Answer
$$L = 2\sqrt 2 \pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {\cos t + \sin t,\cos t - \sin t} \right\rangle ,\,\,\,\,\,{\text{for }}0 \leqslant t \leqslant 2\pi \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {\cos t + \sin t,\cos t - \sin t} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {\cos t - \sin t, - \sin t - \cos t} \right\rangle \cr
& {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& \,{\text{for }}0 \leqslant t \leqslant 2\pi \to a = 0{\text{ and }}b = 2\pi .{\text{ then}} \cr
& L = \int_0^{2\pi } {\left| {\left\langle {\cos t - \sin t, - \sin t - \cos t} \right\rangle } \right|} dt \cr
& L = \int_0^{2\pi } {\sqrt {{{\left( {\cos t - \sin t} \right)}^2} + {{\left( { - \sin t - \cos t} \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^{2\pi } {\sqrt {{{\left( {\cos t - \sin t} \right)}^2} + {{\left( {\sin t + \cos t} \right)}^2}} } dt \cr
& {\text{expanding}} \cr
& L = \int_0^{2\pi } {\sqrt {{{\cos }^2}t - 2\cos t\sin t + {{\sin }^2}t + {{\sin }^2}t + 2\cos t\sin t + {{\cos }^2}t} } dt \cr
& L = \int_0^{2\pi } {\sqrt {{{\cos }^2}t + {{\sin }^2}t + {{\sin }^2}t + {{\cos }^2}t} } dt \cr
& L = \int_0^{2\pi } {\sqrt {2{{\cos }^2}t + 2{{\sin }^2}t} } dt \cr
& L = \int_0^{2\pi } {\sqrt {2\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} } dt \cr
& L = \int_0^{2\pi } {\sqrt 2 } dt \cr
& {\text{integrating}} \cr
& L = \sqrt 2 \left( t \right)_0^{2\pi } \cr
& L = \sqrt 2 \left( {2\pi - 0} \right) \cr
& L = 2\sqrt 2 \pi \cr} $$
