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Motion on a Curve

  1. Aug 6, 2007 #1
    Why is the velocity vector always tangent to the curve?

    P.S.: I know it makes sense!
    But I can't prove it and its driving me crazy!!
  2. jcsd
  3. Aug 6, 2007 #2
    just compare the definitions
    if your curve is given by r(t)
  4. Aug 6, 2007 #3
    In the case your curve is parametrized by [itex]\vec{M}(t)=(x(t),y(t))[/itex], then the tangent vector is given by [itex]\vec{T}(t)=(\dot{x}(t),\dot{y}(t))[/itex]. Now, you can do two things to prove that [itex]\vec{M}[/itex] and [itex]\vec{T}[/itex] are parallel. One is to calculate the normal vector to the curve and then the dot product, proving that they are orthogonal. The second one is to calculate the cross product between the curve and the tangent.

    This is really simple in [itex]\mathbb{R}^2[/itex], so i'll recommend you to prove it for [itex]\mathbb{R}^n[/itex] as well.

    (If you are working with functions in [itex]\mathbb{R}[/itex], then [itex]\vec{M}(t)=(t,f(t))[/itex])
    Last edited: Aug 6, 2007
  5. Aug 6, 2007 #4
    Why would [itex]\vec{M}(t)[/itex] and [itex]\vec{T}(t)[/itex] be parallel to each other? [itex]\vec{M}(t)[/itex] is the position vector correct?
  6. Aug 6, 2007 #5


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    It is my guess that you're trying to prove that the velocity vector is always tangent to the curve, but you don't have a clear idea (i.e. a definition!) of what it means for a vector to be tangent to a curve.

    Like you noted, it makes sense that the velocity is tangent to the curve. So that is how we decide to define "tangency to the curve" (see matness). We'll say that some vector is tangent to the curve at some point if that vector is parallel to the derivative at that point.

    With that definition, your problem is more than trivial.
  7. Aug 6, 2007 #6
    OH Yes! I didn't even think about how tangent vectors were defined! How foolish of me... :rofl:
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