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Math Help - Polar coordinates

  1. #1
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    Polar coordinates

    The polar equation of a curve C is r^2 = 2 sin 2Ѳ.
    (a) Find a Cartesian equation for C.
    (b) Show that C could be represented parametrically by the equations

    x =(2t)/(1 + t^4), y = (2t^3)/(1+t^4)
    where t is a parameter.

    Iíve done (a). But how can I solve (b)?
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  2. #2
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    Quote Originally Posted by geton View Post
    The polar equation of a curve C is r^2 = 2 sin 2Ѳ.
    (a) Find a Cartesian equation for C.
    (b) Show that C could be represented parametrically by the equations

    x =(2t)/(1 + t^4), y = (2t^3)/(1+t^4)
    where t is a parameter.

    I’ve done (a). But how can I solve (b)?
    So you know that \left( x^2 + y^2 \right)^2 = 4xy. (Here's some advice: The answer to (a) is needed in order to answer (b). So next time, to save a poor sap like me from spending 10 seconds re-inventing the wheel, be a sport and include your answer).

    The easiest approach is to sub the parametric expressions into the cartesian equation and show that you get an identity:

    Consider: x^2 + y^2 = \frac{4t^2}{(1 + t^4)^2} + \frac{4t^6}{(1 + t^4)^2} = \frac{4t^2 (1 + t^4)}{(1 + t^4)^2} = \frac{4t^2}{1 + t^4}.


    Therefore LHS = (x^2 + y^2)^2 = \frac{16t^4}{\left( 1 + t^4 \right)^2}.


    RHS = 4xy = 4 \left( \frac{2t}{1 + t^4} \right) \left( \frac{2t^3}{1+t^4} \right) = \frac{16t^4}{\left( 1+t^4 \right)^2} = \, LHS.

    Q.E.D.
    Last edited by mr fantastic; January 27th 2008 at 03:41 AM. Reason: Fixed the several careless typos .....
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