Converting parametric to cartesian

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Homework Help Overview

The discussion revolves around converting parametric equations to Cartesian form, specifically the equations x = (1/2)cos(θ) and y = 2sin(θ). The context is calculus, with participants exploring the relationship between the parameters and their Cartesian counterparts.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various attempts to eliminate the parameter θ, including manipulating the equations and considering trigonometric identities. There is a focus on whether the resulting shape is a circle, with some participants questioning this assumption.

Discussion Status

Multiple approaches are being explored, with some participants offering guidance on rearranging the equations and applying trigonometric identities. There is no explicit consensus on the final form of the Cartesian equation, but productive suggestions have been made.

Contextual Notes

Participants note the challenge of dealing with different coefficients in the parametric equations, which complicates the elimination of θ. There is also mention of the lack of an answer key in the textbook, contributing to uncertainty in verifying results.

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(This is actually a calculus problem, not a physics one, but physics is based on calculus, so I hope it's fine)

1. Homework Statement

Eliminate the parameter to find the Cartesian equation of x = (1/2)cos(θ) y = 2sin(θ)

Homework Equations


x^2 + y^2 = 1 (eq of circle)

The Attempt at a Solution


First approach: x^2 + y^2 = (1/4)cos^2(θ) + 4sin^2(θ) = ?
I can't get rid of θ because the constant preceding cosine and sine are not equal.

2nd try: y/x = 2sin(θ)/(0.5cos(θ)) = 4 * (sin(θ)/cos(θ)) = y/4x = tan(θ), so θ = arctan(y/4x)
But θ is still there. I need an answer in x and y.

Thank you for your help.
 
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It is useful to note that

##sin(arctan(x))~=~\frac{x}{\sqrt{1+x^2}}##

and

##cos(arctan(x))~=~\frac{1}{\sqrt{1+x^2}}##
 
Do not assume it is a circle. It is not.

Arrange the equations so as cosθ and sinθ are alone on one side of the equations: cosθ= ? sinθ = ?
Take the square of both equations and add them together.
 
showzen said:
It is useful to note that

##sin(arctan(x))~=~\frac{x}{\sqrt{1+x^2}}##

and

##cos(arctan(x))~=~\frac{1}{\sqrt{1+x^2}}##
Thank you
 
ehild said:
Do not assume it is a circle. It is not.

Arrange the equations so as cosθ and sinθ are alone on one side of the equations: cosθ= ? sinθ = ?
Take the square of both equations and add them together.
Following your method, I got 4x^2 + (y^2)/4 = 1
My textbook doesn't have an answer key, but I am going to assume that this is the correct answer. Thank you very much
 

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