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Parametric equations

Hopefully this is the right place for this

Hey I would like some help with these questions.



1.) x=t/(2t-1) y=t/(t+3)
2.) x=sin2theta y=costheta.3
.)x=sintheta +2costheta y=2sintheta +costheta


For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer
Reply 1
Avatar for Notnek
Notnek
21
Original post by Sylarus
Hopefully this is the right place for this

Hey I would like some help with these questions.



1.) x=t/(2t-1) y=t/(t+3)
2.) x=sin2theta y=costheta.3
.)x=sintheta +2costheta y=2sintheta +costheta


For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer

1) and 2) are correct.

For 3), try squaring x and y and also consider their product.
Original post by Sylarus
Hopefully this is the right place for this

Hey I would like some help with these questions.



1.) x=t/(2t-1) y=t/(t+3)
2.) x=sin2theta y=costheta.3
.)x=sintheta +2costheta y=2sintheta +costheta


For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer


Agree with your first two.

For the third one, obtain expressions for cos⁡θ\cos\theta and sin⁡θ\sin\theta, in terms of x,y, and use a well known identity to eliminate theta.

Edit: Nothing for 50 minutes and then two at once!
Reply 3
Avatar for Sylarus
Sylarus
OP
If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

Costheta= y-2sintheta
Sintheta= x-2costheta

Then squaring I got this big messy expansion which I can't seem to get the Thetas out of :frown:
Reply 4
Avatar for Notnek
Notnek
21
Original post by Sylarus
If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

Costheta= y-2sintheta
Sintheta= x-2costheta

Then squaring I got this big messy expansion which I can't seem to get the Thetas out of :frown:

Ghostwalker may have to advise you here because I'm not sure.

An alternative method (as mentioned in my last post) : Compare x2+y2x^2+y^2 and xyxy.
Original post by Sylarus
If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

Costheta= y-2sintheta
Sintheta= x-2costheta

Then squaring I got this big messy expansion which I can't seem to get the Thetas out of :frown:


You haven't got cos/sin in terms of x,y.

Using the two equations, you can eliminate sin, and get cos as a function of x,y.
Similarly you can eliminate cos, and get sin as a function of x,y.

Then apply the identity.
Reply 6
Avatar for Notnek
Notnek
21
Original post by ghostwalker
You haven't got cos/sin in terms of x,y.

Using the two equations, you can eliminate sin, and get cos as a function of x,y.
Similarly you can eliminate cos, and get sin as a function of x,y.

Then apply the identity.

Yes that's a much better way. I should have spotted it.
Reply 7
Avatar for Sylarus
Sylarus
OP
K so after some rearranging got it to (2x-y)^2+(2y-x)^2=1
Original post by Sylarus
K so after some rearranging got it to (2x-y)^2+(2y-x)^2=1


Close. But it doesn't equal 1.

As a check you can choose a value for the parameter, theta = 0 say. Sub into your original equations, and check if the x,y values that come out satisfy your cartesian equation.

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