Answer
$0,1$
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, $r'(t)=-\sin t i+\cos t j+k$ and $r''(t)=-\cos ti-\sin tj$
$|r'(t)|=\sqrt{(-\sin t)^2+(\cos t)^+1^2}=\sqrt{2}$
$r'(t) \cdot r''(t)=[1i+(2t-2)j] \cdot [2j]=\sin t \cos t-\sin t \cos t$
and
$r'(t) \times r''(t)=[-\sin t i+\cos t j+k] \times [-\cos ti-\sin tj]=\sin t i-\cos tj+k$
From equation (1), we have
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{\sin t \cos t-\sin t \cos t}{\sqrt{2}}=0$
From equation (2), we have
$a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}=\dfrac{\sin t i-\cos tj+k}{\sqrt{2}}=1$
