Tangent

2011 6 26
4 个回答
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0 个投票

Hi,
I have x and y values which represent some sort of sin wave but I don't know the function. something like x = [1.190984579 1.15776987 1.12455516 1.096085409 1.067615658 1.034400949 1.003558719 0.975088968 0.944246738 0.911032028 0.877817319 0.846975089 0.825622776]
y = [0.526690391 0.517200474 0.519572954 0.533807829 0.550415184 0.56227758 0.552787663 0.536180308 0.521945433 0.521945433 0.524317912 0.517200474 0.495848161]
I want to draw tangent line on the point I specify. Please help me do it.
Thanks.

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 采纳的回答

Andrew Newell
Andrew Newell 2011-6-26

1 个投票

You'll need to fit a function to the data and then take its derivative. First, the fit:
plot(x,y,'.'); hold on
n = 10;
[p,~,mu] = polyfit(x,y,n); % degree 10 polynomial fit, centered and scaled
% Plot the fit with the data
xfit = linspace(min(x),max(x),100);
yfit = polyval(p,xfit,[],mu);
plot(xfit,yfit,'r')
Now you can use a numerical derivative to calculate the tangent. I recommend downloading the package Adaptive Robust Numerical Differentiation package from the File Exchange. Then you can do the following:
idx = input(['Enter index in the range ',num2str([1 length(x)]),':']);
x0 = x(idx);
f = @(x) polyval(p,x,[],mu);
df = derivest(f,x0); % Here is the derivative
% Calculate the end points of a tangent line
xt = [x0-0.05 x0+0.05];
yt = f(x0) + (xt-x0)*df;
plot(xt,yt)
(Edited to accept an index number instead of an x value).

更多回答（3 个）

Judah S
Judah S 2011-6-26

0 个投票

Thanks Andrew.
Kindly check my this approach.
function tangent1
close all;
clear all;
clc;
format long;
x = [1.190984579 1.15776987 1.12455516 1.096085409 1.067615658 1.034400949 1.003558719 0.975088968 0.944246738 0.911032028 0.877817319 0.846975089 0.825622776];
y = [0.526690391 0.517200474 0.519572954 0.533807829 0.550415184 0.56227758 0.552787663 0.536180308 0.521945433 0.521945433 0.524317912 0.517200474 0.495848161];
idx = input('Enter index of X:');
plot(x,y)
%%Slope at a given point
m=(y(idx+1)-y(idx))/(x(idx+1)-x(idx)) %Equation of forward difference
m1=(y(idx)-y(idx-1))/(x(idx)-x(idx-1)) %Equation of backward difference
m2=(y(idx+1)-y(idx-1))/(x(idx+1)-x(idx-1)) %Average of both, also equation of secant
fprintf('Equation of tangent \n')
fprintf('q = %4.2f*(p-%4.2f)+%4.2f \n',m1,x(idx),y(idx))
p = 0.75:0.01:1.6;
q = m1*(p-x(idx))+ y(idx);
hold on;
plot(p,q,'r')
end
Do you think it's correct?
Thanks,

5 个评论
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Andrew Newell
Andrew Newell 2011-6-26
It depends on what you are trying to do. To me, these points look like a coarse sampling of a smooth function. If so, you're computing a secant, not a tangent. That's why I suggested fitting the function.
A minor point: the function fails if you choose the starting point or end point of the curve.
Judah S
Judah S 2011-6-26
Thanks for the comment.
Yes it fails because of the m1. It probably won't if I use m.
Appreciate your help.
:-)
Andrew Newell
Andrew Newell 2011-6-26
Actually, m2 works pretty well.
Judah S
Judah S 2011-6-26
Your code is awesome. This is exactly what I needed.
I was considering interpolation or extrapolation to choose points since I was using indexes.
derivest is really cool.
Andrew Newell
Andrew Newell 2011-6-26
Glad you like it! I agree, derivest is very cool.

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Judah S
Judah S 2011-6-30

0 个投票

Hi Andrew,
I am curious to know if I want to measure angle between tangent line and X or Y axis, how can I do that?
Thanks.

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