Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone tell me whether or not my work and answer below are correct? This is question 13.3.48 in Stewart Calculus 7th edition.

Here is the problem definition:
"Find the vectors $\vec T, \vec N, \vec B$ at the given point. $\vec r(t) = (\cos t)i +(\sin t)j + (ln \cos t)k$ , (1,0,0)"

Here is my work:
Note: $t=0 \text{ because } \cos 0 = 1, \sin 0 = 0,\text{ and }\ln[ \cos 0] = \ln(1)=0 $

$\vec r'(t) = (-\sin t)i + (\cos t)j + (-\tan t)k $

$|\vec r'(t)| = \sec t$

$\vec T(t)= (-\sin t \cos t)i + ((\cos t)^2)j + (-\sin t)k$

$\vec T'(t) = (1-2(\cos t)^2)i -(2 \sin t \cos t)j -(\cos t)k $

$|\vec T'(t)| = \sqrt{1+(\cos t)^2} $

$\vec N(t) = {\vec T'(t)\over|\vec T'(t)|} = {(1-2(\cos t)^2)\over \sqrt{1+(\cos t)^2}}i + ({(-2(\sin t)(\cos t))\over \sqrt{1+(\cos t)^2}})j + {(-\cos t)\over \sqrt{1+(\cos t)^2}}k $

$\vec B(t) = \vec T(t) \times \vec N(t) = \begin{vmatrix} i & j & k \\ -\sin t \cos t & \cos^2(t) & -\sin t \\ {1-2(\cos t)^2\over \sqrt{1+(\cos t)^2}} & {-2(\sin t)(\cos t)\over \sqrt{1+(\cos t)^2}} & {-\cos t\over \sqrt{1+(\cos t)^2}} \end{vmatrix}$

Thus, for t=0 and point (1,0,0) we have:

$\vec T(0) = j $
$\vec N(0) = ({-\sqrt2\over 2})i + ({-\sqrt2\over 2})k $
$\vec B(0) = ({-\sqrt2\over 2})i + ({\sqrt2\over 2})k $

share|cite|improve this question
You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. – joriki Sep 21 '12 at 22:08
Everything up to "abs(r'(t)) = sec t" seems OK (if by that you mean $|r'(t)|=\sec t$). But I don't see how you arrived at the expression for $T(t)$ in the next line. Shouldn't you just divide $r'(t)$ by $|r'(t)|$ do get $T(t)$? And $|T(t)|$ should be $1$, not $\sqrt{1+\cos^2t}$, no? – joriki Sep 21 '12 at 22:15
Strings like "sin" are interpreted as concatenations of variable names and are therefore italicized. To get the right font and spacing for function names like $\sin$, you can use the predefined commands like \sin, or if you need a function name like $\operatorname{erf}$ for which there's no predefined command, you can use \operatorname{erf}. To format square roots, use e.g. \sqrt{1+x^2} to get $\sqrt{1+x^2}$. – joriki Sep 21 '12 at 22:25
To include text like "omitting this because I don't know how to format it" in a formula, you can use \text{the text}. – joriki Sep 21 '12 at 22:41
@joriki, Thank you. I fixed the omission and cleaned up the formatting enough that it now accurately represents my work in a more easily readable manner. I may make more formatting changes later when I have time. Can you tell yet whether I got this correct? If not, where any math error might be? Thank you. – CodeMed Sep 21 '12 at 22:52
up vote 1 down vote accepted

I went through your work and I couldn't find any more errors.

share|cite|improve this answer
Thank you very much for all of your insights. Thank you in particular for helping me start to learn the syntax for proper formatting. +1 – CodeMed Sep 22 '12 at 3:12
@CodeMed: You're welcome! – joriki Sep 22 '12 at 3:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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