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  1. #1
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    Smile Question

    How to find the Asymptote of the Curve-

    2a/r= 1 + 2sin(\theta)<br />

    I guess I posted this in the right section?
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  2. #2
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    Plz help me
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  3. #3
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    Quote Originally Posted by cnpranav View Post
    How to find the Asymptote of the Curve-

    2a/r= 1 + 2sin(\theta) <br />

    I guess I posted this in the right section?


    You should have stated what r is. That's probably why you didn't recieve any help right away. When I looked at the problem, I couldn't tell what the curve is. However, I'm going to assume that r is a variable that depends on \theta, and then I can solve this equation for r to obtain:

    r=\frac{2a}{1+2sin(\theta)}

    So you can think of r as a function of \theta. In this situation, there is vertical asymptote to the curve. This occurs when:

    1+2sin(\theta)=0

    sin(\theta)=-\frac{1}{2}

    This occurs when \theta=\frac{7\pi}{6} and \theta=\frac{11\pi}{6} on the interval [0,2\pi]

    There are really an infinite number of asymptotes to this curve since the sine function is periodic.

    Two of the asymptotes are just the verticle lines x=\frac{7\pi}{6} and x=\frac{11\pi}{6}, really I should say theta, but the x-axis is usually represented by x even when we are using angles. So you could just replace theta with x in all of the work above.
    Last edited by adkinsjr; October 25th 2009 at 01:00 PM.
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