Obtain x and y variables for a given value of z which is a function of both
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Zhenteng Shen
on 22 May 2020
Commented: Ameer Hamza
on 22 May 2020
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)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/298163/image.png)
2 Comments
Stephen23
on 22 May 2020
"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?
Accepted Answer
Ameer Hamza
on 22 May 2020
Edited: Ameer Hamza
on 22 May 2020
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);
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