How to get the intersection points of a line and a curve which was fit to data?

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Khanh il 24 Ott 2014

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https://it.mathworks.com/matlabcentral/answers/159912-how-to-get-the-intersection-points-of-a-line-and-a-curve-which-was-fit-to-data

Modificato: Hussein Qenawy il 13 Apr 2019

Hi,

I have a line and a curve that was fit to a data. I also get Coefficients of Equation of the Curve, but don't know how to find its equation to make two equations equal to find the points of the tangency. Could someone give me some recommends?

Here is my code:

clc
array=[515 525 561 600 632 700 761 800 900 1000 1014 1750;
    0 150 300 394 450 540 600 631 696 745 750 865];
x=linspace(array(1,1),array(1,end),101)
y=interp1(array(1,:),array(2,:),x,'pchip')
x=transpose(x)
y=transpose(y)
%
f=fit(y,x,'pchip')
a=coeffvalues(f)
plot(f,y,x)
hold on
% Equation of line that pass through origin
x1=0:1000;
slope=tan(51.5*pi/180);
y1=slope*x1
plot(x1,y1)

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Andrei Bobrov il 24 Ott 2014

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https://it.mathworks.com/matlabcentral/answers/159912-how-to-get-the-intersection-points-of-a-line-and-a-curve-which-was-fit-to-data#answer_156434

Modificato: Andrei Bobrov il 25 Ott 2014

one way with Curve Fitting Toolbox

array=[515 525 561 600 632 700 761 800 900 1000 1014 1750;
         0 150 300 394 450 540 600 631 696  745  750  865];
x = array([2 1],:)';
f = fit(x(:,1),x(:,2),'cubicinterp');
df = fit(x(:,1),differentiate(f,x(:,1)),'cubicinterp');
xx = fzero(@(x)f(x)/x - df(x),[1 750]);
x1 = linspace(x(1,1),x(end,1),300);
plot(x1,f(x1),x1,x1*df(xx),xx,f(xx),'ro');

well, more

f = fit(x(:,1),x(:,2),'cubicinterp');
df = fit(x(:,1),differentiate(f,x(:,1)),'cubicinterp');
k = tand(10);
xx = fzero(@(x)df(x) - k,[1 x(end,1)]);

you line: y = k*x + b

b = f(xx) - k*xx;
x1 = linspace(x(1,1),x(end,1),300);
plot(x1,f(x1),x1,k*x1 + b,xx,f(xx),'ro');

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Khanh il 27 Ott 2014

Wow. Perfect. Thank you so much.

Hussein Qenawy il 13 Apr 2019

Modificato: Hussein Qenawy il 13 Apr 2019

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