Chain rule with finding equation of line tangent

**Truthbetold** · November 20th 2007, 07:21 PM

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

**earboth** · November 20th 2007, 08:24 PM

Originally Posted by Truthbetold

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.

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