%Example data
x=0:0.1:10;
y = 0.230 + -0.046*x + -0.208*x.^2 ;
plot(x,y);
%Find the tangent and normals at all points
dy = [0 diff(y)./diff(x)];
py = -1./dy;
%Choose a point
n = 60;
X = x(n);
Y = y(n);
hold on
plot(X, Y, 'or')
%Find the equation of the straight line normal to that point. You can do this in one step (yn = py(n)*(x - X) + Y) but I've done it in two to illustrate where this comes from
c = Y - py(n)*X;
yn = py(n)*x + c;
plot(x, yn, 'g')
how to calculate equation to a perpendicular to polynomial line
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26-06-2022 - |
Pergunta
I have a polynomial y = 0.230 + -0.046*x + -0.208*x^2
.
I want to calculate the perpendicular to this line cutting another line at (X,Y).
Solução 2
Outras dicas
An alternative is to compute the analytical result which is not terribly difficult. (you could use the symbolic toolbox for that but the NN sitting on your head will do):
%Example data
x=0:0.1:10;
y = 0.230 + -0.046*x + -0.208*x.^2 ;
plot(x,y);
%Find the tangent and normals at all points (*edited*)
slope = (-0.046 + -2*0.208*x);
py = -1./slope; % <-- modified from Dan's expression
% to use analytical derivative
%Choose a point
n = 60;
X = x(n);
Y = y(n);
hold on
plot(X, Y, 'or')
% Copying @Dan: Find the equation of the straight line normal to that point. You can do this in one step (yn = py(n)*(x - X) + Y) but I've done it in two to illustrate where this comes from
c = Y - py(n)*X;
yn = py(n)*x + c;
plot(x, yn, 'g')
axis tight equal
Using axis equal
is also a good idea in this example to see that you really have orthogonal curves.
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