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Homework Help: Determining Domain and Range of Cartesian Equations

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Eliminate the parameter to find the Cartesian equation of the curve.

    Problem #1:
    x = sin t, y = csc t, 0 < t < ∏ / 2

    Problem #2:
    x = sin θ, y = cos 2θ

    2. Relevant equations
    What's shown above is what's listed in the book. However, the authors felt compelled to include the domain (for the the second problem) and range (for the first problem) in the answers section in the back of the book which are as follows:

    Problem #1:
    {y|y > 1}

    Problem #2:
    {x|- 1 ≤ x ≤ 1}

    I'm not sure how they got that (as it has been a year since I finished precal) and I don't think my professor requires me to know how to get the domain and range for these problems, but I'd like to know how just in case I ever come across something like this again.

    3. The attempt at a solution
    If it makes any difference, here are the Cartesian equations (and the work it took to get them):

    Problem #1:
    x = sin t
    y = csc t
    y = 1 / sin t
    y = 1 / x

    Problem #2:
    x = sin θ
    y = cos 2θ
    y = cos θ^2 - sin θ^2
    y = 1 - 2sin θ^2
    y = 1 - 2x^2
  2. jcsd
  3. Jun 14, 2012 #2
    1. notice that 0<x<1. Thus y=1/x>1, and approches ∞ ("continuously")
    2. here we have x=sin(θ), so -1≤x≤1.

    These answers are not the complete answer, often, the answers in the back are not necessarily what someone might answer on their homework or test. A complete solution might include what the book said, and anybody else might mention in their version, like the 0<x<1 above. The book never mentions that. Nor does it mention that for part 2, that -1≤y≤1.
  4. Jun 14, 2012 #3


    Staff: Mentor

    Comment on your notation:

    You wrote cos θ^2 and sin θ^2, which would be interpreted as cos(θ2) and sin(θ2), rather than what you really meant, which was [cos(θ)]2 and [sin(θ)]2.

    The last two are usually written as cos2(θ) and sin2(θ).
  5. Jun 29, 2012 #4
    Thank you. Much appreciated.
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