How do I find the phase of any point in the graph below

x=[0:10:91]; str={'Phase 1','Phase 2'}; y=[0.01*(x.^2)]; xt=[30 70]; yt=[60 20]; plot(x,y,'k','LineWidth',5),xlabel('Composition'),ylabel('Temperature');

[Yazan]

Your question is not well-defined. How do you define the phase for your data and what does this data represent?

[Owuraku Amankwah]

The data represents a basic phase diagram. The x values are my composition values and those of y are my temperature values.

[Owuraku Amankwah]

I'm supposed to find the points above and below the curve and also write a code that shows whether any input, that is température and composition, is above or below the curve drawn

Answer 1

[DGM]

Considering that there's only one boundary of interest, it's fairly simple.

x=0:10:91; 
y=0.01*x.^2; 
xt=[30 70]; 
yt=[60 20]; 

str={'Phase 1','Phase 2'}; 
plot(x,y,'k','LineWidth',5); hold on;
plot(xt,yt,'*')
xlabel('Composition');
ylabel('Temperature');

isabovecurve = yt>(0.01*xt.^2)

That will give you a logical vector describing where each point is WRT the boundary curve.

If you don't like repeating the expression, you could do something like:

x=0:10:91; 
fy = @(xx) 0.01*xx.^2; % define y as a function
y=fy(x); % and just evaluate it
xt=[30 70]; 
yt=[60 20]; 

% do all the plotting stuff

isabovecurve = yt>fy(xt) % and evaluate it again with different points

If your problem is more complicated than this example (i.e. more boundaries), then this probably isn't going to help much.

[Owuraku Amankwah]

Any other alternatives will be helpful

[Owuraku Amankwah]

How do I do so for a problem with more boundaries