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How to find the slope of a tangent on a point on a nonlinear curve?

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Aqua22
Aqua22 2015 年 4 月 15 日
コメント済み: John D'Errico 2018 年 11 月 10 日
I have a set of data points x and y. I am plotting these on a (x,y) graph. The result is a nonlinear curve. On each and every point on the curve, tangents can be drawn and the slopes for every tangent will be different. I want to know, how I can I draw these tangents and find their slope. Please guide. I am new to MATLAB.

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John D'Errico
John D'Errico 2015 年 4 月 15 日
編集済み: John D'Errico 2015 年 4 月 15 日
You cannot find a slope until you define a curve. Simple points are not a curve. You have connected the points in your mind, so you "see" a curve. But all you have are points.
So you need to find a function that interpolates the points. A spline will do. Then differentiate the spline to get the slope.
You will probably want to use tools like spline (to fit a spline), fnder (differentiate it), and fnval (evaluate the derivative).
If these points form a completely general multi-valued relationship, such that there are multiple values of y for a given point x, you can still form a spline model, but it will take an extra step or two to do the work. An example of such a curve is a circle, where a direct spline fit using spline will fail.
  7 件のコメント
Deepak sharma
Deepak sharma 2018 年 11 月 9 日
can u please share the whole code by taking some random data points
John D'Errico
John D'Errico 2018 年 11 月 10 日
I'm a bit perplexed. I already did share the code. I.e., use pchip, then fnder, then fnval. Other tools might apply in the scenario where you don't want to use pchip, or lack the curve fitting toolbox.

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その他の回答 (1 件)

farzad
farzad 2015 年 4 月 15 日
I think you just can define a straight line between each of the two point and simply find the slope of that line
  1 件のコメント
Aqua22
Aqua22 2015 年 4 月 15 日
@farzad: I can not define a straight line between two points on the curve to find slope as the slope changes at every point. Its a nonlinear curve. Further, I need to find the tangent at the particular point and then find its slope. Tangent at a particular point on the curve is unique and hence its slope.

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