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Cartesian and Parametric

  1. May 31, 2007 #1
    Okay, I was doing 3D modelling. To save space I used vector functions to render terrain. Anyway, I came up with 3 parametric equations - each a function of an axis: e.g.: x=4t, y=5t+6, z=7t-9. How can you convert this into a Cartesian Equation?:confused:
     
  2. jcsd
  3. May 31, 2007 #2
    What do x, y and z stand for?
     
  4. May 31, 2007 #3
    Not sure what you mean by a "Cartesian Equation".
    In your case, your are looking for a nonhomogeneous linear system of 2 equations in 3 unknowns.

    EDIT.

    Basic theorem from linear algebra:
    Every linear manifold is a solution set of a nonhomogeneous linear system of equations.

    In your example, the function-locus parametrically defined by f(t) = (4t,5t+6,7t-9) can be considered a linear manifold.
    It is the range of f.
    It is the line of intersection of the two planes determined by the system.
     
    Last edited: May 31, 2007
  5. May 31, 2007 #4
    z = 5t + 6 + 2t - 15
     
  6. May 31, 2007 #5
    z(t)=7t-9
    x(t)=4t
    y(t)=5t+6
    i.e. it is like a vector
    I was wondering if there is a way to transfer this into a Cartesian equation. I mean if the z did not exist, I would just solve the 2 equations for t and then equate them: x/4=(y-6)5. With that extra function, how would one go about creating a single cartesian equation?
     
  7. May 31, 2007 #6

    Read the edit in post 3.
     
  8. May 31, 2007 #7
    There is no cartesian equation for 3d vectors. Think about it: if a single relationship included three variables, x, y, z, then any of these variable could not be determined by the value of simply 1 other variable. In fact, such a relationship represents a surface, not a curve or line thereof.
     
    Last edited: May 31, 2007
  9. Jun 1, 2007 #8
    Exactly what I wished to confirm...
     
  10. Feb 21, 2011 #9
    Actually, there is a Cartesian form.

    Given a position vector A <a1,a2,a3>
    And a direction vector B <b1,b2,b3>

    The Cartesian form of a line in 3 dimensional space is:

    (x - a1) / b1 = (y - a2) / b2 = (z - a3) / b3

    This can be determined from two points (C,D) by generating the direction vector like so:
    A = C - D

    And then picking one of the points as the position vector.
    See:
    http://www.netcomuk.co.uk/~jenolive/vect17.html

    And I believe you can generate a Cartesian form of a line in N dimensions..
     
  11. May 3, 2011 #10
    Some people refer to the Cartesian Equation Form as the
    Symmetric Form

    JFYI
     
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