How to plot a 3 dimensional matrix against its last independent variables?

suppose that I have a variable 'u' which depends on 3 independent variables 'x', 'y' and 'z'.

If i,j,k indicate x,y,z steps respectively, like suppose that

n=10;

x=linspace(0,1,n);y=linspace(0,1,n);z=linspace(0,1,n);

for i=1:n
  for j=1:n
      for k=1:n
          u(i,j,k)=x(i)+y(j)+z(k);
      end
  end
end

I want to see the variation of 'u' with 'z'. How can I do that?

I know it will be a straight line in this case but how to plot the graph as plot command plots graphs by considering only first 2 independent variables.

[John Chilleri]

How you define time here? Is time simply the length of u or something else?

[Udit Srivastava]

I am sorry for my typing error. Its 'z' not time.

[John Chilleri]

Would a plot of u vs z not suffice? Since you would see how u changes with z?

[John BG]

Udit

wouldn't it be easier to start with a concise definition of u=f(x,y,z)?

is it possible for you to code, if only approximately, the function you write about?

[Udit Srivastava]

for a fixed value of x and y, using plot(z,u(3,5,:)) gives the following error.

Error using plot Data cannot have more than 2 dimensions.

Answer 1

[John Chilleri]

Hello,

A simple solution to get around your problem is:

for i = 1:size(u,3)
   uplot(i) = u(3,5,i);
end
plot(z,uplot)

I believe that the reason it is encountering trouble is because you can think of a 3d array as sheets of 2d arrays, and although it can call x and y across a sheet, it would need to call one z value from each sheet ("2 dimensions") which it can't do, I would take this explanation with a grain of salt.

Hope this helps!

[Udit Srivastava]

Works great. Thanks!

One thing I would like to know how size(u,3) is helping. I mean size(u,3) should give 10 X 10 but it gives 10. Why so?