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Math Help - polar equations

  1. #1
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    polar equations

    find the points of horizontal and vertical tangency (if any) to the polar curve.
    r=1-sin theta

    i got horizontal tangent at (1/2, pi/6) (2, 3pi/2) (1/2, 5pi/6)
    and vertical tangent at (3/2, -pi/6) (3/2, 7pi/6)

    can anyone check my work and see if those are correct? i am not sure about the vertical tangent points...
    thanks a bunch
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  2. #2
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    Hello, Holly!

    Your answers are correct.

    But there is another vertical tangent at the cusp: . \left(0,\tfrac{\pi}{2}\right)

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  3. #3
    MHF Contributor Calculus26's Avatar
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    Your results are correct , however you'd probably list

    (3/2,-pi/6) as (3/2, 11pi/6)

    Don't know the uncertainty you have on the VTs

    the demominator of dy/dx = 2sin^2(t)-sin(t)- 1 = 0

    (2sin(t)+1)(sin(t)-1) = 0

    sin(t) = -1/2 which yields 7pi/6 , 11pi/6


    or sin(t) = 1 but this is the cusp at pi/2
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  4. #4
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    Thanks, I had the point at pi/2 for the horizontal tangent but I threw it out because I got pi/2 for the vertical tangent too and I thought you had to get rid of it for some reason??
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  5. #5
    MHF Contributor Calculus26's Avatar
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    A technical and subtle but important point. because you have 0/0 at pi/2

    for dy/dx the lim as theta -> pi/2+ = -inf

    lim as theta ->pi/2- = inf

    So it is a cusp and not a vertical tangent
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  6. #6
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    thanks! and its not a horizontal tangent either? just a cusp?
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  7. #7
    MHF Contributor Calculus26's Avatar
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    Yes -- just a cusp

    Think about the cardioid 1 + cos(theta)

    at pi you wouldn't have a horizontal tangent but a cusp.

    Typically with polar graphs 0/0 yields a cusp

    Another good example is y = x^(2/3) where you have a cusp at x = 0

    since y ' = 2/3/{x^(1/3)} as you approach 0+ you get inf but as you approach 0- you get - inf

    But for y = x^ (1/3) you have a vertical tangent since

    y ' = 1/3/{x^(2/3)} you get infinity as x approaches 0 either from the left or right
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  8. #8
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    thanks! that was a good example easy to understand
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