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Let $\gamma : I \rightarrow R^3 $ be a unit speed curve such that $\kappa >0$ and $\tau \neq 0 $. Suppose that trace of $\gamma$ is on the sphere with center equal to $\rho_0$ and radius $r>0$.

a) Show that $\gamma - \rho_0 = -\rho N - \rho_0' \sigma B $, where $\rho= 1/ \kappa$ and $\sigma = 1 / \tau $

Can anyone give me a hint?

Alejandra Ramirez
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    First of all, do we agree that the curve $\gamma$ is traced on the sphere $S(\rho_0,r)$ ? Then I don't understand the meaning you attach to the product $\rho_0 \sigma B$ (a point times a real times times the binormal vector)... Finally, could you tell at which page of Do Carmo's book you have this formula ? – Jean Marie Oct 06 '16 at 05:22
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    Yes, we are agree. And the product was wrong. And can't tell you, so I have to edit that – Alejandra Ramirez Oct 06 '16 at 10:50

1 Answers1

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HINTS: First of all, note that you're told that $\gamma$ is arclength parametrized (speed $1$). This means that you can use the Frenet formulas without any adjustments. Next, you can write $\gamma(s)-\rho_0 = \lambda(s)T(s)+\mu(s)N(s)+\nu(s)B(s)$ for some smooth functions $\lambda,\mu,\nu$. What does $\|\gamma-\rho_0\|=\text{constant}$ tell you? Last, as always in this game, differentiate, use the Frenet formulas, and use the fact that $T(s),N(s),B(s)$ is a basis for $\Bbb R^3$.

Ted Shifrin
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