# Calculus

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 its velocity v(t) is given by v(t)= t/(1+t^2).

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.

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1. dx/dt = t/(1+t^2)
that will be maximum when the derivative d^2x/dt^2 = 0
d^2x/dt^2 = [(1+t^2)1 -t(2t) ](1+t^2)^2
0 when numerator is zero
0 = 1 + t^2 - 2 t^2
0 = 1 - t^2
t^2 = 1
so max or min at t = 1
we know it is max because v decreases with big x

a ) so max v = 1/2 at t = 1

integrate that mess and put in initial condition of x = 5 at t = 0
x =int t dt/(t^2+1) = C+ (1/2)ln(t^2+1)
when t = 0, x = 5
5 = C + (1/2) ln(1) = C + 0 = C
so
x = 5 + (1/2) ln(t^2+1)
put in t = 6
x = 5 + .5(3.61) = 6.81

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2. c) as t --->oo v ---->1/t = 0

d) 500 = 5 + .5 ln (t^2+1)
495 (2) = ln(t^2+1)
990 = ln (t^2+1)

e^(990) = t^2 + 1
oh, my, calculator overflow :)

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3. Thank you!

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