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Math Help - Chain rule with finding equation of line tangent

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    Chain rule with finding equation of line tangent

    Find the equation of the line tangent to the curve at the point defined by the given value x.

    x= sin (2pi* t), y= cos (2pi*t),  t= -1/6

    I have no idea how to do this beyond the fact the it needs the chain rule and at some time (probably the end) I have to plug in t as -1/4
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  2. #2
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    Quote Originally Posted by Truthbetold View Post
    Find the equation of the line tangent to the curve at the point defined by the given value x.

    x= sin (2pi* t), y= cos (2pi*t),  t= -1/6

    I have no idea how to do this beyond the fact the it needs the chain rule and at some time (probably the end) I have to plug in t as -1/4
    Hello,

    substitute t by -\frac16 to get the coordinates of the tangent point: T\left(-\frac{\sqrt{3}}{2} , \frac12  \right)

    The slope of the tangent can be calculated by:

    m =\frac{\frac{d(y)}{dt}}{\frac{d(x)}{dt}} . With your problem you get:

    m = \frac{- 2 \pi \cdot \sin(2 \pi \cdot t)}{2\pi \cdot \cos(2 \pi \cdot t)} . Plug in t=-\frac16 and you have m = \sqrt{3}
    Now us point-slope-formula to calculate the equation of the tangent. I've got:

    y = \sqrt{3} \cdot x +2

    I've attached a sketch of the graph and the tangent.
    Attached Thumbnails Attached Thumbnails Chain rule with finding equation of line tangent-krstang_parametr.gif  
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