Calculus particle moves along x axis problem?

A particle moves along the x axis with position at time t given by x(t)= e^ (-t)sint for 0<t<2π

a. find the time t at which the particle is farthest to the left. Justify answer

b. find the value of the constant A for which x(t) satisfies the equation Ax"(t) + x'(t) + x(t) = 0 for 0<t<2π

Comments

  • a)

    x(t)= e^ (-t) sin t

    x'(t) = -e^(-t) sin t + e^(-t) cos t = 0

    e^(-t) ( cos t - sin t) = 0

    cos t = sin t

    divide both sides by sin t

    tan t = 1

    t = PI/4

    t = 5PI/4

    x''(t) = e^(-t) sin t -e^(-t) cos t - e^(-t) cos t - e^(-t) sin t = -2 e^(-t) cos t

    When t = PI/4, x''(t) = -0.644794 < 0, so x has a maximum

    When t=5PI/4, x''(t) = 0.027864 > 0, so x has a minimum

    Farthest left when t=PI/4

    b)

    x(t)= e^ (-t) sin t

    x'(t) = -e^(-t) sin t + e^(-t) cos t

    x''(t) = e^(-t) sin t -e^(-t) cos t - e^(-t) cos t - e^(-t) sin t = -2 e^(-t) cos t

    Ax"(t) + x'(t) + x(t) = A [ -2 e^(-t) cos t] + ( -e^(-t) sin t + e^(-t) cos t) + e^ (-t) sin t = 0

    -2 A e^(-t) cos t + e^(-t) cos t = 0

    e^(-t) cos t ( -2A+1) = 0

    A=1/2

  • the equation you provide is one dimensional; this suggests that there is not any replace in y or z displacement, meaning there at the instant are not any forces interior the y or z instructions interior the x direction, we come across the acceleration interior the x direction by skill of taking the 2d spinoff of x(t); this provides us a=-18t+8 so the rigidity appearing in this particle is F=0.15kg x (-18t+8) at a time t

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