$Finding Binormal vector from the derivative of the Normal and Tangent.$

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$Finding Binormal vector from the derivative of the Normal and Tangent.$

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 $B(t)$?

Multivariable Calculus

Miguel Gomes

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Daniel90

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The bounty is a bit low.

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In general, there is no way to determine the binormal vector at all times without some additional assumptions on the curve. Does the curve have constant curvature and torsion?

If you just want to find $B(0)$:

If so, one may use the Frenet-Serret formulae. Since $T' = \kappa N$, and $N$ is of unit norm, then $\kappa = 1/\sqrt{5}$ and $N = -(1,-1,0)/\sqrt{2}$.

Taking the dot product of $T$ and $N'$ and using the second formula yields $T \cdot N' = -\kappa$. Since $T$ is perpendicular to $T'$ and unit norm, we have three constraints which yield $T(0) = (0, 0, -1)$.
Then $B(0) = (N'(0) + \kappa T(0))/\tau = 0$.

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$Finding Binormal vector from the derivative of the Normal and Tangent.$

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