vfc-a-20

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Vector Functions and Curves One variable functions Question An object travels on the curve of intersection of the cylinders y = -x 2 and z = x 2 with increasing x. When the particle is at (1, -1, 1), it has a speed of 9cm/s which is increasing at a rate of 3cm/s 2 . Given that all distances are in cm, find the velocity and acceleration of the object at that point. Answer r = xi - x 2 j + x 2 k v = dx dt (i - 2xj +2xk ) a = d 2 x dt 2 (i - 2xj +2xk )+ ˆ dx dt ! 2 (-2j +2k ). |v | = dx dt 1+4x 4 +4x 4 = 1+8x 4 dx dt , as x is increasing. At (1, -1, 1), x = 1 and |v | = 9, dx dt = 3. at that point v =3i - 6j +6k . Now d dt |v | = 1+8x 4 d 2 dt 2 + 16x 3 1+8x 4 ˆ dx dt ! 2 . When x = 1, the left side is 3. 3 ˆ d 2 x dt 2 ! + 48 = 3 and d 2 x dt 2 = 15 at that point. Acceleration at that point a = -15(i - 2j +2k ) + 9(-2j +2k ) = -15i + 12j - 12k 1

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VFC-A-20

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  • Vector Functions and Curves

    One variable functions

    Question

    An object travels on the curve of intersection of the cylinders y = x2 andz = x2 with increasing x. When the particle is at (1,1, 1), it has a speed of9cm/s which is increasing at a rate of 3cm/s2. Given that all distances arein cm, find the velocity and acceleration of the object at that point.Answer

    r = xi x2j + x2kv =

    dx

    dt(i 2xj + 2xk)

    a =d2x

    dt2(i 2xj + 2xk) +

    (dx

    dt

    )2(2j + 2k).

    |v| =

    dxdt1 + 4x4 + 4x4

    =1 + 8x4

    dx

    dt,

    as x is increasing.At (1,1, 1), x = 1 and |v| = 9, dx

    dt= 3.

    at that pointv = 3i 6j + 6k

    .Now

    d

    dt|v| =

    1 + 8x4

    d2

    dt2+

    16x31 + 8x4

    (dx

    dt

    )2.

    When x = 1, the left side is 3.

    3

    (d2x

    dt2

    )+ 48 = 3

    andd2x

    dt2= 15

    at that point. Acceleration at that point

    a = 15(i 2j + 2k) + 9(2j + 2k)= 15i + 12j 12k

    1