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March 25, 2017

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4. A particle starts at the point (5, 0) at t = 0 and moves along the x-axis in such a way that at time t > 0 the velocity is given by v(t)=t/(1+t^2)
a. Determine the maximum velocity attained by the particle. Justify your answer.
b. Determine the position of the particle at t = 6.
c. Find the limiting value of the velocity as t increases without bound.
d. Does the particle ever pass the point (500,0)? Explain.

  • calculus - ,

    dv/dt = [ (1+t^2) -t(2t) ] / (1+t^2)^2
    = (1-t^2) /(1+t^2)^2
    zero when
    t = + or - 1
    when t = +1
    v = 1/2
    when t = -1 that was before we started
    so v max is 1/2 at t = 1

    integrate dx = v dt to get x(6), remember to add 5
    (1/2)ln(t^2+1) from t = 0 to 6
    it is 0 when t = 0
    so
    x = 5 + .5 ln 37 when x = 6


    looks like 1/t for large t, so approaches 0

    sure it reaches all +x
    because as t ---> oo
    x = 5 + ln (oo)

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