Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.5 - Tangential and Normal Components of Acceleration - Exercises 13.5 - Page 771: 6

$a=2 \ T +\sqrt 2 \ N$

$v(t)=\dfrac{dr}{dt}= e^t (\cos t -\sin t)i+e^t(\sin t+ \cos t ) j+\sqrt 2 e^t k \implies |v(t)|=e^t \sqrt {(\cos t -\sin t)^2+(\sin t+\cos t)^2+2}= e^t \sqrt {1+1+2}=2e^t$ and $a(t)=\dfrac{d \ v(t)}{dt}= 2e^t$ $|a(0)|= 2e^{0}=2$ Now, $a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {(\sqrt 6)^2 -(2)^2}=\sqrt 2 $ So, $a=a_T T+a_{N}=2 \ T +\sqrt 2 \ N$