Answer
The first positive value of ${t_0}$ is ${t_0} = \frac{{3\pi }}{4}$.
The speed at ${t_0} = \frac{{3\pi }}{4}$ is
$\frac{{ds}}{{dt}}{|_{t = 3\pi /4}} = {{\rm{e}}^{ - 3\pi /4}}\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
$x\left( t \right) = {{\rm{e}}^{ - t}}\cos t$, ${\ \ }$ $x'\left( t \right) = - {{\rm{e}}^{ - t}}\cos t - {{\rm{e}}^{ - t}}\sin t$,
$y\left( t \right) = {{\rm{e}}^{ - t}}\sin t$, ${\ \ }$ $y'\left( t \right) = - {{\rm{e}}^{ - t}}\sin t + {{\rm{e}}^{ - t}}\cos t$.
Using Eq. (8) of Section 12.1, the slope of the tangent line is
$\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{ - {{\rm{e}}^{ - t}}\sin t + {{\rm{e}}^{ - t}}\cos t}}{{ - {{\rm{e}}^{ - t}}\cos t - {{\rm{e}}^{ - t}}\sin t}} = \frac{{ - \sin t + \cos t}}{{ - \cos t - \sin t}}$
$\frac{{dy}}{{dx}} = \frac{{\sin t - \cos t}}{{\sin t + \cos t}}$
The tangent line is vertical if $\frac{{dy}}{{dx}}$ is infinite. This occurs when $\sin t + \cos t = 0$. So, $\tan t = - 1$.
The solutions are $t = - \frac{\pi }{4} + \pi n$, for $n = 0, \pm 1, \pm 2, \pm 3,...$
The first positive value of ${t_0}$ is ${t_0} = \frac{{3\pi }}{4}$.
By Theorem 2 of Section 12.2, the speed along $c\left(t\right)$ is
$\frac{{ds}}{{dt}} = \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}} $
$\frac{{ds}}{{dt}} = \sqrt {{{\rm{e}}^{ - 2t}}{{\left( { - \cos t - \sin t} \right)}^2} + {{\rm{e}}^{ - 2t}}{{\left( { - \sin t + \cos t} \right)}^2}} $
$\frac{{ds}}{{dt}} = {{\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} $
$\frac{{ds}}{{dt}} = {{\rm{e}}^{ - t}}\sqrt 2 $
The speed of the curve at ${t_0} = \frac{{3\pi }}{4}$ is
$\frac{{ds}}{{dt}}{|_{t = 3\pi /4}} = {{\rm{e}}^{ - 3\pi /4}}\sqrt 2 $
