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tmn50

  • 3 years ago

let's say we have a function F(y(t),t)=y*G(t) the partial derevative of F with respect to t is y*dG/dt or dy/dt*G+dG/dt*y I would really apreciate an explanation

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  1. Mashy
    • 3 years ago
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    option 2

  2. Mashy
    • 3 years ago
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    since both are function of t!

  3. abb0t
    • 3 years ago
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    \[F(y(t), t) = y \times G(t)\] ?

  4. tmn50
    • 3 years ago
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    yes

  5. abb0t
    • 3 years ago
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    \[\frac{ ∂G }{ ∂t }\]

  6. tmn50
    • 3 years ago
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    yes

  7. Mashy
    • 3 years ago
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    if both G and Y are function of t.. then you have to use the product rule!

  8. tmn50
    • 3 years ago
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    mashy our teacher used the first one in determining the solution of a differential equation of first order that's what makes me crazy

  9. Mashy
    • 3 years ago
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    no he mentioned its a function of time see the question!!

  10. tmn50
    • 3 years ago
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    y is a function of time

  11. Mashy
    • 3 years ago
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    Then Y must be independent of t... !!

  12. abb0t
    • 3 years ago
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    stop yelling @ me!! Ur hurting my feelings, bro :'(

  13. Mashy
    • 3 years ago
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    awww.. come here you !!

  14. tmn50
    • 3 years ago
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    -_-

  15. Mashy
    • 3 years ago
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    wow i get a medol and abbot gets none.. yay yay :P

  16. tmn50
    • 3 years ago
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    who the hell gave u a medal i think your answer is wrong man here we're talking about partial derevative which is diffrent from derevating everything with respect to t

  17. wio
    • 3 years ago
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    Is it the partial derivative of \[ F(u, t) \]Or of \[ F(y(t),t) \]?

  18. tmn50
    • 3 years ago
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    the second one bro partial derevative with respect to t

  19. wio
    • 3 years ago
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    Well it was ambiguous, and \(F(u, t) \neq F(y(t),t)\)

  20. tmn50
    • 3 years ago
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    and so ?

  21. Mashy
    • 3 years ago
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    if both functions are time dependent.. then really there is no difference between partial and normal derivatives.. HELL why would you even DO a partial derivative doesn't even make any sense :-/

  22. tmn50
    • 3 years ago
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    -_- it makes sense when u'r solving a diferential equation of first order first you find F(y,t) then you derevative partially with respect to y or t and force it to be equal to the equation you have left so yeh u need a partial derevative

  23. Mashy
    • 3 years ago
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    give me a differential equation!!

  24. wio
    • 3 years ago
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    Like @Mashy is saying, \(F(y(t),t)\) doesn't have a partial derivative because it is not a multiple variable function. It's the OUTPUT of \(F(u, t)\) which happens to be a single variable function.

  25. abb0t
    • 3 years ago
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    well, that escalated quickly.

  26. Mashy
    • 3 years ago
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    lol :D

  27. tmn50
    • 3 years ago
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    M(y,t)dt+N(y,t)dy=0 where M=df/dt and N=df/dy

  28. abb0t
    • 3 years ago
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    Whoa, now we're doing ODE's?

  29. Mashy
    • 3 years ago
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    thats called as EXACT FORM right?

  30. abb0t
    • 3 years ago
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    Correct, Mashy :)

  31. tmn50
    • 3 years ago
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    wio i didnt catch up with what you said i mean here we got 2 variable well not two variables y is a function of time and hell

  32. Mashy
    • 3 years ago
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    well then in that.. case.. i have forgetting how to do it :D

  33. abb0t
    • 3 years ago
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    This is escalating too damn quickly.

  34. tmn50
    • 3 years ago
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    it's like i have holes in mathematics so when i advance i fall into some of them

  35. Mashy
    • 3 years ago
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    holes in mathematics.. lol funny :D

  36. wio
    • 3 years ago
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    Can you give us \(M, N\)?

  37. Mashy
    • 3 years ago
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    he already mentioned.. df/dy and df/dt!

  38. abb0t
    • 3 years ago
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    Well, you want to find the PD for either M or N first. I think it's ∂M/∂x and ∂N/∂y ?

  39. tmn50
    • 3 years ago
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    it doesn't matter what M and N are the only restriction is that M and N are the partial derevatives of a certain function F(y,t)

  40. Mashy
    • 3 years ago
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    wow i did this 4 years back!! ..

  41. Mashy
    • 3 years ago
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    so i don't remember much :-/

  42. tmn50
    • 3 years ago
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    sadly

  43. abb0t
    • 3 years ago
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    Well, you have to check that they are exact first for this to work.

  44. abb0t
    • 3 years ago
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    Otherwise, you are wasting time trying to do furhter work than is necessary.

  45. wio
    • 3 years ago
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    \[ \begin{array}{rcl} M(y,t)dt+N(y,t)dy &=& 0 \\ M(y,t)dt &=& -N(y,t)dy \end{array} \] I would try integrating \(M\) with respect to \(t\) and then differentiating with respect to \(y\).

  46. tmn50
    • 3 years ago
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    wait can i post a picture ?

  47. tmn50
    • 3 years ago
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    yup i can wait

  48. tmn50
    • 3 years ago
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  49. Mashy
    • 3 years ago
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    really?!? of yourself?? we are in the middle of math here!

  50. tmn50
    • 3 years ago
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    here check tht

  51. tmn50
    • 3 years ago
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    i'm too handsome :p

  52. Mashy
    • 3 years ago
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    so i have heard :P

  53. Mashy
    • 3 years ago
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    i mean .. i did.. just now :P.. from you!!

  54. tmn50
    • 3 years ago
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    so any idea about what i posted?

  55. abb0t
    • 3 years ago
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    What it looks like you're trying to do in that step is find a integrating factor to make the solution exact?

  56. tmn50
    • 3 years ago
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    yup

  57. tmn50
    • 3 years ago
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    wait wait wait we found the integrating factor we're in the middle of finding F

  58. tmn50
    • 3 years ago
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    see where's the arrow pointing

  59. abb0t
    • 3 years ago
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    I'm not quite sure what you did there, I'm used to doing it a different method..

  60. tmn50
    • 3 years ago
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    tht's the teacher's doing not me :3

  61. tmn50
    • 3 years ago
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    oh well i'm off now i'll check again after half an hour or so thanks ppl

  62. tmn50
    • 3 years ago
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    i'm back people any news,

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