swin2013
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.
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swin2013
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@zepdrix
swin2013
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i know i must take the antiderivative
zepdrix
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\[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.
swin2013
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yea i did u sub for 1 + t^2
swin2013
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i meant +
swin2013
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i got 1/u du
zepdrix
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\[\large \int\limits \frac{1}{u}du\]Yep, sounds good.
zepdrix
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Woops, should be a factor of 1/2 in front I think.
swin2013
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and i should get u = 1 and u = 10?
swin2013
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oh yea 1/2 (1/u) du
zepdrix
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No we don't want to add limits to our integral.
swin2013
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no?? :O
zepdrix
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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.
swin2013
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oh wait... i do the t(3) = t(0) + the integral
zepdrix
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So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O
swin2013
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yes!
zepdrix
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Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]
swin2013
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OHHHHH
swin2013
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well i did something similar... but i put t(0) -_-
zepdrix
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heh :D
swin2013
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so i plug 3 for t?
zepdrix
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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.
swin2013
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i gotsss 3.848
swin2013
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idk if that's right.... scary
zepdrix
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Hmm I got 6.15
swin2013
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:((((((
zepdrix
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\[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]
swin2013
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oh i subt 5 -_-