如何計算方程到垂直於多項式的線

Question

我有一個多項式y = 0.230 + -0.046*x + -0.208*x^2。 我想計算與此線垂直切割另一條線（X，Y）。

Answer 1

%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') 

Answer 2

另一種方法是計算分析結果，這並不是非常困難。 （你可以用象徵性的工具箱如此，但NN坐在你的頭會做）：

%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 

使用axis equal也是在這個例子中是一個好主意，看看你真的有正交曲線。