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How to derive a Parametric Equation?

  1. Apr 22, 2009 #1
    I've been looking everywhere for a tutorial or lesson on Parametric equations and can't find one that shows how to derive the actual equations. Either they show how to eliminate the parameter or sketch the graph. Let's say you had a function like y = x^2 + x - 3

    What would you do to find the Parametric equation? Thanks.
     
  2. jcsd
  3. Apr 22, 2009 #2

    jgens

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    Gold Member

    How about these parametric functions.

    y = t^2 + t - 3
    x = t
     
  4. Apr 22, 2009 #3
    Yeah, I've seen a lot of that. That's pretty clear cut but there was one that given as an example that was so confusing. The original function was y = -x^2/72 + x

    Then they said from this we obtain the parametric equations:

    x = 24(sqrt2)(t)
    y = -16t^2 + 24(sqrt2)(t)

    I'm pretty lost on to how they got those ones.
     
  5. Apr 22, 2009 #4

    jgens

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    When you're working with functions, I'm fairly certain that you can always just replace the x's with t's in the parameter for y and then use the parameter x = t.
     
  6. Apr 22, 2009 #5
    There is no unique parametrization of a given graph. For example, note that mapping t to 4t does not change the resulting points (x, y) that make up the graph. If the graph is the path of a particle, some physics texts will say that you have simply changed the speed at which the point is moving along the graph. Indeed, changing t to f(t) for any bijective function f (over the domain required for the parametrization to cover the graph) gives exactly the same graph. For example, consider the unit semicircle above the x-axis centered at the origin. The parametrizations (x, y) = (cos(t), sin(t)) for t in [0, Pi] and (x, y) = (t, Sqrt[1 - t^2]) for t in [-1, 1] both cover it.
     
  7. Apr 22, 2009 #6
    All they did for y was replace x in the equation with 24(sqrt2)(t). You can do whatever you want to x, as long as you then make sure y and x satisfy the equation.
     
  8. Apr 22, 2009 #7
    Yeah, I understand that part. Maybe I'm not being clear. The whole example is about a projectile that is shot at an angle of 45 degrees and a initial velocity of 48 ft/sec. Then they say it follows the path given by y = -x^2/72 + x

    They go on and say how this equation does not give all the information possible. Then how we need to introduce a third variable t for time. Then they by writing x and y as functions of t we get:

    x = 24(sqrt2)(t)
    y = -16t^2 + 24(sqrt2)(t)

    That's where I'm confused, they don't explain how they got to that point (Mathematically).
     
  9. Apr 22, 2009 #8
    They are using the additional physical information to get the horizontal speed of the particle, so that they can replace x with x(t). Being fired at a 45 degree angle, the horizontal and vertical components of the velocity are the same, as vx = 48*cos(45) = 48*(Sqrt(2)/2) = 24sqrt(2), etc. Splitting a vector into horizontal and vertical components should have been covered in a previous chapter.
    Using the kinematic equation for constant velocity (since there is no horizontal acceleration), we get x = vx*t = 24sqrt(2)*t.
     
  10. Apr 22, 2009 #9
    Yay, thanks.
     
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