# Given the vector r(t) = { cosT, sinT, ln (CosT) } and point (1, 0, 0) find vectors T, N and B at that point.Vector T is the unit tangent vector, so the derivative r(t) is needed.

Given the vector $$r(t) = { \cos T, \sin T, \ln ( \cos T) }$$ and point (1, 0, 0) find vectors T, N and B at that point.

Vector T is the unit tangent vector, so the derivative r(t) is needed.

Vector N is the normal unit vector, and the equation for it uses the derivative of T(t). 

The B vector is the binormal vector, which is a crossproduct of T and N.

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Ezra Herbert

The derivative of the vector is,
$$r'(t)=\{-\sin 5, \cos t, -\tan t \}$$
The unit tangent vector, T at the given point is obtained as follows.
$$T(t)=\frac{r'(t)}{||r'(t)||}$$
$$T(t)=\frac{\langle-\sin t,\cos t,-\tan t\rangle}{\sqrt{(-\sin t)^2+(\cos t)^2+(-\tan t }^2}$$
$$T(t)=\langle-\frac{\sin t}{\sec t},\frac{cos t}{sec t},-\frac{tan t}{\sec t}\rangle$$
$$T(t)|_{1,0,0}=\langle-0.45,1,0\rangle$$
The derivative of the unit tangent vector, T is,
$$\displaystyle{T}'{\left({t}\right)}={\left\langle-\frac{{ \cos{{t}}- \tan{{t}} \sin{{t}}}}{ \sec{{t}}},- \sin{{2}}{t},- \cos{{t}}\right\rangle}$$
The normal unit vector, N at the given point is obtained as follows.
$$\displaystyle{N}{\left({t}\right)}=\frac{{{T}'{\left({t}\right)}}}{{{\left|{\left|{T}'{\left({t}\right)}\right|}\right|}}}$$
$$\displaystyle{N}{\left({t}\right)}=\frac{{{\left\langle-\frac{{ \cos{{t}}- \tan{{t}} \sin{{t}}}}{ \sec{{t}}},- \sin{{2}}{t},- \cos{{t}}\right\rangle}}}{{\frac{{\sqrt{{{\left(- \cos{{t}}+ \tan{{t}} \sin{{t}}\right)}^{2}+{{\sin}^{2}{\left({2}{t}\right)}}{{\sec}^{2}{\left({t}\right)}}+{1}}}}}{ \sec{{\left({t}\right)}}}}}$$
$$\displaystyle{N}{\left({t}\right)}{|}_{{{1},{0},{0}}}={\left\langle{0.366},{0},-{0.70}\right\rangle}$$
The binormal vector, B at the given point is obtained as follows.
$$B(t)=T(t) xx N(t)$$
$$\displaystyle{B}{\left({t}\right)}{|}_{{{1},{0},{0}}}={\left[\begin{matrix}{i}&{j}&{k}\\-{0.45}&{1}&{0}\\{0.366}&{0}&-{0.70}\end{matrix}\right]}$$
$$=(-0.70)i-(0.315)j-(0.366)k$$