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    C has polar equation:

    r = a (1 + cos theta), theta between pi and - pi.

    The point on C with polar coordinates (r, theta) has cartesian coordinates (x, y). Find the min. value of y.

    -------------------------------------------------------------------------------------------------------------------------

    So, what I did is:

    y = a (1 + cos theta) sin theta
    y = a/2 ( 2 sin theta + sin 2theta)

    I know the min y comes out when theta = - pi/3

    But I did it by trial and error, and it's pretty time consuming and may not be the correct method. Is there any other way to do it?



    P.S. Sorry, never used LaTex.
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    (Original post by Alpha510)
    C has polar equation:

    r = a (1 + cos theta), theta between pi and - pi.

    The point on C with polar coordinates (r, theta) has cartesian coordinates (x, y). Find the min. value of y.

    -------------------------------------------------------------------------------------------------------------------------

    So, what I did is:

    y = a (1 + cos theta) sin theta
    y = a/2 ( 2 sin theta + sin 2theta)

    I know the min y comes out when theta = - pi/3
    .... and when theta=+pi/3, -pi or +pi



    But I did it by trial and error, and it's pretty time consuming and may not be the correct method. Is there any other way to do it?



    P.S. Sorry, never used LaTex.[/QUOTE]

    It seems to be the most simple

    \frac{dy}{d \theta}=0

    from this
    \cos \theta=\frac{1}{2} \rightarrow \sin \theta=\frac{\sqrt{3}}{2} for \theta=\frac{\pi}{3} and \sin \theta=-\frac{\sqrt{3}}{2} for \theta=-\frac{\pi}{3}
    And the other root
    \cos \theta =-1 \rightarrow \sin \theta =0

    when you resubstitute use these sin \theta values.
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    (Original post by ztibor)

    It seems to be the most simple

    \frac{dy}{d \theta}=0

    from this
    \cos \theta=\frac{1}{2} \rightarrow \sin \theta=\frac{\sqrt{3}}{2} for \theta=\frac{\pi}{3} and \sin \theta=-\frac{\sqrt{3}}{2} for \theta=-\frac{\pi}{3}
    And the other root
    \cos \theta =-1 \rightarrow \sin \theta =0

    when you resubstitute use these sin \theta values.
    Ah yes, yes, yes. That's elementary. Shameful. I'm stressing too much.

    Thanks a lot.

    (Original post by ztibor)
    .... and when theta=+pi/3, -pi or +pi
    Btw, -pi or pi don't give the min value. I drew the curves and tried to compare distances.
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    I think you can still do it your way

    y = a/2 ( 2 sin theta + sin 2theta)

    Dy/dtheta= 2cos theta + 2cos2theta
    Dy/dtheta=0

    0 =2 cos theta+ 2*(2cos^2 -1)
    0=2 cos^2 theta + cos theta -1

    Then you can factorise and find the theta values
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    (Original post by freakynerdlol)
    I think you can still do it your way

    y = a/2 ( 2 sin theta + sin 2theta)

    Dy/dtheta= 2cos theta + 2cos2theta
    Dy/dtheta=0

    0 =2 cos theta+ 2*(2cos^2 -1)
    0=2 cos^2 theta + cos theta -1

    Then you can factorise and find the theta values
    Yes, I got that. Thank you.
 
 
 
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