How to plot this implicit function? (It's one parameter also changes)

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Binayak Behera
Binayak Behera 2022 年 9 月 11 日
編集済み: John D'Errico 2022 年 9 月 11 日
Here Φ also varies.
First i got that it is an implicit function. So, fimplicit should be used. But i'm unable to take Φ varied.

回答 (2 件)

Torsten 2022 年 9 月 11 日
編集済み: Torsten 2022 年 9 月 11 日
So you want several implicit plots in different colors for different values of phi ?
Use "fimplicit" in a loop using "hold on" before the loop.
phi = [pi/4 3*pi/8 pi/2 5*pi/8 3*pi/4];
hold on
for i=1:numel(phi)
fun = @(x,y) y.^4 + y.^2.*(2*x*tan(phi(i))-1) + x.^2.*sec(phi(i))^2;
hold off

John D'Errico
John D'Errico 2022 年 9 月 11 日
編集済み: John D'Errico 2022 年 9 月 11 日
The vertical axis indicates the value of phi.
F3 = @(x,y,phi) y.^4 + y.^2.*(2*x.*tan(phi)-1) + x.^2.*sec(phi).^2;
xlabel x
ylabel y
zlabel phi
So at any fixed phi, you will get a simple path through the (x,y) plane, indicating the solution locus for that value of phi.
And for a fixed set of values of phi, you can use fimplicit to get the curves as shown by @Torsten.


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