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    A curve is defined by the polar equation  r=1+\cos \theta,  0 \leqslant \theta < 2\pi
    The tangents to the curve when r=1 meet at a single point.
    Find the area of the region that lies inside the triangle formed by the tangents and the y axis, but outside the area bounded by the curve and the y axis.
    I got the area as 3-3π/4
    Is this correct??
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    (Original post by Ano123)
    A curve is defined by the polar equation  r=1+\cos \theta,  0 \leqslant \theta < 2\pi
    The tangents to the curve when r=1 meet at a single point.
    Find the area of the region that lies inside the triangle formed by the tangents and the y axis, but outside the area bounded by the curve and the y axis.
    I got the area as 3-3π/4
    Is this correct??
    Zacken
    16Characters....
    EricPiphany
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    (Original post by Ano123)
    A curve is defined by the polar equation  r=1+\cos \theta,  0 \leqslant \theta < 2\pi
    The tangents to the curve when r=1 meet at a single point.
    Find the area of the region that lies inside the triangle formed by the tangents and the y axis, but outside the area bounded by the curve and the y axis.
    I got the area as 3-3π/4
    Is this correct??
    I also got this.
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    (Original post by 16Characters....)
    I also got this.
    It's a nice question don't you think?
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    (Original post by Ano123)
    It's a nice question don't you think?
    It was a nice question. Tested quite a few different ideas but was not too long and it "flowed", unlike some questions which just force concepts together.
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    got tagged here by teeEm. Finally maths I can actually do :P
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    A vertical line touches the curve r=a(1+\cos(\theta)) at two points. Show that the area enclosed between the line and the curve is equal to \diaplaystyle a^2\left(\frac{15\sqrt{3}}{16}-\frac{\pi}{2}\right).
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    (Original post by EricPiphany)
    A vertical line touches the curve r=a(1+\cos(\theta)) at two points. Show that the area enclosed between the line and the curve is equal to \diaplaystyle a^2\left(\frac{15\sqrt{3}}{16}-\frac{\pi}{2}\right).
    OMG, I actually managed to do this. :ahee:

    x = r\cos \theta = ar\cos \theta(1 + \cos\theta) \Rightarrow x' = 0 \iff \sin \theta (1 + 2\cos \theta) = 0

    The two values that concern us are: \theta = \frac{4\pi}{3} and \theta = \frac{2\pi}{3}

    The are bounded in the polar curve (using symmetry) is then:

    \displaystyle 

\begin{equation*}a^2\int_{2\pi/3}^{\pi} (1 + \cos \theta)^2 \, \mathrm{d}\theta = \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right)\end{equation*}

    Now draw the lines corresponding to the two arguments and we get a triangle that has area

    \displaystyle 

\begin{equation*}\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2} \times \sin \frac{2\pi}{3} = \frac{a^2\sqrt{3}}{16}\end{equat  ion*}

    So the area required is given by:

    \displaystyle

\begin{equation*} \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right) - \frac{a^2\sqrt{3}}{16} = \frac{a^2}{8}\left(\frac{15\sqrt  {3}}{16} - \frac{\pi}{2}\right)\end{equatio  n*}

    as required.
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    (Original post by Zacken)
    OMG, I actually managed to do this. :ahee:

    x = r\cos \theta = ar\cos \theta(1 + \cos\theta) \Rightarrow x' = 0 \iff \sin \theta (1 + 2\cos \theta) = 0

    The two values that concern us are: \theta = \frac{4\pi}{3} and \theta = \frac{2\pi}{3}

    The are bounded in the polar curve (using symmetry) is then:

    \displaystyle 

\begin{equation*}a^2\int_{2\pi/3}^{\pi} (1 + \cos \theta)^2 \, \mathrm{d}\theta = \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right)\end{equation*}

    Now draw the lines corresponding to the two arguments and we get a triangle that has area

    \displaystyle 

\begin{equation*}\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2} \times \sin \frac{2\pi}{3} = \frac{a^2\sqrt{3}}{16}\end{equat  ion*}

    So the area required is given by:

    \displaystyle

\begin{equation*} \frac{a^2}{8}\left(4\pi - 7\sqrt{3}\right) - \frac{a^2\sqrt{3}}{16} = \frac{a^2}{8}\left(\frac{15\sqrt  {3}}{16} - \frac{\pi}{2}\right)\end{equatio  n*}

    as required.
    Very good, and extra credit for latexing it up
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    (Original post by EricPiphany)
    Very good, and extra credit for latexing it up
    Thanks for this! I learnt quite a bit about polar shiz from doing it.
 
 
 
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