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Thread: calc 3, motion

  1. #1
    Sep 2009

    calc 3, motion

    Consider the motion of a particle along a helix given by r(t) = sinti + costj + (t^2 - 3t + 2)k, where the k component measures the height in meters above ground and t >= 0.
    (a) Does the particle ever move downward?
    (b) Does the particle ever stop moving?
    (c) At what times does it reach a position 12 meters above the ground?
    (d) What is the velocity of the particle when it is 12 meters above the ground?

    no idea what to do here, please help
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  2. #2
    MHF Contributor Matt Westwood's Avatar
    Jul 2008
    Reading, UK
    a) For this question you're only interested in the k component as that measures the up-and-down-ness.

    If the particle is moving downwards, it means its vertical velocity (rate of change of the k component w.r.t. t) is negative.

    So, differentiate the k component w.r.t. t (that is, the $\displaystyle t^2 - 3t + 2$ and see whether what you get can ever be negative.

    b) For it to stop moving, the components of i, j and k must all have zero velocity.

    So differentiate all components w.r.t. t and see if there's a value of t that will make all of these zero at the same time. My suggestion is that I rather think there isn't.

    c) Simply solve the equation in t that puts the k component equal to 12. You'll have a quadratic to solve.

    d) Plug the value of t into tha equation you got in a) and see what you get.
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