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September 30, 2014

September 30, 2014

Posted by **Laurie** on Thursday, September 22, 2011 at 9:34pm.

- calculus -
**Steve**, Friday, September 23, 2011 at 3:29pmNow, the slope of the tangent line at any point (x,y) on the curve is just y'(x) = cosx

At x=pi, y' = -1 and y=3

So, we want the line through (pi,3) with slope -1.

y-3 = -(x-pi) = pi-x

y = -x + 3+pi

The slope of the normal line is -(1/-1) = 1

y-3 = (x-pi)

y = x + 3-pi

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