Necessary condition to a curve lying on a sphere I'm reading by myself the classical Do Carmo's differential geometry book and I couldn't solve this question (page 25):

The hint on the back of the book says:




My attempt
$(|\alpha(s)|^2)'=0\implies (\langle \alpha(s),\alpha(s)\rangle)'=0\implies  \langle\alpha'(s),\alpha(s)\rangle=0$
Differentiating again the last equation above, we have: $\langle\alpha''(s),\alpha(s)\rangle=-1$, in other words $\langle k(s)n(s),\alpha(s)\rangle=-1$
Differentiating a third time, we have:
$\langle k'(s)n(s)+k(s)n'(s),\alpha(s)\rangle+\langle k(s)n(s),\alpha'(s)\rangle=0\implies \langle k'(s)n(s)+k(s)n'(s),\alpha(s)\rangle=0$
So my last result was 

$$\langle k'(s)n(s)+k(s)n'(s),\alpha(s)\rangle=0$$

I'm stuck there, any help would be very useful.
EDIT
Using the Frenet-serret formulas for the $n'$ in my last result above gave me: 

$$\langle b(s),\alpha(s)\rangle=\frac{1}{R'(s)\tau(s)}$$ 

However, I couldn't prove $\alpha(s)=-Rn+R'Tb$ using this fact. I'm stuck again.
 A: If you substitute the Frenet-Serret equation $\displaystyle n'(s)=-k(s)\alpha'(s)+\tau(s) b(s)$ into your third equation, and also substitute your second equation $n(s)\cdot\alpha(s)=-\frac{1}{k(s)},$ you're almost there. Finally just express $\alpha^2$ in terms of the Frenet-Serret frame.
Edit: Here it is step by step.


*

*$\alpha^2 = r^2$

*Take derivative: $\alpha\cdot\alpha' = 0.$

*Take derivative: $(\alpha')^2 + \alpha\cdot\alpha''=0.$

*Take the derivative: $3\alpha'\cdot\alpha'' + \alpha\cdot\alpha''' = 0.$

*Arc length is by definition the parametrization which makes $\alpha'$ the unit tangent vector, so $(\alpha')^2=1$ in step 3 or $\alpha'\cdot\alpha'' = 0 $ in step 4. Leaving $\alpha\cdot\alpha''' = 0.$

*$\alpha''=kn,$ so now step 3 can be written $\alpha\cdot n=-\frac{1}{k}.$

*Also $\alpha'''=k'n+kn'.$

*From steps 5 and 7, $k'\alpha\cdot n+k\alpha\cdot n'=0.$

*By the Frenet-Serret formula $n'=-k\alpha'+\tau b,$ so we now have $k'\alpha\cdot n+k\tau\alpha\cdot b=0.$

*By step 6, this is $-\frac{k'}{k}+k\tau(\alpha\cdot b)=0$ or $\alpha\cdot b=\frac{k'}{k^2\tau}.$

*Writing $\alpha$ in the Frenet-Serret frame, $\alpha = (\alpha\cdot\alpha')\alpha' + (\alpha\cdot n)n + (\alpha\cdot b)b,$ we see that $\alpha^2=r^2=(\alpha\cdot n)^2+(\alpha\cdot b)^2$ by the Pythagorean theorem.

*So $\alpha = -\frac{n}{k} + \frac{k'b}{k^2\tau}$ and $r^2=\left(-\frac{1}{k}\right)^2+\left(\frac{k'}{k^2\tau}\right)^2=\left(\frac{1}{k}\right)^2+\left(\left(\frac{1}{k}\right)'\right)^2\left(\frac{1}{\tau}\right)^2,$ which was to be proved.

