How to find the slope of the curve at x=0 and y=0

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NITHIN XAVIER
NITHIN XAVIER am 4 Aug. 2019
Bearbeitet: Star Strider am 4 Aug. 2019
I wanted to find the slope of the curve, where is cross the x-axis and y-axis (there are four points), I am a beginner in Matlab, It would be very helpful if someone can help me.
Thanks in advance.
IV curve.jpg
  2 Kommentare
Walter Roberson
Walter Roberson am 4 Aug. 2019
Do you have the equations, and do you have the symbolic toolbox? If so then evaluate the derivatives at the appropriate points.
darova
darova am 4 Aug. 2019
I want to help you

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Antworten (1)

Star Strider
Star Strider am 4 Aug. 2019
Bearbeitet: Star Strider am 4 Aug. 2019
Since you did not share your data, I created my own.
This should get you started:
x = linspace(-0.5, 1.5); % Create Data
y1 = 20*exp(0.8*x)-25; % Create Data
y2 = 15*exp(1.2*x)-35; % Create Data
dv = [-1 0 1]*1E-7;
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Approximate Zero-Crossing Indices Of Argument Vector
y1zx = zci(y1); % Zero-Crossing Indices
dy1dx = gradient(y1,(x(2)-x(1))); % Numerical Derivative
y1y0x = interp1([y1(y1zx(1)-1),y1(y1zx(1)+1)],[x(y1zx(1)-1),x(y1zx(1)+1)],0); % Precise Zero-Crossing
dy1y0 = interp1([x(y1zx(1)-1),x(y1zx(1)+1)],[dy1dx(y1zx(1)-1),dy1dx(y1zx(1)+1)],y1y0x); % ‘y1’ Slope At y=0 ***
y2zx = zci(y2); % Zero-Crossing Indices
dy2dx = gradient(y2,(x(2)-x(1))); % Numerical Derivative
y2y0x = interp1([y2(y2zx(1)-1),y2(y2zx(1)+1)],[x(y2zx(1)-1),x(y2zx(1)+1)],0); % Precise Zero-Crossing
dy2y0 = interp1([x(y2zx(1)-1),x(y2zx(1)+1)],[dy2dx(y2zx(1)-1),dy2dx(y2zx(1)+1)],y2y0x); % ‘y2’ Slope At y=0 ***
x0 = find(x>=0, 1, 'first');
dy1x0 = interp1([x(x0-1),x(x0+1)],[dy1dx(x0-1),dy1dx(x0+1)],0); % ‘y1’ Slope At x=0 ***
dy2x0 = interp1([x(x0-1),x(x0+1)],[dy2dx(x0-1),dy2dx(x0+1)],0); % ‘y2’ Slope At x=0 ***
figure
plot(x, y1, x, y2)
grid
txtcely0 = sprintfc('\\leftarrow Slope = %.3f', [dy1y0, dy2y0]);
txtcelx0 = sprintfc('Slope = %.3f \\rightarrow', [dy1x0, dy2x0]);
text([y1y0x, y2y0x],[0 0], txtcely0, 'HorizontalAlignment','left', 'Rotation',-45)
text([0, 0],[y1(x0) y2(x0)], txtcelx0, 'HorizontalAlignment','right')
The slope results are signified by ‘***’ at the end of the comments.
EDIT — Added plot & text Calls.

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