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Math Help - Vertical asymptote in polar?

  1. #1
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    Vertical asymptote in polar?

    Show that the polar curve r = 4 + 2sec(theta) has the line x = 2 as a vertical asymptote by showing that lim (r -> inf) x = 2

    I don't understand what its asking for me to do, and how I should do it.

    I have, however, shown that r = r + 4 without using limits, but it specifically asks for a limit.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    Show that the polar curve r = 4 + 2sec(theta) has the line x = 2 as a vertical asymptote by showing that lim (r -> inf) x = 2

    I don't understand what its asking for me to do, and how I should do it.

    I have, however, shown that r = r + 4 without using limits, but it specifically asks for a limit.
    Convert it to rectangular co-ordinates and evaluate the limit?
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  3. #3
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    I don't understand. Perhaps, I have forgotten how to do limits.

    I would really appreciate a step by step answer, even though it seems that I just want the answer, I learn best from seeing an example. (Its not in my text book either)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    I don't understand. Perhaps, I have forgotten how to do limits.

    I would really appreciate a step by step answer, even though it seems that I just want the answer, I learn best from seeing an example. (Its not in my text book either)
    Did you convert it to rectangular co-ordinates first? IF so show me and I will help with the limit
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  5. #5
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    Does this look right?

    1 = 4/(sqrt(x^2 + y^2)) + 2/x
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    Does this look right?

    1 = 4/(sqrt(x^2 + y^2)) + 2/x
    Yeah it does...I am pretty sure...now 1=\frac{4}{\sqrt{x^2+y^2}}+\frac{2}{x}
    How would you solve that for a limit?
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  7. #7
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    Well, I'd like you to help me understand limits again. It's been a while since we solved for limits in class. So, does lim (r to inf) x = 2 mean that r approaches inf when x is = to 2? And if that is the case, how do I "solve" this because I am unsure how to show it. I'd think that we can plug in x = 2 but what would that do?
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    So, does lim (r to inf) x = 2 mean that r approaches inf when x is = to 2?
    No, it means that x approaches 2 as r approaches +\infty. To evaluate this limit you need to have an expression of x which only depends on r, acknowledged that x=r\cos \theta and r=4+\frac{2}{\cos \theta}.
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  9. #9
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    OMG, thank you guys! I think I got it now. Basically, 2/(r - 4) = cos(Theta). x = r(2/(r - 4)). Lim r -> inf of 2r/(r - 4). Using L'Hospital, you get 2/1 = 2. Did I do it right?!
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    OMG, thank you guys! I think I got it now. Basically, 2/(r - 4) = cos(Theta). x = r(2/(r - 4)). Lim r -> inf of 2r/(r - 4). Using L'Hospital, you get 2/1 = 2. Did I do it right?!
    good job
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  11. #11
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    As much as I hate asking another similar question, this one seems harder to solve. r = sinTtanT show that it has the line x = 1 as a vertical asymptote. I assume its done the same way but I dont see how to change r solely to x as flyign squirrel described
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  12. #12
    Super Member flyingsquirrel's Avatar
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    The idea is the same, except that me use x(T) and y(T) instead : \left\{ \begin{array}{l}x(T)=r\cos T=\sin^2T\\y(T)=r\sin T=\frac{\sin^3T}{\cos T}\end{array}\right.

    As you are told that the asymptote is the line x=1, you can find for which T_0 it will be reached solving \lim_{T\to T_0}x(T)=1. (here, the limit is useless because x(T_0) exists but it will not always be true) Then, check that \lim_{T\to T_0}y(T)=\pm \infty and conclude.
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