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Math Help - tangent line on curve is negative of expected

  1. #1
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    tangent line on curve is negative of expected

    Hi. I think I've reversed a sign somewhere, but for the life of me I can't seem to find it.

    I have these parametric equations (using t to represent theta):

    x = 2 - 3\,cos\,t
    y = 3 + 2\,sin\,t

    I've got to find the equation of the tangent line at various given points on the curve. So, I want to create a general equation first that I can plug the points into. First, I solve for t:

    y = 3 + 2\,sin\,t
    t = \arcsin{\frac{y-3}{2}}

    Then I get the slope:

    \frac{dy/dt}{dx/dt} = \frac{3\,sin\,t}{2\,cos\,t} = \frac{3}{2}\,tan\,t

    Now, the intercept in the tangent line equation will be

    B = y_{1} - (\frac{3}{2}\,tan\,(\arcsin{\frac{y_{1}-3}{2}}))\,x_{1}

    Giving me

    y = (\frac{3}{2}\,tan\,(\arcsin{\frac{y_{1}-3}{2}}))\,x + y_{1} - (\frac{3}{2}\,tan\,(\arcsin{\frac{y_{1}-3}{2}}))\,x_{1}

    or

    y = \frac{3}{2} (tan \circ arcsin) (\frac{y_{1}-3}{2}}) (x - x_{1}) + y_{1}

    Great. But if I plug in the first coordinate I'm given

    (x_{1}, y_{1}) = (\frac{4 + 3\sqrt{3}}{2}, 2)

    I eventually get

    \frac{-\sqrt{3}}{2}\,x + \sqrt{3} + \frac{17}{4}

    A tangent line with a negative slope! Yet when I look diagram provided in the book itself, the tangent line at this point clearly has a positive slope.
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  2. #2
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    Re: tangent line on curve is negative of expected

    Hey infraRed.

    Given your parametric model, you have:

    dy/dt = 2cos(t) and
    dx/dt = 3sin(t) which means

    dy/dx = dy/dt / dx/dt = 2/3 * cot(t)
    Thanks from infraRed
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  3. #3
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    Re: tangent line on curve is negative of expected

    Wow. I think I'm going to shoot myself now.
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