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neilski_2003

2nd derivatives?

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Hi there guys, I have worked out the equation for the path of a particle to be y = 3sin(pi*x/L) so to get the first derivative i have used the chain rule and come up with the following (which may need checking) given that y = 3sin(pi*x/L) k = pi*x/L therefore: y = 3sin (k) and that gives me dy/dx = Cos(k)*pi/L So now i want to work out d2y/dx2 and i'm getting some funky stuff. I suppose the chain rule still applies but now i hold two and differentiate one so i'd get something like d2y/dx2 = 3 (-sin(pi*x/L)*(pi/L))(Cos(pi/L)*(pi/L))(Cos(pi*x/L)*(-L)) I'm not sure if thats right - or if it can be simplified - if it is correct can someone give me a shout - if not can someone point to my error - ta very much Neil

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You got the first one nearly right. You forgot the constant at the beginning though. So correctly the first derivative would be 3*cos(x*PI/L)*(PI/L)

I don't get how you got that for the second derivative? Anyway, if I'm right the second derivative would be -3*sin(x*PI/L)*(PI/L)^2

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2nd derivative of a*sin(bx) is always -a*b^2*sin(bx). Similar for cosine. It's related to the property that the points of inflection in a sinusoidal wave are located where the function is minimized, and the curvature is maximized where the function is maximized.

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My grasp of the 2nd derivatives and the chain rule is skethcy at best so i'll go with you on that one - seeing as looking at it makes sense (what i was doing was using 3 things in my chain rule and getting myself muddled!)

Cheers

Neil

[Edited by - neilski_2003 on April 25, 2006 6:12:26 AM]

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Maybe you should refresh your maths skills a bit :) (no offense meant)
Maybe I should do that too, because I don't know much about the radius of curvature, besides what it is... :)

So, to get to the poin: you cannot cancel the (PI/L)^2 there, because
you have 1 + (something)*(PI/L)^2 there, so you would have to divide the '1' too, not to mention the whole expression is on the power of 3/2, which introduces further problems. I hope I did make some sense there. If not, just ask away, and I'll try to clarify...

Frankly, I don't really see a way to simplify this, but I might be missing something.

By the way, this isn't a maths homework we're solving here, is it? Maybe you could tell us what you need this fancy stuff for (maybe there's a more simple way of doing it?).

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Hi,

I realised that when checking my work this morning and have since remedied that.

Thanks for your help thus far - if you have any more questions for me fell free to ask.

I have sent you a private message.

Cheers again

Neil

[Edited by - neilski_2003 on April 25, 2006 6:28:40 AM]

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I don't have a clue as of how to extract L from that equation, but then again, I'm not that good at maths, so maybe someone here can enlighten us both :)

As for your thinking, it's correct, just one thing: if (a=v^2/p) then p=(v^2/a), not (a/v^2) but I'm sure that was only oversighted by you.

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...
For circular motion:

F= ma = m(v^2)/r
a = v^2/r


Therefore

.7*9.8m/s^2 = (22m/s)^2/r
r = (484 m^2/s^2)/(6.86 m/s^2)
r = 70.559 m

r being the radius of the circle that you are travelling in.

Now what the hell is L? You never told us what L is supposed to be ;-)

Unless L is the thing in your first equation... Then I don't want to solve that :-P

[edit]
Although...
pi/L is the angular velocity...

w(angular velocity) = v/r

v = 22m/s
r = 70.55m

w = pi/L
so
pi/L = 70.55m/22m/s
pi/L = 3.2/s
L = 10.06

So im guesing here you have a different equation of motion since Ididn't come up with the same answer. That or I made a typo somewhere.

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Hi there,

Its quite difficult to explain because i have a diagram so i won't bother you with it i'll go with what i think and leave it at that.

Cheers

Neil

[Edited by - neilski_2003 on April 25, 2006 6:13:38 AM]

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