Solve systems of equations graphically

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Rytis Lekstutis
Rytis Lekstutis2021-11-1
评论: Walter Roberson ,2021-11-1
Hello, i've searched many webpages and yet can't find a specific and easy answer, so im writting here and hoping to get the answer. How do i plot them to see where the curves crosses graphically?
There is the system:

回答(3 个)

Bjorn Gustavsson
Bjorn Gustavsson 2021-11-1
You can start investigating the surfaces of the functions f1(x,y) = x.^2+y.^2-29 and f2(x,y) = y.^2-4*x.^2-9 with something like this:
[x,y] = meshgrid(-10:0.01:10,-10:0.01:10);
surf(x,y,x.^2+y.^2-29,ones(size(x))),shading flat
hold on
surf(x,y,-4*x.^2+y.^2-9),shading flat
contour(x,y,x.^2+y.^2-29,[0 0],'r'),shading flat
contour(x,y,-4*x.^2+y.^2-9,[0 0],'b'),shading flat
axis([-10 10 -10 10 -100 50])
view(-37.5,70)
Then from that point you can go whatever direction you see fit. You can for example use this nice tool found on the FEX: fast-and-robust-curve-intersections.
HTH

Walter Roberson
Walter Roberson 2021-11-1
syms x y
fimplicit([x^2 + y^2 == 29, y^2 - 4*x^2 == 9], [-8 8 -8 8])
  1 个评论
Rytis Lekstutis
Rytis Lekstutis 2021-11-1
Thank you for the response. One problem is that this fimplicit function doesn't work on Octave and i must do it on Octave. Is there any other easy way to do it?

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Dave B
Dave B 2021-11-1
I think you're asking how to plot these two implicit functions: you can do that with fimplicit. To see where they cross, I suppose I'd probably make the axis limits a bit bigger, and a grid might be useful?
fimplicit(@(x,y)x.^2+y.^2-29)
hold on
fimplicit(@(x,y)y.^2-4*x.^2-9)
axis([-10 10 -10 10])
grid on
  2 个评论
Walter Roberson
Walter Roberson 2021-11-1
Octave is off-topic here. You can find Octave discussion lists mentioned at https://savannah.gnu.org/mail/?group=octave
(Sorry, but the purpose of Free Software Foundation projects such as Octave is to ruin the market for proprietary software. FSF believes that softwar ecompanies can and should survive solely by people hiring them to write software or to answer questions. So, if you have a question about Octave, you should be paying someone to answer it for you.)

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