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The Wildest point on an ellipse

  1. Apr 11, 2008 #1
    The motivation behind my question stems from my own curiosity. There was recently a post in this forum titled "The Widest Point on an ellipse" (or something to that effect). In any event, I misread the title, as "The wildest". I got to thinking, and remembered from vector calculus there existed a formula to measure the severity of a curve at any point.

    The formula is as follows:
    [tex]K(t) = \frac{||r\prime(t) \times r\prime\prime(t)||}{||r\prime(t)||^3}[/tex]

    Where r(t) is our curve in parametric form.

    I wanted to use this to figure out in general , what points on an ellipse are the wildest, ie at which points can we find a maximum value for K?

    To keep things simple I assumed the ellipse I would look at would be soley in the xy plane. the equation I decided on is the following:

    [tex]r(t) =\left( \begin{array}a a\cos(t) \\ b \sin(t) \\0 \end{array}
    \right) [/tex]

    Now I am running into a problem when trying to calulate K(t)

    The numerator actually works out very nice:
    [tex] \pm 2ab[/tex]

    the denominator I can't figure out. I get stuck at this:
    [tex](\sqrt{a^2\sin(t)^2 + b^2 \cos(t)^2})^3[/tex]

    Anyone see a way to reduce this?
    Last edited: Apr 11, 2008
  2. jcsd
  3. Apr 11, 2008 #2


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    Why do you need to simplify it? Just differentiate and solve for the maximum.
  4. Apr 11, 2008 #3


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    Staff Emeritus
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    The "wildest" points on an ellipse, the points with largest curvature are the points at the ends of the major axis. The points with the smallest curvature are the points at the ends of the minor axis.
  5. Apr 11, 2008 #4

    I was going to mention that this was my hypothesis, but I wanted to figure it out for myself.
  6. Apr 11, 2008 #5
    Ah ha! Of course...

    Thanks, man.
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