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Math Help - derivatives of vector functions

  1. #1
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    derivatives of vector functions

    first bit done stuck on second part:

    The equations of motion of a point particle in space are given by

    1. x(t) = 2t^3 - 3, y(t)= -3t^3, z(t)= 4t^3 - 1

    2. x(t)= asin(wt), y(t) = acos(wt), z(t) = t

    In each case, calculate the velocity vector v and the acceleration vector a. Also, calculate the absolute values of these vectors.

    Now the bit I can't do is,

    what are the curves along which the particle moves?
    (find these values for an arbitraryt, with a and w being fixed parameter)
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  2. #2
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    Quote Originally Posted by sonia1 View Post
    first bit done stuck on second part:

    The equations of motion of a point particle in space are given by

    1. x(t) = 2t^3 - 3, y(t)= -3t^3, z(t)= 4t^3 - 1

    2. x(t)= asin(wt), y(t) = acos(wt), z(t) = t

    In each case, calculate the velocity vector v and the acceleration vector a. Also, calculate the absolute values of these vectors.

    Now the bit I can't do is,

    what are the curves along which the particle moves?
    (find these values for an arbitraryt, with a and w being fixed parameter)
    x = 2t^3 - 3 \Rightarrow t^3 = \frac{x+3}{2} .... (1)

    y = -3t^3 \Rightarrow t^3 = - \frac{y}{3} .... (2)

    z = 4t^3 - 1 \Rightarrow t^3 = \frac{z+1}{4} .... (3)

    From (1), (2) and (3) it follows that the path is a line with cartesian equation \frac{x+3}{2} = - \frac{y}{3} = \frac{z+1}{4}


    x = a \sin (\omega t) \Rightarrow x^2 = a^2 \sin^2 (\omega t) .... (1)

    y = a \cos (\omega t) \Rightarrow y^2 = a^2 \cos^2 (\omega t) .... (2)

    (1) + (2): x^2 + y^2 = a^2.

    You have a spiral with a circular cross-section centred along the z-axis.
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  3. #3
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    could you please show me a sketch of what it actually looks like
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