swin2013 Group Title 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. one year ago one year ago

1. swin2013

@zepdrix

2. swin2013

i know i must take the antiderivative

3. zepdrix

$\large s(t)=\int\limits v(t) dt$Having trouble with the integral? It looks like another nice easy U substitution.

4. swin2013

yea i did u sub for 1 + t^2

5. swin2013

i meant +

6. swin2013

i got 1/u du

7. zepdrix

$\large \int\limits \frac{1}{u}du$Yep, sounds good.

8. zepdrix

Woops, should be a factor of 1/2 in front I think.

9. swin2013

and i should get u = 1 and u = 10?

10. swin2013

oh yea 1/2 (1/u) du

11. zepdrix

No we don't want to add limits to our integral.

12. swin2013

no?? :O

13. 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.

14. swin2013

oh wait... i do the t(3) = t(0) + the integral

15. zepdrix

So the integral gave you something like...$\large s(t)=\frac{1}{2}\ln(1+t^2)+C$Yes? :O

16. swin2013

yes!

17. zepdrix

Then just plug in the Initial Condition to solve for C!$\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C$

18. swin2013

OHHHHH

19. swin2013

well i did something similar... but i put t(0) -_-

20. zepdrix

heh :D

21. swin2013

so i plug 3 for t?

22. 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.

23. swin2013

i gotsss 3.848

24. swin2013

idk if that's right.... scary

25. zepdrix

Hmm I got 6.15

26. swin2013

:((((((

27. zepdrix

$\large C=5$$\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5$

28. swin2013

oh i subt 5 -_-