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Math Help - Find equation of the tangent line

  1. #1
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    Find equation of the tangent line

    Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. c(t)=cost, cost + sin^2(t) t = pi
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  2. #2
    Member Mentia's Avatar
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    Well right off there is a problem because the curve is at an endpoint when t = pi, so the slope isn't defined. However, I suppose we can take a limit as t approaches pi.

    What is the location of the point? We have:

    x = \cos (t)
    y = \cos (t) + \sin ^2(t)

    Then for t = \pi we have:
    x = -1
    y = -1

    So what is the slope of the curve at that point? We have

    c'(t) = - \sin (t), - \sin (t) + 2 \cos (t) \sin (t)

    So the slope of a line passing through that point would be:

    \frac{- \sin (t) + 2 \cos (t) \sin (t)}{- \sin (t)}

    For t = \pi we have:

    Slope =  \lim_{t \rightarrow \pi } \frac{- \sin (t) + 2 \cos (t) \sin (t)}{- \sin (t)} = 3


    Oooookay, so whats the equation of a line passing through (-1,-1) with slope 3?

    y = mx + b ---> -1 = -3 + b ----> b = 2

    Then our equation of the line is:

    y = 3x + 2
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