Obtain x and y variables for a given value of z which is a function of both

Hello all, thanks for looking.

For example if I have composite material consisting of concentrations of A and B and their resulting measured stress value is Z (in picture below this is the matrix in green). Now given the value of z, how do I find the value of A and B which correspond to it?

Usually the interp2 would work if I know A and B and want to find Z value, but now I want to do the opposite. The z value is beyond the range of results too, so may need to be extrapolated.

I currently have the script below whcih plots a surface sketch but I want to essentially find x and y values that correspond to a specific z value. I assigned the A concentration as x, B concentration as y and the stress value of thir combination z.

So as an exmaple, I want to know x and y values that would make z=0.4.

Many thanks for your help!

x=[2.5; 5]

y=[1; 1.5]

z=[4.9 5.3; 13.8 14.4]

surf(x,y,z)

[Stephen23]

"Now given the value of z, how do I find the value of A and B which correspond to it? "

Even if your surface is nicely continuous there can be zero, one, some finite number, or infinite solutions to this. Consider your example data, for example if we look for values where z=10 then this means all x and y values on the line from (2.5,1.2865) to (5,1.2582), i.e. infinite points. Which of those infinite points do you want as the output?

[Zhenteng Shen]

I see, I completely understand what you are saying now, Ameer demonstrated it graphically for me in my other question. Thank you for taking your time to comment.

Answer 1

[Ameer Hamza]

As mentioned in the comment to your other question, if you try to extrapolate the surface that there is not a unique solution corresponding to a given z value. The following shows one of the ways; however, you can note that the solution is different every time you run it. It is because each initial guess to fsolve() will lead to a different solution.

x=[2.5; 5];
y=[1; 1.5];
z=[4.9 5.3; 13.8 14.4];

[X,Y] = meshgrid(x,y);
mdl = scatteredInterpolant(X(:), Y(:), z(:));
fun = @(x) mdl(x(1),x(2));
sol = fsolve(@(x) fun(x)-0.4, rand(1,2));
x_sol = sol(1);
y_sol = sol(2);

[Zhenteng Shen]

Hello again, sorry I wrote this question because I didn't want to trouble you too much on the previous post as you hadn't commented yet on the second part of original question. But I understand what you mean now.

For the code you posted (pasted below) which works perfectly for me, I assume I replace the 0.4 after fun(x)- with whatever value of z i want right to get x_sol and y_sol?

sol = fsolve(@(x) fun(x)-0.4, rand(1,2));

Thank you very much again for your patience and help!

[Ameer Hamza]

Correct, you just need to replace 0.4 with the z value of your choice. However, note that the solution is not unique if you use rand(1,2) as the staring point.

[Zhenteng Shen]

Perfect! Thanks a lot! :D

[Zhenteng Shen]

Hi Ameer, just an update I have found out which starting points to use in place of rand(1,2) for fsolve so will create a 1x2 matrix of the starting x value and y value to use. In this case your answer is exactly what I was looking for!!! I can't believe it, THANK YOU SO MUCH AGAIN!!!

[Ameer Hamza]

I am glad to be of help!