Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

sharathe

  • 4 years ago

Can someone help me with this initial value problem? d^2y over dt^2 = 2/t^3. Given is: dr/dt = t=1 and that whole thing = 1. Another Given is: r(1) = 1. I have no idea how to go about this problem after finding the antiderivative for the d^2r over dx^2.

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

    so you never meant to say anything about y right?

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

    Sorry meant r not y.

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

    \[r''=2t^{-3}\]

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

    So you said you integrated both sides: \[r'=2 \cdot \frac{t^{-3+1}}{-3+1}+C_1=2 \cdot \frac{t^{-2}}{-2}+C_1=-1 t^{-2}+C_1\] => \[r'=\frac{-1}{t^2}+C_1\]

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

    So we are given r'(1)=1?

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

    Just r(1)=1

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

    dr/dt = t=1? r'=t=1 I look at that is when t=1 ,r '=1

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

    \[1=\frac{-1}{1^2}+C_1 \] Solve for C_1

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

    where does the r' come from?

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

    Oh! so another way to look at d^2r/dt^2 is r''?

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

    yep yep

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

    Okay! Also just to clarify, when substituting t and r, just plug in one in newly acquired derivative where r and t are respectively? If r(1)=1 and t(1)=2, then plug in respectively and then ... what would happen?

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

    well if you are given r'(a)=b then you use the function for r' If t=a then r'=b ------------------------ and if you are given r(c)=d then use the function r If t=c, then r=d You will use both of this conditions to find your constants

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

    So, since r(1)=1, then t(1)=2, then r of t = 1? Somehow I confused myself...

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

    t(1)=2 makes no sense

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

    I'm sorry to be so blunt

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

    No, it makes no sense to me either.

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

    You are given r'(1)=1 and r(1)=1

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

    |dw:1332451923490:dw|

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

    r'(1)=1 says when t=1 r'=1

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

    \[1=\frac{-1}{1^2}+C_1 \] The reason why I wrote this

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

    I replaced t with 1 and I replaced r' with 1

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

    That makes sense...what happens to the 2 of t(1)=2 tho? how would I solve for C then? Or is it just replace r' with 1 and t with 1?

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

    Sorry for all these questions...

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

    I don;t understand where you are seeing t(1)=2?

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

    I added it after getting the orignal problem in the "what happens after" part...cause the next problem was basically this problem, but instead of t(1)=1, it was t(2)=1...Sorry if this was all confusing! I don't quite get how to solve for C when the r' and t equal different numbers but have the same number plugged in...

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

    @sharathe you're confused between the input t = 1, output r' ( 1) = 1 From these input and output => constant C1 = 2

  28. 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