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!

  • $\begingroup$ 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} $. $\endgroup$ – Berci Feb 15 '13 at 1:59
  • $\begingroup$ Why do you post the same question two times? math.stackexchange.com/questions/304320/… $\endgroup$ – zaarcis Feb 15 '13 at 2:00
  • $\begingroup$ because i didn't get any help $\endgroup$ – user62336 Feb 15 '13 at 2:05
  • $\begingroup$ @Berci The OP has it right; simplify what you have... $\endgroup$ – David Mitra Feb 15 '13 at 2:11
  • 1
    $\begingroup$ 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 ${\bf N}(t)={\bf F}(t)/ | {\bf F}(t)|$. $\endgroup$ – 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.)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.