anonymous
  • anonymous
A particle moves along a line so that at any time t> (or equal to) 0 its velocity is given by v(t) = t/1+t^2. At time t = 0, the position of the particle is s(0) = 5. Determine the position of the particle at t = 3.
Biology
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
i know i must take the antiderivative
zepdrix
  • zepdrix
\[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.

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anonymous
  • anonymous
yea i did u sub for 1 + t^2
anonymous
  • anonymous
i meant +
anonymous
  • anonymous
i got 1/u du
zepdrix
  • zepdrix
\[\large \int\limits \frac{1}{u}du\]Yep, sounds good.
zepdrix
  • zepdrix
Woops, should be a factor of 1/2 in front I think.
anonymous
  • anonymous
and i should get u = 1 and u = 10?
anonymous
  • anonymous
oh yea 1/2 (1/u) du
zepdrix
  • zepdrix
No we don't want to add limits to our integral.
anonymous
  • anonymous
no?? :O
zepdrix
  • zepdrix
Evaluate the `Indefinite Integral` to get a function of position, s(t). It will include an unknown constant +C. We can use our `Initial Conditions` to solve for C.
anonymous
  • anonymous
oh wait... i do the t(3) = t(0) + the integral
zepdrix
  • zepdrix
So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O
anonymous
  • anonymous
yes!
zepdrix
  • zepdrix
Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]
anonymous
  • anonymous
OHHHHH
anonymous
  • anonymous
well i did something similar... but i put t(0) -_-
zepdrix
  • zepdrix
heh :D
anonymous
  • anonymous
so i plug 3 for t?
zepdrix
  • zepdrix
After you've determined the value of C, yes, the final step is asking for s(3). So you would plug in t=3 and solve.
anonymous
  • anonymous
i gotsss 3.848
anonymous
  • anonymous
idk if that's right.... scary
zepdrix
  • zepdrix
Hmm I got 6.15
anonymous
  • anonymous
:((((((
zepdrix
  • zepdrix
\[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]
anonymous
  • anonymous
oh i subt 5 -_-

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