Finding Binormal vector from the derivative of the Normal and Tangent.
Knowing that C is a regular, smooth curve parameterized by arc length.
It's $T'(0) = \frac{-1}{\sqrt10}(1,-1,0)$ and it's $N'(0) = \frac{-1}{\sqrt5}(0,0,-1)$ find the formula for $B(t)$ for all t. I have absolutely no idea what to do, I know that doing the cross product of $T'(0)$ with $N'(0)$ should give me $B'(0)$ but how do I go from there to $B(t)$?
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 1211 views
- $4.00
Related Questions
- Compute $\iint_D \frac{dx dy}{\sqrt{1+x+2y}}$ on $D=[0,1]\times [0,1]$
- Find the coordinates of the point $(1,1,1)$ in Spherical coordinates
- Finding the arc length of a path between two points
- Line Integral
- Show that $\int_\Omega \Delta f g = \int_\Omega f \Delta g$ for appropriate boundary conditions on $f$ or $g$
- Vector fields, integrals, and Green's Theorem
- Does $\sum_{n=2}^{\infty}\frac{\sin n}{n \ln n}$ converge or diverge?
- Prove that $f$ is a diffeomorphism $C^∞$, that maps... (More inside)
The bounty is a bit low.