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\lhead{Math 4441 Activity 10}
\rhead{<your name(s)>}

\begin{document}
\title{Math 4441 Activity 10}
\author{<your name(s)>}
\date{November 9, 2022}
\maketitle
\vspace{5em}

\setcounter{section}{10}


\begin{exercise}\label{activity-10:sphere}
Consider the point $p=(1,0,0)$ on the unit sphere $S^2\subset\mathbb{R}^3$ and the function $f\colon S^2\to\mathbb{R}$ defined by $f(x,y,z):=z$.  Then
\[
D_{\vec{v}_1}f = 1
\quad\text{and}\quad
D_{\vec{v}_2}f = 0,
\]
where $\vec{v}_1=(0,0,1)^T\in T_pS^2$ and $\vec{v}_2=(0,1,0)^T\in T_pS^2$.  Draw a picture and write some complete sentences explaining why these equations are true.
\end{exercise}
\begin{solution}
<your solution>
\end{solution}

\begin{exercise}\label{activity-10:scalars}
Use the definition of the directional derivative to prove that
\[
D_{c\,\vec{v}}f = c\,D_{\vec{v}}f,
\]
for any tangent vector $\vec{v}\in T_p\mathrm{im}(\vec{x})$, differentiable function $f\colon\mathrm{im}(\vec{x})\to\mathbb{R}^n$, and scalar $c\in\mathbb{R}$.
\end{exercise}

\begin{lemma}\label{activity-10:downstairs-expression}
Consider a simple surface $\vec{x}\colon U\to\mathbb{R}^3$, a point $p=\vec{x}(u^1_0,u^2_0)$ on the surface, and a tangent vector $\vec{v}\in T_p\mathrm{im}(\vec{x})$.  If $f\colon\mathrm{im}(\vec{x})\to\mathbb{R}^n$ is differentiable at $p$, then we have
\[
D_{\vec{v}}f = \left.\left(V^1\,\dfrac{\partial(f\circ\vec{x})}{\partial u^1} + V^2\,\dfrac{\partial(f\circ\vec{x})}{\partial u^2}\right)\right\vert_{(u^1_0,u^2_0)},
\]
where $\vec{v}=V^1\vec{x}_1 + V^2\vec{x}_2$.
\end{lemma}

\begin{exercise}
Prove Lemma~\ref{activity-10:downstairs-expression}.
\end{exercise}

\begin{exercise}
Use Lemma~\ref{activity-10:downstairs-expression} to show that
\[
D_{\vec{v}+\vec{w}}f = D_{\vec{v}}f + D_{\vec{w}}f,
\]
for any $\vec{v},\vec{w}\in T_p\mathrm{im}(\vec{x})$ and differentiable function $f\colon\mathrm{im}(\vec{x})\to\mathbb{R}^n$.  Combined with Exercise~\ref{activity-10:scalars}, this tells us that the directional derivative is \textbf{linear} in the vector argument.  Why is this fact harder than Exercise~\ref{activity-10:scalars} to prove from the definition?
\end{exercise}

\begin{exercise}
Fix a point $p\in\mathrm{im}(\vec{x})$ and a vector $\vec{v}\in T_p\mathrm{im}(\vec{x})$.  Use Lemma~\ref{activity-10:downstairs-expression} to show that
\[
D_{\vec{v}}(c\,f + d\,g) = c\, D_{\vec{v}}f + d\,D_{\vec{v}}g,
\]
for any differentiable functions $f,g\colon\mathrm{im}(\vec{x})\to\mathbb{R}^n$ and scalars $c,d\in\mathbb{R}$.
\end{exercise}

\begin{exercise}
Fix a point $p\in\mathrm{im}(\vec{x})$ and a vector $\vec{v}\in T_p\mathrm{im}(\vec{x})$.  Use Lemma~\ref{activity-10:downstairs-expression} to show that
\begin{equation}\label{activity-10:inner-product-rule}
D_{\vec{v}}\langle\vec{X},\vec{Y}\rangle = \left\langle D_{\vec{v}}\vec{X},\vec{Y}\right\rangle + \left\langle \vec{X},D_{\vec{v}}\vec{Y}\right\rangle,
\end{equation}
for any differentiable vector fields $\vec{X},\vec{Y}\colon\mathrm{im}(\vec{x})\to\mathbb{R}^3$.
\end{exercise}

\begin{exercise}
Consider a surface curve $\vec{\alpha}=\vec{x}\circ\vec{\alpha}_U$ and a vector field $\vec{V}$ defined along $\vec{\alpha}$.  Remember that we call a vector field $\vec{V}$ \textbf{parallel} along $\vec{\alpha}$ if it doesn't appear to be changing as we move along $\vec{\alpha}$ --- from the perspective of an inhabitant of the surface.  Write down an equation to determine whether or not $\vec{V}$ is parallel using the covariant derivative.  Write a sentence or two explaining why you wrote the equation you did.
\end{exercise}

\end{document}