Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Exercises - Page 735: 33

We obtain $s\left( t \right) = \sqrt {17} {{\rm{e}}^t}$ and $\kappa \left( t \right) = \frac{4}{{s\left( t \right)}}$. We show that $R\left( t \right) = \frac{1}{4}s\left( t \right)$, where $R\left( t \right)$ is the radius of curvature. Hence, the radius of curvature is proportional to $s\left( t \right)$.

Part 1. The arc length function We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t,{{\rm{e}}^t}\sin 4t} \right)$. The derivative is ${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}\cos 4t - 4{{\rm{e}}^t}\sin 4t,{{\rm{e}}^t}\sin 4t + 4{{\rm{e}}^t}\cos 4t} \right)$ ${\bf{r}}'\left( t \right) = {{\rm{e}}^t}\left( {\cos 4t - 4\sin 4t,\sin 4t + 4\cos 4t} \right)$ So, $||{\bf{r}}'\left( t \right)|{|^2} = {\bf{r}}'\left( t \right)\cdot{\bf{r}}'\left( t \right)$ $||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left( {{{\left( {\cos 4t - 4\sin 4t} \right)}^2} + {{\left( {\sin 4t + 4\cos 4t} \right)}^2}} \right)$ $||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}({\cos ^2}4t - 8\cos 4t\sin 4t + 16{\sin ^2}4t$ $ + {\sin ^2}4t + 8\sin 4t\cos 4t + 16{\cos ^2}4t)$ $||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left( {1 + 16} \right) = 17{{\rm{e}}^{2t}}$ $||{\bf{r}}'\left( t \right)|| = \sqrt {17} {{\rm{e}}^t}$ Evaluate $s\left( t \right) = \mathop \smallint \limits_{ - \infty }^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$ $s\left( t \right) = \sqrt {17} \mathop \smallint \limits_{ - \infty }^t {{\rm{e}}^u}{\rm{d}}u = \sqrt {17} {{\rm{e}}^u}|_{ - \infty }^t$ $s\left( t \right) = \sqrt {17} {{\rm{e}}^t}$ Part 2. The curvature Now, we have $x\left( t \right) = {{\rm{e}}^t}\cos 4t$ and $y\left( t \right) = {{\rm{e}}^t}\sin 4t$. So, the derivatives are $x'\left( t \right) = {{\rm{e}}^t}\cos 4t - 4{{\rm{e}}^t}\sin 4t$ ${\ \ \ \ \ }$ $ = {{\rm{e}}^t}\left( {\cos 4t - 4\sin 4t} \right)$ $x{\rm{''}}\left( t \right) = {{\rm{e}}^t}\cos 4t - 4{{\rm{e}}^t}\sin 4t - 4{{\rm{e}}^t}\sin 4t - 16{{\rm{e}}^t}\cos 4t$ ${\ \ \ \ \ }$ $ = - 15{{\rm{e}}^t}\cos 4t - 8{{\rm{e}}^t}\sin 4t$ ${\ \ \ \ \ }$ $ = - {{\rm{e}}^t}\left( {15\cos 4t + 8\sin 4t} \right)$ $y'\left( t \right) = {{\rm{e}}^t}\sin 4t + 4{{\rm{e}}^t}\cos 4t$ ${\ \ \ \ \ }$ $ = {{\rm{e}}^t}\left( {\sin 4t + 4\cos 4t} \right)$ $y{\rm{''}}\left( t \right) = {{\rm{e}}^t}\sin 4t + 4{{\rm{e}}^t}\cos 4t + 4{{\rm{e}}^t}\cos 4t - 16{{\rm{e}}^t}\sin 4t$ ${\ \ \ \ \ }$ $ = - 15{{\rm{e}}^t}\sin 4t + 8{{\rm{e}}^t}\cos 4t$ ${\ \ \ \ \ }$ $ = - {{\rm{e}}^t}\left( {15\sin 4t - 8\cos 4t} \right)$ 1. Evaluate $x'\left( t \right)y{\rm{''}}\left( t \right)$ $x'\left( t \right)y{\rm{''}}\left( t \right) = - {{\rm{e}}^{2t}}\left( {\cos 4t - 4\sin 4t} \right)\left( {15\sin 4t - 8\cos 4t} \right)$ $ = - {{\rm{e}}^{2t}}\left( {15\cos 4t\sin 4t - 60{{\sin }^2}4t - 8{{\cos }^2}4t + 32\sin 4t\cos 4t} \right)$ $ = - {{\rm{e}}^{2t}}\left( {47\cos 4t\sin 4t - 60{{\sin }^2}4t - 8{{\cos }^2}4t} \right)$ 2. Evaluate $x{\rm{''}}\left( t \right)y'\left( t \right)$ $x{\rm{''}}\left( t \right)y'\left( t \right) = - {{\rm{e}}^{2t}}\left( {15\cos 4t + 8\sin 4t} \right)\left( {\sin 4t + 4\cos 4t} \right)$ $ = - {{\rm{e}}^{2t}}\left( {15\cos 4t\sin 4t + 8{{\sin }^2}4t + 60{{\cos }^2}4t + 32\sin 4t\cos 4t} \right)$ $ = - {{\rm{e}}^{2t}}\left( {47\cos 4t\sin 4t + 8{{\sin }^2}4t + 60{{\cos }^2}4t} \right)$ 3. Evaluate $x'{\left( t \right)^2} + y'{\left( t \right)^2}$ $x'{\left( t \right)^2} + y'{\left( t \right)^2} = {{\rm{e}}^{2t}}{\left( {\cos 4t - 4\sin 4t} \right)^2}$ $ + {{\rm{e}}^{2t}}{\left( {\sin 4t + 4\cos 4t} \right)^2}$ $x'{\left( t \right)^2} + y'{\left( t \right)^2} = {{\rm{e}}^{2t}}\left( {{{\cos }^2}4t - 8\cos 4t\sin 4t + 16{{\sin }^2}4t} \right)$ $ + {{\rm{e}}^{2t}}\left( {{{\sin }^2}4t + 8\sin 4t\cos 4t + 16{{\cos }^2}4t} \right)$ $x'{\left( t \right)^2} + y'{\left( t \right)^2} = {{\rm{e}}^{2t}}\left( {1 + 16} \right) = 17{{\rm{e}}^{2t}}$ Substituting the corresponding parts in Eq. (11) of Exercise 28: $\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$ gives $\kappa \left( t \right) = \frac{{\left| { - {{\rm{e}}^{2t}}\left( { - 60 - 8} \right)} \right|}}{{{{\left( {17{{\rm{e}}^{2t}}} \right)}^{3/2}}}} = \frac{{68}}{{{{17}^{3/2}}{{\rm{e}}^t}}} = \frac{4}{{\sqrt {17} {{\rm{e}}^t}}}$ Since $s\left( t \right) = \sqrt {17} {{\rm{e}}^t}$, so $\kappa \left( t \right) = \frac{4}{{s\left( t \right)}}$. By definition, the radius of curvature $R\left( t \right)$ is given by $R\left( t \right) = \frac{1}{{\kappa \left( t \right)}}$. So, $R\left( t \right) = \frac{1}{4}s\left( t \right)$. Hence, the radius of curvature is proportional to $s\left( t \right)$.