calculus

Two particles move along the x-axis. For 0≤t≤6, the position of particle P is p(t)=2cos(t*π/4), while the position of particle R is r(t)=t^3-6t^2+9t+3.
For 0≤t≤6 find all the times which the two particles travel in opposite directions.

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  1. P is moving to the right when dp/dt > 0
    R is moving to the left when dr/dt < 0

    dp/dt = -pi/2 sin(t*pi/4)
    dq/dt = 3t^2 - 12t + 9 = 3(t^2-4t+3) = 3(t-1)(t-3)

    now it should be fairly easy. what do you get?

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  2. what values do i plug in for t? do i have to guess and check?

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  3. when is -pi/2 sin(t*pi/4) < 0?
    when is sinx > 0?
    for 0 < x < pi
    so, we need 0 < t*pi/4 < pi
    0 < t < 4
    in that interval, sin(t*pi/4) > 0, so -pi/2 sin(t*pi/4) < 0

    similarly, dq/dt > 0 for t<1 or t>3

    so, on 0<t<1 and 3<t<4
    dp/dt < 0 and dq/dt > 0, so P and Q are moving in opposite directions.

    solve the other way to see when P is moving right and Q is moving left.

    don't guess and check. solve the inequalities. That's just algebra, not calculus.

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