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How to plot an Implicit function with certain conditions

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im trying to plot in MATLAB the red implicit function given these 3 conditions, in DESMOS it's super easy but unfortunately i have no idea how to pull it off in MATLAB, i've searched the internet but my luck fell short, would appreciate the help.

Accepted Answer

Pratyush Swain
Pratyush Swain on 17 Apr 2024
Hi Shai,
To plot this implicit functions you need to leverage the fimplicit function in MATLAB.Here is an demonstrated example on how we can use this function for your usecase:
warning('off','all');
% Plotting the function using fimplicit
% NOTE: Here I have arbitrarily chosen input values for x and y
% x range --> 1<=x<=2.5, y range --> 0<=y<=1.5
fimplicit(@fun, [1, 2.5, 0, 1.5]);
title('Implicit Function with Conditions');
xlabel('x');
ylabel('y');
% Callback for Impicit to calculate the values, Here x and y are retreived as vector of input values
function values = fun(x, y)
% Initialize output array to NaN %
values = NaN(size(x));
% Define the condition for y to avoid division by zero %
valid = y ~= 0;
% Calculate the expression (2x/y)*(1-1/2y) under valid condition %
% NOTE: Here the ./ operater ensures division occurs on all valid
% elements of vector x
expression = (2*x(valid)./y(valid)).*(1-1./(2*y(valid)));
% Apply all conditions and obtain a final logical vector
condition = x > 1.166 & expression > 0 & expression < pi/2 & y < 1;
% Apply the implicit function where conditions are met
values(condition) = (2*x(condition)./y(condition)) - tan(expression(condition));
end
I have referred to this thread to form a solution for this usecase: https://www.mathworks.com/matlabcentral/answers/1756760-how-to-plot-implicit-function-with-conditions
For more information on fimplicit function, please refer to https://www.mathworks.com/help/matlab/ref/fimplicit.html
  2 Comments
Shai Zipori
Shai Zipori on 17 Apr 2024
unfortunately, the plot that's being displayed isn't appropriate (when increasing the x range beyond 2.5, it displays crushed line) could you help me with getting the points (coordinates) of given x as input (lets say 1.166:0.01:5.1) and y as output that fullfills these constraints?
0<((2*x)/(y))*(1-((1)/(2*y)))<(pi/2)
y<1
and suits this (same) equation?
((2*x)/(y))=tan(((2*x)/(y))*(1-((1)/(2*y))))
unfortunately, i just can't seem to grasp how to use the logical vectors.
Thank you so much for your help!
Mathieu NOE
Mathieu NOE on 18 Apr 2024
you can improve the result by increasing the 'MeshDensity' factor :
h = fimplicit(@fun, [1, 5, 0.5, 1.1], 'MeshDensity',300);
then you get rid of the waves and NaNs
% Plotting the function using fimplicit
% NOTE: Here I have arbitrarily chosen input values for x and y
% x range --> 1<=x<=5, y range --> 0.5<=y<=1.1
h = fimplicit(@fun, [1, 5, 0.5, 1.1], 'MeshDensity',300);
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
title('Implicit Function with Conditions');
xlabel('x');
ylabel('y');
% if you need to access to the x & y data
x = h.XData;
y = h.YData;
figure
plot(x,y);
title('Implicit Function with Conditions');
xlabel('x');
ylabel('y');
% Callback for Impicit to calculate the values, Here x and y are retreived as vector of input values
function values = fun(x, y)
% Initialize output array to NaN %
values = NaN(size(x));
% Define the condition for y to avoid division by zero %
valid = y ~= 0;
% Calculate the expression (2x/y)*(1-1/2y) under valid condition %
% NOTE: Here the ./ operater ensures division occurs on all valid
% elements of vector x
expression = (2*x(valid)./y(valid)).*(1-1./(2*y(valid)));
% Apply all conditions and obtain a final logical vector
condition = x > 1.166 & expression > 0 & expression < pi/2 & y < 1;
% Apply the implicit function where conditions are met
values(condition) = (2*x(condition)./y(condition)) - tan(expression(condition));
end

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More Answers (1)

Mathieu NOE
Mathieu NOE on 18 Apr 2024
I have to say I don't do much with implicit function problems , tried using fsolve and fminbnd but was lacking how to implement the conditions
now, a poor's man solution is to create a x, y grid , evaluate your function, and find the x,y points where you function is minimal
% solving implicit function
% (2*x./y) - tan((2*x./y).*(1-1./(2*y))) = 0;
x = linspace(1.167,5,1000); % create x array with condition x>1.166
y = linspace(0.5,1,1000);
% create a X Y meshgrid and evaluate function
[X,Y] = meshgrid(x,y);
C = (2*X./Y).*(1-1./(2*Y));
fun = (2*X./Y) - tan(C); % this is your implicit function
% apply condition (C>0 & C<pi/2)
ind = (C>0 & C<pi/2);
Z = NaN(size(fun));
Z(ind) = fun(ind); % keep only valid fun values according to condition (C>0 & C<pi/2)
% plot function Z=f(x,y) to show minimum line (is what we are looking for)
figure
h = imagesc(x,y,log(abs(Z)));
colorbar('vert')
set(gca,'YDir','normal')
set(h, 'AlphaData', 1-isnan(Z))
% find x,y of minimum line
for ci = 1:numel(x)
zz = Z(:,ci);
% find y coordinate to get minimum z value
[val,ind] = min(abs(zz));
if ~isempty(ind)
xc(ci) = x(ci);
yc(ci) = y(ind);
end
end
figure
plot(xc,yc);
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