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Matlab Second Derivative Ploting Problem

  1. Dec 5, 2004 #1
    Hello, this is my first post in this site, so here goes:

    Im working on a Matlab program. I have to calculate the second derivative of a given equation. The problem is that plotting the derivative is necessary, and that's where I'm stuck.

    Here's what I've got so far:

    % Evaluar f(x) y f'(x) usando diferencias hacia atras.
    %
    x = -10:0.1:10;
    f = x.^3 - 5*x.^2 + 2*x + 8;
    df = diff(f);
    dx = diff(x);
    df_dx = df./dx;
    xd = x(2:length(x));
    d2f = diff(f,2);
    d2x = diff(f,2);
    d2f_d2x = d2f./d2x;
    %
    % Graficar lo anterior.
    %
    subplot(2,1,1),...
    plot(x,f,xd,df_dx,xd,d2f_d2x),...
    title('Derivada de un polinomio de quinto grado'),...
    xlabel('x'),grid,...
    ylabel('f(x)'),...
    axis([-4,5,-500,1500]);

    On the internet I found out that diff(f,2) would calculate the second diff, I guess thats true. Ignore the spanish text :P.

    When I try to run the program, the following appears:

    ??? Error using ==> plot
    Vectors must be the same lengths.

    Error in ==> C:\MATLAB6p5\work\segunda_diferencial.m
    On line 16 ==> plot(x,f,xd,df_dx,xd,d2f_d2x),...

    Hope you can help.

    Thanks!
     
  2. jcsd
  3. Dec 6, 2004 #2

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    De donde eres? Casi no se nota que hablas español.


    Well, the command line "diff" shortens one dimension the array:

    [tex] diff (\overline{x})=[x_2-x_1, x_3-x_2,....x_n-x_{n-1}] [/tex]

    you will obtain a n-1 array dimension.

    You should interpolate the last component from interior ones, in order to have n components.

    With diff2 is the same, surely you will obtain n-2 components.
     
  4. Dec 6, 2004 #3
    Gracias! Soy del norte de México.

    In fact, I can graph the first diff, though what you say does make sense, as with every consecutive derivative one constant term will dissappear.

    Interpolate? As with 'interp1'? The last component? d2f_d2x in my code?

    Casi no se nada de Matlab :(

    Thanks for that, I will try it out, hope I understood you...
     
  5. Dec 6, 2004 #4
    Ok, I got it! Thanks though.
     
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