# 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
on 25 Dec 2023

Answered: Sulaymon Eshkabilov
on 25 Dec 2023

[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 Comments

Dyuman Joshi
on 25 Dec 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.

### Accepted Answer

Sulaymon Eshkabilov
on 25 Dec 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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