Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.5 - Tangential and Normal Components of Acceleration - Exercises 13.5 - Page 771: 5

$a=0 \ T +2 \ N$

$v(t)=\dfrac{dr}{dt}= 2ti+(1+t^2) j+(1-t^2) k \implies |v(t)|=\sqrt {(2t)^2+(1+t^4+2t^2)^2+(1+t^4-2t^2)}=\sqrt {2t^4+4t^2+2}$ and $a(t)=\dfrac{d \ v(t)}{dt}= \dfrac{4t^3}{\sqrt {2t^4+4t^2+2}} $ $|a(0)|= \dfrac{4(0)^3}{\sqrt {2(0)^4+4(0)^2+2}} = 0$ Now, $a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {(2)^2 -(0)^2}=2 $ So, $a=a_T T+a_{N}=0 \ T +2 \ N$