + ln(b*(x^2)
ln(y) = 2ln(a) + ln(b*(x^2))
I can see that the intercept is 2ln(a), but I'm not sure how I would read a slope from ln(b*(x^2))

 

[technician]

I used your numbers and got something very similar! Where did you get the data, was it from measurements you made?

 

[Cade]

The data was given to me as part of the question by my instructor.

Could also you please also take a look at my second question in post #6?

 

[technician]

What was post #6? I am not certain how to find it

 

[Cade]

The numbers are marked on the right side of a post's header, opposite from the time and date of its posting.
https://www.physicsforums.com/showpost.php?p=3618807&postcount=6

 

[technician]

Got it (a bit on the slow side!)
You can't do this:
y = a + a*b*(x^2)
ln(y) = ln(a) + ln(a*b*(x^2))
You cannot add the ln of 2 numbers added together.When you add ln it means you are dealing with multiplication. You did this right in your post #1
For this equation I would leave it as:
y = a + abx^2
Try plotting y against x^2... what will be the gradient and what will be the intercept?

 

[Cade]

When I added, I was thinking of ln(x) + ln(x) = ln(x*x) = 2*ln(x)

D'oh... if the intercept is a and gradient is ab, then I can find b as gradient/a, thanks.

 

[technician]

you have got it
Cheers

 

[sophiecentaur]

That graph I showed is R against 1/(r^2). My problem is that it is very curved, so I don't have a best-fit line.

Sorry. I was being dumb. You are right. The graph looks so wrong that I assumed you'd not actually plotted it right. But the values certainly give that odd curve.

I think the data must be wrong - unless it relates to something that can get hot for thin samples and not following Ohms Law. You'll have to ask your tutor, I think.