Unit Vectors Question

[mooshupork34]

If anyone could explain how the following is done, it would be greatly appreciated!:

Find the unit tangent vector T and unit normal vector N for the given curve at the indicated point: r(t)=<t, t^2, t^3> at (1,1,1)

r(t) = <t, t^2, t^3>

Now, I've found T by the following:
v(t) = <1, 2t, 3t^2>
v(1) = <1, 2, 3>
|v(1)| = sqrt(14)
T = <1/sqrt(14), 2/sqrt(14), 3/sqrt(14)>.

But I was wondering if anyone could explain how to find N.

 

[galactus]

\(\displaystyle \L\\r'(t)=i+2tj+3t^{2}k\)

\(\displaystyle \L\\||r'(t)||=\sqrt{1^{2}+(2t)^{2}+(3t^{2})^{2}}\)

\(\displaystyle \L\\r'(1)=i+2j+3k\)

\(\displaystyle \L\\||r'(1)||=\sqrt{14}\)

\(\displaystyle \L\\T(t)=\frac{1}{\sqrt{9t^{4}+4t^{2}+1}}i+\frac{2t}{\sqrt{9t^{4}+4t^{2}+1}}j+\frac{3t^{2}}{\sqrt{9t^{4}+4t^{2}+1}}k\)

\(\displaystyle \L\\T(1)=\frac{r'(1)}{||r'(1)||}=\frac{1}{\sqrt{14}}i+\frac{2}{\sqrt{14}}j+\frac{3}{\sqrt{14}}k\)

\(\displaystyle \L\\T'(t)=\frac{-2t(9t^{2}+2)}{(9t^{4}+4t^{2}+1)^{\frac{3}{2}}}i-\frac{2(9t^{4}-1)}{(9t^{4}+4t^{2}+1)^{\frac{3}{2}}}j+\frac{6t(2t^{2}+1)}{(9t^{4}+4t^{2}+1)^{\frac{3}{2}}}k\)

\(\displaystyle \L\\T'(1)=\frac{-11\sqrt{14}}{98}i-\frac{8\sqrt{14}}{98}j+\frac{9\sqrt{14}}{98}k\)

\(\displaystyle \L\\||T'(t)||=\frac{2\sqrt{9t^{4}+9t^{2}+1}}{9t^{4}+4t^{2}+1}\)

\(\displaystyle \L\\||T'(1)||=\frac{\sqrt{19}}{7}\)

\(\displaystyle \L\\N(t)=\frac{T'(t)}{||T'(t)||}=\)

Now, you find N(t), OK?.