Answer
$\kappa \left( 1 \right) = \frac{1}{{2\sqrt 2 }}$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\ln t,t} \right)$. So, $\left( {x\left( t \right),y\left( t \right)} \right) = \left( {\ln t,t} \right)$.
$x'\left( t \right) = \frac{1}{t}$, ${\ \ \ }$ $y'\left( t \right) = 1$
$x{\rm{''}}\left( t \right) = - \frac{1}{{{t^2}}}$, ${\ \ \ }$ $y{\rm{''}}\left( t \right) = 0$
By Eq. (11) of Exercise 28 (Section 14.4), the curvature of a plane curve ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$ is given by
(11) ${\ \ \ \ }$ $\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
Thus,
$\kappa \left( t \right) = \frac{{\left| { - \frac{1}{{{t^2}}}} \right|}}{{{{\left( {\frac{1}{{{t^2}}} + 1} \right)}^{3/2}}}} = \frac{1}{{{t^2}}}\frac{{{t^3}}}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}} = \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}$
$\kappa \left( 1 \right) = \frac{1}{{2\sqrt 2 }}$
