Answer
$\dfrac{4+18t^2}{\sqrt{4+9t^2}},\dfrac{6t^2}{\sqrt{4+9t^2}}$
Work Step by Step
Tangential acceleration component can be calculated as:
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}$
Normal acceleration component can be calculated as:
$a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}$
From the question, $r'(t)= 2ti+3t^2 j$ and $r''(t)=2i+6tj$
This yields, $|r'(t)|=\sqrt{(2t)^2+(3t^2)}=t\sqrt{4+9t^2}$
Also, $r'(t) \cdot r''(t)=[2ti+3t^2 j] \cdot [2i+6tj]=4+18t^2$
Also, $r'(t) \times r''(t)=[2ti+3t^2 j] \times [2i+6tj]=6t^2 k$
Plug all the calculated in their respective formulas.
Therefore,
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4+18t^2}{\sqrt{4+9t^2}}$
Plug all the calculated in their respective formulas.
$a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{6t^2}{\sqrt{4+9t^2}}$
