I have generated a plot from t=0->1, and I wish to append another plot from t=1->2. How can I do this?
Thanks in advance.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Will this do what you want? g1=Plot[1/2Sin[2Pi t],{t,0,1}] and then g2=Plot[1/3Cos[2Pi t],{t,1,2}] followed by Show[{g1,g2},PlotRange->{{0,2},{-1,1}}] $\endgroup$– BillCommented Jan 26, 2015 at 1:23
-
1$\begingroup$ Not exactly, both plots are over the same domain, 0->1. I need to show that the end of the first plot connects to the beginning of the second plot. Such as plotting Sin[x] on x=0->1 and then appending another Sin[x] x=0->1 but placing it on x=1->2 so it looks as 2 periods of Sin[x]. $\endgroup$– gKirklandCommented Jan 26, 2015 at 1:44
-
$\begingroup$ I never would have guessed this from your original wording. OK, use t from 1 to 2 (to satisfy Plot), but use t-1 as the argument for your second function (to satisfy the 0 to 1 domain of your second function.) Thus my example code is only changed by substituting g2=Plot[1/3Cos[2Pi (t-1)],{t,1,2}]. Think your old Algebra 1 class with translation of functions. $\endgroup$– BillCommented Jan 26, 2015 at 6:18
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Perhaps Translate?
p1 = Plot[1/2 Sin[2 Pi t], {t, 0, 1}]
Show[#, Graphics[Translate[#[[1]], {1, 0}]], PlotRange -> All] &@p1
