Cartesian Equation for Parametric Curve: x(t)=3sin(2t), y=4cos(2t)

[will_lansing]

Homework Statement

Give a Cartesian equation for the parametric curve x(t)=3sin(2t) and y=4cos(2t)

Homework Equations

The Attempt at a Solution

I'm not sure if I'm doing this right
since x^2+y^2=1

I thought

sin^2(2t)+cos^2(2t)=1 should be the right answer
am i wrong
so how do you go about converting parametric curve to a cartesian equation

 

[Avodyne]

You want an equation that involves x and y only, and not t.

And I can't tell if sin2(t) is supposed to mean sin^2(t) or sin(2t). Either way, can you express cos(2t) in terms of sin2(t)? Once you do that, you're basically done.

 

[unplebeian]

If you have taken a calculus III course at all (looks like you are taking one right now)the questions are reversed and stated as parameterize the following. If you think in a reverse way, you may get some insight.

Look up ellipse in the form of ...oops I'm not supposed to give out the answer!

 

[HallsofIvy]

Homework Statement

Give a Cartesian equation for the parametric curve x(t)=3sin2(t) and y=4cos(2t)

Do you mean x= 3sin(2t)?

Homework Equations

The Attempt at a Solution

I'm not sure if I'm doing this right
since x^2+y^2=1

Where did you get that?

I thought

sin^2(2t)+cos^2(2t)=1 should be the right answer
am i wrong
so how do you go about converting parametric curve to a cartesian equation

No, sin^2(2t)+cos^2(2t)=1 is a good start but it is not the "answer"!

Assuming you meant x= 3 sin(2t) then x/3= sin(2t). If y= 4 sin(2t) then y/4= sin(2t). Now, what do you get if you square both sides of those equations and then add?

 

[will_lansing]

sorry i meant 3sin(2t)

okay so if you are supposed to square both side of the equation you should get
x^2/9=sin^2(2t)
how did you get y=4sint(2t)?
but if you were to square that you would get y^2/16=sin^2(2t)
then if you add the two equations
x^2/9=y^2/16
x^2/9-y^2/16=0
is that what you mean?

 

[malty]

sorry i meant 3sin(2t)

okay so if you are supposed to square both side of the equation you should get
x^2/9=sin^2(2t)
how did you get y=4sint(2t)?
QUOTE]

It was just an error... he meant y=4cos (2t)


Ok so you have x=3sin (2t) and y=4cos(2t) if we let 2t= \theta

Does it help? Remember \cos^2 {\theta} +\sin^2{\theta}=1