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Reparametrizing a space curve in terms of arclength:

  1. May 15, 2004 #1
    Hello folks! I'd be happy for any help you could give me. Thanks! :)

    First though I'd like to make clear that I want to solve this problem myself, even if something seems obvious to you and you feel there is no harm in making a certain logical leap - please take the time to consider whether or not it's a step which I should be working out myself. I have very little background in Calc. and it took me a while just to place this problem when digging through my old texts from uni, but I would like to understand what the concepts I'm dealing with are.

    Here is the original question: (Everything else that follows are notes that I have tracked down on the net to help me understand it.)

    This was all that was presented. Is this a matrix then? The spacing does not seem to indicate something as simple as f(t)=(5-3t); g(t)=(5-2t); h(t)=(-4-2t); but I don't want to make any assumptions.

    I'm really quite at a loss as to where to start. In simple terms it seems that I really haven't gotten anywhere yet and must still:

    1. Figure out what ( 5 -3 t )i + ( 5 -2 t )j + ( -4 -2 t )k represent in a way that I can understand what I'm working with.
    2. Find the arclength.
    3. Reparametrize.

    I'm assuming the relationship between the point given and the curve is one where by the question is supposed to be that much similar. Something akin to t=0 or something?

    Other notes I picked up and which I'm still digesting:
    A vector function of a single variable is much like a parametrized curve,
    (where i, j, k, are the vectors (1,0,0), (0,1,0), and (0,0,1), respectively)

    The functions fi are the component functions and much of calculus of vector functions of one variable is done by components, for example,

    A curve is simply defined as a vector function of a single variable:

    http://planetmath.org/encyclopedia/SpaceCurve.html [Broken]
    http://rsp.math.brandeis.edu/Generi.../Documentation/DocumentationPages/Curves.html (Documentation for 3D-XplorMath)

    Thanks for any help you can give! I'm quite in the dark. :confused:
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. May 15, 2004 #2


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    I'll take a whack at it.

    Call arc length s. s is a function of the parameter t. Let s(t) = 0 at the point (5,5,-4), i.e. s(0)=0.


    ds = sqrt[(dx)^2 + (dy)^2 + (dz)^2]
    = sqrt[(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2] dt
    = sqrt[(-3)^2 + (-2)^2 + (-2)^2] dt

    So s(t) = sqrt[17] t.

    Solving for t as a function of s, t(s) = s/sqrt[17].


    r(s)= (5-3s/sqrt[17])i + (5-2s/sqrt[17])j + (-4-2s/sqrt[17])k.

    Am I even on the right track? I haven't tried to do this sort of thing before.
  4. May 15, 2004 #3


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    You have the right answer, Janitor.
    1. Hint:
    Try to find out why your curve is a straight line.
  5. May 15, 2004 #4
    Would I be crazy to try to attempt to do row-reduced form like this?

    Code (Text):

     5 -3 t
     5 -2 t
    -4 -2 t
     0 -5  0     R1-R2
     0  1  0     R2-R1
     1 -5  2t    R3+R1
     1  0  2t    R3-R1,  R3 <-> R1
     0  1  0
     0  0  0     R1+5R3, R1 <-> R3
  6. May 15, 2004 #5
    Your matrix doesn't have any meaning.

    Why are you trying to use matrices to solve this problem?

  7. May 15, 2004 #6
    ? When are matrices used anyways? I dun see a real system of equations here...
  8. May 15, 2004 #7
    That's what I'm trying to understand. I'm not a math student. I work as a receptionist/secretary in a physio clinic. Approaching this problem I'm trying not to let what little math I did learn all blend together. If you tell me I can drop what little I remember from linear, I will - but by talking about parameters and such it seemed to tie right in with it.

    So is ( 5 -3 t ) actually (5-3t) ? Or is it a vector ( 5 -3 t)?

    If so then where does i = <1,0,0>, j= <0,1,0>, k= <0,0,1> come in?

    Does it become r(t) = (5) + (-2) + (t) ?
  9. May 15, 2004 #8
    r is given in the problem statement.

    r(t) = (5-3t)i + (5-2t)j + (-4-2t)k

    Do you understand what a vector function is?

    Last edited: May 15, 2004
  10. May 16, 2004 #9


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    I see now where you got lost in the first place:
    The "spacing" between a number and "t" has absolutely no significance whatsoever!!
    You are absolutely correct in your assumption that:
    f(t)=5-3t, g(t)=(5-2t), h(t)=(-4-2t).

    Good luck!
  11. May 16, 2004 #10
    Thank you all for your time!

    Ok, thanks a lot then! :smile:

    Sorry everyone for all that confusion. Please correct me anywhere I go wrong, and if there is a proper mathematical way of saying something please let me know. ;)

    So following Janitor like he did it:

    r(t) = (5-3t)i + (5-2t)j + (-4-2t)k

    So given x, y, z in terms of parameter t; t= 0 as starting point; and t>0 as positive acrlength s; the formula is s(t)= int_0^t sqrt[x'^2+y'^2+z'^2]dt.

    Is it because t can grow to be infinitely large that the numbers by themselves have no meaning? For example with (5-3t) in [5/10000] - [(-3*10000)/10000], [5/10000] has almost no meaning so we drop it. So far so good? Or is it because we are starting from the point (5, 5, -4)?

    s(t)= int_0^t sqrt[(-3)^2 + (-2)^2 + (-2)^2] dt
    s(t)= int_0^t sqrt[(-3)^2 + (-2)^2 + (-2)^2] dt
    s(t)= int_0^t sqrt[(9) + (4) + (4)] dt
    s(t)= int_0^t sqrt[17] dt

    So now with arclength s as a function of t, we solve for t as a function of s and substitute that formula into the parametric equation:

    t(s) = s/sqrt[17]
    r(t(s))= (5-3s/sqrt[17])i + (5-2s/sqrt[17])j + (-4-2s/sqrt[17])k

    Is that all?

    I think I still need to understand that point above about whether starting from the point (5, 5, -4) affects things. Is there an upper limit? Are there any specific areas that you would suggest I read up on? :)
  12. May 16, 2004 #11


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    No, absolutely not!!

    First, let us regard the usual notation for an arbitrary point in 3-D:

    An arbitrary point P in 3-D can be written as a 3-tuple: (X,Y,Z).
    (X,Y,Z) is a code that allows us to locate the point P:
    We go a distance X along the x-axis from the origin (0,0,0); we have then reached the intermediate point (X,0,0)

    2) We then travel from (X,0,0) a distance Y along an axis parallell to the y-axis, such that we reach the intermediate point (X,Y,0)

    3) Finally, we reach P by going a distance Z along an axis parallell to the z-axis, such that we reach (X,Y,Z) (i.e., P itself!)

    Now, in explicit vector notation, we have:
    [tex](1,0,0)=\vec{i}, (0,1,0)=\vec{j}, (0,0,1)=\vec{k}[/tex]

    Hence, the arbitrary point P can be represented as the vector [tex]\vec{P}[/tex]:
    (The only reason why I used uppercase letters formerly, was that I there chose to use the lowercase letters in the axis descriptions)

    Now, let us return to a vector representation of a curve C:

    That is, the x-coordinate of a point P on C is given as a function of the single variable "t", and we call that function x(t), for obvious reasons.
    A similar argument holds for the y and z-coordinates of P.

    Now, we will find an expression for the arclength of a curve segment:
    Suppose we have 2 points on C, [tex]\vec{P}_{1},\vec{P}_{2}[/tex], where:

    Note in particular that I have not yet used the fact that I can represent the curve C as a function of "t" at all. This fact will enter a bit later!

    Now, clearly, if we draw the straight line segment/secant L between [tex]\vec{P}_{1},\vec{P}_{2}[/tex], the length of L, [tex]\bigtriangleup{s}[/tex], is given by the Pythagorean theorem:

    Let us now utilize that the curve C may be represented as a function of "t":
    Let [tex]\vec{P}_{1}[/tex] be given at the evaluation [tex]t=t_{1}[/tex], whereas [tex]\vec{P}_{2}[/tex], be given at the evaluation [tex]t=t_{2}[/tex].
    That is:
    [tex]\vec{r}(t_{1})=x(t_{1})\vec{i}+y(t_{1})\vec{j}+z(t_{1})\vec{k}, x(t_{1})=x_{1}, y(t_{1})=y_{1}, z(t_{1})=z_{1}[/tex]
    [tex]\vec{r}(t_{2})=x(t_{2})\vec{i}+y(t_{2})\vec{j}+z(t_{2})\vec{k}, x(t_{2})=x_{2}, y(t_{2})=y_{2}, z(t_{2})=z_{2}[/tex]

    We now introduce:
    and we assume that [tex]\bigtriangleup{t}[/tex] is small.
    Then we have:

    Possibly, you are unfamiliar with the notation [tex]\frac{dx}{dt}\mid_{(t=t_{1})}[/tex].

    This simply means that you are to evaluate the derivative of x(t) (that is the function [tex]\frac{dx}{dt}[/tex]) at the t-value [tex]t_{1}[/tex].

    For example, if [tex]x(t)=t^{2}[/tex], then [tex]\frac{dx}{dt}=2t[/tex] and

    Clearly, similar expressions hold for [tex]\bigtriangleup{y},\bigtriangleup{z}[/tex]

    Therefore we have when we suppress the function evaluation notation [tex]\mid_{(t=t_{1})}[/tex]:
    [tex]\bigtriangleup{s}\approx\sqrt{ (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}}\bigtriangleup{t}[/tex]

    We now proceed to the limit [tex]\bigtriangleup{s}\rightarrow{0}[/tex], and sum together all line segments to find the total arclength S:
    [tex]S\equiv\int_{C}{ds}=\int_{0}^{T}\sqrt{ (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}}dt[/tex],

    where the curve C has been represented by t-values lying between 0 and T.

    Hence, the reason why the constants disappears is that you worh with the derivatives of the functions (the derivative of a constant is zero)
    Last edited: May 16, 2004
  13. May 16, 2004 #12
    Wow, I really appreciate the work you put in to explaining that.

    Thanks again for all your time and effort! :)
  14. May 16, 2004 #13
    Calculus, by arildno, to be published in December. arildno to be made a millionaire by January.

  15. May 16, 2004 #14
    man... so formal... :eek: cookie man could make a practical man edition out of such formality. :approve:
  16. May 16, 2004 #15
    Hey, I'd buy it. And after your first million, don't forget to show your love for the forum and contribute, ok? It's not like you need that much anyway.
  17. May 16, 2004 #16
    Should be a Schaum's guide thing... like arildno's guides... And have a through 3000page explanation for Calculus. :)
  18. May 16, 2004 #17
    It'll be better than the so called book we use.
  19. May 17, 2004 #18
    What do you guys use? Clay tablets?
  20. May 17, 2004 #19


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    Just because I had another boring day, doesn't mean I'm getting rich..
  21. May 17, 2004 #20


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    Well, since I didn't define the integral as the limit of Riemann sums, and didn't present [tex]\epsilon,\delta[/tex]-proofs of my statements, I thought my argument was rather sloppy...
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