MATLAB: How to implement Linear Aproximation of a function

linear approximationlinearizationMATLABmatlab function

How one can translate the linear approximation of the function f(x,y) into MATLAB. The resulting approximation given by :

f(x,y) = f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0)

both x0 and y0 are known but how this can be translated into MATLAB

Best Answer

The attached file pplane8 might help you. This tool helps you to identify nonlinear two variable first order differential equations and find the equilibrium points and draw phase portraits.

Related Solutions

MATLAB: 4th order Runge-Kutta code that can solve for several intial conditions

Here's a version:

% Initial conditions
t0=0;
x0=0:0.2:1;
y0=1:0.2:2;
dt=0.1;
tz=1;
t_range= t0:dt:tz;
X = zeros(numel(x0),numel(t_range));
Y = zeros(numel(y0),numel(t_range));
x_rk=zeros(1,numel(t_range));
y_rk= zeros(1,numel(t_range));
for n = 1:numel(x0)
    x_rk(1)=x0(n);
    y_rk(1)=y0(n);
    for i=1:length(t_range)-1
        [x_rk(i+1), y_rk(i+1)]=rk_method(x_rk(i),y_rk(i),dt);
    end
    X(n,:) = x_rk;
    Y(n,:) = y_rk;
    %% PLots
    subplot(numel(x0)/2,2,n)
    plot (t_range, x_rk,'b-o',t_range, y_rk,'y-o','MarkerSize',5);
    grid
    title([num2str(x0(n)),'   ',num2str(y0(n))])
    legend('x','y')
end
disp('X = '), disp(X)
disp('Y = '), disp(Y)
%% Functions
function [dxdt, dydt]= f(x,y)
dxdt =  5*x-3*y;
dydt = -6*x+2*y;
end 
function [x1, y1]=rk_method(x0,y0,dt)
[k11, k12]=f(x0,y0);     
[k21, k22]=f(x0+0.5*dt*k11,y0+0.5*dt*k12); 
[k31, k32]=f(x0+0.5*dt*k21,y0+0.5*dt*k22); 
[k41, k42]=f(x0+dt*k31, y0+dt*k32); 
x1=x0+dt*((k11/6)+((k21+k31)/3)+(k41/6));
y1=y0+dt*((k12/6)+((k22+k32)/3)+(k42/6));
end

MATLAB: Law of motion through numerical aproximation

Here's a basic integration of the system, along with plots:

% mass, initial time, final time
m = 0.12;
t0 = 0;
tf = 3;
% number of time intervals and resulting spacing dt
N = 1001;
t = linspace(t0,tf,N);
dt = t(2) - t(1);
% vectors for storage of position, velocity, and acceleration
x = zeros(1,N);
xdot = zeros(1,N);
xdoubledot = zeros(1,N);
%initial conditions for position, velocity, and acceleration
x(1) = 0;
xdot(1) = 0;
xdoubledot(1) = (6*sin(4*pi*t(1)-pi/6)-0.5)/m;
% numerical integration
for i = 2:1:N
    x(i) = x(i-1) + xdot(i-1)*dt;
    xdot(i) = xdot(i-1) + xdoubledot(i-1)*dt;
    xdoubledot(i) = (6*sin(4*pi*t(i)-pi/6)-0.5)/m;
end
figure(1)
plot(t,xdoubledot)
title('acceleration')
figure(2)
plot(t,xdot)
title('velocity')
figure(3)
plot(t,x)
title('position')

I left the final time at 3 seconds so you can get a better picture of what's going on. I encourage you to play around with the values of t0 and tf a little bit.

Notice that you have a constant term (-0.5) added to a sinusoidal term. What do you expect the results of this "offset" to be?

Best Answer

Related Solutions

MATLAB: 4th order Runge-Kutta code that can solve for several intial conditions

MATLAB: Law of motion through numerical aproximation

Related Question