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Parameterization of x^(2/3)+y^(2/3)=a^(2/3)

  1. Jun 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Parametrize the curve x^(2/3) + y^(2/3) = a^(2/3) in the standard counterclockwise sense.

    2. Relevant equations

    x^(2/3) + y^(2/3) = a^(2/3)
    Any trig identity... I was thinking cos(x)^2 + sin(x)^2 = 1


    3. The attempt at a solution

    Because it has to be parameterized in the standard counterclockwise sense, I know I have to use a cosine.

    The shape of the graph is a diamond-like shape. (NOT a diamond, though.) At first I let

    x = a(cos[t])^(3/2), so that
    y = a(sin[t])^(3/2)

    *if my substitutions and simplifications are correct... Maybe my problem lies in here?

    My results are
    t
    x
    y​

    0
    a
    0
    pi/2
    0
    a
    pi
    nonreal answer
    0
    3*pi /2
    0
    nonreal answer

    Instead of the nonreal answers, it should be -a. Does any one know how I can get around it?

    Thank you in appreciation for your help, and I'd just like to say to the people that help everyone else, You guys are awesome. I've seen your answers before, and you're really helping a lot of people out! (Hopefully me!)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 23, 2007 #2
    What do you get on the right when you substitute x = a(cos[t])^(3/2) and y = a(sin[t])^(3/2) on the LHS?

    And welcome to the forum. :)
     
  4. Jun 23, 2007 #3
    When I substitute the x = a(cos[t])^(3/2) and y = a(sin[t])^(3/2) into the Left Hand Side of the equation, I get:

    (a(cos[t])^(3/2))^(2/3) + (a(sin[t])^(3/2))^(2/3) = a^(2/3)
    a^(2/3)(cos[t]) + a^(2/3)(sin[t]) = a^(2/3)
    cos[t] + sin[t] = 1
    or also
    cos[t]^2 + sin[t]^2 = 1, which is the trigonometic identity, which proves that when x = a(cos[t])^(3/2) and y = a(sin[t]^(3/2) the equation is still correct.

    I'm hoping if I keep typing, the answer will just come to me. So far, no good.
    This doesn't change that I get nonreal answers when pi < t < 2pi.

    cos[pi] = -1, but when it is raised to the power of (3/2), it becomes an unreal answer. By "unreal" I mean nonreal. I really appreciate the hint, but I don't think I got it. I'll keep working on it and hope you or someone else replies with another hint! :D Thank you! :!!)
     
  5. Jun 23, 2007 #4
    What else am I missing here? I've tried rearranging the equation, and I figure I have to raise everything to a power so that the cos[x] and sin[x] aren't raised to the power of (3/2). It has to be an even number power so that cos[pi] will remain -1. Any other hints?
     
  6. Jun 23, 2007 #5

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    how does cos[t] + sin[t] = 1 become cos[t]^2 + sin[t]^2 = 1 ??

    "cos[t] + sin[t] = 1
    or also
    cos[t]^2 + sin[t]^2 = 1"

    And is the curve:
    x^(2/3) + y^(2/3) = a^(2/3)
    meant to be one positive turn? i.e connected curve?

    What values of x and y are allowed?
     
  7. Jun 23, 2007 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You WANT sin2 x+ cos2 x= 1 so you DON'T WANT that 1/2 power in x= (a sin x)3/2 and y= (a cos x)3/2!
     
  8. Jun 23, 2007 #7
    " how does cos[t] + sin[t] = 1 become cos[t]^2 + sin[t]^2 = 1 ??

    "cos[t] + sin[t] = 1
    or also
    cos[t]^2 + sin[t]^2 = 1""

    It doesn't, ha ha ha. I've been working (thinking about) this question for the past 2 weeks, and I guess I just made up some trig identities. Wouldn't it be nice if cos[t] + sin[t] = 1?

    And yes, the curve x^(2/3 + y^(2/3) = a^(2/3) is a connected curve, and to parameterize it, it is meant to go one complete cycle, t goes from 0 to 2pi.
     
  9. Jun 23, 2007 #8
    Hey guys! I got it! I know it took me a while, and you guys were probably just rolling your eyes at me, but I GOT IT! Yaaaay! Hooray for me! Calculus on a Saturday night!

    I just want to give my heartfelt thanks to you who have helped me. It means so much to me that you would reply so quickly to my post. :) :)
    Enjoy the rest of your weekend!!!
     
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