0
$\begingroup$

For the curve given by: $r(t)= [\sin(t) - t\cos(t), \cos(t) + t\sin(t), 6t^2 + 2]$

Solve for the Unit Normal Vector N(t).

I was successfully able to solve the Unit Tangent Vector $T(t)$ as $r'(t)/|r'(t)|$. The solution is $$ T(t)=\frac{(t\sin(t), t\cos(t), 12(t))}{\sqrt{t^2sin(t)^2+t^2\cos(t)^2+144t^2}}. $$

Now to solve for the Normal Vector, it should be $N(t)=T'(t)/|T'(t)|$. It is my understanding that $T'(t)=r''(t)=(t\cos(t)+\sin(t), \cos(t)-t\sin(t), 12)$.

My final solution then is: $$ N(t) = \frac{(t\cos(t)+\sin(t), \cos(t)-t\sin(t), 12)}{\sqrt{(t\cos(t)+\sin(t))^2+(\cos(t)-t\sin(t))^2+144}} $$ But that is not correct. Any help?

EDIT:

After some suggestions i was able to factor out a $ t^2 $ from the radical on the bottom resulting in the following for T(t) (please check my math): $$ T(t) = \frac{(sin(t), cos(t), 12)}{\sqrt{145}}$$.

Using this to solve for N(t) I get:

$N(t)=T'(t)/|T'(t)|=\frac{(cos(t), -sin(t), 0)}{sqrt(145)}$

But this is still not correct? I feel like i am close but i am not seeing where the problem is. Any more help?

Thanks!

$\endgroup$
  • $\begingroup$ The unit normal is defined to be ${\bf N}(t) = {\bf T}'(t)/|{\bf T}'(t)|$. Your mistake is in thinking ${\bf T}'(t) = {\bf r}''(t)$. This usually is not the case. You'll need to differentiate ${\bf T}(t)$ (unfortunately since it's kind of nasty). :( $\endgroup$ – Bill Cook Jan 22 '14 at 1:18
  • $\begingroup$ Ah yes that is what i was assuming, i had 2 problems before in which that was the case. Apparently it was just a coincidence... $\endgroup$ – Riley Jan 22 '14 at 1:21
0
$\begingroup$

Hint: Your $T$ is just about correct, except for the typo 1144 which should be 144. Now note that $\cos^2(t)+\sin^2(t)=1$ and factor out a $t^2$ from the square root in the denominator. This will put a $t$ downstairs that will cancel and make $T(t)$ nice.

Finally, as noted by one of the comments, you should really take the derivative of $T(t)$, after it is simplified of course, to get $N=T'/|T'|$. You should find $|T'|$ constant.

$\endgroup$
  • $\begingroup$ So I factor out the t^2 from the bottom, and take it out of the radical as well resulting in (after simplification): (sin(t), cos(t), 12)/sqrt(145)? $\endgroup$ – Riley Jan 22 '14 at 1:33
  • $\begingroup$ That looks about right. $\endgroup$ – abnry Jan 22 '14 at 1:41
0
$\begingroup$

Attention this derivate this product: $t.sin(x)$ or $t.cos(t)$:

$r'(t) = \frac{d}{dt}[sin(t)-tcos(t), cos(t)+tsin(t), 6t^2+2] = [cos(t) - (cos(t) + t.(-sin(t))\,\,,\,\,-sin(t) + (sin(t) + tcos(t))\,\,,\,\,12t] = \\ \\ = [ tsin(t), tcos(t), 12t]$.

And

$|r'(t)| = \sqrt{t^2sin^2(t) + t^2cos^2(t) + 144t^2}$.

You can simplifier this vector $T(t) = \large\frac{(tsin(t),tcos(t),12t)}{|t|\sqrt{145}}$ and apply $N(t) = T'(t)/|T'(t)|$

EDIT: it is wrong (forget |t| below) ===> $N(t) = \large\frac{(tcos(t) + sin(t), cos(t)-tsin(t), 12)}{\sqrt{145}}$. It is exactly your answer in your question.

The function $|t|$ the point $t=0$ this function doesn't have derivate. But t>0 (|t|)' = 1 and t<0 (|t|)' = -1. You need to analyze these cases.

$\endgroup$
  • $\begingroup$ Yes i'm pretty sure thats what i already have, thank you. $\endgroup$ – Riley Jan 22 '14 at 1:32
  • 1
    $\begingroup$ Ok I see where you are going with that... Do the t's up on the top cancel as well? $\endgroup$ – Riley Jan 22 '14 at 1:54
  • $\begingroup$ @Riley I forgot this detail, sorry. You should consider $T(t) = \frac{(sin(t),cos(t),12)}{\sqrt{145}}$ if $t>0$ because $\sqrt{t^2sin^2t+t^2cos^2+144t^2} = |t|\sqrt{145}\neq t\sqrt{145}$. $\endgroup$ – miguel747 Jan 22 '14 at 13:20

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.