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Math Help - The unit sphere in R^3 is arcwise connected

  1. #1
    Senior Member Pinkk's Avatar
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    The unit sphere in R^3 is arcwise connected

    So I attempted to find the intersection of the unit sphere with a plane that contains any two arbitrary points a, b on the sphere and the origin, but I only came up with a really convoluted expression that I am not even sure can be used to find a function f: [0,1] \to \mathbb{R}^{3} such that f(0) = a, f(1) = b and for all t\in [0,1], f(t) is on the unit sphere. Any help would be appreciated.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Pinkk View Post
    So I attempted to find the intersection of the unit sphere with a plane that contains any two arbitrary points a, b on the sphere and the origin, but I only came up with a really convoluted expression that I am not even sure can be used to find a function f: [0,1] \to \mathbb{R}^{3} such that f(0) = a, f(1) = b and for all t\in [0,1], f(t) is on the unit sphere. Any help would be appreciated.
    You can parametrize the sphere with two angles \vartheta, \varphi so that you get x=r\sin\vartheta \cos\varphi, y = r\sin \vartheta \sin\varphi, z=r\cos\vartheta.
    This is a map g:\, [0;2\pi)\times [-\pi/2;\pi/2]\to \in\mathbb{R}^3, (\varphi,\vartheta)\mapsto (x,y,z).
    Assuming that you have coordinates (\varphi_a,\vartheta_a), (\varphi_b,\vartheta_b) for two given points a,b on the unit sphere, you next parametrize the line segment connecting the points (\varphi_a,\vartheta_a), (\varphi_b,\vartheta_b) that lie in the rectangular area of the domain of these two parameters. This gives you the map h:\, [0;1]\to [0;2\pi)\times [-\pi/2;\pi/2].
    Now a parametrization of the arc connecting a and b on the sphere is f\,: [0;1]\to \mathbb{R}^3, t\mapsto g(h(t)), I think.
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  3. #3
    Senior Member Pinkk's Avatar
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    What would a parameterization of the line segment look like, simply h(t) = (\varphi_a,\vartheta_a) + t(\varphi_b - \varphi_a,\vartheta_b - \vartheta_a)?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Pinkk View Post
    What would a parameterization of the line segment look like, simply h(t) = (\varphi_a,\vartheta_a) + t(\varphi_b - \varphi_a,\vartheta_b - \vartheta_a)?
    Sure, why not? Since the domain of the parametrization of the sphere by g:\, (\varphi,\vartheta)\to\mathbb{R}^3 is a rectangle and thus convex, that parametrization h(t) of the line segment is sure to give you a line segment that remains completely within the domain of g.
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