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December 20, 2014

December 20, 2014

Posted by **Anonymous** on Friday, September 16, 2011 at 6:29am.

- calculus -
**Damon**, Friday, September 16, 2011 at 6:40amThe slope of f(x) at any point on the circle will be perpendicular to the radius (tangent to circle perpendicular to radius at intersection).

so the slope we are looking for = -1/f')x)

f' = (1/2)(36-x^2)^-(1/2) (-2x)

= -x/sqrt(36-x^2)

the negative inverse of that is

(1/x)sqrt(36-x^2)

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