Answer
$T(0)=\dfrac{2}{3}i-\dfrac{2}{3}j+\dfrac{1}{3}k$;
$r''(0)=4i+4j+4k$;
and
$ r'(t) \cdot r''(t) =12e^{4t}-8e^{-4t}+12te^{4t}+8t^2e^{4t}$
Work Step by Step
Given: $r(t)=\lt e^{2t},e^{-2t},te^{2t} \gt $
In order to find $r'(t)$ we will have to take the derivative of $r(t)$ with respect to $t$.
Thus, $r'(t)=2e^{2t}i-2e^{-2t}j+(e^{2t}+2te^{2t})k$
As we know $T(t)=\dfrac{r'(t)}{|r'(t)|}$
Then, $T(0)=\dfrac{r'(0)}{|r'(0)|}$
$r'(0)=2e^{2(0)}i-2e^{-2(0)}j+(e^{2(0)}+2(0)e^{2(0)})k=2i-2j+k$
and
$|r'(0)|=\sqrt {4+4+1}=3$
So, $T(0)=\dfrac{2i-2j+k}{3}=\dfrac{2}{3}i-\dfrac{2}{3}j+\dfrac{1}{3}k$
$r''(t)=4e^{2t}i+4e^{-2t}j+(2e^{2t}+2e^2t+4te^{2t})k$
$\implies r''(0)=4e^{2(0)}i+4e^{-2(0)}j+(2e^{2(0)}+2e^2(0)+4te^{2(0)})k=4i+4j+4k$
$ r'(t) \cdot r''(t) =(2e^{2t}i-2e^{-2t}j+(e^{2t}+2te^{2t})k) \cdot 4e^{2t}i+4e^{-2t}j+(2e^{2t}+2e^2t+4te^{2t})k)=12e^{4t}-8e^{-4t}+12te^{4t}+8t^2e^{4t}$
Hence,
$T(0)=\dfrac{2}{3}i-\dfrac{2}{3}j+\dfrac{1}{3}k$;
$r''(0)=4i+4j+4k$;
and
$ r'(t) \cdot r''(t) =12e^{4t}-8e^{-4t}+12te^{4t}+8t^2e^{4t}$
