# Parametric Representation of an equation.

**URL:** <https://boards.straightdope.com/t/parametric-representation-of-an-equation/111155>\
**Category:** Factual Questions\
**Created:** [May 28, 2002, 11:18am UTC](https://boards.straightdope.com/t/parametric-representation-of-an-equation/111155 "2002-05-28T11:18:18Z")\
**Posts on this page:** 4\
**Page:** 1

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**Author:** ![Phlip](https://avatars.discourse-cdn.com/v4/letter/p/ed8c4c/32.png) [@Phlip](https://boards.straightdope.com/u/Phlip)\
**Post date:** [May 28, 2002, 11:18am UTC](https://boards.straightdope.com/t/parametric-representation-of-an-equation/111155/1 "2002-05-28T11:18:18Z")

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What is the proper procedure for changing an equation from x,y format to parametric?

For example the equation x^4+y^4=1 . Parametrically I am thinking that x=cos (t) and y=sin(t)? Is this correct? The shape of this thing is kind of a rounded square if you know what I mean.

I am just guessing at the paramtric representaion of this based off of the shape. But what is the formal procedure? What steps would I take this thru to reach the parametric values of the x and y components?

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**Author:** ![Phlip](https://avatars.discourse-cdn.com/v4/letter/p/ed8c4c/32.png) [@Phlip](https://boards.straightdope.com/u/Phlip)\
**Post date:** [May 28, 2002, 11:40am UTC](https://boards.straightdope.com/t/parametric-representation-of-an-equation/111155/2 "2002-05-28T11:40:12Z")

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or is x=cos^2(t) and y=sin^2(t)?

I am equating this to the trig function cos^2+sin^2=1

is there a more reliable way to do this?

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**Author:** ![Jabba](https://avatars.discourse-cdn.com/v4/letter/j/b5e925/32.png) [@Jabba](https://boards.straightdope.com/u/Jabba)\
**Post date:** [May 28, 2002, 12:50pm UTC](https://boards.straightdope.com/t/parametric-representation-of-an-equation/111155/3 "2002-05-28T12:50:06Z")

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As a first point, in general curves do not have unique parameterisation. To take a simple example, consider the straight-line segment joining (0,0) to (1,1). We could parameterise this as (t,t) (0\<=t\<=1); as (t[sup]2[/sup],t[sup]2[/sup]) (0\<=t\<=1); or as (sin t, sin t) (0\<=t\<=pi/2) among many others.

For your specific example, no ( sin t,cos t) won’t work; this is a parameterisation of x[sup]2[/sup] + y[sup]2[/sup] = 1. You could perhaps use ( (sin t)[sup]1/2[/sup],(cos t)[sup]1/2[/sup]) (0\<=t\<=pi/2) for the region of the curve where x\>0,y\>0. Judicious insertion of minus signs will get you the other three quadrants.

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**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\
**Post date:** [May 28, 2002, 7:11pm UTC](https://boards.straightdope.com/t/parametric-representation-of-an-equation/111155/4 "2002-05-28T19:11:29Z")

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One way to get a paremetrization (not the only one, and probably not the most useful one) is to use one of your variables as the parameter. For instance, you have x[sup]4[/sup] + y[sup]4[/sup] = 1, which means that x = (1 - y[sup]4[/sup])sup[/sup]. Then you can take the parametrization to be x = (1 - t[sup]4[/sup])sup[/sup], y = t. Note that this will only work for the half right of the y axis; for the other half, stick on a minus sign.