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Reparametrize the cure in terms of arc length instead of time 
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#1
Jul1405, 11:43 PM

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Yikes, im really starting to spam this place up!
On the subject of curvature, it says that we can reparametrize the cure in terms of arc length instead of time. If we have time,t, and arc length s(t), we can write it as t=t(s). It seems to me that this is NOT true in general, if you cannot factor t from s(t), then there is no way relate t in terms of t(s). So under what conditons is this true, and could you show a proof that shows that these conditions will insure reparametrization? Secondly, the curvature is defined as the rate of change of the unit tangent vector with respect to arc length. I cannot see why they would choose this definition of curvature. What reasoning is behind using this as the definition? Thanks, Cyrus ADDITION:Since not all functions can be parametrized in terms of arc length, that means that not all fuctions have cuvature, but all functions DO have arc length. Is that correct? 


#2
Jul1505, 07:04 AM

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To answer your 2nd question, if you think about it, the more it changes of a short distance the more 'curvy' it is. Which is why it is called curvature, but the reason we have it is because its a useful property.



#3
Jul1905, 09:11 AM

P: 4,780

I think the definition of curvature makes more sense to me now. If we look as to how fast the tangent vector changes with respect to distance in terms of arc length, we can get a feel for how curved the function is. If dT/ds is zero, then its not a curve, but a line. If dT/ds is a constant, then we should expect a circle. If dT/ds is anything else, we should expect to see a nonuniform curve of varing radius in any arbitrary direction.
Could anyone help me out with the first part to my question? 


#4
Jul1905, 12:55 PM

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Reparametrize the cure in terms of arc length instead of time
http://mathworld.wolfram.com/RadiusofCurvature.html IIRC, I intuitively used it in a mechanics problem. A particle was constrained to move on a parabola (y=something*x^2) subject to gravity pointing down and was thus oscillating. I wanted to know the centripetal force (or normal force, can't remember) when the particle was at the vertex. I know for circular motion it's mv^2/R. So I used this and put for R the radius of curvature of the parabola at the vertex, which was correct. :) One reason I can think of why they used the change of unit tangent wrt arc length definition is that by doing so the radius of curvature will be certainly independent of the parametrization, which is what you want. For the first part, I'm pretty sure the parametric equations need to be injective (save some exceptions). From this you can show the arc length of a curve is independent from its parametrization. So there's a oneone correspondence between a point in the parameterdomain and a point on the curve. That's why you can invert s(t). 


#5
Jul1905, 11:17 PM

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#6
Jul2005, 02:32 PM

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Are you kidding? A function (say [itex]f:X \to Y[/itex])is invertible iff it is bijective. That is: it is injective (or onetoone) and surjective (onto). So to every [itex]x\in X[/itex] there corresponds one and precisely one [itex]y\in Y[/itex] (namely f(x)).



#7
Jul2005, 03:08 PM

P: 4,780

Actually, I was serious. Ok, thats fine. Lets say I am given some big messy formula. Is there an easy way I could test to see if it meets the conditions to be invertible? Graphing it would be one way. So would manipulating the function, or would that depend greatly on that particular function and what tricks one could play to be able to factor it, if at all possible.
I was thinking along the lines of something like, if the function is continuous on this region and x,y,z conditions are met, then the function is invertible. Some sorta proof to show invertability that runs along those lines. (if such a thing even exists?) 


#8
Jul2105, 02:19 AM

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Sorry, I forgot this isn't usually taught in physics classes.
Usually only injectivity is important, since surjectivity is obtained by limiting the codomain to the range of the function f:X> f(X) So you have to prove that the following holds: [tex]f(x)=f(y) \to x=y[/tex] for all x,y in X. In the case of one variable (f:R>R for example) you can sometimes check injectivity by solving y=f(x) for x in terms of y. Example: [tex]y=\frac{x2}{x+6} \Rightarrow x=\frac{26y}{y1}[/tex] so f(x)=(x2)/(x+6) is injective. It seems for the case of curves [itex]\vec r(t)[/itex] is not always injective, since it can cross itself without problems. In this case there is more than one value of t corresponding to a particular point on the curve. It is easy to see that s(t) (arc length as function of time) is injective. I guess this is intuitively clear. Also, from the definition of arc length you can see the integrand is always positive: [tex]s(t)=\int_{t_0}^{t}\sqrt{(x'(t')^2+(y'(t'))^2}dt'[/tex] and for smooth curves we have [itex]r'(t)\not= 0[/itex] so s(t) is always increasing. (I always wondered why [itex]r'(t)\not=0[/itex] was important. I guess this gives one reason.) So you can invert s(t) into t(s). 


#9
Jul2105, 08:22 AM

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PF Gold
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A differentiable function has an inverse as long as its derivative is not 0. If you are thinking of the parameter as "time" so that a particle is at distance s(t) from the starting point at time t, then you can inverts s(t) to t(s) as long as the particle does not come to a stop.
By the way, some texts use a different but equivalent definition of curvature. If φ(s) is the angle the tangent to the curve makes with the xaxis at s (so that tan(φ)= df/dx) then the curvature at s is dφ/ds. That is, it tells how fast the angle is changing: how fast the curve is curving. 


#10
Jul2105, 11:31 AM

P: 701

[tex]f(x) = x^2, x \epsilon \mathbb{R}[/tex] this function does not have an inverse since the domain of [itex]f^{1}(x) = x^{1/2}[/itex] will not cover all of the reals...and we can see that it's derivative at x = 0 is in fact zero. but take [tex]f(x) = x^3, x \epsilon \mathbb{R}[/tex] whose inverse does exist and yet the derivative at x = 0 is also zero. 


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