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I want to make a function that plots 3D quadratic surface

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Maryam Abdelhamid
Maryam Abdelhamid am 9 Mai 2020
Kommentiert: Walter Roberson am 18 Mai 2020
I want to creat a function that returns a graph of hyperbolic paraboloid just by entring the coefficient of x^2,y^2 and z
function hyperbolicparaboloid(A,B,C)
A=input('enter coeffecient of x^2');
B=input('enter coeffecient of y^2');
C=input('enter coeffecient of z');
%y^2/b^2-x^2/a^2=z/c
%where B=1/b^2, A=1/a^2, C=1/c
X=linspace(-10,10,100);
Y=linspace(-10,10,100);
Z=linspace(-10,10,100);
[X,Y]=meshgrid(X,Y);
Z=((Y.^2*B)-(X.^2*A))./C;
mesh(X,Y,Z);
view([130,30])
end
Althoug the code is working the function cannot be created
so what is the problem with this function?
  1 Kommentar
Walter Roberson
Walter Roberson am 9 Mai 2020
What is the point of accepting three input parameters and then promptly ignoring them? You should either have your function not accept any parameters or else you should have your function use the A, B, C values passed in.
Z=linspace(-10,10,100);
That statement is not productive: you overwrite the Z you create there.

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Antworten (1)

rajat aggarwal
rajat aggarwal am 18 Mai 2020
Bearbeitet: rajat aggarwal am 18 Mai 2020
You can use ezplot to draw hyperbolic paraboloid
following links can also help
clc;
clear all;
[X,Y,Z] = meshgrid(-10:0.5:10,-10:0.5:10,-10:0.5:10);
a=1;
b=1;
c=1;
V = X.^2/a^2 + Y.^2/b^2 - Z.^2/c^2;
p=patch(isosurface(X,Y,Z,V,1)); % This is the key step. It involves getting the part of the volume corresponding to the surface defined by the equation
set(p,'FaceColor','red','EdgeColor','none');
daspect([1 1 1])
view(3);
camlight
  1 Kommentar
Walter Roberson
Walter Roberson am 18 Mai 2020
Note that ezplot() is not recommended anymore. It uses older technology to create the plot. fplot() is a better choice in most cases now.

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