# Locating point where slope becomes most steep

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A on 19 Jun 2012
Commented: GOPEE Ajit Kumar on 25 Feb 2017
Hello, I am trying to extract data from this graph, generated from the image to the right of it:
As you can tell from the graph, the slope starts to decrease most rapidly at about 1000 pixels.
My code will ask for a restriction zone, so it is not a problem to only investigate between say 800 pixels and 1100 pixels. However, I am not sure how to locate the point where the slope begins to decrease most rapidly.
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Image Analyst on 20 Jun 2012
Do you really need the x location of the steep part? Or can you just threshold the y values to find it? For example, maybe you can get the x like this: x= find(signal > 0.3, 1, 'last') so you find the x location where the signal exceed some threshold, like 0.3.
What if you had a steep part that was in the range y = [0.2 to 0.5] but you had a steeper one somewhere down in the range of y= [0 to 0.2]? Are you sure you'd want the lower amplitude signal just because it's steeper?

the cyclist on 19 Jun 2012
As Ryan comments, you "just take the derivative", but that is a gross simplification. Taking the numerical derivative of noisy discrete data is a tricky business.
I would suggest taking a look at the following FEX submission, and carefully reading the description:

Ayodele Oladeji on 12 Jan 2015
Edited: Ayodele Oladeji on 12 Jan 2015
Perform the following steps
• Use a low pass filter to smoothen the data points (smooth function could be helpful)
• Find the gradient of all points on the data and calculate the slope= (( diff(y_data)/diff(x_data))
• Depending on what you define as "most rapidly", you can apply the function find to your slope data and track index as follows: find(slope >= steep_threshold).
GOPEE Ajit Kumar on 25 Feb 2017
ok but how do you calculate the gradient of all points on the data if I have curve in a binary image and that I do not know the equation of the curve? Thanks

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