Answer
$4e^{2t},2e^{t}$
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, $r'(t)= i+2e^tj+2e^{2t}k$ and $r''(t)=2e^tj+4e^{2t}k$
This yields,$|r'(t)|=\sqrt{(1)^2+(2e^t)^2+(2e^{2t})^2}=1+2e^{2t}$
Thus, $r'(t) \cdot r''(t)=[i+2e^tj+2e^{2t}k] \cdot [2e^tj+4e^{2t}k]=4e^{2t}(1+2e^t)$
Also, $|r'(t) \times r''(t)|=[i+2e^tj+2e^{2t}k] \times [2e^tj+4e^{2t}k]=2e^{t}(1+2e^t)$
After simplifications, we get $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4e^{2t}(1+2e^t)}{1+2e^{2t}}$
or, $=4e^{2t}$
Now, $a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}$
or, $=\dfrac{2e^{t}(1+2e^t)}{1+2e^{2t}}$
or, $=2e^{t}$
