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Curvature is defined as

  1. Feb 3, 2005 #1
    A quick question. In stewart, the curvature is defined as:

    [tex] \kappa = \abs(\frac{dT}{ds}) [/tex]

    and it says that:

    "The curvature is easier to compute if it is expressed in terms of the parameter t instead of s, so we can use the Chain Rule to write:

    [tex] \frac{dT}{dt}=\frac{dT}{ds} \frac{ds}{dt} [/tex]
    and
    [tex] \kappa= \abs(\frac{dT}{ds}) = \abs(\frac{dT/dt}{ds/dt}) [/tex].

    I dont see how they used the chain rule on that problem, normally, it is T(t).
    So T is a function of time, (t) directly. Are they saying that they replaced the variable t, in terms of the varaible s. S, which is arclength, is a function of t. So I suppose the rewrote t, as a function of s instead. The same as saying, [tex] y=x^2 [/tex] or you could also say, [tex] x=+/-\sqrt(y) [/tex].
    And they pluged this in where (t) used to be in T(t). So now it is writen as:
    T(s), but s(t), so it is like saying T(s(t)) ?

    Also, did they just solve the first fraction dT/dt =dT/ds ds/dt by dividing both sides by ds/dt in order to get, dT/ds= (dT/dt)/(ds/dt). Is it correct to multiple and divide by ds/dt like that?


    so when you do the chain rule you get, dT/dt = dT/ds * ds/dt ?

    On another note, im taking dynamics now, and it is quite amazing how I looked at three books, physics 1, dynamics, and calc 3, and found three different ways of finding the acceleration vector, the dynamics and physics book do almost the same approach with diagrams of vectors, and the math book quite nicely does the same thing without every using a picture. Neat, I think.
     
    Last edited: Feb 3, 2005
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  3. Feb 4, 2005 #2

    dextercioby

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    Yes,that's the advantage of using differentials.U can ALMOST multiply & divide through them at will.There are some "miserable" exceptions,of course,but in this case,no...

    Daniel.
     
  4. Feb 4, 2005 #3
    Let me ask this again, becuase I see something I may have done wrong.

    Normally the tangent is in terms of t, T(t).

    Now, we also know that arc length is a function of time, t as well:

    s=s(t).

    But if we reparametrize the tangent in terms of s, instead of t, that means we have to rewrite arc length as t in terms of s, and subsitute that in for t. In other words, we have to chagne s=s(t), into t=t(s). And we can plug THIS back in for t in the tangent.

    So the tangent should now become T(t(s)). Right?

    So now its a function of s. So when we do the chain rule, how come it is not written as:

    [tex] \frac{dT}{ds} = \frac{dT}{dt}\frac{dt}{ds} [/tex]

    They both give the same anwser if you move around the ds/dt term, but I think the way the book arrived to the anwser is incorrect. s is not a function of t, i.e s != s(t), because it is the other way around, t is a function of s, t=t(s). Which is why i strongly disagree with their use of the chain rule.

    dT/dt= dT/ds ds/dt, nonono!? :biggrin: How can you say ds/dt like that, s is no longer a function of t, s is the variable now, and t is the dependent variable on s.




    Thanks for your help.
     
    Last edited: Feb 4, 2005
  5. Feb 5, 2005 #4

    dextercioby

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    It doesn't really matter for well-behaved functions (diffeomorphisms) if it's s(t) or viceversa.That's another advantage of the chain rule.

    Daniel.
     
  6. Feb 5, 2005 #5
    Im sorry, I see what your saying, but I dont understand what you mean by that.
    How does it make sense to talk of dT/ds ds/dt, when s is not a function of t? I agree that s=s(t) for arc length, but when you reparametirize the tangent vector in terms of s, you have to change from s=s(t) into t=t(s), if you want to replace the t's with s's. So then ds/dt becomes meaningless, no?
     
  7. Feb 5, 2005 #6

    dextercioby

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    Sure.But it's all about reparametrization.For example,if u have the curve s=s(t),then the curvature will be
    [tex] K(t)=K(s(t)) [/tex]
    If u chose another parameter "u",for example
    [tex]K(u)=K(s(u)) [/tex]

    Daniel.
     
  8. Feb 5, 2005 #7

    arildno

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    To be pedantic:
    Let "s" and "t" be related through:
    [tex]s=f(t), t=f^{-1}(s)[/tex]
    Expressed in terms of "s", the tangent vector may be called [tex]\vec{T}_{s}(s)[/tex]
    wheras expressed in terms of "t", we have [tex]\vec{T}_{t}(t)[/tex]
    The two functions [tex]\vec{T}_{s},\vec{T_{t}}[/tex] must satisfy the relations:
    [tex]\vec{T}_{s}(f(t))=\vec{T}_{t}(t), \vec{T}_{s}(s)=\vec{T}_{t}(f^{-1}(s))[/tex]¨
    Thus, you have:
    [tex]\frac{d\vec{T}_{s}}{ds}=\frac{d\vec{T}_{t}}{dt}\frac{dt}{ds}, t=f^{-1}(s)[/tex]
    and
    [tex]\frac{d\vec{T}_{t}}{dt}=\frac{d\vec{T}_{s}}{ds}\frac{ds}{dt}, s=f(t)[/tex]
     
    Last edited: Feb 5, 2005
  9. Feb 5, 2005 #8
    But that is incorrect daniel, you cant just put s(t) in place for t like that. s(t) might be something like [tex] s(t)=t^3 [/tex], so you cant just put in s(t), everywhere you see at t in T(t), because, in this case, s is related to [tex] t^3 [/tex] not t. Thats my dilema.
     
  10. Feb 5, 2005 #9

    arildno

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    Perhaps you should take a closer look at my response.
    Your dilemma is real, and based on a sloppy and ambiguous notation met in textbooks.
    Hopefully, what I've written, clears up a bit.
     
  11. Feb 5, 2005 #10

    dextercioby

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    That's what i call a functional dependence (explicit or/and implicit) and i don't recall claiming that the K function be SURJECTIVE... :wink:

    Daniel.
     
  12. Feb 5, 2005 #11
    :approve: Thanks, you guys are my heros. I wish I was a math TITAN like you guys,
     
  13. Feb 6, 2005 #12

    Ah, I think I see what you mean now. Do you mean that for [tex] t= f^{-1}(s))[/tex], you are calling that t=t(s), which is where the dt/ds comes from?
    and the same applies for the s=f(t), you mean s=s(t), which is where the ds/dt comes from.

    Does this:
    [tex]\vec{T}_{s}(f(t))=\vec{T}_{t}(t) [tex],
    correspond to this:
    [tex]\frac{d\vec{T}_{t}}{dt}=\frac{d\vec{T}_{s}}{ds}\frac{ds}{dt}, s=f(t)[/tex][/QUOTE]

    and this:
    [tex] \vec{T}_{s}(s)=\vec{T}_{t}(f^{-1}(s))[/tex]
    correspond to this:
    [tex]\frac{d\vec{T}_{s}}{ds}=\frac{d\vec{T}_{t}}{dt}\frac{dt}{ds}, t=f^{-1}(s)[/tex]

    I think you put them in reverse order from the equality you had above it, so im not sure if that is the correspondance or not.

    In words, does this say that suppose we know the tangent vector in terms of arclength, s, from the getgo, then we can convert it into terms of t. Or vise versa, if we know the tangent in terms of time, t, we can change it to arclength.

    The thing is, thoughout the book, everything is in terms of t, so it did not seem natural to start with s, and work my way to t. Up until then, everything had been done in terms of t, and we worked our way to s. Maybe thats partly been the problem.
     
  14. Feb 7, 2005 #13

    arildno

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    You've right, I reversed the order, but you seem to have figured it out.

    It was perhaps somewhat unfortunate of me to use the word "tangent vector" here , since the relations I set up, are of course, entirely general, independent of the particular interpretation of it (that is, it holds for any representation of a vector function in either of two variables which can be set in bijective correspondence to each other).
     
  15. Feb 7, 2005 #14
    Got it. My math book wanted to show that dT/dt = (dT/ds)/(ds/dt), and I thought that I could prove it as well, by another method. I defined an inverse function of s as, t=t(s), so then it would be clear that dS/dt=(dT/ds)(dt/ds), and I THOUGHT, that if I made it as ds/dt=(dT/ds)/(ds/dt), I would arrive at the same anwser, since I just put the fraction in the denominator but flipped it, they two are equivalent: dS/dt=(dT/ds)(dt/ds)=ds/dt=(dT/ds)/(ds/dt), but on review that seems like a big NO NO, because dt/ds is NOT a fraction, so I cannot divide by the inverse like that. The reason that the other way, dT/dt, works is because im dividing both sides by the entire fraction. I belive the only way my alternative method would be valid is if I could prove that
    [tex] ds/dt=(dt/ds)^{-1} [/tex], but then, why bother when the first way is so much simpler, and does not require a proof. I should just not worry about writting the tangent in terms of t and converting to s, becuase there may be situations where it is most natural to write the tangent vector in terms of s, and not t, but either way, I should arrive at the same anwser. (provided that the function s=s(t) can be separated to t=t(s) ).
     
  16. Feb 11, 2005 #15
    I just bring this back up from the dead, so that perhaps you can respond to it and give me your thoughts arildno,

    Thanks, cyrus
     
  17. Feb 12, 2005 #16

    arildno

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    All-right:
    Let:
    [tex]s=f(t), t=f^{-1}(s)[/tex]
    f and its inverse necessarily obeys the relations (since they are inverses of each other)
    [tex]f(f^{-1}(s))=s, f^{-1}(f(t))=t[/tex]
    Differentiate, by using the chain rule:
    [tex]\frac{df}{dt}\frac{df^{-1}}{ds}=1,\frac{df^{-1}}{ds}\frac{df}{dt}=1[/tex]
    But suppress now the f-notation, and we have the pleasing result:
    [tex]\frac{ds}{dt}\frac{dt}{ds}=1[/tex]
    which is what you seek.
     
  18. Feb 12, 2005 #17
    I dont understand what you mean by supress the f-notation, or how you got the fraction below, could you explain that please. And also, when we do it this way, we are assuming there is an inverse function that relates s and t, but is it always going to be true in general that s is related to t in a way that they can be separated like that?
     
    Last edited: Feb 12, 2005
  19. Feb 12, 2005 #18

    arildno

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    Let [tex]\vec{r}(t)[/tex] be a parametrization of a curve . Then, the arc-length "s" is given by the integral (when the curves starts at t=0):
    [tex]s=\int_{0}^{t}||\frac{d\vec{r}}{d\tau}||d\tau[/tex]
    (That is, in our case f(t) is the integral at the right-hand side).
    But, note that f'>0, that is f is a strictly increasing function.
    This is sufficient to ensure that an inverse exists.
     
  20. Jul 21, 2005 #19
    Man, this post goes back. Im re-reading this and its making even more sense than the first time I read it. I seem to follow all your work fine. The only part I got stuck was when you said
    "Differentiate, by using the chain rule:
    [tex]\frac{df}{dt}\frac{df^{-1}}{ds}=1,\frac{df^{-1}}{ds}\frac{df}{dt}=1[/tex]"

    where did the =1 come from. That would imply that you just did:

    d/ds *(s)

    Does this notation make sense? It just seems weard for some reason. Typicaly, it is chage of y, WRT change in x, or dy/dx. Rate of change of s with respect to s doesnt make much sense. S doesnt make s change.

    Actually, I gues it does make sense if you call s f(s)=s then you would have
    df(s)/ds, which would be 1.
     
    Last edited: Jul 21, 2005
  21. Jul 22, 2005 #20

    arildno

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    Okay, I'll try to add a bit of words here to my previous posts:
    1. Whenever "t" is regarded as the independent variable, the corresponding s-values is given by s=f(t)
    2.Whenever "s" is regarded as the independent variable, the corresponding t-values is given by [tex]t=f^{-1}(s)[/tex]
    3. f and f^{-1} are inverses of each other, that is:
    [tex]f(f^{-1}(s))=s,f^{-1}(f(t))=t[/tex]
    4.Now, take the first identity under 3., and differentiate both sides with respect to "s".
    Use the chain rule on the left-hand side, whereas the right-hand side reduces to [tex]\frac{d}{ds}s=1[/tex]
     
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