Vector Calculus

Introduction

Curves

Differentiating the Curve

The vector function is differentiable at t if, as \(\delta t \rightarrow 0\), we can write:

\[\mathbf{x}(t+\delta t)- \mathbf{x}(t)= \dot{\mathbf{x}}(t)\delta t + \mathcal{O}(\delta t^2) \]

If the derivative \(\dot{\mathbf{x}}(t)\) exists everywhere then the curve is said to be smooth.

Notations:

1. \(\dot{\mathbf{x}}(t)\): Newton original notation, usually denote differentiation wrt. time
2. \(f'(x)\): Usually used for Differentiation wrt. space

To differentiate vectors with basis vectors \(\mathbf{e}_i\), we can do it component by component:

\[ \mathbf{x}(t) = x^i(t)\mathbf{e}_i \Rightarrow \dot{\mathbf{x}}(t) = \dot{x}^i(t)\mathbf{e}_i \]

Here Leibnitz identities hold for the products of scalar and vector functions:

\[ \begin{align*} \frac{d}{d t}(f\mathbf{g}) &= \frac{d f}{d t}\mathbf{g} + f\frac{d \mathbf{g}}{d t}\\ \frac{d}{d t}(\mathbf{g}\cdot \mathbf{h}) &= \frac{d \mathbf{g}}{d t}\cdot \mathbf{h} + \mathbf{g}\cdot \frac{d \mathbf{h}}{d t}\\ \end{align*} \]

Moreover, if \(\mathbf{g}(t)\) and \(\mathbf{h}(t)\) are vectors in \(\mathbb{R}^3\), we also have cross product identity:

\[ \frac{d}{d t}(\mathbf{g}\times \mathbf{h}) = \frac{d \mathbf{g}}{d t}\times \mathbf{h} + \mathbf{g}\times \frac{d \mathbf{h}}{d t} \]

Tangent Vectors

The direction of the tangent vector \(\dot{\mathbf{x}}(t)\) depends only on the curve \(C\) itself, and not the choice of parameterisation.

The magnitude of the tangent vector \(|\dot{\mathbf{x}}(t)|\) depends on parameterisation.

Example: Consider the curve \(C\) in \(\mathbb{R}^2\):

\[ \mathbf{x}(t) = (t^3, t^3) \]

Here curve C is straight line \(x=y\), you can use any parameterisation here. Current parameterisation gives \(\dot{\mathbf{x}}(t) = 3t^2(1, 1)\). We can find out that the magnitude \(3\sqrt{2}t^2\) vanishes at \(t=0\) and only depends on parameterisation.

A parameterisation is called regular if \(\dot{\mathbf{x}}(t) \neq 0\) for any \(t\).

Piecewise smooth curve: \(C=C_1+C_2+\dots\). A tangent vector exists everywhere except at the cusps where two curves meet.

Arc Length

From the above figure, we can see the distance between two nearby points is:

\[ \delta s = |\delta \mathbf{x}| + \mathcal{O}(|\delta\mathbf{x}|^2) = |\dot{\mathbf{x}}\delta t| + \mathcal{O}(\delta t^2) \]

We then have

\[ \frac{ds}{dt} = \pm |\frac{d\mathbf{x}}{dt}| = \pm |\dot{\mathbf{x}}| \]

where plus sign is for direction of increasing \(t\) while minus sign is the other way around.

If we pick starting point \(t_0\), then the distance along the curve to any point \(t>t_0\) is given:

\[ s = \int_{t_0}^t d\hat{t}|\dot{\mathbf{x}}(\hat{t})| \]

This distance is called arc length and it does not depend on the choice of parameterisation.

Proof: Assume we choose another parameterisation of curve: \(\tau(t)\). The chain rule tells us:

\[ \frac{d\mathbf{x}}{dt} = \frac{d\mathbf{x}}{d\tau} \frac{d\tau}{dt} \]

The arc length is then become:

\[ s = \int_{t_0}^t d\hat{t}|\dot{\mathbf{x}}(\hat{t})| = \int_{\tau_0}^\tau d\hat{\tau}\frac{d\hat{t}}{d\hat{\tau}}|\frac{d\mathbf{x}}{d\hat{\tau}} \frac{d\hat{\tau}}{dt}| = \int_{\tau_0}^\tau d\hat{\tau}|\frac{d\mathbf{x}}{d\hat{\tau}} | \]

Thus, the arc length is independent of the choice of parameterisation of the curve. The arc length can be natural choice of parameterisation, i.e., \(\mathbf{x}(s)\) with correponding tangent vector \(d\mathbf{x}/ds\). Here \(|d\mathbf{x}/ds|=1\) (See above for why)

Curvature and Torison

Curvature represents magnitude of the acceleration of the curve with respect to arc length:

\[ \kappa(s) = |\frac{d^2\mathbf{x}}{ds^2}| \]

Note: We have already known that tangent vector \(\mathbf{t}=d\mathbf{x}/ds\) has unit length \(|\mathbf{t}|=1\)

Consider an example of a circule of radius \(R\):

\[ \mathbf{x}(t) = (R\cos t, R\sin t) \]

The arc length would be

\[ \frac{ds}{dt} = |\frac{d\mathbf{x}}{dt}| = R \rightarrow s = Rt \]

Then, we get \(\mathbf{x}(s) = (R\cos (s/R), R\sin (s/R))\), the curvature would be:

\[ \kappa = |\frac{d^2\mathbf{x}}{ds^2}| = \frac{1}{R} \]

As \(R\rightarrow \infty\), the circule would become a straight line.

There is a unit vector associated to curvature (magnitude), i.e. Principal normal:

\[ \mathbf{n} = \frac{1}{\kappa} \frac{d^2\mathbf{x}}{ds^2} = \frac{1}{\kappa} \frac{d\mathbf{t}}{ds} \]

The factor of \(1/\kappa\) ensures that \(|\mathbf{n}|=1\). The principal normal \(\mathbf{n}\) is perpendicular to the tangent vector \(\mathbf{t}\). Proof:

\[ \mathbf{n} \cdot \mathbf{t} = \frac{1}{\kappa} \frac{d\mathbf{t}}{ds} \mathbf{t} = \frac{1}{2\kappa}\frac{d(\mathbf{t} \cdot \mathbf{t})}{ds} = 0 \]

This means \(\mathbf{t}\) and \(\mathbf{n}\) defines a plane, known as osculating plane. Drawing a circle at point \(s\), whose curvature matches \(\kappa(s)\). This is called osculating circle.

We can also define a binormal here in \(\mathbb{R}^3\):

\[ \mathbf{b} = \mathbf{t}\times \mathbf{n} \]

Since we know \(\mathbf{t}\cdot \mathbf{b}=0\), we can differentiate it:

\[ 0 = \frac{d\mathbf{t}}{ds} \cdot \mathbf{b} + \mathbf{t} \cdot \frac{d\mathbf{b}}{ds} = \kappa \mathbf{n} \cdot \mathbf{b} + \mathbf{t} \cdot \frac{d\mathbf{b}}{ds} = \mathbf{t} \cdot \frac{d\mathbf{b}}{ds} \]

Now we know \(\mathbf{t} \cdot \frac{d\mathbf{b}}{ds}=0\) and \(\mathbf{b} \cdot \frac{d\mathbf{b}}{ds}=0\). Thus, \(\frac{d\mathbf{b}}{ds}\) is orthogonal to both \(\mathbf{b}\) and \(\mathbf{t}\), which means it must be parallel to \(\mathbf{n}\).

The torsion \(\tau(s)\) is defined as a measure of how the binormal changes:

\[ \frac{d\mathbf{b}}{ds} = -\tau(s)\mathbf{n} \]

NOTE:

1. Curvature: how much the curve fails to be a straight line
2. Torsion: how much the curve fails to be planar

Frenet-Serret Equations

\[ \begin{align*} \frac{d\mathbf{t}}{d s} &= \kappa\mathbf{n}\\ \frac{d\mathbf{b}}{d s} &= -\tau \mathbf{n}\\ \frac{d\mathbf{n}}{d s} &= \tau \mathbf{b}-\kappa\mathbf{t}\\ \end{align*} \]

Line Integrals

Scalar Fields

A scalar field is a map:

\[ \phi: \mathbb{R}^n \rightarrow \mathbb{R} \]

Given a parameterised curve \(C\) in \(\mathbb{R}^n\), we can put them together \(\phi (\mathbf{x}(t))\), which is a composite map \(\mathbb{R} \rightarrow \mathbb{R}\). To make the integration be independent to parameterisation of the curve, we can use arc length \(s\). We can integrate from point \(\mathbf{a}\) to point \(\mathbf{b}\), with \(\mathbf{x}(s_a)=\mathbf{a}\) and \(\mathbf{x}(s_b)=\mathbf{b}\) and \(s_a < s_b\) by defining the line integral:

\[ \int_C \phi ds = \int_{s_a}^{s_b} \phi (\mathbf{x}(s)) ds \]

Note: Line integral \(\int_C ds\) is always positive as there is no directional information in the integral.

Given a parameterised curve \(\mathbf{x}(t)\) with \(\mathbf{x}(t_a)=\mathbf{a}\) and \(\mathbf{x}(t_b)=\mathbf{b}\) (\(t_b > t_a\) i.e. \(ds/dt = + |\dot{\mathbf{x}}(t)|\)), we can obtain:

\[ \int_C \phi ds = \int_{t_a}^{t_b} \phi (\mathbf{x}(t))\frac{ds}{dt}dt = \int_{t_a}^{t_b} \phi (\mathbf{x}(t))|\dot{\mathbf{x}}(t)|dt \]

Here line integral comes with length of tangent vector \(\dot{\mathbf{x}}\) and thus independent of the choice of parameterisation.

Vector Fields

A vector field is a map:

\[ \mathbf{F}: \mathbb{R}^n \rightarrow \mathbb{R}^n \]

The line integral of a vector field \(\mathbf{F}\) along \(C\) is defined to be

\[ \int_C \mathbf{F} (\mathbf{x}) \cdot d\mathbf{x} = \int_{t_a}^{t_b} \mathbf{F}(\mathbf{x}(t)) \cdot \dot{\mathbf{x}}(t) dt \]

Here no matter \(t_a > t_b\) or \(t_a < t_b\), we always have the same equatiion as only the orientation \(\dot{\mathbf{x}}(t)\) determines the limits.

Example

Consider the vector field in \(\mathbb{R}^3\):

\[ \mathbf{F}(\mathbf{x}) = (xe^y, z^2, xy) \]

We consider two curve which we perfom integration. For curve \(C_1:\mathbf{x}(t) = (t, t^2, t^3)\), we have

\[ \mathbf{F}(\mathbf{x}(t)) = (te^{t^2}, t^6, t^3) \]

Thus, the line integral is:

\[ \begin{aligned} \int_{C_1} \mathbf{F} \cdot d\mathbf{x} &= \int_0^1 dt \mathbf{F}\cdot \dot{\mathbf{x}} \quad \text{Note here} \quad \dot{\mathbf{x}} = (1, 2t, 3t^2)\\ &= \int_0^1 dt (te^{t^2}+2t^7+3t^5) \\ &= \frac{1}{4}(1+2e) \end{aligned} \]

For curve \(C_2: \mathbf{x}(t) = (t, t, t)\), we have;

\[ \int_{C_2} \mathbf{F} \cdot d\mathbf{x} =\int_0^1 dt \mathbf{F}\cdot \dot{\mathbf{x}} = \int_0^1 dt (te^t + 2t^2) = \frac{5}{3} \]

More Curves, more integrals

For line integral around a close curve (loop), i.e., the start/end points are the same, we have

\[ \oint_c \mathbf{F}\cdot d\mathbf{x} \]

This quantity is called the circulation of \(\mathbf{F}\) around \(C\).

For line integral on curve \(C\) that can decompose into a number of piecewise smooth curves \(C = C_1 + C_2 + \dots\), the line integral is

\[ \int_C \mathbf{F} \cdot d\mathbf{x} = \int_{C_1} \mathbf{F} \cdot d\mathbf{x} + \int_{C_2} \mathbf{F} \cdot d\mathbf{x} + \dots \]

For curve \(-C\), we can think of it as the same curve \(C\) but with opposite orientation:

\[ \int_{-C} \mathbf{F} \cdot d\mathbf{x} = - \int_{C} \mathbf{F} \cdot d\mathbf{x} \]

Conservative Fields

Gradient

Reference:

https://www.damtp.cam.ac.uk/user/tong/vc.html