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sharathe

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.

  • 2 years ago
  • 2 years ago

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  1. freckles
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    so you never meant to say anything about y right?

    • 2 years ago
  2. sharathe
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    Sorry meant r not y.

    • 2 years ago
  3. freckles
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    \[r''=2t^{-3}\]

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

    • 2 years ago
  5. freckles
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    So we are given r'(1)=1?

    • 2 years ago
  6. sharathe
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    Just r(1)=1

    • 2 years ago
  7. freckles
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    dr/dt = t=1? r'=t=1 I look at that is when t=1 ,r '=1

    • 2 years ago
  8. freckles
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    \[1=\frac{-1}{1^2}+C_1 \] Solve for C_1

    • 2 years ago
  9. sharathe
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    where does the r' come from?

    • 2 years ago
  10. sharathe
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    Oh! so another way to look at d^2r/dt^2 is r''?

    • 2 years ago
  11. freckles
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    yep yep

    • 2 years ago
  12. sharathe
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    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?

    • 2 years ago
  13. freckles
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    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

    • 2 years ago
  14. sharathe
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    So, since r(1)=1, then t(1)=2, then r of t = 1? Somehow I confused myself...

    • 2 years ago
  15. freckles
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    t(1)=2 makes no sense

    • 2 years ago
  16. freckles
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    I'm sorry to be so blunt

    • 2 years ago
  17. sharathe
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    No, it makes no sense to me either.

    • 2 years ago
  18. freckles
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    You are given r'(1)=1 and r(1)=1

    • 2 years ago
  19. freckles
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    |dw:1332451923490:dw|

    • 2 years ago
  20. freckles
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    r'(1)=1 says when t=1 r'=1

    • 2 years ago
  21. freckles
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    \[1=\frac{-1}{1^2}+C_1 \] The reason why I wrote this

    • 2 years ago
  22. freckles
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    I replaced t with 1 and I replaced r' with 1

    • 2 years ago
  23. sharathe
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    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?

    • 2 years ago
  24. sharathe
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    Sorry for all these questions...

    • 2 years ago
  25. freckles
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    I don;t understand where you are seeing t(1)=2?

    • 2 years ago
  26. sharathe
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    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...

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

    • 2 years ago
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