1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Curvature math problem

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data

    I am trying to calculate the curvature of a curve given by the position function:

    [itex]\vec{r}(t)= sin(t) \vec{i} + 2 \ cos (t) \ vec{j}[/itex]

    The correct answer must be:

    [itex]\kappa (t) = \frac{2}{(cos^2(t) + 4 \ sin^2 (t))^{3/2}}[/itex]

    I tried several times but I can't arrive at this answer. :confused:


    2. Relevant equations

    Curvature is given by:

    [itex]\kappa(t) = \frac{||T'(t)||}{||r'(t)||}[/itex]

    Where

    [itex]T (t) = \frac{r'(t)}{||r'(t)||}[/itex]

    3. The attempt at a solution

    [itex]r'(t) = \left\langle cos(t), \ -2 sin(t) \right\rangle[/itex]

    [itex]||r'(t)|| = \sqrt{cos^2(t) + (2 \ sin(t))^2} = \sqrt{3 \ sin^2t+1}[/itex]

    Therefore

    [itex]T = \frac{cos(t)}{\sqrt{3 \ sin^2 (t) +1}} \ \vec{i}, \ \frac{-2 \ sin(t)}{\sqrt{3 \ sin^2 (t) +1}} \vec{j}[/itex]

    [itex]\frac{dT}{dt} = \frac{-4 \ sin(t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{i} + \frac{-2 \cos (t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{j}[/itex]

    [itex]||T'(t)|| = \sqrt{\left( \frac{-4 \ sin(t)}{3 \ sin^2(t) + 1)^{3/2}} \right)^2 + \left( \frac{-2 \cos (t)}{(3 \ sin^2(t) + 1)^{3/2}} \right)^2}[/itex]

    [itex]= \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}[/itex]

    Putting this in the equation given

    [itex]\kappa = \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}. \frac{1}{\sqrt{3 \ sin^2t+1}}[/itex]

    But I can't see how this can be simplified to get to the correct answer. I appreciate any guidance.
     
    Last edited: Apr 25, 2012
  2. jcsd
  3. Apr 25, 2012 #2
    Re: Curvature

    You dropped +1 from the bottom, while calculating ||T'||. The top equals [itex]12sin^2(t)+4 = 4(3sin^2(t)+1)[/itex] and it all works out nicely.
     
  4. Apr 25, 2012 #3
    Re: Curvature

    Thank you very much! I got it! :)
     
    Last edited: Apr 25, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Curvature math problem
  1. Curvature problem (Replies: 5)

  2. Curvature problem (Replies: 2)

Loading...