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Showing the derivative of a vector is orthogonal to the vector

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/6j8W6.jpg
    I'm trying to understand that example in the text. I can imagine a curve on a sphere having the derivative vector being orthogonal to the position vector. What I don't understand is, how does "if a curve lies on a sphere with center the origin" mean the same thing as "if |r(t)|=c (constant)"?

    2. Relevant equations
    The problem is I don't understand why the statement says "Show that if |r(t)|=c ..."
    Doesn't the example mean that every derivative vector of a curve is orthogonal to the position vector since |r(t)|=c for all t as long as the curve is continuous? (thinking it in terms of sqrt(x^2+y^2+z^2))
    When is |r(t)| not equal to a constant?
     
  2. jcsd
  3. Aug 23, 2011 #2
    If |r(t)| is a constant, then you are essentially talking about a radius. Think about what the definition of |r(t)| is (it looks like you have an idea in the relevant equations section). If it is not equal to a constant, then the position vector and the derivative are not necessarily orthogonal. It is easy to find an example of this (for example, r(t) = <t,t>).
     
  4. Aug 23, 2011 #3

    Dick

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    I don't quite get your question. |r(t)|=c means the position r(t) is at a constant distance of c from the origin at all times. If it's always at a distance of c from the origin, then it lies on a sphere of radius c around the origin. For a general curve, the distance from the origin will vary with time.
     
  5. Aug 23, 2011 #4
    How is |r(t)| not equal to a constant for r(t)=<t,t>?

    Thanks for replying!
     
  6. Aug 23, 2011 #5
    Ohh, I get it now. I wasn't thinking of it the way you have it in the bold.

    Thanks!
     
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