3D plot Between one known and two unknown parameter.

For a given equation, A.^2+B.^2=R

I need to plot between A,B and R where R is known(R lies between 1 to 10 for 100 resolutions) and A and B is not known.

Thanks in advance.

[Nathan Hardenberg]

Not quite sure if this is what you want. But this plots the resulting circles, in this case continuously R = [1,10]

fimplicit3(@(A,B,R) A.^2 + B.^2 - R)
axis equal
zlim([1,10]); xlim([-4,4]); ylim([-4,4])

[MADHVI]

Thanks, but this is a plot for A,B and the function. I need plot for A,B and R.

In 2D, it will definitely give a circle.

[Nathan Hardenberg]

hmm... I'm still not sure what you want. If you do want to have three "inputs" A,B and R there is no function anymore. It can be either true or false.

Example:

A = 1; B = 1; R = 2;
A^2 + B^2 == R      % is true

ans = logical
   1

A = 1; B = 1; R = 1;
A^2 + B^2 == R      % is false

ans = logical
   0

You could only plot the "true"-datapoints. But this would result in the same plot as above, if I'm not mistaken

[MADHVI]

Thank you.

Answer 1

[Dyuman Joshi]

"In 2D, it will definitely give a circle."

I assume you want to obtain all the circles corresponding to values in R in the same plot.

figure
hold on
%R lies between 1 to 10 for 100 resolutions
for R = linspace(1,10,100)
    fimplicit(@(x, y) x.^2+y.^2-R)
end
grid on
hold off

Edit - You can also achieve this from the code mentioned in the comments by changing the view. But note that the resolution of R will not be same as you mentioned and as in the code above.

figure
fimplicit3(@(A,B,R) A.^2 + B.^2 - R)
axis equal
zlim([1,10]); xlim([-4,4]); ylim([-4,4])
view(2)

[MADHVI]

Thank you.