Chapter 12 - Vector-Valued Functions - 12.5 Exercises - Page 861: 54

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

$$\eqalign{ & y = {e^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[ {{e^x}} \right] \cr & y'\left( x \right) = {e^x} \cr & y''\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}} \right] \cr & y''\left( x \right) = {e^x} \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| {{e^x}} \right|}}{{{{\left( {1 + {{\left( {{e^x}} \right)}^2}} \right)}^{3/2}}}} \cr & K = \frac{{{e^x}}}{{{{\left( {1 + {e^{2x}}} \right)}^{3/2}}}} \cr & \left( {\bf{a}} \right){\text{ Find the point where }}K{\text{ is maximum}} \cr & K' = \frac{{{{\left( {1 + {e^{2x}}} \right)}^{3/2}}{e^x} - {e^x}\left( {\frac{3}{2}} \right){{\left( {1 + {e^{2x}}} \right)}^{1/2}}\left( {2{e^{2x}}} \right)}}{{{{\left( {1 + {e^{2x}}} \right)}^3}}} \cr & K' = \frac{{\left( {1 + {e^{2x}}} \right){e^x} - {e^x}\left( {\frac{3}{2}} \right)\left( {2{e^{2x}}} \right)}}{{{{\left( {1 + {e^{2x}}} \right)}^{5/2}}}} \cr & K' = \frac{{{e^x} + 2{e^{3x}} - 3{e^{3x}}}}{{{{\left( {1 + {e^{2x}}} \right)}^{5/2}}}} \cr & K' = \frac{{{e^x} - {e^{3x}}}}{{{{\left( {1 + {e^{2x}}} \right)}^{5/2}}}} \cr & K' = 0 \cr & {e^x} - {e^{3x}} = 0 \cr & {e^x}\left( {1 - {e^{2x}}} \right) = 0 \cr & 1 - {e^{2x}} = 0 \cr & {e^{2x}} = 1 \cr & x = 0 \cr & K{\text{ is maximum when }}x = 0,{\text{ at the point }}\left( {0,1} \right) \cr & \cr & \left( {\bf{b}} \right){\text{ Find }}\mathop {\lim }\limits_{x \to \infty } K \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{{{{\left( {1 + {e^{2x}}} \right)}^{3/2}}}} = 0 \cr & \cr & {\text{Summary}} \cr & \left( {\bf{a}} \right)K{\text{ is maximum when }}x = 0,{\text{ at the point }}\left( {0,1} \right) \cr & \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to \infty } K = 0 \cr} $$