Answer
$0,1$
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 $r'(t)=-\sin t i+\cos t j+k$ and $r''(t)=-\cos ti-\sin tj$
This yields, $|r'(t)|=\sqrt{(-\sin t)^2+(\cos t)^+1^2}=\sqrt{2}$
and $r'(t) \cdot r''(t)=[1i+(2t-2)j] \cdot [2j]=\sin t \cos t-\sin t \cos t$
Also, $r'(t) \times r''(t)=[-\sin t i+\cos t j+k] \times [-\cos ti-\sin tj]=\sin t i-\cos tj+k$
Find the values.
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}$
or, $=\dfrac{\sin t \cos t-\sin t \cos t}{\sqrt{2}}$
or, $=0$
Now, $a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}$
or, $=\dfrac{\sin t i-\cos tj+k}{\sqrt{2}}$
or, $=1$
