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  1. #1
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    Smile Question

    How to find the Asymptote of the Curve-

    $\displaystyle 2a/r= 1 + 2sin(\theta)
    $

    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-

    $\displaystyle 2a/r= 1 + 2sin(\theta)$$\displaystyle
    $

    I guess I posted this in the right section?


    You should have stated what $\displaystyle r$ 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 $\displaystyle r$ is a variable that depends on $\displaystyle \theta$, and then I can solve this equation for $\displaystyle r$ to obtain:

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

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

    $\displaystyle 1+2sin(\theta)=0$

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

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

    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 $\displaystyle x=\frac{7\pi}{6}$ and $\displaystyle x=\frac{11\pi}{6}$, 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; Oct 25th 2009 at 01:00 PM.
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