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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

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    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!)
     
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