Answer
$\dfrac{4+18t^2}{\sqrt{4+9t^2}},\dfrac{6t^2}{\sqrt{4+9t^2}}$
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, $r'(t)= 2ti+3t^2 j$ and $r''(t)=2i+6tj$
$|r'(t)|=\sqrt{(2t)^2+(3t^2)}=t\sqrt{4+9t^2}$
$r'(t) \cdot r''(t)=[2ti+3t^2 j] \cdot [2i+6tj]=4+18t^2$
and
$r'(t) \times r''(t)=[2ti+3t^2 j] \times [2i+6tj]=6t^2 k$
From equation (1), we have
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4+18t^2}{\sqrt{4+9t^2}}$
From equation (2), we have
$a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{6t^2}{\sqrt{4+9t^2}}$
