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  1. #1
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    Question vector problems

    1] how to find the unit tangent vector T(t)at the indicated point of the vector function r(t)=(e^(8t) cos(t)) i + (e^(8t)sin(t)) j + (e^(8t)) k ,what is T(pi/2) ?

    >> i am not sure, but i kind of think that it should be derivative and divided by the magnitude.. but somehow it cannot be done, please help!!!


    And also 2] how to find parametric equations for the tangent line at the point (cos(3pi/6),sin(3pi/6),(3pi/6)) on the curve x = cos t, y = sin t ,z = t ?

    >> i totally get lost from the question.. can someone please help?
    Last edited by iwonder; September 11th 2008 at 09:36 PM.
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  2. #2
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    Quote Originally Posted by iwonder View Post
    1] how to find the unit tangent vector T(t)at the indicated point of the vector function r(t)=(e^(8t) cos(t)) i + (e^(8t)sin(t)) j + (e^(8t)) k ,what is T(pi/2) ?

    >> i am not sure, but i kind of think that it should be derivative and divided by the magnitude.. but somehow it cannot be done, please help!!!

    [snip]
    A vector tangent to the curve defined by r(t) is T(t) = \frac{dr}{dt}. Substitute t = \frac{\pi}{2} and divide the resulting vector by its magnitude.
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  3. #3
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    Quote Originally Posted by iwonder View Post
    [snip]
    And also 2] how to find parametric equations for the tangent line at the point (cos(3pi/6),sin(3pi/6),(3pi/6)) on the curve x = cos t, y = sin t ,z = t ?

    >> i totally get lost from the question.. can someone please help?
    A vector tangent to the curve has i, j and k components given respectively by \frac{dx}{dt} = - \sin t, \frac{dy}{dt} = \cos t, \frac{dz}{dt} = 1.

    So at t = \frac{3 \pi}{6} = \frac{\pi}{2} a vector in the direction of the tangent is ......

    And a point on the tangent is obviously \left( \cos \frac{\pi}{2} = 0, \, \sin \frac{\pi}{2} = 1, \, \frac{\pi}{2} \right).

    So you have a vector in the direction of the line and a point on the line. You should know how to get the parametric equations of the line from these two things.
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