Thm 3.3: Suppose F is a continuous vector field. F = ∇f iff F has path independent line integrals. Moreover if C is a piecewise C^{1} curve, then

∫_{C} F · ds = f(B) − f(A)

where A is the initial point of C and B is the terminal point of C. 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 = ∇f for f ∈ C^{2} iff ∇ × F = 0 for all x ∈ R. Suppose F = (M(x, y), N(x, y)). Then ∇ × F = (\frac{∂N}{∂x} − \frac{∂M}{∂y})k

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

Parametrized curves: Ex: f : [0, 2π] → 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, 2π) (but not 1:1 on boundary of [0, 2π]

Thus the image of f = {(cos(t), sin(t)) | t ∈ R} is a curve in R^{2}. A parametrization of the image of f is

x(t) = cos(t), y(t) = sin(t).

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

The graph of f = {(t, f(t)) = (t, cos(t), sin(t)) | t ∈ R} is also a curve in R^{3}. A parametrization of the graph of f is

x(t) = t, y(t) = cos(t), z(t) = sin(t).

7.1 Parametrized surfaces Ex: f(s, t) = [0, 2π] × R → 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.

The graph of f is also a 2-dimensional surface (in R^{5}), but we will focus on the image of f.

Defn: Suppose X : D → R^{n}, D ⊂ R^{2}. Fix t_{0} ∈ R. The s-coordinate curve at t = t_{0} is the image of the map c_{1}(s) = X(s, t_{0}). Fix s_{0} ∈ R. The t-coordinate curve at s = s_{0} is the image of the map c_{2}(t) = X(s_{0}, t).

Suppose X(s, t) differentiable.

Let T_{s}(s_{0}, t_{0}) = \frac{∂X}{∂s} (s_{0}, t_{0}) = tangent vector to the s-coordinate curve X(s, t_{0})

Let T_{t}(s_{0}, t_{0}) = \frac{∂X}{∂t} (s_{0}, t_{0}) = tangent vector to the t-coordinate curve X(s_{0}, t)

Thus T_{s} and T_{t} are tangent to the surface X(D)

A normal to this surface is

Defn: A parametrized surface S = X(D) is 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})×T_{t}(s_{0}, t_{0}) ≠ 0.

If S is smooth at every point in D, then the surface S is smooth.

If S is a smooth parametrized surface, then N = T_{s} × T_{t} is the standard normal vector arising from the parametrization of X.

Let V be a finite-dimensional vector space over R. The dual of V = V^{∗} = {f : V → R | f linear } Note V^{∗} is a vector space. The elements of V^{∗} are called covectors. If e_{1}, ..., e_{n} basis for V, then w_{1}, ..., w_{n} basis for V^{∗} where

w_{i} : V → R where w_{i}(e_{j}) = δ_{ij} = \{ \begin{matrix}1 & i = j \\ 0 & i ≠ j \end{matrix}

dim V = dim V^{∗} Let F_{∗} : V → W be a linear map between vector spaces The dual map map is F^{∗} : W^{∗} → V^{∗}, F (g) = g ∘ F. F_{∗} is injective implies F^{∗} injective F_{∗} is surjective implies F^{∗} surjective (G_{∗} ∘ F_{∗})^{∗} = F^{∗} ∘ G^{∗}. d : V → (V^{∗})^{∗}, d(v) = h where h : V^{∗} → R, h(f) = f(v). Thus (V^{∗})^{∗} is naturally isomorphic to V. Defn: The dual of T_{p}M = T_{p}^{∗}M is the cotangent space to M at p. If \frac{∂}{∂x_{1}} , ..., \frac{∂}{∂x_{m}} is a basis for T_{p}M, then the dual basis will be denoted dx_{1}, ..., dx_{m}.

B is bilinear if B(cv_{1} + dv_{2}, w) = cB(v_{1}, w) + dB(v_{2}, w) 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}) = B(v_{1}, w_{1} + w_{2}) + B(v_{2}, w_{1} + w_{2}) = B(v_{1}, w_{1}) + B(v_{1}, w_{2}) + B(v_{2}, w_{1}) + B(v_{2}, w_{2})

B is linear if B((v_{1}, w_{1}) + (v_{2}, w_{2})) = B((v_{1}, w_{1})) + B((v_{2}, w_{2})) B(c(v_{1}, w_{1})) = cB((v_{1}, w_{1})