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?


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?


  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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

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


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

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.

  • $\begingroup$ Yes i'm pretty sure thats what i already have, thank you. $\endgroup$ – Riley Jan 22 '14 at 1:32
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    $\begingroup$ 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

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