Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cartesian equation

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Give a Cartesian equation for the parametric curve x(t)=3sin(2t) and y=4cos(2t)


    2. Relevant equations



    3. 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
     
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    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.
     
  4. Dec 11, 2007 #3
    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!
     
  5. Dec 11, 2007 #4

    HallsofIvy

    User Avatar
    Science Advisor

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


    Where did you get that?

    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?
     
  6. Dec 11, 2007 #5
    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?
     
  7. Dec 11, 2007 #6

    malty

    User Avatar
    Gold Member

     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook