A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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?

  • This Question is Closed
  1. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *is a solution

  5. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  9. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you may be reading to much into the problem :)

  10. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    is y=tan(t) a solution?

  15. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no, y'=sec^2t

  16. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  17. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ?

  18. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ?

  19. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, no

  20. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  21. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  24. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  26. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  27. TuringTest
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right!

  28. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but arctan(0)=0

  29. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  30. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    mmhmm

  31. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    C=0 so y=arctant

  32. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  34. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thank you so much

  36. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.