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Assume $f:(a,b) \to \mathbb R^3$ is an admissible unit speed curve (hence $f^{\prime} \times f^{\prime\prime}$ is never zero)

If $f$ lies on the sphere with center $a$ and radius $r$ prove that

$f = a - (1/\kappa) \mathbf N - (1/\kappa)^{\prime} (1/\tau) \mathbf B$

With $|f^{\prime}| = 1$,

$\mathbf T = f^{\prime}$

$\mathbf N = f^{\prime\prime}$

$\mathbf B = \mathbf T \times \mathbf N$

$\mathbf T^{\prime} = \kappa \mathbf N$

$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$

$\mathbf B^{\prime} = -\tau\mathbf N$

I've been using the hint that since $f$ lies on the sphere then

$(f-a) \cdot (f-a) = r^2$

And trying to differentiate it (three times) to get what I want but I'm not getting very far. I start with

$(f-a) \cdot (f-a) = r^2$


$2\mathbf T \cdot (f-a) = 0$


$2 + 2\kappa\mathbf N \cdot (f-a) = 0$


$\mathbf N \cdot \mathbf T + (-\kappa\mathbf T + \tau\mathbf B + 2(\kappa^{\prime}/\kappa)\mathbf N) \cdot (f-a) = 0$

Then I'm not sure where to go.

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The $\TeX$ goes wonky for some reason when you stick them too close to paragraphs... if you can't put them within a sentence, separate them by newlines. – J. M. Dec 3 '10 at 2:00
up vote 4 down vote accepted

For each $t\in(a,b)$, $\{T(t),N(t),B(t)\}$ forms an orthonormal basis for $\mathbb{R}^3$, so you can express $f(t)-a$ in terms of this basis as $f(t)-a=c_1(t)T(t)+c_2(t)N(t)+c_3(t)B(t)$, and the coefficients are obtained by taking the dot products with the corresponding vectors. Thus the result of differentiating once shows that $c_1(t)\equiv 0$. Solving for $(f-a)\cdot N$ after differentiating twice shows that $c_2(t)=-1/\kappa(t)$. After differentiating three times, notice that $N$ and $T$ are orthogonal, and use what you already know from the previous steps to solve for $(f-a)\cdot B$ and obtain $c_3(t)$. You will want to use the fact that $(1/\kappa)'=-\kappa'/\kappa^2$ to simplify the final coefficient to the given form.

Also note that it is not correct that $f''=N$.

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