![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/220325/image.png)
Tangents to the curves
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Yash Runwal
am 20 Mai 2019
Kommentiert: Star Strider
am 21 Mai 2019
Hello, I have plotted a few figures as shown below. Now i would like to plot the tangents to each of these curves. How can i do that?
I have searched on the forum but could not get any definitive answer.
The code for this graph is divided in 2 functions as shown below:
function [x,y_val] = plot_trial(Z)
g = 9.8; K = 0.3; % g is the gravitational accleration
y = @(V_d) (1/(2*g*K^2))*(V_d.^2/Z^2);
x = linspace(0,1,100);
y_val = y(x);
end
clc; clear; close all;
% Use the other function 'plot_trial()'
[v_d1, y_val1] = plot_trial(1);
[~,y_val2]= plot_trial(2);
[~,y_val3]= plot_trial(3);
figure
hold on
plot(v_d1, y_val1,v_d1,y_val2,v_d1,y_val3)
legend('Z=1','Z=2','Z=3')
title('Der Tank (Konstante Querschnittsfläche)')
xlabel('dV1/dt')
ylabel('Y')
![tank_trial.PNG](https://www.mathworks.com/matlabcentral/answers/uploaded_files/220321/tank_trial.png)
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Star Strider
am 20 Mai 2019
Bearbeitet: Star Strider
am 20 Mai 2019
At what point do you want to plot the tangent?
The usual approach is to use the gradient function to find the slope, then use that and the value of the function at that point to calculate the intercept. The slope and intercept then define the tangent line at that point.
Example —
x = linspace(0, 5);
y = x.^2;
h = x(2)-x(1);
dydx = gradient(y, h);
xi = 2.1; % Choose An ‘x’ Value
yi = interp1(x, y, xi); % Corresponding ‘y’ Value
dydxi = interp1(x, dydx, xi); % Derivative At ‘x’
intcpt = yi - dydxi*xi; % Calculate Y-Intercept
figure
plot(x, y, '-b')
hold on
plot(x, dydxi*x+intcpt, '-r')
hold off
grid
legend('Data','Tangent', 'Location','NW')
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/220325/image.png)
EDIT —
To clarify my approach, I choose to use interp1 and gradient simply because it allows the desired ‘x’ value to be any value between the limits of the x-axis.
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