Modifying an equation to plot a straight line.


by thatguythere
Tags: equation, line, modifying, plot, straight
thatguythere
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#1
Feb10-13, 11:47 PM
P: 91
1. The problem statement, all variables and given/known data
Show all the steps required to modify the equation so that the plot yields a straight line.
N/N0=[itex]e\ [/itex]-ux

This equation demonstrates the fraction radiation absorbed by a material, where "N0" is the number of incident photons from the radioactive source without any absorbed introduced, "N" is the number of transmitted photons, "u" is the absorption coefficient of the absorber (units m-1) and x is the thickness of the absorber.

2. Relevant equations



3. The attempt at a solution
Subsequently I am instructed to graph lnN/N0 vs "x". Therefore I am assuming N/N0 will be my "y" value in the equation of the line, whereas "x" (the thickness) will be my "x" value. I tried something along these lines, but have no idea if I am even in the ballpark.

N/N0=e-ux
lnN/N0=-ux

That seems to give me lnN/N0 as "y", -u as "m", x as "x" and 0 as "b". Please give me some thoughts. Thank you.
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tms
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#2
Feb11-13, 12:38 AM
P: 501
Well, do you get a straight line when you plot it? If you do, you're done.
thatguythere
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#3
Feb11-13, 12:45 AM
P: 91
The problem is that I have no data to plug into the equation and verify it because this is a question that I need to have answered before we do the experiment.

ehild
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#4
Feb11-13, 12:46 AM
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Modifying an equation to plot a straight line.


Quote Quote by thatguythere View Post
1. The problem statement, all variables and given/known data
Show all the steps required to modify the equation so that the plot yields a straight line.
N/N0=[itex]e\ [/itex]-ux

This equation demonstrates the fraction radiation absorbed by a material, where "N0" is the number of incident photons from the radioactive source without any absorbed introduced, "N" is the number of transmitted photons, "u" is the absorption coefficient of the absorber (units m-1) and x is the thickness of the absorber.

2. Relevant equations



3. The attempt at a solution
Subsequently I am instructed to graph lnN/N0 vs "x". Therefore I am assuming N/N0 will be my "y" value in the equation of the line, whereas "x" (the thickness) will be my "x" value. I tried something along these lines, but have no idea if I am even in the ballpark.

N/N0=e-ux
lnN/N0=-ux

That seems to give me lnN/N0 as "y", -u as "m", x as "x" and 0 as "b". Please give me some thoughts. Thank you.
It is all right, but you have to write N/N0 in parentheses.

ln(N/N0)=-ux

ehild
tms
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#5
Feb11-13, 01:23 AM
P: 501
Quote Quote by thatguythere View Post
The problem is that I have no data to plug into the equation and verify it because this is a question that I need to have answered before we do the experiment.
Just let x run from 0 to the total thickness.
Redbelly98
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#6
Feb11-13, 03:26 PM
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Quote Quote by tms View Post
Just let x run from 0 to the total thickness.
That's not a bad suggestion. Just plug into Excel or other graphing program, and look at the graph. You'd need to make numbers up for u and N0, but it would show whether you can expect a straight line.

Quote Quote by ehild View Post
It is all right, but you have to write N/N0 in parentheses.

ln(N/N0)=-ux

ehild
Exactly.
tms
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#7
Feb11-13, 03:36 PM
P: 501
Quote Quote by Redbelly98 View Post
That's not a bad suggestion. Just plug into Excel or other graphing program, and look at the graph. You'd need to make numbers up for u and N0, but it would show whether you can expect a straight line.
In my universe, all constants equal 1.
CompuChip
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#8
Feb11-13, 03:44 PM
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Quote Quote by thatguythere View Post
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That seems to give me lnN/N0 as "y", -u as "m", x as "x" and 0 as "b". Please give me some thoughts. Thank you.
That sounds good. This is what is called a logarithmic plot (or a log-linear plot, to indicate that one of the axes is logarithmic and the other is just linear). An exponential function such as the one you have for N, will become a straight line in such a plot. This is particularly useful since for a relatively small range of x, the y-values usually go through a wide range of values and taking the logarithm keeps it all a bit more easily viewable.

Note that you can also use the property ln(a/b) = ln(a) - ln(b) to rewrite the equation to
ln(N) = - u x + ln(N0)
in which case you will get a log-linear plot for N and your starting value "b" will be the initial value N0 (although on your logarithmic y-axis, you will actually plot ln(N0).

[edit]Here is another example of a log-plot:

Note how equal distances on the y-axis correspond to multiplications instead of additions, in other words, they've plotted the log10 of the actual quantity.

(Googled it from http://www.eyephysics.com/tdf/models.htm)


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