I want to get a scaled vector field $\vec F=(x^2-x)i+(y^2-y)j$. The Fig should be like the attached Fig:
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Atom
il 25 Dic 2023
Risposto: Sulaymon Eshkabilov
il 25 Dic 2023
I want to get a scaled vector field $\vec F=(x^2-x)i+(y^2-y)j$. The Fig should be like the attached Fig:
[x,y] = meshgrid(-2:.16:2,-2:.16:2);
Fx = (x.*x-x);
Fy=y.*y-y;
figure;
quiver(x,y,Fx,Fy,'k','linewidth',1.2)
hold on
x=-2:0.01:2;
y=1-x;
plot(x,y,'k','linewidth',2);
Note that we have $div \vec F>0$ for $x+y>1$, $div \vec F<0$ for $x+y<1$ and $div \vec F=0$ on the line $x+y=1$ .
3 Commenti
Dyuman Joshi
il 25 Dic 2023
What is the vector field scaled to?
The arrows of the vector field in the reference image appear to be of the same length.
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Sulaymon Eshkabilov
il 25 Dic 2023
Is this what you are trying to obtain:
[x, y] = meshgrid(-2:0.25:2); % x and y values
F_x = 1*(x.^2 - x); % The vector field of x
F_y = 1*(y.^2 - y); % The vector field of y
F = F_x+F_y;
% Normalize the vectors
magnitude = sqrt(F_x.^2 + F_y.^2);
F_x_Nor = F_x ./ magnitude;
F_y_Nor = F_y ./ magnitude;
% Plot the scaled vector field
quiver(x, y, F_x_Nor, F_y_Nor);
hold on
X=-2:0.1:2;
Y=1-X;
plot(X,Y,'k','linewidth',2);
xlabel('x');
ylabel('y');
title('Scaled Vector Field: F = (x^2 - x) + (y^2 - y)');
axis([-2 2 -2 2])
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