# Thread: Find vectors T, N, and B at the given point

1. ## Find vectors T, N, and B at the given point

Find the vectors T, N, and B at the given point.

r(t) = {cos(t), sin(t), ln(cos(t))}, (1,0,0)
I got T(t)= r'(t)/|r'(t)|= (-sint, cost, -sint/cost)/square root(1+tan^2t) = (-sint,cost,-tant)=(0,1,0) because t=0.

then for N(t)= T'(t)/|T'(t)|=(-cost,-sint,-sec^2)/ sqaure root(1+sec^4t)=(-cost/sqaure root 2,-sint/sqaure root 2, sec^2t/sqaure root 2)

I am just wondering if my math is correct or if I went wrong somewhere.

B(t)= T(t)xN(t)= -1/sqaure root2[(sect+sinttant)i+(sint(sec^2t + 1)j -k]
=(-1/sqaure root2,0,1/sqaure root 2)

Did I solve this correctly?

2. Looks like you found the correct answer, only

$\mathbf{N}=\frac{\mathbf{T'}}{|\mathbf{T'}|}=\frac {1}{\sqrt{1+\sec^4 t}}\left<-\cos t,-\sin t,-\sec^2 t\right>.$

Therefore,

$\mathbf{B}=\mathbf{T}\times\mathbf{N}=\frac{|\cos t|}{\sqrt{1+\sec^4 t}}
\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-\sin t & \cos t & -\tan t \\
-\cos t & -\sin t & -\sec^2 t
\end{array}\right|.$

3. if t=0 wouldn't 1/sqaure root(1+sec^4t)= 1/sqaure root 2?

And where did you get |cost| from?

4. Originally Posted by andreaspohle
if t=0 wouldn't 1/sqaure root(1+sec^4t)= 1/sqaure root 2?
Correct.

And where did you get |cost| from?
It is the value of $|\mathbf{r'}(t)|^{-1}$ as shown by

$\frac{1}{\sqrt{1+\tan^2 t}}=\frac{1}{\sqrt{\sec^2 t}}=\frac{1}{|\sec t|}=|\cos t|.$