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Homework Help: Parameterization of Arc Length Function

  1. Aug 20, 2003 #1
    I'm a little confused by the following in my textbook:

    Arc Length Function of a curve, 's', is defined by:

    s(t) = [inte]|r'(u)|du =
    [inte][squ]((dx/du)^2 + (dy/du)^2 + (dz/du)^2)du

    integrate both sides and you get ds/dt = |r'(t)|.

    Arc length is independent of the parameterization that's used, is that why there appears to be no rhyme or reason to the interchangeability of 'u' and 't' in the equations above?

    Also along the same lines, curvature of a curve is defined as:
    k = |dT/ds| where T is the unit tangent vector.
    The curvature is easier to compute if it is expressed in terms of the parameter 't' instead of 's', so using the Chain Rule the book gets:

    dT/ds = (dT/ds)(ds/dt)??????

    I guess I'm a little confused by the parameterization or the chain rule because I'm not seeing how this came about.

    Can someone please explain?

    Thanks much.
  2. jcsd
  3. Aug 20, 2003 #2


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    The bounds on your integral should be from 0 to t, which explain from where it comes.

    And dT/ds = dT/dt * dt/ds
    or alternatively,
    dT/dt = dT/ds * ds/dt
  4. Aug 20, 2003 #3
    Actually the bounds in the book were given as 'a' to 't'.

    dT/dt is the derivative of the function T(t) with respect to 't' or is it the function T(s(t))? this is where I'm confused.
  5. Aug 20, 2003 #4
    When I took vector calc, the professor simply said that since we integrate with respect to time from t=a (usually where a=0) to t=b, we want to use an arbitrary character to denote time in the integral...for notation purposes...

    in your book, they use the character u...in my book, they use w...in class, we used the greek letter τ...

    Basically, if we took the original position vector r(t), and substituted t for the s we obtained from the arclength formula, we have position in terms of distance traveled, r(s).

    We obtain the unit tangent vector, r'(s), by taking the derivative of the position vector (in terms of arclength) with respect to the arc length, dr/ds. This is the main benefit of arclength parametrization...

    Then we can calculate the curvature, κ(s), which is the length of the rate of change of the unit tangent vector with respect to s (basically the deviation of the curve from the tangent), by taking the magnitude of the derivative of the unit tangent vector, r'(s) with respect to s.
    Last edited: Aug 20, 2003
  6. Aug 21, 2003 #5


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    Are you asking whether dT/dt is the derivative or the function?
    That should be obvious: dT/dt MEANS derivative.

    If you are asking whether dT/dt means derivative of T(t) with respect to t or derivative of T(s(t)) with respect to t, the answer is "it doesn't matter". If T is a function of s (T(s)) and s is a function if t (s(t)), then "T(t)" is just shorthand for "T(s(t))"- they mean exactly the same thing.
  7. Aug 21, 2003 #6
    Thanks, but can you go step by step with either representation of the function -- T(t) or T(s(t)) -- and show me exactly how dT/dt = dT/ds*ds/dt.

    This is probably very fundamental, but while I understand how to apply the chain rule to a specific equation (eg. cos2t), I just can't follow the how dT/dt becomes dT/ds*ds/dt.

  8. Aug 21, 2003 #7
    I think I don't really understand the question...

    are we looking for curvature? what are we given? r(t)? r(s)?

    if we're given r(s), the problem's done in my mind...take the second derivative with respect to s, then take it's magnitude...easy right? that's why it's advantageous to parametrize the curve with arclength...

    if we're given r(t), there are a few more steps...but just a few...

    we need to know the arclength, s...

    let's say r(t)=(cos(2t), sin(2t), t)

    dr/dt = (-2sin(2t), 2cos(2t), 1)

    |dr/dt| = √(4sin^2(2t) + 4cos^2(2t) +1)
    = √(4(sin^2(2t) + cos^2(2t)) +1)
    = √5

    S = ∫|dr/dt|dτ = (√5)τ (from 0 to t) = t√5

    therefore, r(s) = (cos(2S/√5), sin(2S/√5), S/√5)

    take dr/ds for the unit tangent vector, T (you can test this by noting that its magnitude = 1)

    curvature, κ = |dT/ds| = |d^r/ds^2|

    so you wanted to know how dT/dt = dT/ds * ds/dt?

    remember that T(t) is just T(s(t)), so you still need to know arclength in terms of time...

    if you know arclength in terms of time, s(t), and you know the unit tangent vector in terms of length traveled, T(s), it's plug and chug...

    honestly...although I see where we would want to know curvature in terms of time for certain applications, I don't see how it could be seen as "easier" to calculate...

    can anybody fill me in on a way to calculate curvature without arclength, or without being initially given the position with respect to arclength? I don't have any notes on that whatsoever...I didn't consider it possible
  9. Aug 25, 2003 #8


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    That IS the chain rule. You appear to be asking for a proof of the chain rule. You can find that in any good calculus book.
  10. Aug 25, 2003 #9
    I found it.

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