Chapter 12 - Section 12.5 - Tangential and Normal Components of Acceleration - Exercises - Page 669: 5

$a= 0 \ T +2 \ N$

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