Finding tangent to a curve

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Faisal 2023-1-23

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https://ww2.mathworks.cn/matlabcentral/answers/1898890-finding-tangent-to-a-curve

评论： Mathieu NOE 2023-1-24

I have some data in x and y form. The data gives a plot as shown in the figure,

I want to draw tangent line on this curve, I am new to Matlab and would like someone to answer here.

Thank you!

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Mathieu NOE 2023-1-23

在 MATLAB Online 中打开

basocally no big deal , simply a few points I think the code could be improved :

no need to fit a polynomial curve here , everything can be done directly on the experiimental data
always make sure that you don't have too much rounding on excel data , this can create issue in diff / gradient computation
I thought maybe the tangent must be computed automatically at certain x/y point but the user is requested to entre the x index (??)
endpoints of the tangent are computed automatically

issue with x rounding : see what you get when you plot(diff(x))

my suggestion so far :

close all;
clear all;
clc;
file='C_2.xlsx';
% [x] = xlsread(file,1,'C2:C5092');
% [y] = xlsread(file,1,'A2:A5092');
data = xlsread(file);
y = data(:,1);
x = data(:,2);
% NB : x is not constant sampling (check with plot(diff(x)) you will see resolution issue)
xnew = linspace(min(x),max(x),300); % we really don't need much more than 300 points on this curve
ynew = interp1(x,y,xnew,'linear');
% overwritte data (my personnal choice)
x = xnew;
y = ynew;
clear xnew ynew
plot(x,y,'-*');
hold on
% slope (gradient)
dx = mean(diff(x));
slop = gradient(y,dx);
idx = input(['Enter index in the range ',num2str([1 length(x)]),':']);
x0 = x(idx);
y0 = y(idx);
df = slop(idx);
% Calculate the end points of a tangent line
xt = [max(0,x0-y0/df)  x0+(max(y)-y0)/df ];
yt = y0 + (xt-x0)*df;
plot(x0,y0,'dr',xt,yt,'r','LineWidth',2.5);