Values not populating in table for variable

I have the following code:

clc
clear all
close all
%define some initial variables
theta = pi/4;
R = 0.0006;
w = 1.318*R;
format longg
%calculate the vector length, p, to the edge of the ellipse 
phi=0:0.01:2*pi;
p = (w*sin(theta)^2*cos(phi)+sqrt(R^2*sin(phi).^2+R^2*sin(theta)^2*cos(phi).^2-w^2*sin(theta)^2*sin(phi).^2))/(sin(phi).^2+sin(theta)^2*cos(phi).^2);
x = p.*cos(phi);
y = p.*sin(phi);

When I run this and look at the results in Workspace, the tables holding the values of x and y each have 629 columns with values. However, the table holding the values of p only has one column with one value.

It looks like Matlab is populating the x and y table using only the one value of p and then multiplying it by looping phi to yield the 629 entries. What I need is for Matlab to change "p" with each value of phi and then calculate the corresponding values of x and y. How do I get Matlab to calculate all of the values of p based on the preceding line of code which is:

phi=0:0.01:2*pi;

Answer 1

[Dyuman Joshi]

%clc
%clear all
%close all
%define some initial variables
theta = pi/4;
R = 0.0006;
w = 1.318*R;
format longg
%calculate the vector length, p, to the edge of the ellipse 
phi=0:0.01:2*pi;

There's a element-wise dot symbol missing between the numerator and denominator division sign/operator (scroll to right) -

%                                                                                                            v
p = (w*sin(theta)^2*cos(phi)+sqrt(R^2*sin(phi).^2+R^2*sin(theta)^2*cos(phi).^2-w^2*sin(theta)^2*sin(phi).^2))./(sin(phi).^2+sin(theta)^2*cos(phi).^2);
x = p.*cos(phi);
y = p.*sin(phi);
whos

  Name       Size             Bytes  Class     Attributes

  R          1x1                  8  double              
  p          1x629             5032  double              
  phi        1x629             5032  double              
  theta      1x1                  8  double              
  w          1x1                  8  double              
  x          1x629             5032  double              
  y          1x629             5032  double              

Now as you can see, p (along with x and y) is also same size as phi (1x629).

[Robert Demyanovich]

That did the trick. Thanks!