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Math Help - Point on a CSC graph

  1. #1
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    Point on a CSC graph

    Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

    0 is less than or equal to x which is less than or equal to 2*pi

    Thanks
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    Quote Originally Posted by Mr_Green View Post
    Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

    0 is less than or equal to x which is less than or equal to 2*pi

    Thanks
    3y - 2x = 4\implies y = (2/3)x+(4/3).

    To be parallel it means the derivative must be equal to 2/3. Now find all point with derivative 2/3.
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  3. #3
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    so thh derivate of -csc is =

    So I set this equal to 2/3?
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    Quote Originally Posted by Mr_Green View Post
    so thh derivate of -csc is =

    So I set this equal to 2/3?
    Yes.
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    so i get

    pi/3

    Is this correct? Is this the only point?
    Last edited by Mr_Green; October 4th 2007 at 03:39 AM.
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  6. #6
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    i found that out by plugging into my calculator and using the intersect ability on the graph. Is there anything mathematical that I could do to solve it?
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    Quote Originally Posted by Mr_Green View Post
    Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

    0 is less than or equal to x which is less than or equal to 2*pi

    Thanks
    Quote Originally Posted by Mr_Green View Post
    so i get

    pi/3

    Is this correct? Is this the only point?
    y = -csc(x) = -\frac{1}{sin(x)}

    \frac{dy}{dx} = - \left ( - \frac{cos(x)}{sin^2(x)} \right ) = \frac{cos(x)}{sin^2(x)}

    So we need to solve
    \frac{cos(x)}{sin^2(x)} = \frac{2}{3}

    3cos(x) = 2sin^2(x) = 2(1 - cos^2(x))

    2cos^2(x) + 3cos(x) - 2 = 0

    Let z = cos(x). Then this equation becomes:
    2z^2 + 3z - 2 = 0

    This has solutions:
    z = -2 or z = \frac{1}{2}

    Thus
    cos(x) = -2 <-- Impossible for real x.
    or
    cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5 \pi}{3}

    So there are two answers.

    -Dan
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    Last edited by Mr_Green; October 5th 2007 at 10:52 AM.
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