Calculus

Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3.

1. For 0 is less than or equal to t is less than or equal to 6, find all times t during which particle R is moving to the left. 2. for 0 is less than or equal to t is less than or equal to 6, find all times t during which the two particles travel in opposite directions. 3. Find the acceleration of particles P at time t=3. Is particle P speeding up, slowing down, or doing neither at time t=3? Explain your reasoning. 4. Show that during the interval (1,3), there must be at least one instant when the particle R must have a velocity of -2.

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  1. 1. p '(t) = -2(π/4) sin(π/4t) , which is the velocity
    to move to the left, p '(t) < 0 or negative, so
    sin (π/4t) must be positive.
    the period of sin(π/4t) is 2π/(π/4) = 8
    so for the first 4 seconds the curve is above the x-axis or it is positive
    so for 0 ≤ t ≤ 6, the particle is moving to the left for the first 4 seconds, then to the right for the next 2 seconds.

    ARGGGG , it asked for particle R and I found it for P, Oh well, maybe we can use the above later on

    r ' (t) = 3t^2 - 12t + 9
    = 3(t-1)(t-3)
    zeros are t = 1 and t = 3
    so r'(t) is positive for t < 1 or t > 3 and is negative for 1 < t < 3

    make a sketch of both derivatives on the same grid
    look at when both curves are on the same side of the x-axis ---> going same direction
    when both curves are on opposite sides of the x-axis --> same direction

    From 0 to 1, R is moving to the right,
    from 1 to 3, R is moving to the left
    from 3 to 6 , R is moving to the right again.

    So from 0 to 1, they are moving in the opposite directions
    From 1 to 3, they are moving in the same direction , (both v's are negative
    from 3 to 4 , they are moving in opposites
    from 4 to 6 , they are moving in same directions

    3. r '' (t) = 6t - 12
    r ''(3) = 18-12 which is positive, so r ' (t) or the velocity of R is increasing.

    4. when is 3t^2 - 12t + 9 = -2
    3t^2 - 12t + 11 = 0
    t = (12 ± √12)/6 = appr 2.577 or 1.423

    both t = 2.577 and t = 1.423 fall in the interval 1 ≤ t ≤ 3

    YEAHHH

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    posted by Reiny

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