Tangents to the curves

22 次查看(过去 30 天)
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

采纳的回答

Star Strider
Star Strider 2019-5-20
编辑:Star Strider 2019-5-20
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')
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.
  4 个评论
Yash Runwal
Yash Runwal 2019-5-21
Thank You Guys, I will try this and first try to understand the code. This helps a lot. If i have some doubts i will comment here.
Thank you again.
Star Strider
Star Strider 2019-5-21
As always, our pleasure!

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Programming 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by