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

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)

Answer 1

[Andrei Bobrov]

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');

[Khanh]

Hi Andrei,

Your code worked so great except one thing the tangent line which I desired had the slope is tan(10*pi/180). Sorry about my question. It was a little wrong. I would like to determine the tangent line which has the slope is tan(10*pi/180) . The tangent line will have the equation y=tan(10*pi/180)*x+b . The equation of the curve is y=f. So I need to solve the equation df=tan(10*pi/180) to find x coordinate of the point of tangency. I use function

x_tangency=solve(df-tan(10*pi/180)).

But it didn't work. Could you please take a look?

[Andrei Bobrov]

see code after "well"

[Chad Greene]

Side note, tan(10*pi/180) is the same thing as tand(10).

[Andrei Bobrov]

Hi Chad! I agree with you.

[Khanh]

Hi Andrei,

Really thank you for your help. It worked so perfectly. Could I ask you one more question? How can I determine two intersection points as the following picture? The equation of the line: y=tand(30)*x+400 and the curve as above. I tried to solve the equation: xx=fzero(@(x) f(x)-tand(30)*x-400,[x(end,1)]). Then I just get one intersection point instead of two.

[Andrei Bobrov]

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');
f1 = @(x)tand(30)*x+400;
xx = x(1):x(end,1);
[~,ii] = min(f(xx(:)) - f1(xx(:)));
x1 = fzero(@(x)f(x) - f1(x),[0,xx(ii)]);
x2 = fzero(@(x)f(x) - f1(x),[xx(ii),xx(end)]);
plot(xx(:),[f(xx(:)),f1(xx(:))],[x1,x2],f([x1,x2]),'ro');

[Khanh]

Wow. Perfect. Thank you so much.

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