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\begin{document}

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\begin{pro}{(\S 2.3: 10)}{}
{\red{Spherical curves}}. Let $\alpha$ be a unit-speed curve with $\kappa>0, \tau\neq 0$.
\begin{enumerate}[label={\rm (\alph*)}]
\item If $\alpha$ lies on a sphere of center $\cc$ and radius $r$, show that
\[
\alpha-\cc=-\rho\, N-\rho'\sigma B,
\]
where $\rho=1/\kappa$ and $\sigma=1/\tau$. Thus $r^2=\rho^2+(\rho'\sigma)^2$.
\item Conversely, if $r^2=\rho^2+(\rho'\sigma)^2$ has constant value $r^2$ and $\rho'\neq 0$, show that
$\alpha$ lies on a sphere of radius $r$.
\end{enumerate}


\end{pro}

\begin{proof} Assume that $\alpha$ is on the sphere of radius $r$. Then we have $\|\alpha-\cc\|^2=r^2$. Taking the derivative, we get
\[
(\alpha-\cc)\cdot\alpha'=(\alpha-\cc)\cdot T=0.
\]
Thus we can write $\alpha-\cc$ as a linear combination of $N$ and $B$:
\[
\alpha-\cc=aN+bB,
\]
where $a,b$ are functions of $t$. Taking the derivative again and using the Frenet formulas, we get
\[
T=\alpha'=a'N+aN'+b'B+bB'=a'N+a(-\kappa T+\tau B)+b'B-b\tau N.
\]
From the above equation, we get the following equations
\[
1+a\kappa =0,\qquad a'-b\tau=0,\qquad a\tau+b'=0.
\]
We therefore have
\[
a=-\rho,\qquad b=-\rho'\sigma.
\]
Thus
\[
\|\alpha-\cc\|^2=\rho^2+(\rho'\sigma)^2=r^2.
\]

Conversely,  if
\[
\alpha-\cc=-\rho\, N-\rho'\sigma B,
\]
then
\[
\|\alpha-\cc\|^2=\rho^2+(\rho'\sigma)^2=r^2.
\]
so $\alpha$ is on the sphere of radius $r$.

\end{proof}
\begin{pro}{(\S 2.3: 11)}{}
Let $\beta,\bar\beta: I\to\R^3$  be unit-speed curves with nonvanishing curvature and
torsion. If $T=\bar T$,  then $\beta$  and $\bar \beta$ are parallel (Ex. 10 of Sec. 2). If
$B=\bar B$, prove that $\bar\beta$ is parallel to either $\beta$ or the curve $s\mapsto -\beta(s)$.
\end{pro}

\begin{proof}
We consider $(\beta(s)-\bar\beta(s))'=T-\bar T=0$. Thus $\beta(s)-\bar\beta(s)=\cc$ is a constant. Thus $\beta$ and $\bar\beta$ are parallel. 

If $B=\bar B$, then $B'=\bar B'$ and hence
\[
-\tau N=-\bar\tau \bar N.
\]
Thus since both $N$ and $\bar N$ are unit vectors, we must have
\[
N=\pm\bar N, \quad \tau=\pm\bar\tau.
\]
Thus we have
\[
-\kappa T+\tau B=N'=\bar N'=-\bar \kappa \bar T+\bar\tau \bar B.
\]
From the above, we conclude that $T=\bar T$. Thus $\beta$ is either parallel to $\bar\beta$ or $-\bar\beta$.
\end{proof}

\begin{pro}{(\S 2.4: 1)}{}
Express the curvature and torsion of the curve $\alpha(t) = (\cosh t, \sinh t, t)$
in terms of arc length $s$ measured from $t = 0$.
\end{pro}

\begin{proof} We have 
\begin{align*}
&\alpha'=(\sinh t,\cosh t,1),\\
&\alpha''=(\cosh t,\sinh t, 0),\\
&\alpha'''=(\sinh t,\cosh t,0).
\end{align*}
Thus we have
\begin{align*}
&\|\alpha'\|=\sqrt 2\cosh t,\\
&
\alpha'\times\alpha''=(-\sinh t,\cosh t, -1),\\
&\|\alpha'\times\alpha''\|=\sqrt 2\cosh t,\\
& (\alpha'\times\alpha'')\cdot\alpha'''=1.
\end{align*}
We then have
\[
\kappa=\tau=\frac{1}{2\cosh^2 t}.
\]
In order to find the arc-length reparametrization, we solve the differential equation
\[
s'(t)=\|\alpha'(t)\|=\sqrt 2\cosh t,\quad s(0)=0.
\]
We then have 
\[
s(t)=\sqrt 2\sinh t.
\]
Therefore 
\[
\kappa(s)=\tau(s)=\frac{1}{2\cosh^2 t}=\frac{1}{2+s^2}.
\]
\end{proof}

\begin{pro}{(\S 2.4: 4)}{}
Show that the curvature of a regular curve in $\R^3$ is given by
\[
\kappa^2\nu^4=\|\alpha''\|^2-(d\nu/dt)^2.
\]
\end{pro}

\begin{proof} Since $\nu=ds/dt$, we have
\[
\frac{ d\nu}{dt}=\frac{d}{dt}\sqrt{\alpha'\cdot\alpha'}
\]

\end{proof}

\begin{pro}{(\S 2.4: 5)}{}
If $\alpha$ is a curve with constant speed $c > 0$, show that
\begin{align*}
&
T=\alpha'/c,\qquad N=\alpha''/\|\alpha''\|,\qquad B=\alpha'\times \alpha''/(c\|\alpha''\|),\\
&\kappa=\frac{\|\alpha''\|}{c^2},\qquad \tau=\frac{\alpha'\times \alpha''\cdot\alpha'''}{c^2\|\alpha''\|^2}.
\end{align*}
\end{pro}
\begin{pro}{(\S 2.4: 6)}{}
(a). If $\alpha$ is a cylindrical helix, prove that its unit vector $\uu$ (Thm. 4.5) is
\[
\uu=\frac{\tau}{\sqrt{\kappa^2+\tau^2}}\,T+\frac{\kappa}{\sqrt{\kappa^2+\tau^2}}\,B
\]
and the coefficients here are $\cos\theta$ and $\sin\theta$ (for $\theta$ as in Def. 4.5).

(b). Check (a) for the cylindrical helix in Example 4.2 of Chapter 1.
\end{pro}

\begin{pro}{(\S 2.4: 12)}{}
If $\alpha(t) = (x(t), y(t))$ is a regular curve in $\R^2$, show that its plane curvature
(Ex. 8 of Sec. 3) is given by
\[
\tilde\kappa =\frac{\alpha''\cdot J(\alpha')}{\nu^3}=\frac{x'y''-x''y'}{((x')^2+(y')^2)^{3/2}}.
\]
\end{pro}

\begin{pro}{(\S 2.5: 1)}{}
Consider the tangent vector $\vv = (1, -1, 2)$ at the point $\p = (1, 3, -1)$.
Compute $\nabla_\vv W$ directly from the definition, where
\begin{enumerate}[label={\rm (\alph*)}]
\item $W=x^2U_1+yU_2$,
\item $W=xU_1+x^1U_2-z^2U_3$.
\end{enumerate}
\end{pro}

\begin{pro}{(\S 2.5: 3)}{}
If $W$ is a vector field with constant length $\|W\|$,  prove that for any vector
field $V$, the covariant derivative $\nabla_VW$ is everywhere orthogonal to $W$.
\end{pro}


\end{document}











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