Unit Tangent and Unit Normal Vectors -- Calculus III Question

Solution 1:

After using the product rule: $$ {\bf T}'(t)= {1\over 2e^{2t}+2e^{-2t}} \bigl< 0, 4e^{2t}, 4 e^{-2t} \bigr> -{4e^{2t}-4e^{-2t}\over (2e^{2t}+2e^{-2t})^2 } \bigl< 2\sqrt2, 2e^{2t}, -2 e^{-2t} \bigr> . $$

By definition, the direction of the unit normal vector is the direction of the vector ${\bf T'}$. To simplify things when finding the unit normal, you can multiply ${\bf T}'(t)$ by a positive scalar. This will give a vector in the same direction as that of $\bf N$; multiplying a vector by a positive number does not change its direction. (Said differently ${{\bf T}(t)\over |{\bf T}(t)|} = {|a|{\bf T}(t)\over |a{\bf T}(t)|} $ for any nonzero $a$.)

Once we have our direction vector, divide by its length to get ${\bf N}(t)$.

So, let's multiply ${\bf T}'(t)$ by $(2e^{2t}+2e^{-2t})^2/2$. This gives the vector $$ {\bf F}(t)={ (e^{2t}+e^{-2t})} \bigl< 0, 4e^{2t}, 4 e^{-2t} \bigr> -({2e^{2t}-2e^{-2t}}) \bigl< 2\sqrt2, 2e^{2t},- 2 e^{-2t} \bigr> $$ which is a bit easier to deal with.

After finding $|{\bf F}(t)|$, you can compute ${\bf N}(t) ={{\bf F}(t)\over |{\bf F}(t) | }$.

(Note that when finding the curvature, you need to find $|{\bf T}'(t)|$ proper.)