Answer
$s = \sqrt 2 \left( {{{\rm{e}}^b} - {{\rm{e}}^a}} \right)$
Work Step by Step
We have
$x\left( t \right) = {{\rm{e}}^t}\cos t$, ${\ \ \ }$ $x'\left( t \right) = {{\rm{e}}^t}\cos t - {{\rm{e}}^t}\sin t$,
$y\left( t \right) = {{\rm{e}}^t}\sin t$, ${\ \ \ }$ $y'\left( t \right) = {{\rm{e}}^t}\sin t + {{\rm{e}}^t}\cos t$.
Let $a \le t \le b$ be a definite interval. Using Eq. (3) of Theorem 1, the arc length is
$s = \mathop \smallint \limits_a^b \sqrt {{{\left( {{{\rm{e}}^t}\cos t - {{\rm{e}}^t}\sin t} \right)}^2} + {{\left( {{{\rm{e}}^t}\sin t + {{\rm{e}}^t}\cos t} \right)}^2}} {\rm{d}}t$
$ = \mathop \smallint \limits_a^b \sqrt {{{\rm{e}}^{2t}}{{\cos }^2}t - 2{{\rm{e}}^{2t}}\cos t\sin t + {{\rm{e}}^{2t}}{{\sin }^2}t + {{\rm{e}}^{2t}}{{\sin }^2}t + 2{{\rm{e}}^{2t}}\sin t\cos t + {{\rm{e}}^{2t}}{{\cos }^2}t} {\rm{d}}t$
$ = \mathop \smallint \limits_a^b \sqrt {2{{\rm{e}}^{2t}}} {\rm{d}}t = \sqrt 2 \mathop \smallint \limits_a^b {{\rm{e}}^t}{\rm{d}}t$
So,
$s = \sqrt 2 {{\rm{e}}^t}|_a^b = \sqrt 2 \left( {{{\rm{e}}^b} - {{\rm{e}}^a}} \right)$
