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Plot multivariable function,can't get the plot right ?

Asked by Huy Nguyen on 22 May 2019
Latest activity Commented on by Huy Nguyen on 23 May 2019
I need to plot a multivariable (x,y) function in matlab and find its critical points. I plotted it but can't seem to get the correct plot as shown in the picture.
[x,y] = meshgrid(-5:.2:5);
f=3.*x.*exp(y)-x.^3-exp(3.*y);
fa1 = gradient(f, 0.2, 0.2); % Derivative
zv = contour(x,y,fa1, [0; 0]); % Critical Points
figure(1)
surf(x,y,f)
hold on
plot3(zv(1,2:end), zv(2,2:end), zeros(1,size(zv,2)-1), 'r', 'LineWidth',2)
hold off

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1 Answer

Answer by Are Mjaavatten on 23 May 2019
 Accepted Answer

The main problem with your plot is simply that the large values of f for y > 0.5 dwarf the variations at lower values. Changing the plotting range solves this problem, I also rotate the axes to make the graph more similar to the original.
[x,y] = meshgrid(-1.5:.1:1.5,-2:0.1:0.5);
f=3.*x.*exp(y)-x.^3-exp(3.*y);
figure(1)
surf(x,y,f)
xlabel x
ylabel y
view(136,3)
The critical points are where both partial derivatives are zero. You can visually find one solution by plotting the zero contours for bot in the same graph:
[fa1,fa2] = gradient(f, 0.2, 0.2); % Derivative
figure(1);
contour(x,y,fa1, [0; 0]);
hold on;
contour(x,y,fa2, [0; 0]); % maximum where df/dx = df/dy = 0
grid on % we find one such point at (1,0]
hold off
You can also find analytical expressions for the partial derivatives and solve using fsolve from the optimization toolbox. Here I use an anonymous function J. z(1) is x and z(2) is y.
J = @(z) [3*exp(z(2))-3*z(1)^2;3*z(1)*exp(z(2))-3*exp(3*z(2))]
zc = fsolve(J,[2;0])

  1 Comment

Thank you so much kind sir

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