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Basic vector function questions

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    True or False:
    a. if k(t)=o, the curve is a straight line
    b. if the magnitude of r(t)=1 for all t then r'(t) is orthogonalo to r(t)
    c. different parametrizations of the same curve result in identical tangent vectors at a given point


    2. Relevant equations



    3. The attempt at a solution
    a. k(t) is the curvature, which means T'(t) is zero, but what does T'(t)= 0 tells me?
    b. if the magnitude of r(t)=1, is the r(t) just a circle/sphere? if r(t) is a sphere, is r'(t) orthogonal to r(t)?
    c. i think its true, but i dont know why
     
  2. jcsd
  3. Oct 2, 2009 #2

    LCKurtz

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    "k(t) is the curvature, which means T'(t) is zero, but what does T'(t)= 0 tells me?"

    It tells you that T(t) is a constant vector. What does that tell you?

    If r(t) is a curve, how could it be a sphere?
    Does a curve on the surface of a sphere have to be a circle?
     
  4. Oct 2, 2009 #3
    what about the last question? i think it is right, but i dont know why
     
  5. Oct 2, 2009 #4

    Dick

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    For b, the magnitude of r(t) is sqrt(r(t).r(t))=1. DIfferentiate that expression. Squaring both sides first makes it a little easier. c is false. Let r(t)=(t,0) and r(t)=(-t,0). What's the tangent vector in each case?
     
    Last edited: Oct 2, 2009
  6. Oct 2, 2009 #5
    r(t)=(t,0) and r(t)=(-t,0) wont give you the same cure. i think (-t,0) is not the reparametrization for (t,0)
     
  7. Oct 2, 2009 #6

    Dick

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    r(t)=(t,0) and r(t)=(-t,0) for t in R are both parametrizations of the curve y=0 in the x-y plane. So for that matter is (t^3,0). Unless you have a much more specific notion of 'reparametrization' in mind. And by 'tangent' do you mean the unit tangent or just r'(t)?
     
    Last edited: Oct 2, 2009
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