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Polar coordinates (trig question)

  1. Dec 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Given r = 2tan(theta)sec(theta)

    Find cos(theta) then use inverse key to find sec(theta)

    The answer given in the solution guide is y = 1/2 x^2

    Attempt at solution
    Since tan = sin/cos and sec = 1/cos
    We have r = 2sin/cos * 1/cos
    So rcos^2 = 2sin
    rcos^2 is defined as x^2 so x^2 = 2sin

    Unless 2 sin is defined as 2 y, I don't see how we can get to y = 1/2x^2??
  2. jcsd
  3. Dec 29, 2008 #2


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    Gold Member

    I do not understand where you got this part from:

    In polar coordinates,

    Substitute your expression for r into these two equations. Solve the first one in such a way that you can plug your result into the second one to solve for y in terms of x.
  4. Dec 29, 2008 #3


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    Homework Helper

    r cos^2 isn't the same as x^2. [tex]r^2 \cos^2 \theta = x^2[/tex].

    For this problem just use the equations for converting between rectangular and polar coordinates, using the expression for cos theta and sine theta as hinted and substituting them into the problem equation.
  5. Dec 29, 2008 #4
    Okay, so:

    r = 2sin/cos * 1/cos

    rcos = 2sin * 1/cos

    x = 2sin * 1/cos
    x = 2tan

    Where does the x^2 come from and where does the y come from?
  6. Dec 29, 2008 #5


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    Don't worry about where the x^2 comes from, that is part of the answer and is a result of you solving the problem correctly.

    You know that [tex]y=r\sin\theta[/tex] by definition. Find y as a function of theta only.

    Then you can eliminate theta when you find y as a function of x only.
  7. Dec 29, 2008 #6
    r = 2tan(theta)sec(theta)
    r = 2sin(theta)/cos(theta) * 1/cos(theta)
    rcos^2 = 2sin(theta)
    r^2cos^2 = r2sin(theta)
    x^2 = 2y
    y = 1/2(x^2)

    It seems like this works too. Thanks!! I haven't done trig in like forever...
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