vfc-a-19

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Vector Functions and Curves One variable functions Question An object travels on the curve given by the parameterization r =3ui + 3u 2 j +2u 3 k . Given that the object has a constant speed of 6 and that u is increasing, find the velocity and acceleration of the object at the point (3, 3, 2). Answer r = 3ui +3u 2 j +2u 3 k v = du dt (3i +6uj +6u 2 k ) a = d 2 u dt 2 (3i +6uj +6u 2 k ) + ˆ du dt ! 2 (6j + 12uk ) As the speed of the object of 6 and u is increasing, 6= |v| = 3 du dt 1+4u 2 +4u 4 = 3(1 + 2u 2 ) du dt du dt = 2 1+2u 2 , and d 2 u dt 2 = -2 (1 + 2u 2 ) 2 4u du dt = -16u (1 + 2u 2 ) 3 It can be seen that the object is as (3, 3, 2) when u = 1. At this point du dt =2/3 and d 2 u dt 2 = -16/27. v = 2 3 (3i +6uj +6u 2 k ) = 2i +4j +4k a = - 16 27 (3i +6j +6k )+ 2 3 2 (6j + 12k ) = 8 9 (-2i - j +2k ) 1

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VFC-A-19

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  • Vector Functions and Curves

    One variable functions

    Question

    An object travels on the curve given by the parameterization r = 3ui +3u2j + 2u3k. Given that the object has a constant speed of 6 and that uis increasing, find the velocity and acceleration of the object at the point(3, 3, 2).Answer

    r = 3ui + 3u2j + 2u3k

    v =du

    dt(3i + 6uj + 6u2k)

    a =d2u

    dt2(3i + 6uj + 6u2k)

    +

    (du

    dt

    )2(6j + 12uk)

    As the speed of the object of 6 and u is increasing,

    6 = |v| = 3dudt

    1 + 4u2 + 4u4

    = 3(1 + 2u2)du

    dt

    du

    dt=

    2

    1 + 2u2, and

    d2u

    dt2=

    2(1 + 2u2)2

    4udu

    dt

    =16u

    (1 + 2u2)3

    It can be seen that the object is as (3, 3, 2) when u = 1.At this point du

    dt= 2/3 and d

    2u

    dt2= 16/27.

    v =

    2

    3(3i + 6uj + 6u2k)

    = 2i + 4j + 4k

    a = 1627(3i + 6j + 6k) +

    (2

    3

    )2(6j + 12k)

    =8

    9(2i j + 2k)

    1