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Math Help - Trajectory of an arrow

  1. #1
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    Trajectory of an arrow

    An arrow shot from a bow travels along a curved trajectory given by the set of points satisfying


     P(x,y) = (0,22) + r(t)

    where r(t) = <36\sqrt{2}t , (36t - 6t^2)>

    a. find the unit tangent vector T to this curve as a function of t
    b. find unit nomrla vector N to this curve as function of t
    c. taking {T,N} as an orthonormal basis defininf the orientation of the arrow, find the angle through which the arrow rotates from t=3 to t=4
    d. using this angle, or otherwise, find a transformation matrix that maps the basis at time t=3 to the basis at time t=4


    k so r'(t) = <36\sqrt{2} , (36 - 12t)

    so a. T = \frac{<36\sqrt{2} , (36 - 12t)>}{\sqrt{3888-864t+144t^2}}

    b. N=\frac{<12t-36 , 36\sqrt{2}>}{\sqrt{3888-864t+144t^2}}

    c. i tried evaluating the values
    T(3) = \frac{<36\sqrt{2},0>}{\sqrt{2592}}
    N(3)= \frac{<0,36\sqrt{2}>}{\sqrt{2592}}

    T(4) = \frac{<36\sqrt{2},-12>}{\sqrt{2736}}
    N(4) = \frac{<-12,36\sqrt{2}>}{\sqrt{2736}}

    but i don't know if this is usefull or where to go from there.

    d. No idea.
    Last edited by linalg123; June 15th 2013 at 02:26 AM.
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  2. #2
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    Re: Trajectory of an arrow

    The angle through which the arrow rotates is the angle between the tangent vector evaluated at t=3 and t= 4. And you can find that from \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta). Since these are unit tangent vectors, that is just cos(\theta)= \vec{u}\cdot\vec{v}.

    A linear transformation in two dimensions can be represented as a two by two matrix. The linear transformation that maps \begin{bmatrix}x_1 & y_1\end{bmatrix} to \begin{bmatrix}x_2 & y_2\end{bmatrix} is the matrix, \begin{bmatrix}a & b \\ c & d\end{bmatrix} satisfying \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x_1 \\ y_1\end{bmatrix}= \begin{bmatrix}x_2 \\ y_2\end{bmatrix}. That reduces to the two equatons a x_1+ by_1= x_2 and cx_1+ dy_1= y_2. Solve those equations for a, b, c, d (using your values for x_1, y_1, x_2, y_2, of course).
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  3. #3
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    Re: Trajectory of an arrow

    so for the angle i did

    cos(\theta) = \frac{36\sqrt{2}*36\sqrt{2}}{\sqrt{2592}\sqrt{2736  }}

    and theta = inverse of that but it said that was wrong?
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  4. #4
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    Re: Trajectory of an arrow

    Please, do not use 'unattached' pronouns! What does "it" refer to?
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