Parametric Equations and cartesian equation

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Homework Help Overview

The discussion revolves around finding the Cartesian equation corresponding to given parametric equations, specifically x = sin(2πt) and y = cos(2πt), with the parameter t ranging from 0 to 1. The context involves understanding the relationship between parametric and Cartesian forms of equations in trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the use of the identity sin²(θ) + cos²(θ) = 1 to derive the Cartesian equation. Questions arise regarding the role of the parameter t and whether x and y can be considered independent of t.

Discussion Status

The discussion is active, with participants clarifying the parametric equations and addressing typographical errors. Some guidance has been provided regarding the relationship between the parameter t and the resulting Cartesian equation, with acknowledgment of the circular nature of the curve as t varies.

Contextual Notes

There is a noted concern about the implications of the parameter t in the solution, as well as the potential for miscommunication due to typographical errors in the equations presented.

courtrigrad
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(1)If you are given the parametric equations [itex]x = sin(2\pi\t)[/itex] [itex]y = cos(2\pi\t)[/itex] and [itex]0\leq t\leq 1[/itex] how would you find the cartesian equation for a curve that contains the parametrized curve?

Using the identity [itex]\sin^{2}\theta + cos^{2}\theta = 1[/itex] would it be [itex]x^{2} + y^{2} = 1[/itex]?

Thanks
 
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Sorry... What does anything have to do with t? Isn't t part of the problem? If so, should it not be part of the solution as well?
 
Are you sure x and y are independent of t?? If so, the cartsian equation is just the point (0,1)
 
come on its a typo... [itex]x = \sin(2\pi t ), y = \cos(2\pi t )[/itex]

thanks
 
plugpoint, you were the one who made the typo- "Sorry, it was a typo" would be better than "Come on its a typo"!

Yes, you are correct, since [itex]sin^2(2\pi t)+ cos^2(2\pi t)[/itex].
You should also note that, as t goes from 0 to 1, [itex]2\pi t[/itex] goes from 0 to [itex]2\pi[/itex] so this would be exactly once around the circle.
 
sorry about that. I was actually saying that to myself, because I was annoyed that I always make typos with Latex. Sorry To Tsar and EnumaFish. And thank you HallsofIvy for helping me

:smile:
 

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