Finding exact point on the surface

Hi all,

I have a 3D surface obtained from the equation y = f(x, z). I'm wondering how I can determine the exact value of z when I have the values of x and y. It's challenging to find the equation for z = g(x, y), which is why I created the surface based on x and z.

Essentially, I have a rectangular region on the x-y plane, and I need to project it onto the surface. To achieve this, I need to calculate the exact z values for each corner of the rectangle based on the given x and y values.

I appreciate any assistance with this.

Thank you!

[Torsten]

Given x and y, is there exactly one value of z with y = f(x,z) or could there be several or none of them ?

[M]

It could be two, I mean surface has two sheets, so I want the value on the lower sheets.

[Torsten]

Use "fsolve" to solve y - f(x,z) = 0 with known x and y and unknown z.

Answer 1

[Matt J]

z = fzero(@(z)y-f(x,z),  [z1,z2])

[M]

Thanks for the answer.

It doesn't work and give me this error:

Error using fzero>localFirstFcnEval

FZERO cannot continue because user-supplied function_handle ==>

@(c)h-(-0.4.*A.*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf))) failed with the error

below.

Undefined function 'times' for input arguments of type 'function_handle'.

Error in fzero (line 245)

fa = localFirstFcnEval(FunFcn,FunFcnIn,a,varargin{:});

[Matt J]

We cannot see the complete code that produced this message. It is advisable for you to run the code here.

[M]

I changed it a little bit and again gives another error:

gamma=5.5;
T=1/(gamma*40);
kh=0.1;
p=0.09;
delta=0.1;
ktau=0.04;
Kc=0.2; 
Khat=0.000015;
Kp=0.3;
kb=0.4;
Vs=T*0.9;
v_pmm=T*0.07;
alph0=T*0.003;
alph1=T*0.01;
Ke=14;
ks=0.2;
Kf=T*40;
kplc=0.11;
ki=2;
tmax=200/T;

e=0.0016;
vss=Vs/e;
K=(Khat)/ktau^4;
alpha0=delta.*(alph0)/ktau^4;
alpha1=delta.*(alph1)/ktau^4;
v_pm=delta.*(v_pmm)/ktau^4;
tmaxhat=tmax*ktau^4;
%[c,ct]=meshgrid(0:0.01:10);


A=@(ct,c) (-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=@(ct,c) (-0.4.*A.*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf)));

% Given values of ct and h
ct = 0.32;
h = 0.18;

% Define a function for the equation to solve
equation_to_solve = @(c) h(ct, c) - h;

% Initial guess for c
c0 = 0.010663;

% Use fzero 
z_optimized = fzero(equation_to_solve, c0);

Error using fzero>localFirstFcnEvalError
FZERO cannot continue because user-supplied function_handle ==> @(c)h(ct,c)-h failed with the error below.

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in fzero (line 302)
localFirstFcnEvalError(FunFcn,FunFcnIn,ME);

disp(['The optimized z value is approximately z = ', num2str(z_optimized)]);

[Dyuman Joshi]

You are over-writing the function handle h with the numerical value 0.18 and then trying to use it as a function handle.

Also, it's a better idea to define constants before defining function handles.

gamma=5.5;
T=1/(gamma*40);
kh=0.1;
p=0.09;
delta=0.1;
ktau=0.04;
Kc=0.2; 
Khat=0.000015;
Kp=0.3;
kb=0.4;
Vs=T*0.9;
v_pmm=T*0.07;
alph0=T*0.003;
alph1=T*0.01;
Ke=14;
ks=0.2;
Kf=T*40;
kplc=0.11;
ki=2;
tmax=200/T;

e=0.0016;
vss=Vs/e;
K=(Khat)/ktau^4;
alpha0=delta.*(alph0)/ktau^4;
alpha1=delta.*(alph1)/ktau^4;
v_pm=delta.*(v_pmm)/ktau^4;
tmaxhat=tmax*ktau^4;
%[c,ct]=meshgrid(0:0.01:10);


%Renamed to constant to h0
% Given values of ct and h0
ct = 0.32;
h0 = 0.18;

A=@(c) (-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=@(c) (-0.4.*A(c).*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf)));


% Define a function for the equation to solve
equation_to_solve = @(c) h(c) - h0;

% Initial guess for c
c0 = 0.010663;

% Use fzero 
z_optimized = fzero(equation_to_solve, c0);

disp(['The optimized z value is approximately z = ', num2str(z_optimized)]);

The optimized z value is approximately z = -0.2563

[M]

Thanks for the answer, so it is not the correct one that I want because z which is c should be larger than zero. Do you think is there a way that I can solve this problem? because I am sure that for those value of ct and h there should be a positive c value on the surface

[Dyuman Joshi]

Change the initial guess

gamma=5.5;
T=1/(gamma*40);
kh=0.1;
p=0.09;
delta=0.1;
ktau=0.04;
Kc=0.2; 
Khat=0.000015;
Kp=0.3;
kb=0.4;
Vs=T*0.9;
v_pmm=T*0.07;
alph0=T*0.003;
alph1=T*0.01;
Ke=14;
ks=0.2;
Kf=T*40;
kplc=0.11;
ki=2;
tmax=200/T;

e=0.0016;
vss=Vs/e;
K=(Khat)/ktau^4;
alpha0=delta.*(alph0)/ktau^4;
alpha1=delta.*(alph1)/ktau^4;
v_pm=delta.*(v_pmm)/ktau^4;
tmaxhat=tmax*ktau^4;
%[c,ct]=meshgrid(0:0.01:10);


%Renamed to constant to h0
% Given values of ct and h0
ct = 0.32;
h0 = 0.18;

A=@(c) (-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=@(c) (-0.4.*A(c).*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf)));


% Define a function for the equation to solve
equation_to_solve = @(c) h(c) - h0;


% Initial guess for c
c0 = 0.010663;

% Use fzero 
z_optimized = fzero(equation_to_solve, 0.5);

disp(['The optimized z value is approximately z = ', num2str(z_optimized)]);

The optimized z value is approximately z = 0.2563

[M]

Thanks, it works, but I don't know how to accept your answer since it's in a comment, not seperated answer.

Could you please look at this question as well, it is similar to this but the point is that I have line there not points.

https://au.mathworks.com/matlabcentral/answers/2020676-how-do-i-project-a-rectangular-plane-onto-the-surface