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Math Help - Curvature

  1. #1
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    Curvature

    1.) Find the unit tangent vector to the curve at the following points:

    r(t) = <3*cos(t), 2*sin(t)>, t = 0, t = -Pi/2, t = Pi/2

    2.) Sketch this curve (the one above) with the addition of the following vectors: r(0), T(0), r(Pi/2), and T(Pi/2)

    Note: in the above r and T are both vectors.
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  2. #2
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    Hello, Ideasman!

    The first one is very straight-forward. .Where is your difficulty?


    1) Find the unit tangent vector to the curve at the following points:

    . . r(t) .= .< 3搾os(t), 2新in(t) >, .t = 0, -π/2, π/2

    The tangent vector is: .r'(t) .= .< -3新in(t), 2搾os(t) >

    At t = 0: .r'(0) .= .< -3新in(0), 2搾os(0) > .= .< 0, 2 >
    . . The unit vector is: .< 0, 1 >

    At t = -π/2: .r'(-π/2) .= .< -3新in(-π/2), 2搾os(-π/2) > .= .< 3, 0 >
    . . The unit vector is: .< 1, 0 >

    At t = π/2 .r'(π/2) .= .< -3新in(π/2), 2搾os(π/2) > .= .< -3, 0 >
    . . The unit vector is: .< -1, 0 >


    2.) Sketch this curve (the one above) with the addition of the following vectors:
    . . r(0), T(0), r(π/2), and T(π/2)

    To identify this curve, eliminate the parameter.

    . . x .= .3搾os(t) . . x/3 .= .cos(t) .[1]

    . . y .= .3新in(t) . . y/2 .= .sin(t) .[2]


    Square [1]: .x/9 .= .cos(t)

    Square [2]: .y/4 .= .sin(t)

    Add: .x/9 + y/4 .= .sin(t) + cos(t)

    . . . . . . . . . . . . . . . . . . . x . .y
    And we have the ellipse: . --- + --- .= .1
    . . . . . . . . . . . . . . . . . . . 9 . . .4

    I assume you can find those four vectors.

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