2nd derivatives?
Author
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
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
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
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.
Author
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]
Cheers
Neil
[Edited by - neilski_2003 on April 25, 2006 6:12:26 AM]
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?).
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?).
Author
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]
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]
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.
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.
...
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.
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.
Author
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]
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]
This topic is closed to new replies.
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