Calculus (Related Rates)

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The position of a particle moving in a straight line is given by s(t) = (e^(-t))(cos(5t)) for t>0, where t is in seconds. If the particle changes direction at time T seconds, then T must satisfy the equation:
cos(5T) = 0
5T = arctan(-1/5)
5e^(-t) sin(5t) = 0
tan(5T) = -1/5
cos(5T) = 5

I know that a change in direction will be marked by a change from positive to negative or vice versa, but I don't understand the equations the question gives me. Could someone please talk me through this process to find the right answer?

  • Calculus (Related Rates) -

    you are correct, as far as you go. When it changes from pos to neg, it will be zero.

    Note that the particle changes direction, not position. So, its velocity changes sign. The velocity is given by the derivative.

    s = e^-t cos5t
    s' = e^-t (-cos5t - 5sin5t)

    so, since e^-t is always positive, we need

    -cos5t - 5sin5t = 0
    tan 5t = -1/5

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