Real-valued function: If f : R^{n} → R, then Df = ∇f = (\frac{∂f}{∂x_{1}} ....\frac{∂f}{∂x_{n}})

Ch 3: Path in R^{n}:

If x : I ⊂ R → R^{n}, then Dx = [x′(t)]^{T} If x represents position,

v(t) = x′(t) = velocity vector and a(t) = v′(t) = x′′(t) = acceleration vector. Speed of path = ||x′(t)|| If x : [a, b] → R^{n}, then

length of path = L(x) = ∫_{a}^{b} ||x′(t)||dt Arclength parameter = s : [a, b] → [0, L(x)],

s(t) = ∫_{a}^{t} ||x′(u)||du Example: x : [10, 100] → R^{2}, x(t) = (t, 4t) velocity vector = v(t) = x′(t) = (1, 4) and acceleration vector = a(t) = v′(t) = x′′(t) = (0, 0) Speed = ||x′(t)|| = ||(1, 4)|| = \sqrt{1^{2} + 4^{2}} = \sqrt{17} length of path = L(x) = ∫_{10}^{100} ||x′(t)||dt = ∫_{10}^{100} \sqrt{17}dt

= \sqrt{17}t|_{10}^{100} = \sqrt{17}(100 − 10) = 90\sqrt{17} Arclength parameter = s : [10, 100] → [0, 90\sqrt{17}],

s(t) = ∫_{10}^{t} \sqrt{17}du = \sqrt{17}u|_{10}^{t} = \sqrt{17}(t − 10) Thus s : [10, 100] → [0, 90\sqrt{17}],

s(t) = t\sqrt{17} − 10\sqrt{17} Change of parametrization: Solve for t to find s^{−1}: t = \frac{s+10\sqrt{17}}{\sqrt{17}}. Let the function s^{−1} be denoted by t Hence t : [0, 90\sqrt{17}] → [10, 100], t(s) = \frac{s+10\sqrt{17}}{\sqrt{17}} y : [0, 90\sqrt{17}] → R^{2}, y(s) = (x ∘ t)(s)

y(s) = x(t(s)) = x(\frac{s+10\sqrt{17}}{\sqrt{17}}) = (\frac{s+10\sqrt{17}}{\sqrt{17}}, 4(\frac{s+10\sqrt{17}}{\sqrt{17}})) Thus the reparametrization of the path x using the arclength parameter s is y : [0, 90\sqrt{17}] → R^{2}, y(s) = (\frac{s+10\sqrt{17}}{\sqrt{17}}, 4(\frac{s+10\sqrt{17}}{\sqrt{17}})) Observe ||y′(s)|| = 1 and the length of the path traveled by time s is ∫_{0}^{s} ||y′(u)||du = s Unit tangent vector: T(t) = \frac{x′(t)}{||x′(t)||}

Note T(t) is perpendicular to T′(t)

outline of proof: Use \frac{d}{dt}||T(t)||^{2} = \frac{d}{dt}(1)

to show T(t) · T′(t) = 0

Let unit normal vector = N = \frac{T′(t)}{||T′(t)||}

Let unit binormal vector B = T × N.

Note B is a unit vector.

T(t), N(t), B(t) is a set of mutually orthogonal unit vectors called the moving frame (or Frenet frame).

Curvature κ = ||\frac{dT}{ds}|| = \frac{|| \frac{dT}{ds} ||}{\frac{ds}{dt}}

Torsion τ: ||\frac{dB}{ds}|| = \frac{|| \frac{dB}{ds} ||}{\frac{ds}{dt}} = τN

Intrinsic quantity: does NOT depend on parametrization:

Ex: T, N, B, κ, τ, length of path.

Extrinsic quantity: depends on parametrization: Ex: speed, velocity, acceleration