How do I convert the rectangular equation x+y = 1 to parametric equations?

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Homework Help Overview

The discussion revolves around converting the rectangular equation x+y=1 into parametric equations, with a focus on generating three equations for x, y, and z. The subject area includes parametric equations and surface area calculations using double integrals.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore various parametrizations, with some suggesting letting x=t as a starting point. Others express confusion over the infinite number of possible parametrizations and the specific requirements of their textbook.

Discussion Status

Participants are actively discussing different approaches to parametrization, with some guidance provided on selecting parametrizations that simplify integration. There is acknowledgment of the flexibility in choosing parametrizations, though no consensus has been reached on a specific method.

Contextual Notes

Some participants mention the context of their textbook, which includes specific examples of parametrizations for different equations, indicating that there may be preferred methods for their assignments.

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Convert x+y = 1 to parametric equations.

I know that z = v, but after that I'm stuck.
 
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Wait, z=v? What's that got to do with anything?

A simple way to do this is let x=t.
 
I need to convert it into three parametric equations for x, y, and z
 
There are lots of parametrizations.

Which one you would want to use depends on what you want to do with it.
 
It's in the section of my book dealing with finding surface area by using double integrals.
 
its the infinite amount of parametrizations that throws me
 
You typically pick one that cancels out factors in your integration.
 
ok, but my book wanted me to do it a certain way.

like when the function was x^2 + y^2 = 1

z = V
x = cosU
y = sinU

and then for another one they wanted

z = V
x = sinUcosV
y = sinUsinV
 
Last edited:
Right, because they come out nice.

The point is that the parametization doesn't matter. The double integral will always evaluate to the same value.

(t,1-t,z) will probably work fine for you.
 

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