How to define which variable is x and which variable is y axis?
5 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have an equation like "2*c^2 - 4*b = 0" with b ranging from 0 to 10. I want to plot a graph of b verses c with b on the x axis. How would I tell the program to plot b on the x axis and c on the y axis?
0 commentaires
Réponses (4)
Mischa Kim
le 7 Oct 2014
Modifié(e) : Mischa Kim
le 7 Oct 2014
John, simply use
plot(b,c)
The first argument is the x the second the y axis. Or, if you want to show both solution branches
b = 0:0.1:10;
c = sqrt(2*b);
plot(b,c,b,-c)
0 commentaires
Star Strider
le 7 Oct 2014
First, a bit of algebra to create ‘c’ as a function of ‘b’, then it’s simply another plot:
b = linspace(0,10);
c = [-sqrt(2*b); sqrt(2*b)];
figure(1)
plot(b, c)
grid
4 commentaires
Star Strider
le 7 Oct 2014
My pleasure!
The sincerest expression of appreciation here on MATLAB Answers is to Accept the Answer that most closely solves your problem.
Andrew Reibold
le 7 Oct 2014
Modifié(e) : Andrew Reibold
le 7 Oct 2014
Alternatively, PayPal is accepted too xD
Matt J
le 7 Oct 2014
Modifié(e) : Matt J
le 7 Oct 2014
f = @(V,b)V-(pi+atan(b./(1-b)).^(1./2)+atan((b+2).*(1-b)).^(1./2))./(1-b).^(1./2);
ezplot(f,[0,1.1,15,100]);
3 commentaires
Mohammad Abouali
le 7 Oct 2014
Modifié(e) : Mohammad Abouali
le 7 Oct 2014
Yes, with x being V and y being b. so xmin=0 xmax=15
and again put b=1.1 and check what your V turns out to be.
ff=@(b) (pi+atan(b./(1-b)).^(1./2)+atan((b+2).*(1-b)).^(1./2))./(1-b).^(1./2);
V would be a complex number
He wants 0<V<15 on x axis
Matt J
le 7 Oct 2014
Modifié(e) : Matt J
le 7 Oct 2014
Right you are (although I think it is the real part and not the norm that is being plotted). Here's a modification that fixes that issue,
function doPlot
ezplot(@(V,b) f(V,b),[0,15,0,1]);
function fval = f(V,b)
fval = V-(pi+atan(b./(1-b)).^(1./2)+atan((b+2).*(1-b)).^(1./2))./(1-b).^(1./2);
fval=fval/(imag(fval)==0);
end
end
Voir également
Catégories
En savoir plus sur Axis Labels dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!