Curvature — Question 7

Question 7 – Solution

Strategy Use the graph formula $\kappa=|y''|/(1+y'^2)^{3/2}$ and a hyperbolic identity.

Calculation Since $y'=\sinh x$, $y''=\cosh x$, and $1+\sinh^2x=\cosh^2x$, $$\boxed{\kappa(x)=\frac{\cosh x}{\cosh^3x}=\operatorname{sech}^2x}, \qquad \boxed{\rho(x)=\cosh^2x}.$$

Analysis Curvature is largest at $x=0$, where $\kappa=1$. As $|x|\to\infty$, $\kappa\to 0$ and $\rho\to\infty$, so the arms become locally less sharply bent.