Answer
$r'(t)=\lt 1,2t,3t^2 \gt$,
$T(1)=\lt \dfrac{1}{\sqrt {14}},\dfrac{2}{\sqrt {14}},\dfrac{3}{\sqrt {14}} \gt$,
$r''(t)=\lt 0,2,6t \gt $,
and $r'(t) \times r''(t) =\lt 6t^2,-6t,2 \gt $
Work Step by Step
Given: $r(t)=\lt t,t^2,t^3 \gt $
In order to find $r'(t)$ we will have to take the derivative of $r(t)$ with respect to $t$.
Thus, $r'(t)=i+2tj+3t^2k$
As we know $T(t)=\dfrac{r'(t)}{|r'(t)|}$
Then, $T(1)=\dfrac{r'(1)}{|r'(1)|}$
$r'(1)=i+2(1)j+3(1)^2k=i+2j+3k$ and
$|r'(t)|=\sqrt {14}$
So, $T(1)=\dfrac{i+2j+3k}{\sqrt {14}}=\dfrac{1}{\sqrt {14}}i+\dfrac{2}{\sqrt {14}}j+\dfrac{3}{\sqrt {14}}k$
$r'(t)=i+2tj+3t^2k \implies r''(t)=2j+6tk$
$r'(t) \times r''(t) =i+2tj+3t^2k \times2j+6tk$
$=6t^2i-6tj+2k$
Hence,
$r'(t)=\lt 1,2t,3t^2 \gt$,
$T(1)=\lt \dfrac{1}{\sqrt {14}},\dfrac{2}{\sqrt {14}},\dfrac{3}{\sqrt {14}} \gt$,
$r''(t)=\lt 0,2,6t \gt $,
and $r'(t) \times r''(t) =\lt 6t^2,-6t,2 \gt $
