How to calculate the slope of a curve

How can I determine the slope of this curve. Thanks for help!

Best regards

[Stephen23]

Where did you get the image from: is it generated from some code in MATLAB, or plotted numeric data, or did someone just email you an image file?

All of these are valid ways of obtaining the curve, but they require different solutions. So you need to help us by actually giving some information about the curve: How is it defined?

[AB ABOULOUARD]

it is generated from code in MATLAB

Answer 1

[Star Strider]

The .fig file has the requisite data, so specific code is possible. The easy way of course would be simply to calculate the analytic derivative and write a function to produce it.

This will calculate and plot it using the gradient function. Experiment with it to get the result you want:

DF = openfig('AB ABOULOUARD fig.fig');
H1 = get(gca, 'Children');                          % Get Handle To Data In ‘.fig’ File
x = get(H1, 'XData');                               % Get ‘x’ Data
y = get(H1, 'YData');                               % Get ‘y’ Data

dx = mean(diff(x));                                 % Find Mean Differece In ‘x’ Values
dy = gradient(y,dx);                                % Calculate Slope Of Data

xq = find((x >= 0.275) & (x <= 0.325));             % Index Of Area Of Interest ‘x’ Values
slope_x = x(xq);                                    % ‘Slope’ ‘x’ Values Of Interest
slope_y = dy(xq);                                   % ‘Slope’ ‘y’ Values Of Interest

figure(2)
plot(x, dy, '-g')
hold on
plot(slope_x, slope_y, '-r', 'LineWidth',1)
hold off
grid
legend('Slope Of All Data', 'Slope Of Data In Region Of Interest', 'Location','W')

[Jessica Humphrey]

Is there a way to get the value of the slope at the specified region of interest?

[Image Analyst]

@Jessica Humphrey, that's what the slope_y variable is in his code.

[Star Strider]

Yes.

There will of course be more than one value unless the slope is constant (and the relation therefore linear in that region).

.

[Jessica Humphrey]

Thank you so much! This is very helpful!