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Parameterised a function

  1. Nov 6, 2006 #1
    Is there a specific method to parameterised a function of scalar and vector?

    I was reading through this website but then, cant understand how it read taht parameterised steps.

    Any guide?

    under double check example)

  2. jcsd
  3. Nov 6, 2006 #2


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    I don't think there is a method that works in all cases. But when the curve is specified as a level curve f(x,y)=0, you can try to solve for y. In that case, just use x as the parameter.

    Exemple: Consider the level curve C={(x,y): y-x²-2x=0. We can solve of y like so: y=x²+2x. In that case, the parametrized curve is r(t)=(t²+2t,t), t [itex]\in \mathbb{R}[/itex].

    In other cases, if the relation between x and y is something familiar you can try to exploit this as illustrated in the following exemple:

    Exemple: Consider the level curve C={(x,y): (x/a)²-(y/b)²=1}. This resembles the identity cosh²(t)-sinh²(t)=1. So set x(t)/a=cosh(t) and y(t)/b=sinh(t), i.e. let r(t)=(acosh(t),bsinh(t)) t [itex]\in \mathbb{R}[/itex] parametrize the level curve.

    N.B. You can convince yourself that this parametrization covers the whole curve because given any value of x, there is a corresponding value of t for which acosh(t)=x and similarly for y. [cosh(t) is surjective on the x-axis and sinh(t) is surjective on the y axis]. See http://en.wikipedia.org/wiki/Image:Sinh_cosh_tanh.svg
  4. Nov 6, 2006 #3


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    No wait, cosh(t) is not surjective on the real line, it only covers (1, infty). So r(t) would only cover part of C.
  5. Nov 7, 2006 #4


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    As quasar987 said, there is no one method of parameterizing a curve. In fact, there exist an infinite number of different parameterizations for any curve.

    For the example given, because the path (From the origin, (0, 0, 0) along the z-axis to (0, 0 1), then along the quarter circle to (0, 1, 0), then along the y-axis to (0, 0, 0) again) is not "smooth" (there are corners at (0,0,0), (0, 1, 0) and (0, 0, 1)), you would break it into three pieces.

    First the line from (0, 0, 0) to (0, 0, 1). At any point on that line, x=y= 0
    An obvious parameterization is to use z itself as parameter: x= 0, y= 0, z= t, with [itex]0\le t\le 1[/itex].
    Second, the quarter circle from (0, 0, 1) to (0, 1, 0). At every point on that circle x= 0 and y2+ z2= 1. A "standard" parameterization for a circle is to use sine and cosine: x= 0, y= sin(t), z= cos(t). Then x= 0 always while y2+ z2= sin2(t)+ cos2(t)= 1 for all t. Of course, it is z= cos(t) and not y because when t= 0, z= cos(0)= 1 and y= sin(0)= 0 as required. When t= [itex]\pi/2[/itex], z= cos([itex]\pi/2[/itex])= 0 and y= sin([itex]\pi/2[/itex])= 1 so [itex]0\le t\le \pi/2[/itex].

    Finally, the line from (0, 1, 0) to (0, 0, 0). Obviously x= z= 0 at every point. We could use y itself as parameter: x= 0, y= t, z= 0 with t going from 1 to 0. Another possibility is x= 0, y= 1- t, z= 0 with t going from 0 to 1.
  6. Nov 7, 2006 #5
    Oh rite, i see.
    Thanks for helps!

    Edit: oh, btw, is there any functions that cant be parameterized?
    Last edited: Nov 7, 2006
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