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I'm trying to solve the following question

Show that for any arc length parameterized curve there is a vector $ω(s)$ that satisfies

$$T'(s) = ω(s) × T (s)$$ $$N'(s) = ω(s) × N(s)$$ $$B'(s) = ω(s) × > B(s)$$

HINT: Consider $ω(s) = a(s)T (s) +b(s)N(s) +c(s)B(s)$ (where $T$, $N$, $B$ are the unit tangent, normal and binormal vectors) and find the coefficients $a$, $b$, $c$ that work.

I managed to get $$a(s) = T(s) \cdot ω(s)$$ $$b(s) = N(s) \cdot ω(s)$$ $$c(s) = B(s) \cdot ω(s)$$

But I don't know how to proceed from this. What direction should I be going in.

Vishwa Iyer
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1 Answers1

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Let me write $\omega(s) = t(s)T(s) + n(s)N(s) + b(s)B(s)$ where $t,n,b$ are scalar functions. Using the Frenet-Serret formulas, the first equation $T'(s) = \omega(s) \times T(s)$ translates into

$$ k(s)N(s) = n(s) (N(s) \times T(s)) + b(s)(B(s) \times T(s)) = -n(s)B(s) + b(s)N(s) $$

which implies that $b(s) = k(s)$ and $n(s) = 0$. The second equation $N'(s) = \omega(s) \times N(s)$ translates into

$$ -k(s)T(s) + \tau(s)B(s) = t(s) (T(s) \times N(s)) + b(s) (B(s) \times N(s)) = t(s)B(s) - b(s)T(s) $$

which implies that $t(s) = \tau(s)$ (and again, that $b(s) = k(s)$). Thus, $\omega$ should be

$$ \omega(s) = \tau(s)T(s) + k(s)B(s). $$

I leave it to you to check that the third equation is also satisfied.

levap
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