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Homework Help: Unit Tangent Vector at a Point

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    r(t) = costi + 2 sint j
    Find the tangent vector r'(t) and the corresponding unit tangent vector u(t) at point P:(.5, 3.5,0)

    2. Relevant equations
    r'(t) = r(t)dt
    u(t) = r'(t) / |r'(t)|

    3. The attempt at a solution

    r'(t) = -sinti + 2costj

    |r'(t)| = [sin2t + 4cos2t].5
    = [1-3cos2t].5

    u(t) = {-sinti + 2costj} / {[1-3cos2t].5}

    I think I am right so far, however I don't know what I am supposed to due with Point P to find the unit tangent vector at that point.

    Thanks in advance for the help.
  2. jcsd
  3. Oct 20, 2008 #2


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    The point P corresponds to a value of the parameter t.
    r'(t) is the tangent vector at the point r(t)
  4. Oct 20, 2008 #3
    So all that is required is to plug the i and j components of point P into both r'(t) and the u(t) equation to "evaluate" them at that point?
  5. Oct 20, 2008 #4


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    How would you plug in the components of one vector into another vector?

    No, the idea is that you plug some t, which corresponds to the point P, into both r'(t) and u(t). You can consider r(t) as describing the position of a particle at time t, and r'(t) its velocity at that time. You can reformulate the question as: "Give the velocity of the particle when it is at P" or, equivalently: "Give the velocity of the particle at that time, at which its position vector is P".
  6. Oct 20, 2008 #5
    Ahh...so you get =>

    P = r(t)
    <.5, 3.5,0> = <cos t, 2sin t, 0>

    => t = 60

    From there evaluate r'(t) and u(t) at t=60.

    Is this correct?
  7. Oct 20, 2008 #6


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    yes. Although you should be thinking "[itex]\pi/3[/itex]" rather than "60" at this point.
  8. Oct 20, 2008 #7
    I know...degrees have always been hard to get out of my head. I need to start thinking in radians.

    Thank you very much for the help!
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