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