Answer
$6t,6$
Work Step by Step
Formula to calculate the tangential acceleration component is:
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}$ ....(1)
and
Formula to calculate the normal acceleration component is:
$a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}$ ....(2)
Thus, $r'(t)=(3-3t^2)i+6tj$ and $r''(t)=-6ti+6j$
$|r'(t)|=\sqrt{(3-3t^2)^2+(6t)^2}=3(t^2+1)$
$r'(t) \cdot r''(t)=[(3-3t^2)i+6tj] \cdot [-6ti+6j]=18t(1+t^2)$
and
$r'(t) \times r''(t)=18(1+t^2)$
From equation (1), we have
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{18t(1+t^2)}{3(t^2+1)}=6t$
From equation (2), we have
$a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}=\dfrac{18(1+t^2)}{3(t^2+1)}=6$
