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How to calculate curvature of a vector in Mathematica.

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    r(t)={(4+cos20t) cost,+(4+cos20t) sint,+0.4sin20t}
    Calculate the curvature of r[t] for 0≤t≤4pi

    2. Relevant equations

    k = | r' x r'' | / | r' |^3

    3. The attempt at a solution

    r[t_]:={4+Cos[20t]*Cos[t],4+Cos[20t]*Sin[t],0.4Sin[20t]}
    k[t_]:=Norm[Cross[r',r'']]/Norm[r']^3
    Plot[k[t],{t,0,4Pi}]



    I don't get any error messages, but the graph is blank.

    http://i.imgur.com/woWlabm.png
    http://i.imgur.com/ptUZcVG.png
     
    Last edited: Feb 28, 2013
  2. jcsd
  3. Feb 28, 2013 #2

    SteamKing

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    Science Advisor
    Homework Helper

    t or t_ in the Plot function?
     
  4. Feb 28, 2013 #3
  5. Feb 28, 2013 #4
    If you have defined a function using the "standard" form, id est, f[x_] := you do not want the x_ on the right hand side of the definition.
    Now to your real problem:
    You forgot to specify that the derivatives are also functions of t :)
    Code (Text):
    r[t_] := {4 + Cos[20 t]*Cos[t], 4 + Cos[20 t]*Sin[t], 0.4 Sin[20 t]}
    k[t_] := Norm[Cross[r'[t], r''[t]]]/Norm[r'[t]]^3
    Plot[k[t], {t, 0, Pi/4}]
    SHBjJ14.png
     
  6. Feb 28, 2013 #5
    Haha, I actually managed to plot it, but I thought it was wrong because I wasn't expecting a huge wave.

    I was also told to calculate the length of the curvature. Do you think this would be an acceptable answer?

    http://i.imgur.com/HHYzFy7.png

    Is there a way to simplify this expression?
     
    Last edited: Feb 28, 2013
  7. Feb 28, 2013 #6
    If I am not mistaken
    [tex]
    \kappa = \left | \frac{f''(x)}{(1 + f'(x)^{2})^{\frac{3}{2}}} \right |
    [/tex]

    is the formula one would use to calculate the curvature.
    Following this we would end up right here:
    JTq0NQi.gif

    Well, in theory you can apply //Simplify, but Mathematica has done this already, so in my opinion - no :)
     
  8. Feb 28, 2013 #7
    Okay, thanks. I have one last question. How can I use the inte\frac{}{}rval <0,4pi> in this equation?



    I'm just a little confused about why this formula, k = |[itex]\frac{r(t)' X r(t)''}{r(t)'^3}[/itex]| gives a different answer to the formula you posted
     
    Last edited: Feb 28, 2013
  9. Feb 28, 2013 #8
    Something like this should do the trick:

    Code (Text):
    Sum[Limit[b, t -> \[Omega]], {\[Omega], 0, 4 Pi}]
    where 'b' is

    JTq0NQi.gif


    Where did you get this formula ?
     
  10. Feb 28, 2013 #9
    Last edited: Feb 28, 2013
  11. Feb 28, 2013 #10
    Well, I am probably using wrong/incorrect formula.
    Sticking to your notebook we get 26.8967 as an answer.
    Can you verify that ?
     
  12. Feb 28, 2013 #11
    I'm not getting a number.
    http://i.imgur.com/mQ4Pfac.png

    I'm guessing I have incorrect syntax.

    Edit:

    I realized i didn't capitalize the 'p' in Pi. [STRIKE]now I"m getting 13k though[/STRIKE] I got 20.39
     
    Last edited: Feb 28, 2013
  13. Mar 1, 2013 #12
    Well, using the simple
    Code (Text):
    Sum[k[t],{t,0,4Pi}]
    I, again, receive 26.8967.
     
  14. Mar 1, 2013 #13
  15. Mar 1, 2013 #14
    You should check the initial settings in r, because I see errors there.
    For example a missing 't' in the first part :)
    Check again the problem and repost what is actually given.
     
  16. Mar 1, 2013 #15
    I can't believe I overlooked that. I also removed the parentheses. Now I'm getting 26.89. Thanks for the help
     
  17. Mar 1, 2013 #16
    You are welcome :)
     
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