How to define which variable is x and which variable is y axis?
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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?
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回答 (4 件)
Mischa Kim
2014 年 10 月 7 日
編集済み: Mischa Kim
2014 年 10 月 7 日
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)
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Star Strider
2014 年 10 月 7 日
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 件のコメント
Star Strider
2014 年 10 月 7 日
My pleasure!
The sincerest expression of appreciation here on MATLAB Answers is to Accept the Answer that most closely solves your problem.
Andrew Reibold
2014 年 10 月 7 日
編集済み: Andrew Reibold
2014 年 10 月 7 日
Alternatively, PayPal is accepted too xD
Matt J
2014 年 10 月 7 日
編集済み: Matt J
2014 年 10 月 7 日
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 件のコメント
Mohammad Abouali
2014 年 10 月 7 日
編集済み: Mohammad Abouali
2014 年 10 月 7 日
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
2014 年 10 月 7 日
編集済み: Matt J
2014 年 10 月 7 日
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
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