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Problems with parametrization

  1. Feb 20, 2004 #1
    Hi all! I'm having some problems with parametrization. I read somewhere that you should locate circles, ellipses, hyperboloides, paraboloides etc and use these elements to express a parametric function.

    But someone must have figured out how to do it! The way I see it, there's nothing logical about
    x^2+y^2-z^2=1 \Leftrightarrow f(t,\varphi)=\left(\sqrt{t^2+1}\cos\varphi,\sqrt{t^2+1}\sin\varphi,t\right)

    Yes, I understand that
    (\sqrt{t^2+1}\cos\varphi)^2 + (\sqrt{t^2+1}\sin\varphi)^2 - t^2 = 1
    but how do you actually get to that conclusion (without plotting the function)?

    Consider this problem (which may be simple):
    Determine the intersecting curve (parametric function) between the surfaces [tex]z^2=x^2+y^2[/tex] and [tex]z=x+y[/tex]. How do you approach that?

    Thanks in advance!
  2. jcsd
  3. Feb 22, 2004 #2


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    I have no idea what is meant by "you should locate circles, ellipses, hyperboloides, paraboloides etc and use these elements to express a parametric function." Certainly standard figures such as circles,ellipses, etc. (hyperboloids and paraboloids are surfaces, not curves) and have standard parameterizations but if your curve, surface, or function doesn't happen to be one, it doesn't help.

    It does help to think about some "standard" parameterizations and perhaps try to adapt them.

    First, you do understand that there exist an infinite number of different parameterizations for any curve, don't you?

    If I were trying to find a parametrization for [tex]x^2+y^2-z^2=1 [/tex], then first thing I would notice is that [tex]x^2+y^2[/tex] and think [tex]sin^2(\theta)+cos^2(\theta)= 1[/tex] (or some people might think "circle!"- it's the same thing.) I see immediately that if I let x= cos(θ) and y= sin(θ), I will have x2+ y2= 1. Now I have to take care of that "-z2" term. It would be nice if I could just set it equal to 0 but I can't!. I might think: If instead I set x= R cos(θ), y= R sin(θ) (Yes, I confess: I wrote "R" instead of "A", for example, because I really thinking "circle"!). Now x2+y2= R2 and I have R2- z2= 1. Okay, how about if I take R2= 1+ z2? That is: set R= √(1+ z2)? It's not illegal to use z itself as a parameter but, since "t" is more common, let z= t and the parametrization becomes

    x= √(1+t2)cos(θ)
    y= √(1+t2)sin(θ)
    z= t.

    Finding the intersection of z2= x2+ y2 and z= x+ y is, in fact, relatively straightforward (in theory at least). Saying that (x,y,z) is on the intersection means that the same values for x,y,z satisfy both equations or that
    z= x2+y2= x+ y.

    The intersection of two surfaces (2 dimensional) is a curve (1 dimensional) so I would expect this to depend on a single variable.

    One thing I could do is think of x2+y2= x+ y as a quadratic equation for x:x2- x+(y2- y)= 0 and use the quadratic formula to find x as a function of y. Use y itself as the parameter (or write y= t) so the solution is a parametric equation for x. We could use z= x+y to then get an equation for z. The only problem with that is that "+/-" in the quadratic formula.

    Actually, you should be able to look at x2+ y2= x+ y and immediately think "circle!" (Your first sentence is making more and more sense, at least for these problems!). A standard way of rewriting an equation for a circle is to complete the square (remember way back when you first learned to do that?). Write this as
    x2-x + y2- y= 0 and think "In order to make both x and y terms perfect squares, I have to add 1/4 to each term". That would of course, add 1/2 to the right:
    x2-x+ 1/4 + y2- y+ 1/4= 1/2 or
    (x- 1/2)2+ (y- 1/2)2= 1/2.

    That's a circle (in the x,y plane) with center at (1/2, 1/2) and radius √(1/2). More importantly I see that if I set
    x= cos(θ)+ 1/2 and y= sin(θ)+ 1/2, then
    (x- 1/2)2+ (y- 1/2)2= cos2(θ)+sin2(θ)= 1.

    To account for the "1/2" on the right side, just multiply each by 1/√(2):

    x= (1/√(2))(cos(θ)+ 1/2)
    y= (1/√(2))(sin(θ)+ 1/2) and
    z= x+ y= (1/√(2)(cos(θ)+ sin(θ)+ 1).
    Last edited by a moderator: Feb 22, 2004
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