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M2 Help Needed

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    r = 2\cos (t) i + 2\sin (t) j + (10-.4t)k

    The perpendicular unit vectors i and j are in the horizontal plane and the unit vector k is directed vertically upwards.

    Write down the first two values of t for which Tom is directly below his starting
    point.

    Here is what I did:
    The i and j components must be 0, so

    2\cos (t) = 0
    t =  \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} .........

    2\sin (t) = 0
    t = 0, \pi, 2\pi, 3\pi, 4\pi .........

    Now I cant get a value of t in which both are 0, let alone a value for which bot are 0 and the K component is negative.

    Help is highly appreciated

    Edit: The MS says that the answers are t = 2pi and t= 4 pi. But putting t=2pi in gives r = 2i + 3.7k, which can't be directly below his starting position as it has moved 2 units east and 3.7 units up ?? And the same problem occurs when t=4 is put in.
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    hint: when t = 0

    i = 2
    j = 0

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    (Original post by a.partridge)
    hint: when t = 0

    i = 2
    j = 0

    But I need i = 0, j = 0 and k to be negative.
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    (Original post by member910132)
    But I need i = 0, j = 0 and k to be negative.
    well no value exists where cos(t) and sin(t) are both 0

    but that doesn't matter because at the starting point (i) and (j) are not both 0.

    the starting point comes from r(0) = ........ you see not the origin!
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    regarding the negative you can see that the solutions must be the first to satisfy the inequality

    (A)t > 10 there A is the coefficient in the expression for the displacement parallel to K
  6. Offline

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    (Original post by a.partridge)
    well no value exists where cos(t) and sin(t) are both 0

    but that doesn't matter because at the starting point (i) and (j) are not both 0.

    the starting point comes from r(0) = ........ you see not the origin!
    Thnx a lot for that

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