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!

Maximum curvature

  1. Sep 25, 2007 #1
    I'm having trouble finding the point of the maximum curvature of the line with parametric equations of: x = 5cos(t), and y = 3sin(t).

    I know the curvature "k" is given by the eq.:

    k = |v X a|/ v^3

    Where v is the derivative of the position vector r = <5cos(t), 3sin(t) > , a is the derivative of v, and "v" is the norm of the velocity vector v.

    I know I first have to find x', x'', y', and y'' to get the vectors v and a, and find the norm of v and cube it.

    After plugging in the values and taking the absolute value of the cross product above, I evaluated when the numerator of k' (the derivative of the curvature equation) is equal to zero. I came up with the values of arctan(3/5), pi, zero, and pi/2. The book says the the maximum curvature occurs at x = plus or minus 5. I'm not sure what I did wrong.

    Edit: Additionally, is there any way to integrate the square root of " (25*t^2) + 9 " dt without using integration tables? Any help is appreciated.
     
    Last edited: Sep 25, 2007
  2. jcsd
  3. Sep 25, 2007 #2
    [tex]k=\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{(\dot{x}^2+\dot{y}^2)^{3/2}}= \frac{15}{(9+16\sin^2t)^{3/2}},[/tex]

    and then the maximum is reached when [itex]9+16\sin^2t[/itex] is minimum, wich is in [itex]0,\pi,2\pi[/itex]. This makes total sence, as your curve is actually an ellipse with major semiaxis in the horizontal line. Now, what is the value of [itex]x[/itex] in such points?


    To do the integral simply do the change of variables [itex]t=\frac{3}{5}\tan x[/itex] so

    [tex]\int \sqrt{25 t^2+9}dt=\frac{9}{5}\int\sec^3 xdx[/tex]

    and then itegrate by parts taking [itex]u=\sec x,\,dv=\sec^2 x dx[/itex].
     
    Last edited: Sep 25, 2007
  4. Sep 25, 2007 #3
    Thank you for the reply. Once you get the equation for "k", don't you need to differentiate to find minimums and maximums?
     
  5. Sep 26, 2007 #4
    That is a way of going, wich will give you the max and the min's (namely [itex]\pi/2,3\pi/2[/itex]). What I did is finding out where the denominator is smaller, wich will give the max of the curvature (since the function [itex]1/x[/itex] is decreasing).
     
  6. Sep 26, 2007 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    After a differentiating the curvature with respect to t, then equation reduces to sin(t)cos(t)= 0. Obviously that can only happen if cos(t)= 0 or sin(t)= 0. If you assume cos(t)= 0, then sin(t) is 1 or -1 and putting those into the formula for curvature makes the curvature 15/5= 3. If you assume sin(t)= 0, then cos(t) is 1 or -1 and putting those into the formula for curvature makes the curvature 15/3= 5. Since 5 is the larger of those, it is the maximum curvature (3 is the minimum curvature). sin(t)= 0, cos(t)= 1 or -1 gives x= 5 or -5.
    (if you got that cos(t)= 0 and so x=0, y= 3 or -3, you may have accidently used |r| rather than |v| in the formula. I did the first time I did the calculation!)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Maximum curvature
  1. Curvature is defined as (Replies: 26)

  2. About curvature (Replies: 0)

  3. Curvature of a Helix (Replies: 1)

  4. Constant curvature (Replies: 1)

  5. Radius of curvature (Replies: 3)

Loading...