# Unit Tangent and Unit Normal Vectors — Calculus III Question

Consider the following vector function. $$r(t) = \left\langle 2t \cdot \sqrt{2}, e^{2t}, e^{-2t}\right\rangle$$

(a) Find the unit tangent and unit normal vectors $T(t)$ and $N(t)$.
$T(t) =$
$N(t) =$

(b) Use this formula to find the curvature.
$κ(t) =$

I am getting bogged down in the math. I know how to calculate the three things but I am having trouble getting the derivative of $T(t)$ after solving for it. I have gotten $T(t)$ to equal $$\frac{1}{2 e^{2t} + 2 e^{-2t}} \left\langle 2 \cdot \sqrt{2}, 2 e^{2t},-2 e^{-2t}\right\rangle$$.

Thank you!

• the length of $\langle 2\cdot\sqrt 2,2e^{2t},-2e^{-2t}\rangle$ is rather $\sqrt{(2\cdot\sqrt 2)^2+(2e^{2t})^2+(-2e^{-2t})^2}$. – Berci Feb 15 '13 at 1:59
• Why do you post the same question two times? math.stackexchange.com/questions/304320/… – zaarcis Feb 15 '13 at 2:00
• because i didn't get any help – user62336 Feb 15 '13 at 2:05
• @Berci The OP has it right; simplify what you have... – David Mitra Feb 15 '13 at 2:11
• You could simply things a bit. If you have $T'(t)=f(t){\bf F}(t)$, then ${\bf F}(t)$ gives the direction (just drop the $f(t)$ term); then ${\bf N}(t)={\bf F}(t)/ | {\bf F}(t)|$. – David Mitra Feb 15 '13 at 2:41

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.)

• I still cannot get an answer from this... help! – user62336 Feb 15 '13 at 4:04