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
- 1292 views
- $4.00
Related Questions
- Convex subset
- Evaluate $\int ...\int_{R_n}dV_n(x_1^2 + x_2^2 + ... + x_n^2)$ , where $n$ and $R_n$ is defined in the body of this question.
- Find the absolute extrema of $f(x,y) = x^2 - xy + y^2$ on $|x| + |y| \leq 1$.
- Optimization of a multi-objective function
- Surface Parameterization
- Show that the line integral $ \oint_C y z d x + x z d y + x y d z$ is zero along any closed contour C .
- Compute $\iint_D \frac{dx dy}{\sqrt{1+x+2y}}$ on $D=[0,1]\times [0,1]$
- Conservative Vector Fields
The bounty is a bit low.