## Calculus: Early Transcendentals 8th Edition

$\dfrac{4+18t^2}{\sqrt{4+9t^2}},\dfrac{6t^2}{\sqrt{4+9t^2}}$
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 $a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{6t^2}{\sqrt{4+9t^2}}$