anonymous 4 years ago show that y=tant satisfies the IVP y'=1/(1+t^2), y(0)=0. what is the largest open interval containing t=0 over which y=tant is a solution?

1. anonymous

i am confused because integral of 1/(1+t^2) is arctant...

2. amistre64

if y=tan(t) IS a solution then itll fit in just nicely

3. anonymous

ok, cuz it actually if a solution...but shouldn't the question have assigned different variables??

4. anonymous

*is a solution

5. amistre64

no, t is t is t in all the derivatives of y

6. anonymous

well how do i find the limit of t now? both y and y' have to considered right? so i can find the largest interval?

7. amistre64

well, tan(t) is that has a zero in it, and is contiuous would be: (-pi/2 , pi/2) if i see this right

8. anonymous

right

9. amistre64

you may be reading to much into the problem :)

10. TuringTest

but y=tan^(-1)t...

11. TuringTest

though the interval -pi/2,pi/2 still makes sense

12. anonymous

i think so too cuzi still don't understand... so limit of t in y' would be neg inf to pos inf but y is restricted so in the end, largest interval is -pi/2 to pi/2?

13. TuringTest

arctan(t) is undefined at $\pi/2+n\pi$

14. amistre64

is y=tan(t) a solution?

15. TuringTest

no, y'=sec^2t

16. anonymous

yes

17. TuringTest

?

18. anonymous

?

19. anonymous

ok, no

20. amistre64

or is it; y=arctan(t) a solution? are you reading it correctly is the quandry

21. TuringTest

http://www.wolframalpha.com/input/?i=%20y'%3D1%2F(1%2Bt%5E2)&t=crmtb01 in case you didn't know the derivative of arctan...

22. anonymous

i do know the derivative that's why i was getting confused. cuz the y in the prolem doesn't equal integral of y'. ok so no, not a solution...so limit IS neg inf to pos inf

23. amistre64

as is, we are stating: $tan'(t)=\frac{1}{1+t^2}$ $sec^2(t)=\frac{1}{1+t^2}$

24. TuringTest

no I said earlier it should be arctan tan is not a solution

25. amistre64

if not a solution; then there is no interval to worry about since no interval is a solution to a nonsolution :)

26. anonymous

right

27. TuringTest

right!

28. anonymous

but arctan(0)=0

29. amistre64

$\frac{dy}{dt}=\frac{1}{1+t^2}$ ${dy}=\frac{1}{1+t^2}dt$ $\int\ ({dy}=\frac{1}{1+t^2}dt)$ $y=tan^{-1}(t)+C_1$

30. anonymous

mmhmm

31. anonymous

C=0 so y=arctant

32. amistre64

when y=0 and t=0; we get: $0=tan^{-1}(0)+C_1$ $0=C_1$ therefore, given the IVP $y=tan^{-1}(t)$

33. anonymous

ok

34. amistre64

this shows that y=tan(t) is NOT a solution to the IVP and is therefore no need to worry about any nonexisting intervals

35. anonymous

ok thank you so much