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Differential geometry

  1. Jun 30, 2014 #1
    whether γ=γ(s):I->R^3 curve parameterized as to arc length (single speed). Assume that γ is the surface sphere centered on the origin (0,0). Prove that the vector t(s) is perpendicular to the radius of the sphere at point γ(s), for each s.


    i know that t(s)=γ΄(s) but i dont know how to continue to prove it , maybe i dont have undrstand the problem on how to show that t(s) is vertical on the radius of sphere
     
  2. jcsd
  3. Jun 30, 2014 #2

    micromass

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    Since ##\gamma(s)## is on the surface of the sphere, you know
    [tex]<\gamma(s),\gamma(s)> = 1[/tex]
    Now differentiate both sides.
     
  4. Jun 30, 2014 #3
    i do not have any other information... for γ(s)
     
  5. Jun 30, 2014 #4
    As was mentioned before, if ##\gamma## is on the surface of a sphere that means that
    \begin{equation*}
    \gamma(t) \cdot \gamma(t) = |\gamma(t)|^2 = \text{ constant}.
    \end{equation*}
    Then you should differentiate both sides with respect to ##t##.
     
  6. Jun 30, 2014 #5
    with this i show that t(s) is perpendicular on the radius of the sphere? we do not use any radius r
     
  7. Jun 30, 2014 #6
    What does the vector ##\gamma(s)## look like?
     
  8. Jun 30, 2014 #7
    i think γ(s) is our curve and vector is t(s)=γ΄(s)
     
  9. Jun 30, 2014 #8
    Well, as you wrote yourself, ##\gamma : [0,1] \rightarrow \mathbb{R}^3##. So ##\gamma(s) \in \mathbb{R}^3##. A point in ##\mathbb{R}^3## is also a vector. Don't you agree?
     
  10. Jun 30, 2014 #9
    yes its my fault , γ(s) is a vector..
     
  11. Jun 30, 2014 #10
    if i differentiate i will find γ'(s)=0
     
  12. Jun 30, 2014 #11
    That is not correct. Differentiate ##\gamma(s) \cdot \gamma(s) = c## again. Remember the product rule.
     
  13. Jun 30, 2014 #12
    2γ(s)γ'(s)=0 but what i will do next?
     
  14. Jun 30, 2014 #13

    micromass

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    So the dot product is zero, what does that mean?
     
  15. Jun 30, 2014 #14
    γ(s) and γ'(s) are perpendicular , but we want t(s) to be perpendicular with radius.
     
  16. Jun 30, 2014 #15
    Think about this.
     
  17. Jun 30, 2014 #16
    i am thinking it but i cant find any answer , its a radious? if yes why?
     
  18. Jun 30, 2014 #17
    Well, ##\gamma(s)## is a point on the surface of the sphere, but it is also a vector, right? What would that vector be? Try drawing it, if you are still unsure.
     
  19. Jun 30, 2014 #18
    and beacause t(s)=γ'(s) and γ(s) ιs radious t(s) is perpendicular on γ(s)?
     
  20. Jul 1, 2014 #19
    Yes.
     
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