Posted by Tiffany on Sunday, March 25, 2012 at 1:28am.
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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