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Thread: Unit Tangent & Unit Normal Vectors

  1. #1
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    Unit Tangent & Unit Normal Vectors

    Find the unit tangent vector.

    r (t) = 4ti

    r'(t) = 4i

    ||r'(t)|| = root {4^2} = root {16} = 4

    So, T (t) = 4i/4 = i.

    Yes? No?

    I now need the unit normal vector.

    I am stuck here.

    I know that T (t) = i (see above).

    I need to find T'(t) and ||T'(t)||.

    What is the derivative of i?
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    Re: Unit Tangent & Unit Normal Vectors

    Hi there,
    So you are close with your result for a unit vector tangent. It is actually:

    T(t) = (4t/(4|t|))i = (t/|t|)i

    The vector you got of simply i is only true when the variable t is positive, when it is negative your vector would point 180 degrees out of phase.

    A normal unit vector will point perpendicular to the curve at all points. Now since i points along the x axis you know that perpendicular to this is the y axis which uses the unit vector j.

    The easiest solution is to simply make the normal unit vector the unit tangent vector at 90 degrees, so....

    N(t) = (t/|t|)j

    I think this will work....good luck,

    Craig
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    Re: Unit Tangent & Unit Normal Vectors

    Quote Originally Posted by clombard1973 View Post
    Hi there,
    So you are close with your result for a unit vector tangent. It is actually:

    T(t) = (4t/(4|t|))i = (t/|t|)i

    The vector you got of simply i is only true when the variable t is positive, when it is negative your vector would point 180 degrees out of phase.

    A normal unit vector will point perpendicular to the curve at all points. Now since i points along the x axis you know that perpendicular to this is the y axis which uses the unit vector j.

    The easiest solution is to simply make the normal unit vector the unit tangent vector at 90 degrees, so....

    N(t) = (t/|t|)j

    I think this will work....good luck,

    Craig
    this completely misses the point that there are infinite normal vectors to a line.
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  4. #4
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    Re: Unit Tangent & Unit Normal Vectors

    Yes, indeed there are....so......
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  5. #5
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    Re: Unit Tangent & Unit Normal Vectors

    I was able to solve the problem.

    Answers:

    T (t) = i

    N (t) = undefined
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  6. #6
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    Re: Unit Tangent & Unit Normal Vectors

    Quote Originally Posted by clombard1973 View Post
    Hi there,
    So you are close with your result for a unit vector tangent. It is actually:

    T(t) = (4t/(4|t|))i = (t/|t|)i

    The vector you got of simply i is only true when the variable t is positive, when it is negative your vector would point 180 degrees out of phase.

    A normal unit vector will point perpendicular to the curve at all points. Now since i points along the x axis you know that perpendicular to this is the y axis which uses the unit vector j.

    The easiest solution is to simply make the normal unit vector the unit tangent vector at 90 degrees, so....

    N(t) = (t/|t|)j

    I think this will work....good luck,

    Craig
    Sorry but this is incorrect.
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