Curvature — Question 8

Question 8 – Solution

Strategy Differentiate twice and reduce the denominator using $1+\tan^2x=\sec^2x$.

Calculation $$y'=-\tan x,\qquad y''=-\sec^2x.$$ Therefore $$\kappa(x)=\frac{\sec^2x}{(1+\tan^2x)^{3/2}} =\frac{\sec^2x}{|\sec x|^3}.$$ On the stated interval, $\cos x>0$ and $\sec x>0$, so $$\boxed{\kappa(x)=\cos x}.$$

Behavior The maximum curvature is $1$ at $x=0$, while $\kappa\to 0$ as the graph approaches either vertical asymptote.