Hi everyone,
Trying to plot a graph unsucssesfully :((
alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Mx=linspace(0,50,0.0001);
Interaction_Curve=(Mx./Mx0).^alpha+(My./My0).^alpha-1
plot(Interaction_Curve,Mx)
Thank You Very much

2 Comments

You didn’t define My and you haven’t used the linspace() properly for Mx.
Shimon Katzman
Shimon Katzman on 24 Dec 2019
Edited: Shimon Katzman on 24 Dec 2019
So what is the correct way?

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 Accepted Answer

One problem is that ‘My’ is missing. Beyond that, the linspace arguments resulted in an empty vector for ‘Mx’.
It might be easier to plot this as an implicit function:
alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
figure
fimplicit(Interaction_Function, [0 50 0 30])
ylim([0 30])
producing:
1plotting a simple Graph - 2019 12 24.png
That seems to produce the sort of plot you want. Make appropriate changes to the second agrument in the fimplicit call to get the result you want.

9 Comments

Hi Star,like always thank you so much.
The thing is that im using 2015a version and it errors me:
Undefined function or variable 'fimplicit'.
i used
ezplot(Interaction_Function, [0 50 0 30])
and now it works well.
Thank you very much Star
My pleasure.
The contour function is another option.
Try this:
alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Mx=linspace(0,50,25);
My=linspace(0,50,25);
[MX,MY] = meshgrid(Mx,My);
figure
contour(Interaction_Function(MX,MY), [0 0])
That should work. The contour function has been around as long as I can remember. It should allow the definition of the single contour at [0 0] in R2015a. (This imay be what the fimpllicit function does as well. It appears to produce the same plot.)
Hi Star,
Small questions for you please,
1)how do i find the intersection point between those two graphs?
2)how do i change the color of one of the graphs?
Thank YOU
alpha=2.2;
Mx0=M_x;
My0=M_y;
Mx = linspace(0,50, 1000);
My = linspace(0,50, 1000);
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
figure
hold on
ezplot(Interaction_Function,[0 40 0 20])
MxdivMy = @(Mx,My) Mx-3*My;
ezplot(MxdivMy, [0 40 0 20])
grid on;
As always, my pleasure!
I do not have ‘M_x’ and ‘M_y’, so I cannot run your latest code.
Oh, im sorry.. Mx0= 34.8262 My0= 15.7563
Not immediately obvious, so it took some time.
Try this:
alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Mx=linspace(0,50,25);
My=linspace(0,50,25);
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
MxdivMy = @(Mx,My) Mx-3*My;
[MX,MY] = meshgrid(Mx,My);
figure
hc1 = contour(Mx, My, Interaction_Function(MX,MY), [0 0])
hc2 = contour(Mx, My, MxdivMy(MX,MY), [0 0])
xy1 = hc1(:,2:end); % ‘Interaction Function’
xy2 = fliplr(hc2(:,2:end)); % ‘MxdivMy’
[xy2u,ix] = unique(xy2(1,:), 'stable'); % Unique ‘MxdivMy’
xy2i = interp1(xy2u(1,:), xy2(2,ix), xy1(1,:)); % Interpolate To Same x-Coordinates
xy2i = [xy1(1,:); xy2i];
xint = interp1(xy1(2,:)-xy2i(2,:), xy1(1,:), 0); % Interpolate To Find X-Intercept
yint = interp1(xy1(1,:), xy1(2,:), xint); % Interpolate To Find Y-Intercept
alpha=2.2;
% Mx0=M_x;
% My0=M_y;
Mx = linspace(0,50, 1000);
My = linspace(0,50, 1000);
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
figure
hold on
ez1 = ezplot(Interaction_Function,[0 40 0 20]);
MxdivMy = @(Mx,My) Mx-3*My;
ez2 = ezplot(MxdivMy, [0 40 0 20]);
plot(xint, yint, 'pr', 'MarkerSize',10, 'MarkerFaceColor','g') % Plot Intercept
grid on;
hold off
That calculates and plots the intercept.
Wow Star. Thank you so much!!
You Are The Best!
As always, my pleasure!
I very much appreciate your compliment!

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More Answers (1)

Try this:
alpha = 2.2;
Mx0 = 34.8262;
My0 = 15.7563;
Mx = linspace(0,50, 1000);
My = linspace(0,50, 1000); % Not sure what My should be!!!
Interaction_Curve = (Mx./Mx0).^alpha+(My./My0).^alpha-1
plot(Mx, Interaction_Curve, 'b-', 'LineWidth', 2)
grid on;
Be sure to define My because I just guessed incorrectly.

2 Comments

Hi, it doesnt plot the right graph :(
I know. Because I don't have the value of the My variable. That's why I asked you to define it. What is it? But doesn't matter since it looks like Star figured it out.

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