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Convert parametric equation into f(x,y,z)

  1. Nov 15, 2011 #1
    I'm attempting to convert the following parametric equation into into one f(x, y, z), and am running into difficulty. (Is it even possible?) Can I get some help?

    Last edited: Nov 15, 2011
  2. jcsd
  3. Nov 15, 2011 #2


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    Hey meiskam and welcome to the forums.

    For this problem, what sticks out to me is to use x^2 + y^2 = (2+cos(3t))^2 and then use the fact that z = SQRT(1 - cos^2(3t)) => SQRT(1 - z^2) = cos(3t) which you can plug in to the RHS which gives you:

    x^2 + y^2 = (2 + SQRT(1 - z^2))^2 which you can expand out using algebra.

    One cautious note though is that you might need to use a particular branch, but apart from that I think this should be ok.
  4. Nov 16, 2011 #3

    I don't understand what you are trying to do.

    Right now, you seem to have a function f(t) that returns a tuple (x,y,z)

    are you trying to put together a function f() to which you are going to pass (x,y,z)? and possibly return t? or what? 'cause I don't think any random choice of x,y,z is going to work.

    or I may be misunderstanding the entire thing.
  5. Nov 16, 2011 #4


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    He is trying to describe an equation that relates x, y, and z in one equation.

    As an example consider x = sin(t), y = cos(t). The equation for this is x^2 + y^2 = 1 from the common trigonometric identity.
  6. Sep 6, 2012 #5
    Does anybody know if there is a general condition or a theorem which determines when it is possible to combine a set of parametric equations into a single one? To begin with, local solvability should be a prerequisite, right? Then if n the number of the parameters and m the number of the parametric equations must at least be n=m-1.
    What else?
  7. Sep 6, 2012 #6


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    Try finding where an inverse mapping exists through the inverse function theorem and along the same lines, consider tensor theory to find the minimum number of parameters for the given the system.

    Once you have an orthogonal basis, then that becomes your parameterization.

    For example in a surface you will have two vectors u and v (let's say its in three dimensional space).

    So you will have u = i*f(x,y) + j*g(x,y) + k*h(x,y) and similarly for v where u . v = 0 and i,j,k are the standard basis vectors for R^3.

    Once you have these two-vectors, that becomes your parameterization, and your surface in this case will be u(x,y) + v(x,y) where u and v are vectors as a function of x and y (and u.v = 0 for all possible u and v given an x and a y).
  8. Sep 6, 2012 #7
    I think what you describe is how to parametrize a geometrical object.

    Does every parametrization leads to an implicit function? For instance a line in three dimensions cannot be described implicitly.

    Moreover I'm not sure what do you mean by that:
  9. Sep 6, 2012 #8


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    Actually I misread your OP: you want to go the other way instead of what I am saying.

    If you want to find an f(x,y,z) just do a substitution.

    Knowing where the inverse function branch cuts are tells you how to divide up the domain.

    Since you are guaranteed to have an inverse function in a particular interval, it means you can invert things like sin(2x) where sin^-1(sin(2x)) = 2x and as a result you can pair up term, collect them together and then get a relationship between all of them.

    The point is to do this kind of thing and get an equation involving all them to equal 0 and that's your equation.

    You don't have to make branch cuts depending on your function, but when you get something really complex where connecting the different parameters is non-trivial, then using branch cuts will guarantee inverses in that region and that makes it a lot easier when you want to consider how to connect things together that don't have an obvious link other than when you take an inverse.
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