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Parametric equations of various shapes

  1. May 5, 2015 #1
    How we can know the parametric equation for any curve?
    Is there some trick?
    Like for parabola ## y^2 =4a x ##
    It has general coordinates## (at^2 , 2at) ##
    It will satisfy the equation but how in first place we know it?
    Also we can have ##(a/t^2, -2a/t) ##, how?
     
  2. jcsd
  3. May 7, 2015 #2
    Can any ideas, discussion or some help be given?
     
  4. May 7, 2015 #3

    HallsofIvy

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    There exist an infinite number of different parametric equations for a curve. The simplest way to get parametric equations for [itex]y= ax^2[/itex] is to use x itself as parameter: If [itex]x= t[/itex] then [itex]y= at^2[/itex]. That can, in fact, be done for any function- if [itex]y= f(x)[/itex] then [itex]x= t[/itex], [itex]y= f(t)[/itex] are parametric equations.

    For non-function curves, we have to be a little more creative. For example, the relation, [itex]x^2+ y^2= a^2[/itex] describe a circle with center at (0, 0) and radius a. We know that [itex]cos^2(t)+ sin^2(t)= 1[/itex] so [itex]a^2 cos^2(t)+ a^2 sin^2(t)= a^2[/itex] so we can take [itex]x= a cos(t)[/itex], [itex]y= a sin(t)[/itex] as parametric equations. Of course, [itex]x= a sin(t)[/itex], [itex]y= a cos(t)[/itex] would work as well.

    For a more general circle, [itex](x- x_0)^2+ (y-y_0)^2= a^2[/itex], still with radius a but now with center at [itex](x_0, y_0)[/itex], with the same analysis as before, we have [itex]x- x_0= a cos(t)[/itex], [itex]y- y_0= a sin(t)[/itex] so [itex]x= a cos(t)+ x_0[/itex], [itex]y= a sin(t)+ y_0[/itex] are parametric equations.

    We can think of [itex]\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1[/itex], the relation describing an ellipse with axes, along the x and y axes of lengths a and b, respectively, as [itex]\left(\frac{x}{a}\right)^2+ \left(\frac{y}{b}\right)^2= 1[/itex] and see that we can take [itex]\frac{x}{a}= cos(t)[/itex], [itex]\frac{y}{b}= sin(t)[/itex] or [itex]x= a cos(t)[/itex], [itex]y= b sin(t)[/itex] as parametric equations for that ellipse.
     
    Last edited: May 7, 2015
  5. May 7, 2015 #4
    Got it, thanks.
     
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