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Astrophysics

  • one year ago

Leibniz notation product rule... @ganeshie8 just be expected to be tagged

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  1. Astrophysics
    • one year ago
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    \[\frac{ d^2y }{ dx^2 }\] I'm trying to get this to equal \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\] so I have \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] which I got from the chain rule but how do I use the product rule here X_X

  2. Astrophysics
    • one year ago
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    |dw:1444614858695:dw|

  3. Astrophysics
    • one year ago
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    @freckles

  4. Kainui
    • one year ago
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    Wait so are we saying y(x) and x(t) is that true?

  5. Astrophysics
    • one year ago
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    Well I'm trying to figure out eulers equation \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\] and it says I need to calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2

  6. Astrophysics
    • one year ago
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    so I can use my substitution

  7. Kainui
    • one year ago
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    Oh ok this makes more sense now.

  8. Kainui
    • one year ago
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    wait are you not just allowed to substitute in \[y=At^n\] and solve?

  9. Astrophysics
    • one year ago
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    No I have to use x = lnt

  10. Astrophysics
    • one year ago
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    and it's telling me to do all that stuff lol

  11. Kainui
    • one year ago
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    Oh alright then

  12. Astrophysics
    • one year ago
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    Just want to know how to use the product rule in leibniz notation dont really care about the problem

  13. Astrophysics
    • one year ago
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    f'g+g'f lol but I can't believe I don't know how to do this

  14. Kainui
    • one year ago
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    \[y(x(t))\] \[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\] \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right)\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \] Try to take it from here.

  15. Kainui
    • one year ago
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    While you do that, I'll do it without logs to see how much faster it is.

  16. Astrophysics
    • one year ago
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    It's the first part, \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \right)\frac{ dx }{ dt }\] there's no changeeee, I was expecting there to be d^2y/dx^2 or something

  17. Kainui
    • one year ago
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    Ultimately they'll get the same answer and this method you're using is more general so it's not without its merits it's just this is a pretty common one to solve.

  18. Kainui
    • one year ago
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    Here's only the derivative part of the first part (notice we'll have a squared dx/dt when we multiply this back in): \[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \frac{dx}{dt} \]

  19. Astrophysics
    • one year ago
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    Ok should it not be \[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

  20. Kainui
    • one year ago
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    If this bothers you instead let: \[\frac{dy}{dx} = u(x(t))\] Then when you do the derivative you have: \[\frac{d}{dt}(u) = \frac{du}{dx} \frac{dx}{dt}\] Then you can look at the du/dx term on its own as the second derivative of y wrt x.

  21. Kainui
    • one year ago
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    Explain your reasoning for this: \[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

  22. Astrophysics
    • one year ago
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    Nvm I'm dumb, it's \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \right) = \frac{ dy }{ dt }\frac{ dt }{ dx }\frac{ d }{ dt } = \frac{ d^2y }{ dt^2 }\frac{ dt }{ dx }\] is this what you mean

  23. Astrophysics
    • one year ago
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    and not dx^2

  24. Kainui
    • one year ago
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    Nope, you're taking the derivatives in a strange order, and this hanging d/dt that's by itself is kind of meaningless here.

  25. Astrophysics
    • one year ago
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    I just used the chain rule, the d/dt is what is confusing me

  26. Kainui
    • one year ago
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    \[y(x(t))\] \[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\] \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \] Looking only at the red part: \[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \frac{dx}{dt} \] make the u substitution I described earlier to get through this part.

  27. Kainui
    • one year ago
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    If there's anything about using the product and chain rule above that confuses you say something now before moving forward.

  28. Astrophysics
    • one year ago
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    Nooo, it's not really the substitution part, I mean \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right)\] it was this in general, so you're just writing \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \frac{ dx }{ dt }\right)\] + this thing again, oh I'm overthinking this, I just realized how simple it actually is

  29. Astrophysics
    • one year ago
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    \[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

  30. Astrophysics
    • one year ago
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    omggggggg kai

  31. Kainui
    • one year ago
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    Where did this come from?

  32. Kainui
    • one year ago
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    Explain why you're trying to use this: \[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

  33. Astrophysics
    • one year ago
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    That's my original q

  34. Kainui
    • one year ago
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    Isn't this what you're trying to solve? \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\]

  35. Astrophysics
    • one year ago
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    \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right) = \frac{ d }{ dx }(\frac{ dy }{ dt })t+\frac{ d }{ dx }\left( t \right)\frac{ dy }{ dt }\]

  36. Astrophysics
    • one year ago
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    I was trying to find d^y/dt^2 in terms of d^2y/dx^2 ahaha

  37. Astrophysics
    • one year ago
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    I think I got it, I sort of just thought about this \[\frac{ d }{ dx }(u*v)=...\] I was just overthinking

  38. Kainui
    • one year ago
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    No stop saying that lol, you can't claim that until you've finished solving the problem xD Right now I'm trying to get us on the path of writing \(\frac{d^2y}{dt^2}\) in terms of x with the chain rule, but I'm not sure where you're going wrong so I'm having difficulty helping you out.

  39. Astrophysics
    • one year ago
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    Oh I thought we were doing it in terms of d^2y/dt^2 the whole time

  40. Kainui
    • one year ago
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    I don't know what you're doing so I don't know

  41. Astrophysics
    • one year ago
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    lmao

  42. Kainui
    • one year ago
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    \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] Explain where you're getting this from, cause I don't know. That doesn't mean it's wrong it just means you need to say words, I can't read your mind.

  43. Astrophysics
    • one year ago
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    All I was doing is finding \[\frac{ d^2y }{ dt^2 }\] in terms of \[\frac{ d^2y }{ dx^2 }\] so I would get \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\]

  44. Kainui
    • one year ago
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    Hmmm explain how you're supposed to solve this: " it says I need to calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2" So I am not seeing this here, so for example, dy/dt in terms of dy/dx would look like this, right: \[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]

  45. Astrophysics
    • one year ago
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    \[\frac{ d^2y }{ dx^2 }\] and started using the chain rule, \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dx } \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt }\frac{ dt }{ dx } \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] and then I was trying to apply chain rule to \[\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\]

  46. Astrophysics
    • one year ago
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    I started with \[\frac{ d^2y }{ dx^2 }\]

  47. Astrophysics
    • one year ago
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    product rule not chain rule* omg

  48. Kainui
    • one year ago
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    So you're writing \[\frac{d^2y}{dx^2}\] in terms of \[\frac{d^2y}{dt^2}\] then? And you're plugging in \[\frac{dt}{dx}=t\] is that right?

  49. Astrophysics
    • one year ago
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    Yes!

  50. Kainui
    • one year ago
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    This is fine then I think, it's just weird cause instead of looking it as y(x(t)) you're looking at it as y(t(x)) which is kind of backwards but in this case works since \(x= \ln t\) which is surjective.

  51. Astrophysics
    • one year ago
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    Haha, I thought that's how you were suppose to do it, I read it on pauls online notes or something, I really have no notes for this, so I was sort of coming up with stuff myself

  52. Astrophysics
    • one year ago
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    I mean the point of getting this is so I can set up my equation as \[\frac{ d^2y }{ dx^2 }+(\alpha - 1)\frac{ dy }{ dx }+\beta y=0\]

  53. Kainui
    • one year ago
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    Right of course, but doing it like this is awkward since time isn't generally thought of as a function of position. And it has no meaning in periodic motion.

  54. Astrophysics
    • one year ago
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    true

  55. Astrophysics
    • one year ago
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    I wasn't really thinking of it that way, haha just some math problem

  56. Astrophysics
    • one year ago
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    And the question stated, let x = lnt and calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2 ahah

  57. Kainui
    • one year ago
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    Here, I'll just walk through how I did it: \[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\] \[\frac{d^2y}{dt^2} = \frac{d^2y}{dx^2} \left( \frac{dx}{dt} \right)^2 + \frac{dy}{dt} \frac{d^2x}{dt^2}\] I guess we can do it now or later, so why not now: \[\frac{dx}{dt} = \frac{1}{t}\] \[\frac{d^2x}{dt^2} = \frac{-1}{t^2}\] And then plugging all this in you get the same simplified differential equation since the \(t^2\) will disappear.

  58. Kainui
    • one year ago
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    I messed up one piece but, oh well should have been a dy/dx on that last term I just noticed

  59. Kainui
    • one year ago
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    At least this way is more straight forward to plug in (well in my opinion). However I wouldn't use this method at all lol. Best way is to notice you have a power next to the derivative which is basically compensating for a lost power by the power rule every derivative. Still pretty easy to spot. \[t^2 y''+\alpha t y' + \beta y = 0\] Now you can just pick \[y=At^n\] and plug it in: \[n(n-1)+\alpha n + \beta = 0\] Now you have a quadratic in n to solve for the two solutions, done!

  60. Astrophysics
    • one year ago
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    Yo that's so simple

  61. Astrophysics
    • one year ago
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    Yeah I already hate this method I have to use

  62. Kainui
    • one year ago
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    Haha use the hate to fuel you through the tedious calculations and before you know it you'll be really fast. I feel like most of this discussion didn't have to happen I just had no idea what you were doing so we wasted a lot of time.

  63. Astrophysics
    • one year ago
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    Yeah haha, well at least I know what I'm doing now X_X

  64. Astrophysics
    • one year ago
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    I just realized how simple this was, I actually just misread something earlier, so was sort of confused haha xD

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