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Curvature of polar curve

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the polar curve r=e^(a*theta), a>0, find the curvature K and determine the limit of K as (a) theta approaches infinity and (b) as a approaches infinity.

    2. Relevant equations
    x=r*cos(theta)
    y=r*sin(theta)
    K=|x'y''-y'x''|/[(x')^2 + (y')^2]^(3/2)

    3. The attempt at a solution
    I've tried converting the polar curve using the first equation, solving for their first and second derivatives, then plugging them into equation 3 but that gets very, very long.

    So next, I apply the properties of limits to the relevant equations. However, I get stuck when I need to find the limit of asin(x), acos(x), sin(x), cos(x) as x approaches infinity. What now?
     
  2. jcsd
  3. Sep 27, 2015 #2

    mfb

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    You know sin(x) and cos(x) are limited. That should be sufficient.

    If asin is the inverse sine, then asin(x) with x to infinity shouldn't occur. If it is a*sin(x), then see above.
     
  4. Sep 27, 2015 #3
    I'm getting 0 for the limit of x', x'', y', y'' as a approaches infinity and when theta approaches infinity.
     
  5. Sep 27, 2015 #4

    mfb

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    A curvature that approaches zero makes sense.
     
  6. Sep 27, 2015 #5
    Yes, but then I'd end up with a 0 in the denominator in the third equation which wouldn't work. However, that would require me to find a work around and solve for the limit some other way so I don't get 0/0. I haven't been able to figure out what I have to do to get around that.

    I am expecting the curvature to be 0 in both cases when a or theta approaches infinity so I'm expecting the numerator to be 0.
     
  7. Sep 27, 2015 #6

    SteamKing

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    Instead of using the radius of curvature formula for cartesian coordinates, why don't you try the equivalent formula for polar coordinates?

    http://mathworld.wolfram.com/RadiusofCurvature.html
     
  8. Sep 27, 2015 #7
    Thanks! That's just what I need.

    I've ended up with R=sqrt((a^2 + 1)e^(2ax)) so K=1/sqrt((a^2 + 1)e^(2ax)). Taking the limit as a approaches infinity I get 0 and taking the limit as x approaches infinity, I also get 0.
     
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