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moongazer

  • 2 years ago

If y^3 = at^2 , then (d^2 y)/dt^2 ?

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  1. moongazer
    • 2 years ago
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    I only got to the part: dy/dt = 2at/3y^2

  2. Yttrium
    • 2 years ago
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    Just derive again, so you can solve the d^2y/dt^2. :))

  3. Yttrium
    • 2 years ago
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    it's like solve for y''

  4. amriju
    • 2 years ago
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    let it remain as 3y^2=2at..differntiate again....using multiplication rule....

  5. moongazer
    • 2 years ago
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    it says the answer is: -2a/9y^2 but I am getting a different answer

  6. moongazer
    • 2 years ago
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    is "a" here constant?

  7. Yttrium
    • 2 years ago
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    Quotient rule, remember?

  8. amriju
    • 2 years ago
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    yo constnt

  9. Yttrium
    • 2 years ago
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    for easier calculation, you can factor out 2a/3 and just derive (t/y^2)

  10. moongazer
    • 2 years ago
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    could you show me the first step for the second derivative? maybe I did something wrong

  11. moongazer
    • 2 years ago
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    i'll try again :)

  12. amriju
    • 2 years ago
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    may be u need to substitute y as (at^2)^2/3

  13. moongazer
    • 2 years ago
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    I got: (6ay^4 - 8a^2 t^2 y)/9y^6 I think I am doing something wrong.

  14. moongazer
    • 2 years ago
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    i'll try again :)

  15. Yttrium
    • 2 years ago
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    \[y' = \frac{ 2at }{ 3y^2 }\], right? Therefore, \[y' = \frac{ 2a }{ 3 } (\frac{ t }{ y^2 })\] \[y'' = \frac{ 2a }{ 3 } [\frac{ y^2 - 2tyy') }{ y^4 }]\] \[y'' = \frac{ 2a }{ 3 } \frac{ y^2 - 2ty(\frac{ 2a }{ 3 })(\frac{ t }{ y^2 }) }{ y^4 }\] Simplifying this equation we will get. \[y'' = \frac{ 2a }{ 3 } [ \frac{ 3y^4 -4at^2y }{ 3y^6 }]\]

  16. Yttrium
    • 2 years ago
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    can you follow @moongazer ?

  17. moongazer
    • 2 years ago
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    That's what I got. I think there is some typo in my answer sheet. Thanks :)

  18. myininaya
    • 2 years ago
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    Well you could simplify

  19. Yttrium
    • 2 years ago
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    @moongazer, it isn't the answer in simplest form. Maybe you can continue this and let us verify if we get the same answers.

  20. moongazer
    • 2 years ago
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    @myininaya could you simply it further? @Yttrium yes, I understood your solution

  21. myininaya
    • 2 years ago
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    Could I? Yes. Can you?

  22. myininaya
    • 2 years ago
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    Eye ball the numerator and the denominator. You should see they share a common factor.

  23. Yttrium
    • 2 years ago
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    For our class, the definition of simplest form is that it is factored out completely. Just factor out some stuff and do cancellation.

  24. moongazer
    • 2 years ago
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    2a(3y^3 - 4at^2) / 9y^5

  25. moongazer
    • 2 years ago
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    that's what I got

  26. myininaya
    • 2 years ago
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    You can also choose to write at^2 as y^3

  27. myininaya
    • 2 years ago
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    Recall our initial equation: y^3=at^2

  28. myininaya
    • 2 years ago
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    I see at^2

  29. myininaya
    • 2 years ago
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    Replace it y^3 then combine like terms

  30. Yttrium
    • 2 years ago
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    @moongazer , do you what myininaya told you?

  31. myininaya
    • 2 years ago
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    @moongazer I'm leaving for tonight but I think Yttrium is still here to help. Goodnight you guys. I think you guys got it from here for sure. :)

  32. moongazer
    • 2 years ago
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    WOW! Thank you very much to both of you. I failed to notice that until you said it. Thanks :) I now got the answer. :)

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