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Thread: 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 $\displaystyle a, b$ 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 $\displaystyle f: [0,1] \to \mathbb{R}^{3}$ such that $\displaystyle f(0) = a, f(1) = b$ and for all $\displaystyle t\in [0,1]$, $\displaystyle f(t)$ 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 $\displaystyle a, b$ 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 $\displaystyle f: [0,1] \to \mathbb{R}^{3}$ such that $\displaystyle f(0) = a, f(1) = b$ and for all $\displaystyle t\in [0,1]$, $\displaystyle f(t)$ is on the unit sphere. Any help would be appreciated.
    You can parametrize the sphere with two angles $\displaystyle \vartheta, \varphi$ so that you get $\displaystyle x=r\sin\vartheta \cos\varphi, y = r\sin \vartheta \sin\varphi, z=r\cos\vartheta$.
    This is a map $\displaystyle g:\, [0;2\pi)\times [-\pi/2;\pi/2]\to \in\mathbb{R}^3, (\varphi,\vartheta)\mapsto (x,y,z)$.
    Assuming that you have coordinates $\displaystyle (\varphi_a,\vartheta_a), (\varphi_b,\vartheta_b)$ for two given points $\displaystyle a,b$ on the unit sphere, you next parametrize the line segment connecting the points $\displaystyle (\varphi_a,\vartheta_a), (\varphi_b,\vartheta_b)$ that lie in the rectangular area of the domain of these two parameters. This gives you the map $\displaystyle h:\, [0;1]\to [0;2\pi)\times [-\pi/2;\pi/2]$.
    Now a parametrization of the arc connecting a and b on the sphere is $\displaystyle f\,: [0;1]\to \mathbb{R}^3, t\mapsto g(h(t))$, 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 $\displaystyle h(t) = (\varphi_a,\vartheta_a) + t(\varphi_b - \varphi_a,\vartheta_b - \vartheta_a)$?
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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 $\displaystyle h(t) = (\varphi_a,\vartheta_a) + t(\varphi_b - \varphi_a,\vartheta_b - \vartheta_a)$?
    Sure, why not? Since the domain of the parametrization of the sphere by $\displaystyle g:\, (\varphi,\vartheta)\to\mathbb{R}^3$ 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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