Chapter 12 - Vector-Valued Functions - 12.5 Exercises - Page 861: 53

$$\eqalign{ & \left( {\bf{a}} \right)K{\text{ is maximum when }}x = \frac{1}{{\sqrt 2 }} \cr & \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to \infty } K = 0 \cr} $$

$$\eqalign{ & y = \ln x \cr & {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr & {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr & y'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x} \right] \cr & y'\left( x \right) = \frac{1}{x} \cr & y''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr & y''\left( x \right) = - \frac{1}{{{x^2}}} \cr & \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr & K = \frac{{\left| { - \frac{1}{{{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{1}{x}} \right)}^2}} \right)}^{3/2}}}} \cr & K = \left| {\frac{{ - 1/{x^2}}}{{{x^3}{{\left( {{x^2} + 1} \right)}^{3/2}}}}} \right| \cr & K = \frac{x}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr & \left( {\bf{a}} \right){\text{ Find the point where }}K{\text{ is maximum}} \cr & {\text{Differentiate using a CAS}} \cr & K' = \frac{{ - 2{x^2} + 1}}{{{{\left( {{x^2} + 1} \right)}^{5/2}}}} \cr & K' = 0 \cr & - 2{x^2} + 1 \cr & x = \pm \frac{1}{{\sqrt 2 }} \cr & \cr & - \frac{1}{{\sqrt 2 }}{\text{ It is not in the domain, then}}{\text{.}} \cr & K{\text{ is maximum when }}x = \frac{1}{{\sqrt 2 }} \cr & \cr & \left( {\bf{b}} \right){\text{ Find }}\mathop {\lim }\limits_{x \to \infty } K \cr & \mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} = 0 \cr & \cr & {\text{Summary}} \cr & \left( {\bf{a}} \right)K{\text{ is maximum when }}x = \frac{1}{{\sqrt 2 }} \cr & \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to \infty } K = 0 \cr} $$