Answer
The curve has finite length for $0 \le t < \infty $.
Its length is $s = \sqrt 2 $
Work Step by Step
Since $c\left( t \right) = \left( {{{\rm{e}}^{ - t}}\cos t,{{\rm{e}}^{ - t}}\sin t} \right)$, we have
$x\left( t \right) = {{\rm{e}}^{ - t}}\cos t$, ${\ \ }$ $x'\left( t \right) = - {{\rm{e}}^{ - t}}\cos t - {{\rm{e}}^{ - t}}\sin t$,
$y\left( t \right) = {{\rm{e}}^{ - t}}\sin t$, ${\ \ }$ $y'\left( t \right) = - {{\rm{e}}^{ - t}}\sin t + {{\rm{e}}^{ - t}}\cos t$.
By Theorem 1 of Section 12.2, the length of the curve for $0 \le t < \infty $ is
$s = \mathop \smallint \limits_0^\infty \sqrt {{{\left( { - {{\rm{e}}^{ - t}}\cos t - {{\rm{e}}^{ - t}}\sin t} \right)}^2} + {{\left( { - {{\rm{e}}^{ - t}}\sin t + {{\rm{e}}^{ - t}}\cos t} \right)}^2}} {\rm{d}}t$
$s = \mathop \smallint \limits_0^\infty \sqrt {{{\rm{e}}^{ - 2t}}{{\left( { - \cos t - \sin t} \right)}^2} + {{\rm{e}}^{ - 2t}}{{\left( { - \sin t + \cos t} \right)}^2}} {\rm{d}}t$
$s = \mathop \smallint \limits_0^\infty {{\rm{e}}^{ - t}}\sqrt {{{\cos }^2}t + 2\cos t\sin t + {{\sin }^2}t + {{\sin }^2}t - 2\sin t\cos t + {{\cos }^2}t} {\rm{d}}t$
$s = \mathop \smallint \limits_0^\infty {{\rm{e}}^{ - t}}\sqrt 2 {\rm{d}}t$
$s = - \sqrt 2 {{\rm{e}}^{ - t}}|_0^\infty = - \sqrt 2 \left( {0 - 1} \right) = \sqrt 2 $
Hence, the curve has finite length for $0 \le t < \infty $.
