- #1
syeh
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Homework Statement
A partcile starts att he point (1,0) and t=0 and moves along the x-axis so that at time t≥0 its velocity v(t) is given by v(t)= [itex]\frac{1+ t}{(1+t^2)}[/itex]
A) Determine the maximum velocity of the particle.
B) Find an expression for the position s(t) of the particle at time t.
C) What is the limiting value of the veolicty as t increases without bound?
D) Determine for which values of t, if any, the particle reaches the point (101,0)
Homework Equations
The Attempt at a Solution
A) To find the maximum velocity:
Checked endpoints: t=0
v(0)=1
Where v'(t)=0: (using chain rule:
v'(t)= [itex]\frac{(1+t^2)(1)-(t)(2t)}{(1+t^2)^2}[/itex]
=[itex]\frac{1+ t^2 - 2t^2}{(1+t^2)^2}[/itex]
0=1+ t^2 - 2t^2
t=0.577
V(.577)=1.433
Where v'(t)=∅
none
So the max. velocity is 1.433
The answers said it was 1.5, but that might just be from rounding
B) I don't know how to find an expression of the position. it starts at (1,0) which is an odd place to start. how to i put that into the equation?
The answer says s(t) = 1+ t + ln√(1+t^2)
How did they get that?!?
C) So I guess limiting value is asking if its approaching a number? Like a limit? How do i find that?