# help with maximum curvature

• November 28th 2008, 07:57 PM
khuezy
help with maximum curvature
What is the maximal curvature of the curve y = ln cos x?
• November 28th 2008, 08:43 PM
whipflip15
Find an expression for the curvature using
Then find the maximum.
• December 1st 2008, 06:58 PM
khuezy
I'm still having
problems.
y' = -sinx/cosx or -tanx
y'' = -secx^2
• December 2nd 2008, 05:56 PM
whipflip15
Put those in to get

$\kappa=\frac{\sec^2 x}{(1+\tan^2 x)^{3/2}}$

Use $1+\tan^2 x=\sec^2 x$

and

$\kappa=\frac{\sec^2 x}{\sec^3 x}$.

So

$\kappa=\frac{1}{\sec x }=\cos x$.