Answer
$\dfrac{4t}{\sqrt{4t^2+10}},\dfrac{2\sqrt{10}}{\sqrt{4t^2+10}}$
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, $r'(t)= i+2t j+3k$ and $r''(t)=2j$
$|r'(t)|=\sqrt{(1)^2+(2t)^+3^2}=\sqrt{10+4t^2}$
$r'(t) \cdot r''(t)=[ i+2t j+3k] \cdot [2j]=4t$
and
$r'(t) \times r''(t)=[ i+2t j+3k] \times [2j]=-6i+2k$
From equation (1), we have
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4t}{\sqrt{10+4t^2}}$
From equation (2), we have
$a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{2\sqrt{10}}{\sqrt{10+4t^2}}$
