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Parametric equation to cartesian

  • #1
109
0

Homework Statement


I have this equation and i need to find the cartesian equation, so i apreciate your help

Homework Equations


X=cost ' y=2sin2t

The Attempt at a Solution


I am usign this [/B]
Sin2t=2costsint
So x+y/2=cost+2costsint
But i dont know what to do after,
I also try to solve that this way
Y/2=sin2t
Y/2=2costsint then sint=y/(4cost)
Y/2=2((x)y/4x)
When i isolate y i get y/y=3x-2
 

Answers and Replies

  • #2
33,103
4,798

Homework Statement


I have this equation and i need to find the cartesian equation, so i apreciate your help

Homework Equations


X=cost ' y=2sin2t

The Attempt at a Solution


I am usign this [/B]
Sin2t=2costsint
So x+y/2=cost+2costsint
But i dont know what to do after,
I also try to solve that this way
Y/2=sin2t
Y/2=2costsint then sint=y/(4cost)
Y/2=2((x)y/4x)
When i isolate y i get y/y=3x-2
So y = 4sin(t)cos(t). What is y/x?

Ideally, you want to get something that involves sin2(t) + cos2(t), which equals 1, and eliminates the parameter t.
 
  • #3
109
0
Thanks for the help i juts figure out how to get rid of that parameter 1-cos alll div by 2
 
  • #4
33,103
4,798
Thanks for the help i juts figure out how to get rid of that parameter 1-cos alll div by 2
I doubt that will work.
 
  • #5
109
0
y=4xsint==>y^2=16x^2(1(cost)^2)=y^2=16x^2(1-x^2)
 
  • #6
109
0
Forget about div for 2
 
  • #7
33,103
4,798
y=4xsint==>y^2=16x^2(1(cost)^2)=y^2=16x^2(1-x^2)
With one correction, that works. Where you have y2 = 16x2(1(cost)^2), it should be y2 = 16x2(1 - cos2(t)). Otherwise, what you ended with is the same as I got.
 
  • #8
109
0
Thanks thats it Sorry pal i have anoter problem i am doing x=sect y=tant
So what i need to do use the 1+tan^2=sec^2
Or use the sin/cos=tan and 1/cos = sec? I can get the solution
 
  • #9
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,508
730
Thanks thats it Sorry pal i have anoter problem i am doing x=sect y=tant
So what i need to do use the 1+tan^2=sec^2
Or use the sin/cos=tan and 1/cos = sec? I can get the solution
Using ##1 + \tan^2\theta = \sec^2\theta## is the obvious choice. But you should post new questions in new threads if you don't want to have your hand slapped by a moderator.
 
  • #10
33,103
4,798
Thanks thats it Sorry pal i have anoter problem i am doing x=sect y=tant
So what i need to do use the 1+tan^2=sec^2
Or use the sin/cos=tan and 1/cos = sec? I can get the solution
This is a new problem, so please start a new thread.
 

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