finding slope of a curve at some specific points

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cagatay yilmaz
cagatay yilmaz 2015 年 4 月 10 日
コメント済み: Caden Armentrout 2021 年 9 月 8 日
Hi
I have a a column data set which is the transverse strain of a composite material. I want to calculate slope of it at some specific points

採用された回答

Star Strider
Star Strider 2015 年 4 月 10 日
Your data appear to be smooth enough that you could fit them with a low-order (about 3 or 4) polynomial with polyfit, dividing ‘Index’ by 1E+5 first. Then use polyder to take the derivative of the polyfit coefficients, and use polyval to evaluate the derivative at the values of ‘Index’ you want.
  6 件のコメント
Ahmed Desoky
Ahmed Desoky 2019 年 3 月 27 日
Thank you Star Strider
Star Strider
Star Strider 2019 年 3 月 27 日
My pleasure.

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その他の回答 (6 件)

Image Analyst
Image Analyst 2015 年 4 月 10 日
How about if you just pass points around the point in question to polyfit and fit a line to them:
coefficients = polyfit(x(index1:index2), y(index1:index2), 1);
slope = coefficients(1);
Or fit a quadratic and get the slope at the middle of the stretch of points you fitted:
coefficients = polyfit(x(index1:index2), y(index1:index2), 2);
slope = 2 * coefficients(1) * x(middleIndex) + coefficients(2);
  3 件のコメント
Image Analyst
Image Analyst 2016 年 2 月 28 日
Chris, post your own question and show your code about how you tried to use interp1() to get uniformly spaced data and then used gradient() and show why that wasn't working. Then someone will probably answer there.
Caden Armentrout
Caden Armentrout 2021 年 9 月 8 日
@Image Analyst Ay yo you da MVP

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Chris McComb
Chris McComb 2015 年 4 月 10 日
You might want to start by looking at MATLAB's gradient function. You could also use simple finite difference formulas, like:
slope(i) = (y(i+1) - y(i-1))/(x(i+1) - x(i-1))

Eirini Gk
Eirini Gk 2016 年 3 月 25 日
Hello, I need also to find some things like that but im new in matlab and i cant understand every statement. i want to make a function that gives the slope in a point in every curve (close curve). I dont have the toolbox to use 'sym' statement because i need to do it just numerically. I think that diff or gradient function would help but i havent understand excaclty how to use it. I have to input F(x,y) (for example x^2+y^2-1 - the circle with radius 1) and a point of it (x0,y0). I saw that with these statement i get a row of numbers. But can somebody explain me excactly input and ouputs in these statement? Thank you.

cagatay yilmaz
cagatay yilmaz 2016 年 3 月 25 日
編集済み: cagatay yilmaz 2016 年 3 月 25 日
let say x your independent variable and y dependent variable
index=(x>=2)&(x<=5) % takes the points where you want to calculate your slope (regression)
[slope,intercept]=polyfit(x(index),y(index),1) % create a tangent line on the your selected data points
I hope this will solve your problem.
  1 件のコメント
Eirini Gk
Eirini Gk 2016 年 3 月 25 日
i dont understand why i have to input an interval. I mean its because of the numerical methods? So when i want a point , lets say (x0,y0)=(2,3) i have to take an area of x=1.5:0.001:2.5 or not?

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ali moshkriz
ali moshkriz 2016 年 12 月 7 日
hey Guys! how can i find the slope of this curve in 3 point that have different slope! please help...
>> A=[0,198.6026,397.2053,993.01346,2376.65783,3968.16977,4566.799,5161.454,6956.4044,9930.1346]; >> B=[0.29835,0.3978,0.467415,0.546975,0.745875,0.975375,1.09395,1.3923,2.56275,4.5288]; >> loglog(A,B);

Image Analyst
Image Analyst 2016 年 12 月 7 日
The slope is deltaB/deltaA. For each point, you will have a slope to the right of the point and a slope to the left of the point. You can take whichever one you want, or even average the slopes on each side if you want. For example, the slopes around element #2:
leftSlope = (B(2)-B(1)) / (A(2)-A(1))
rightSlope = (B(3)-B(2)) / (A(3)-A(2))
averageSlope = (leftSlope + rightSlope) / 2;

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