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Can't plot unsolved function

Posted 11 years ago
I am trying to plot the amplitude squared, A^2, as a function of theta: 
A^2 == 0.000219952 Csc[\[CapitalTheta]]^2 Sin[   5.16831 Sin[\[CapitalTheta]]]^2

When I try the plot comes up with no data, just the axes.

Any suggestions?
POSTED BY: Amy Burnett
3 Replies
Posted 11 years ago
Perhaps this?
A[theta_] := 0.000219952 Csc[theta]^2 Sin[5.16831 Sin[theta]];
Plot[A[theta]^2, {theta, -90 Degree, 90 Degree}]
Does that appear to be doing what you were trying to accomplish?
If not then try explaining where you think I went wrong and we will try it again.

You may want to insert a, PlotRange->{stuff}  just inside that last ] where "stuff" tells Mathematica what y range you want or what x and what y range you want. Look up PlotRange in the help system if the plot without the PlotRange doesn't include the parts you want to see. I also included the Degree because I am guessing from your 90 that you are using degrees for theta.
POSTED BY: Bill Simpson
It is not clear what you are trying to actually plot.  Your expression is not syntactically correct Mathematica code. Perhaps it will be easier to understand if you use the formatting tools for this website's interface to place Mathematica code in its own area rather than in-line.  Also show your plotting expression so that we know what you are trying to attempt.... the more information you supply the easier it will be to figure out your problem.

But one thing that I notice right away is that you are using == rather than = in your assignment to A^2.  But also it is incorrect to assign a value to an expression such as A^2...  you must assign to a atomic parameter.   Thus if your paramter is "f" and it is assgined to an expression as follows (note thex in the arguments of the trig functions)

f = 0.000219952 Csc[x]^2 Sin[5.16831 Sin[x]]^2

you can plot as follows:
Plot[f, {x, 0, 10}]
POSTED BY: David Reiss
Posted 11 years ago
I am trying to Plot[A^2,{theta,-90,90}] but it does not come up.  I need to plot A^2 which depends on the value of theta.
POSTED BY: Amy Burnett
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