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Help with solving parametric equation

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the following parametric curve:

    x=5cos^7(t)

    y=5sin^7(t)

    Write it in cartesian form, giving your answer as an equation of the form F(x,y)=c for some function F and some constant c.

    3. The attempt at a solution

    I know that sin^2(t)+cos^2(t) = 1 but I don't think this will be much help to figure this out and I am also unsure how I would find the constant

    Any help would be much appreciated
     
  2. jcsd
  3. May 10, 2016 #2
    Why not? How are ##x## and ##y## related to ##\sin^2(t)## and ##\cos^2(t)##?
     
  4. May 10, 2016 #3
    would that mean x^2+y^2 = 1

    which would mean

    (5cos^7(t))^2 +(5sin^7(t))^2 = 1


    25cos^9(t) +25sin^9(t) = 1

    But what would I do now?
     
  5. May 10, 2016 #4
    Nope, that is not correct. Try to express ##\sin^2(t)## in terms of ##y## and ##\cos^2(t)## in terms of ##x##.
     
  6. May 10, 2016 #5
    would that mean

    y=1-cos^2(t)

    x=1-sin^2(t)
     
  7. May 10, 2016 #6
    No, why would you think that?
    You already have ##x## and ##y## defined in your original post: ##x = 5 \cos^7(t)## and ##y = 5 \sin^7(t)##.
    Can you manipulate ##x = 5 \cos^7(t)## to get ##\cos^2(t)## in terms of ##x##? And likewise for ##\sin^2(t)## in terms of ##y##.
     
  8. May 10, 2016 #7

    SammyS

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    To start: Solve
    ##x=5\cos^7(t) ##​
    for ##\ \cos(t)\ .##
     
  9. May 10, 2016 #8
    cos(t)=(x/5)^(7/2)

    Is this correct?
     
  10. May 10, 2016 #9

    SammyS

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    No. The exponent is wrong.
     
  11. May 10, 2016 #10

    cos(t)=(x/5)^(1/7)

    What would i do now?
     
  12. May 10, 2016 #11

    SammyS

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    That's better.

    Do similar for y.

    Square each & add.
     
  13. May 10, 2016 #12
    Thanks I got the right answer now!
     
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