Wednesday

April 16, 2014

April 16, 2014

Posted by **sh** on Wednesday, March 24, 2010 at 12:24am.

y=(1/cosx)-2cosx at((pi/3),1)

y'=0+2sinx

slope = √3

y=mx+b

1=(√3)(pi/3)+b

b=(3-√3pi)/3

3√3x-3y+3-√3pi=0

What did I do wrong? Thanks in advance.

- Calculus -
**sh**, Wednesday, March 24, 2010 at 1:20amGot it.

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