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Converting parametric to cartesian

  • Thread starter Calpalned
  • Start date
  • #1
297
6
(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.
 

Answers and Replies

  • #2
34
0
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}}##
 
  • #3
ehild
Homework Helper
15,427
1,827
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.
 
  • #4
297
6
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
 
  • #5
297
6
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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