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Astrophysics
 one year ago
Leibniz notation product rule... @ganeshie8 just be expected to be tagged
Astrophysics
 one year ago
Leibniz notation product rule... @ganeshie8 just be expected to be tagged

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d^2y }{ dx^2 }\] 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444614858695:dw

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Wait so are we saying y(x) and x(t) is that true?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Well 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0so I can use my substitution

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Oh ok this makes more sense now.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2wait are you not just allowed to substitute in \[y=At^n\] and solve?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0No I have to use x = lnt

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0and it's telling me to do all that stuff lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Just want to know how to use the product rule in leibniz notation dont really care about the problem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0f'g+g'f lol but I can't believe I don't know how to do this

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2\[y(x(t))\] \[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\] \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right)\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \] Try to take it from here.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2While you do that, I'll do it without logs to see how much faster it is.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0It'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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ultimately 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.

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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ok should it not be \[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2If 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.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Explain your reasoning for this: \[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Nvm 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Nope, you're taking the derivatives in a strange order, and this hanging d/dt that's by itself is kind of meaningless here.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I just used the chain rule, the d/dt is what is confusing me

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2\[y(x(t))\] \[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\] \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \] 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.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2If there's anything about using the product and chain rule above that confuses you say something now before moving forward.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Nooo, 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Where did this come from?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Explain why you're trying to use this: \[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0That's my original q

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Isn't this what you're trying to solve? \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right) = \frac{ d }{ dx }(\frac{ dy }{ dt })t+\frac{ d }{ dx }\left( t \right)\frac{ dy }{ dt }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I was trying to find d^y/dt^2 in terms of d^2y/dx^2 ahaha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I think I got it, I sort of just thought about this \[\frac{ d }{ dx }(u*v)=...\] I was just overthinking

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2No 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Oh I thought we were doing it in terms of d^2y/dt^2 the whole time

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I don't know what you're doing so I don't know

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0All I was doing is finding \[\frac{ d^2y }{ dt^2 }\] in terms of \[\frac{ d^2y }{ dx^2 }\] so I would get \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Hmmm 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: \[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d^2y }{ dx^2 }\] 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 \[\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I started with \[\frac{ d^2y }{ dx^2 }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0product rule not chain rule* omg

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2So 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?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2This 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I mean the point of getting this is so I can set up my equation as \[\frac{ d^2y }{ dx^2 }+(\alpha  1)\frac{ dy }{ dx }+\beta y=0\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Right 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I wasn't really thinking of it that way, haha just some math problem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0And 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Here, 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.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I messed up one piece but, oh well should have been a dy/dx on that last term I just noticed

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2At 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: \[n(n1)+\alpha n + \beta = 0\] Now you have a quadratic in n to solve for the two solutions, done!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yo that's so simple

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I already hate this method I have to use

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Haha 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah haha, well at least I know what I'm doing now X_X

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I just realized how simple this was, I actually just misread something earlier, so was sort of confused haha xD
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