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Parametrization homework help

  1. Jan 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Our prof talked about arc length as a parameter today and I understand how to do problems associated with it, however I do not fully understand why we do it.

    2. Relevant equations

    In our text, the only relevant reading says: "A curve in the plane or in space can be parametrized in terms of the arc length s."

    Would someone be able to give me a geometric interpretation on why we need to do this parametrization?

  2. jcsd
  3. Jan 22, 2007 #2


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    It is often convenient. Biking and hiking give intuitive real-life examples. IF you hike trails in the mountains, or watch the Tour de France on TV, you'll see a graph of elevation as a function of distance along the route. It is useful to know, e.g., that a steep climb is coming up at mile 2.6 followed by a gentle descent for the next half mile, etc. It is far less convenient to have to look at a topo map that gives elevation at every x,y point (or latitude/longitude), because you have to painstakingly figure out what it is along the trail you'll be taking. Parametrization along the arc or curve or trail is the more appropriate and convenient way to work the problem.
  4. Jan 23, 2007 #3


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    You don't need to. It's often convenient, as marcusl said.

    One formula for arclength is
    [tex]s= \int \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}dt[/tex]
    where x and y are functions of some parameter t.

    On the other hand if we just have y a function of x, we can use the formula
    [tex]s= \int \sqrt{\left(\frac{dy}{dx}\right)^2+ 1}dx[/tex]

    Of course, there exist many interesting curves in which y is not a function of x! Example: the circle.
  5. Jan 23, 2007 #4

    [tex]y^2 + x^2=r^2[/tex] : function of a circle.

    This is actually for wiki not a bad little guide to circles, interesting as it shows all the relationships to 0 the center of the circle and their angles, I just though I'd mention it as it is a good picture of how the angles relate and to which sides. The arc length is of course the distance along the circumference of the circle.

    Probably not relevant, but interesting.

    The curve [tex]\int_0^{ln 2}{\sqrt{1+e^{2x}}dx} => \frac {1}{2} \ln [\sqrt {(1+e^2x)}] - \frac {1}{2} \ln [\sqrt {(1+e^2x)}]+\sqrt {1+e^2x}+C[/tex]
    describes the volume of a toroidal too as I recently found out.

    The definite integrals value is found by placing e as 2. As ln2(e)=2 in this case and then subtracting the lower limit.
    Last edited: Jan 23, 2007
  6. Jan 23, 2007 #5
    great thanks a lot!
  7. Jan 23, 2007 #6


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    For some reason, in my first response I thought you were asking about calculating arc-length when a curve is given by parametric equations. Now I see that you are asking about using arclength as a parameter.

    One point about parameterization in terms of arc length is that it is obvious[/t]! Choose any point at all to correspond to s= 0, choose one direction on the curve to be + and another to be -. Then just literally measure along the curve. The point at distance s from the starting point (in the positive direction) is labeled by parameter s, in the negative direction by parameter -s.

    Another nice thing is that while the vector you get by differentiating the position vector with respect to any parameter is a tangent vector to the curve, if you use arclength, it is the unit tangent vector.
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