\magnification 1900
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\voffset -0.4truein
\vsize 10truein

\def\emph{}

\def\S{\Sigma_{i=1}^n}
\def\Sm{\Sigma_{i=1}^m}

\def\s{{\sigma}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\h{\hskip 10pt}
\def\u{\vskip -8pt}
\def\bh{\hfil \break}

\def\N{{\bf N}}
\def\S{\Sigma_{i=1}^n}

\def\0{{\bf 0}}
\def\Z{{\bf Z}}
\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\a{{\bf a}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\p{{\bf p}}
\def\f{\phi}
\def\s{\sigma}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{ \vskip -5pt }
\def\h{\vskip -5pt \hskip 20pt}
\def\grad{ \nabla }
\def\l{\vskip 5pt
\hrule
\vskip -5pt}



Thm 3.3:  Suppose $F$ is a continuous vector field.

$F = \grad f$ iff $F$ has path independent line integrals. 

Moreover if $C$ is a piecewise $C^1$ curve, then $$\int_C F \cdot ds = f(B) -f(A)$$ where $A$ is the initial point of $C$ and $B$ is the terminal point of $C$.

\l

Thm 3.5.  Suppose $F$ is a $C^1$ vector field and suppose $R =$ the domain of $F$ is simply connected in $\R^2$ or $\R^3$.Then 

$F = \grad f$ for $f \in C^2$ iff $\grad \times F = 0$ for all $x \in R$.

\l

Suppose $F = (M(x, y), N(x, y))$.  Then $\grad \times F = ( {\partial N \over \partial x} - {\partial M \over \partial y} ){\bf k}$

\vfill

Ex: $F(x, y) = (x^3, e^y)$

\vfill
\eject

Parametrized curves:


Ex:  $f:  [0, 2\pi] \rightarrow \R^2$, $f(t) = (cos(t), sin(t))$  

Note this is a function of 1 variable.  Thus 1 degree of freedom.  Hence we obtain 1-dimensional curves.

Note $f$ is 1:1 on $(0, 2\pi)$ (but not 1:1 on boundary of $[0, 2\pi]$

\vfill

Thus the image of $f$ = $\{(cos(t), sin(t)  ) ~|~ t \in \R \}$ is a curve in $\R^2$.



A parametrization of the image of $f$ is 

\centerline{$x(t) = cos(t)$, ~~~$y(t) = sin(t)$.}

\vfill
This curve can also be represented by the level set, $g^{-1}(1)$ where $g(x, y) = x^2 + y^2$

\vfill

The graph of $f$ =  $\{(t, f(t)) = (t, cos(t), sin(t)  ) ~|~ t \in \R \}$ is also a curve in $\R^3$.

A parametrization of the graph of $f$ is 

\centerline{$x(t) = t,~~~ y(t) = cos(t)$, ~~~$z(t) = sin(t)$.}






\vfill
\eject


7.1  Parametrized surfaces

Ex:  $f(s, t) = [0 , 2 \pi] \times \R \rightarrow \R^3$

$f(s, t) = (cos(s), sin(s), t)$

Note this is a function of 2 variables.  Thus 2 degrees of freedom.  Hence the image  is a 2-dimensional surface. 

Note $f$ is 1:1 on the interior of the domain, but not on the boundary.
\vfill
\vfill
The graph of $f$ is also a 2-dimensional surface (in $\R^5$), but we will focus on the image of $f$.


\eject

Defn:  Suppose $X: D \rightarrow \R^n$, $D \subset \R^2$.

Fix $t_0 \in \R$.  The {\it s-coordinate curve at $t = t_0$} is the image of the map $c_1(s) = X(s, t_0)$.


Fix $s_0 \in \R$.  The {\it t-coordinate curve at $s = s_0$} is the image of the map $c_2(t) = X(s_0, t)$.

\vfill

\eject


Suppose $X(s, t)$ differentiable.  


Let $T_s(s_0, t_0) = {\partial X \over \partial s}(s_0, t_0)$ = tangent vector to the $s$-coordinate 

\rightline{curve $X(s, t_0)$}



Let $T_t(s_0, t_0) = {\partial X \over \partial t}(s_0, t_0)$ = tangent vector to the $t$-coordinate 

\rightline{curve $X(s_0, t)$}



Thus $T_s$ and $T_t$ are tangent to the surface  $X(D)$ 

A normal to this surface is  


\vfill

Defn:  A parametrized surface $S = X(D)$ is {\it smooth} at $X(s_0, t_0) $ if $X$ is $C^1$ near $(s_0, t_0)$ and if $N(s_0, t_0) = T_s(s_0, t_0) \times T_t(s_0, t_0) \not= 0$.


If $S$ is smooth at every point in $D$, then the surface $S$ is {\it smooth}.

If $S$ is a smooth parametrized surface, then $N = T_s \times T_t$ is the {\it standard normal vector arising from the parametrization of $X$}.



\eject

Let $V$ be a finite-dimensional vector space over $\R$.

The dual of $V$ = $V^* = \{f:  V \rightarrow \R ~|~ f$ linear $\}$

Note $V^*$ is a vector space.  The elements of $V^*$ are called {\it covectors}.


If $e_1, ..., e_n$ basis for $V$, then $w_1, ..., w_n$ basis for $V^*$ where 
$w_i: V \rightarrow \R$ where $w_i(e_j) = \delta_{ij} = \cases{1 & $i = j$ \cr 0 & $i \not= j$}$

dim $V$ = dim $V^*$

Let $F_*:  V \rightarrow W$ be a linear map between vector spaces \bh
The dual map map is $F^*:  W^* \rightarrow V^*$, $F(g) = g \circ F$.

$F_*$ is injective implies $F^*$ injective 

$F_*$ is surjective implies $F^*$ surjective 


$(G_* \circ F_*)^* = F^* \circ G^*$.


$d:  V \rightarrow (V^*)^*$, $d(v) = h$ where $h: V^* \rightarrow R$, $h(f) = f(v)$.

Thus $ (V^*)^*$ is naturally isomorphic to $V$.

\l

Defn:  The dual of $T_pM= T_p^*M$ is the {\it cotangent space} to $M$ at $p$.

If $ {\partial \over \partial x_1}, ..., {\partial \over \partial x_m}$ is a basis for $T_pM$, then the  dual basis will be denoted $dx_1, ..., dx_m$. 

%%Defn:  The cotangent bundle of M 

%%$B(v_1 + v_2, w) = B(v_1, w) + B(v_2, w)$
%%$B(v, w_1 + w_2) = B(v, w_1) + B(v, w_2)$
\eject

$B$ is bilinear if \bh
$B(cv_1 + dv_2, w) = cB(v_1, w) + dB(v_2, w)$ \bh
$B(v, cw_1 + dw_2) = cB(v, w_1) + dB(v, w_2)$

Thus



$B( (v_1, w_1)  + (v_2, w_2)) = 
B(v_1 + v_2, w_1 + w_2)\bh = B(v_1, w_1 + w_2)  + B(v_2, w_1 + w_2) \bh
= B(v_1, w_1) + B(v_1, w_2)  + B(v_2, w_1)  + B(v_2, w_2) $


$B$ is linear if \bh
 $B( (v_1, w_1)  + (v_2, w_2)) =  B( (v_1, w_1))  + B((v_2, w_2))$ \bh
$B( c(v_1, w_1)) = cB( (v_1, w_1) $   




\end