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Arc length parametrization

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Find an arc length parametrization of the curve r(t) = <e^t(cos t), -e^t(sin t)>, 0 =< t =< pi/2, which has the same orientation and has r(0) as a reference point.

    2. Relevant equations

    s = int[0,t] (||r'(t)||)

    3. The attempt at a solution

    So I found the derivative of r(t), and then normalized it. The problem is that I am left with a value that is negative under a radical, or a function that I simply do not have any idea how to integrate.

    P.S: Sorry, I have no idea how to use that type of script which makes viewing functions more convenient...

    My r'(t) = <(e^t(cos t - sin t)), (-e^t (cos t + sin t))>

    I then normalized the derivative, where I got one of two functions:

    e^t * (sqrt(-4sin t cost t) or sqrt(-e^t) * 2sqrt(sin t cos t)

    But how would one integrate that? using by parts? My teacher claims that there would be nothing extremely difficult to integrate, but this problem seems to be proving otherwise...

    Ps: Sorry, I do not know how to use that script which makes viewing functions more convenient..
     
  2. jcsd
  3. Jul 10, 2010 #2

    LCKurtz

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    Check your work for ||r'(t)||. It should be pretty simple; what did you get?
     
  4. Jul 11, 2010 #3
    Yea Thanks. While typing it out, I realized that I forgot to remove the (-) from the second e^t. I guess then it should reduce down to simply e^t * sqrt(2), which is a fairly easy integral.

    I then solved the s = int[0,t], which I ended up with t = ln(s/sqrt(2) + 1)

    So another question arises, as I probably had messed something up, but where does the limiting of t come into play? Should I instead of integrating from 0 -> t, integrate from 0 -> pi/2?
     
    Last edited: Jul 11, 2010
  5. Jul 11, 2010 #4

    LCKurtz

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    So now substitute your t in terms of s into your original equation. s will range from 0 to the length of the curve. Gotta hit the sack now. Good luck.
     
  6. Jul 11, 2010 #5
    Ok thanks. I am still very unsure why they mention that t goes from 0 -> pi/2

    I watched a youtube video on this subject, and he solves for the arc length, L, but it does not show up anywhere else in the problem. My question is what is the point of solving for L, the definite arc length from 0 -> pi/2 if it is not applied anywhere else?
     
  7. Jul 11, 2010 #6

    LCKurtz

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    A parameterization will give the formula for the curve and the domain, the values of t for which the curve is given. The author could have given you any values for t, he just chose to give you 0 -> pi/2. That, of course, determines the extent of the given curve and the length of the curve. When you change the parameterization to another domain variable such as arc length s, you give a new formula in terms of the new parameter s. To give the same curve, you need to give the specific domain of the new parameter. When you use arc length as the parameter, the ending value is always the length of the curve. 0 ≤ s ≤ L(C). And of course, you can compute L(C) from the original parameterization.
     
  8. Jul 11, 2010 #7
    I guess I am still unclear why he would give me that kind of information when, in terms of the question, it is literally useless. I understand the necessity to know how to find the parametrization in terms of s.

    When you say L(C), that means t, not the length of the curve, say if the length from o -> pi/2 was pi, the integral would still go from t(not) -> t, not t(not) to pi?
     
  9. Jul 11, 2010 #8

    LCKurtz

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    Here's an example that might help you. Consider the parameterization of the quarter circle of radius 1 in the first quadrant:

    [tex]\vec R(t) = \langle \sqrt{1 - t^2},t\rangle,\, 0\le t \le 1[/tex]

    [tex]\vec R'(t) = \langle -\frac{t}{ \sqrt{1 - t^2}},1\rangle[/tex]

    [tex]\frac {ds}{dt}=\sqrt{\frac{t^2}{1-t^2}+1}=\sqrt{\frac 1 {1-t^2}}[/tex]

    So to calculate the arc length from 0 to t:

    [tex] s =\int_0^t\sqrt{\frac 1 {1-t^2}}\,dt = \arcsin(t)|_0^t = \arcsin(t)[/tex]

    Notice if you integrate for t from 0 to 1, you get

    [tex]s = \arcsin{(1)} = \pi/2[/tex], which is the length of this arc.

    Your t can be expressed in terms of s:

    [tex] t = \sin{(s)} \hbox{ and } \sqrt{1 - t^2} = \cos{(s)}[/tex]

    and you can parameterize your curve in terms of s:

    [tex]\vec R(s) = \langle \cos(s), \sin(s)\rangle,\ 0 \le s\le \frac \pi 2[/tex]

    You now have two parameterizations of the same curve, in the same direction, one of which is the arc length parameterization. Hopefully the arc length parameterization is already familiar to you because of s = rθ.
     
  10. Jul 12, 2010 #9
    It makes a little more sense. Thanks for the great help.

    Now I understand the use of the ds/dt, as before I was quite confused on what that meant.

    So when parametritizing the curve, you must also "redo" the value of the interval that t is in? And to do that, one would simply plug in the previous t values into the new function of s?
     
  11. Jul 12, 2010 #10

    LCKurtz

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    ds/dt is the speed at which you move along the curve. Each parameterization has its own formula and domain. As the parameter varies you move along the curve. Two people may walk the same curve at different rates, arrive at different times, yet walk the same distance. Same idea.
     
  12. Jul 12, 2010 #11
    So when I solve for ds/dt, I still use the interval of 0 -> t, regardless of the initial interval that t is restricted to. But, when showing the answer, it is necessary to show that the interval has changed, with respect to the new parametrization.
     
  13. Jul 12, 2010 #12

    LCKurtz

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    I'm not sure what is bothering you. The answer to your original question would be expressed in the form:

    [tex]\vec R(s) = \langle f(s),g(s)\rangle,\, 0 \le s \le \ell (C)[/tex]

    where the f and g formulas you have figured out in terms of s and the length [itex]\ell (C)[/itex] of the curve you can calculate from the usual formula.
     
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