# contours of an implicit function f(x,y,z)=0 in MATLAB

I've already known how to plot a 3d implicit function f(x,y,z)=0 by using the isosurface function. Now I'm curious about how to draw contours of it. Like this one:

f(x,y,z) = sin((x.*z-0.5).^2+2*x.*y.^2-0.1*z) - z.*exp((x-0.5-exp(z-y)).^2+y.^2-0.2*z+3)

## Answers

You could numerically run over Z and look for when the sign changes, creating a matrix that holds the Z value, it's not elegant, but it works.

%Create function to evaluate eq=eval(['@(x,y,z)',vectorize('sin((x*z-0.5)^2+2*x*y^2-0.1*z) - z*exp((x-0.5-exp(z-y))^2+y^2-0.2*z+3)'),';']) %Create grid of x and y values [x,y]=meshgrid(0:0.01:15,-2:0.01:2); %Create dummy to hold the zero transitions foo=zeros(size(x)); %Run over Z and hold the values where the sign changes for i=0:0.001:0.04 aux=eq(x,y,i)>0; foo(aux)=i; end %Contour those values contour(foo)

Edit: I found a slightly more elegant solution using the function scatteredInterpolant

%Create function to evaluate eq=eval(['@(x,y,z)',vectorize('sin((x*z-0.5)^2+2*x*y^2-0.1*z) - z*exp((x-0.5-exp(z-y))^2+y^2-0.2*z+3)'),';']); %Create grid to evaluate volume [xm,ym,zm]=meshgrid(0:0.1:15,-2:0.1:2,-0.01:0.001:0.04); $Find the isosurface s=isosurface(xm,ym,zm,eq(xm,ym,zm), 0); $Use the vertices of the surface to create a interpolated function F=scatteredInterpolant(s.vertices(:,1),s.vertices(:,2),s.vertices(:,3)); %Create a grid to plot [x,y]=meshgrid(0:0.1:15,-2:0.1:2); %Contour this function contour(x,y, F(x,y),30)