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.

  • $\begingroup$ 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. $\endgroup$ Dec 3 '10 at 2:00

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 $f''=N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.