Vectors and Multivariable Calculus Notes

Module Description

A PDF document outlining a course on vectors and multivariable calculus.
Lists chapter titles and page numbers.

Vectors – Revision

Reviews basic vector algebra covered in MATH 10340.
A vector has magnitude and direction.
Point $P$ in space has coordinates $(a1, a2, a_3)$ with respect to a Cartesian frame with origin $O$ .
Distance from $O$ to $P$ is $\sqrt{a1^2 + a2^2 + a_3^2}$ .
$OP$ is a vector with direction from $O$ to $P$ and magnitude $\sqrt{a1^2 + a2^2 + a_3^2}$ .
Vector identified by coordinates; $\vec{OP} \equiv (a1, a2, a_3) \equiv a$ .
Vector addition: $a + b = (a1 + b1, a2 + b2, a3 + b3)$ , consistent with the parallelogram law.
Scalar multiplication: $λ\vec{OP} = λa = λ(a1, a2, a3) = (λa1, λa2, λa3)$ , where $λ ∈ R$ .
Unit vectors (length one) along Cartesian frame directions:
- $\hat{x} = (1, 0, 0)$
- $\hat{y} = (0, 1, 0)$
- $\hat{z} = (0, 0, 1)$
Consistency: $\vec{OP} = a = (a1, a2, a3) = a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1) = a1\hat{x} + a2\hat{y} + a_3\hat{z}$ .

The Dot Product

Definition: For vectors $a = (a1, a2, a3)$ and $b = (b1, b2, b3)$ , the dot product is $a · b = a1b1 + a2b2 + a3b3$ .
Properties:
- Commutative: $a · b = b · a$
- Distributive: $a · (b + c) = a · b + a · c$ and $(a + b) · c = a · c + b · c$
- For all $a, b, c ∈ R^3$ , where $R^3$ denotes all triples $(x, y, z)$ , with $x, y, z$ as real numbers.
Length (magnitude) of a vector: $mag(a) = \sqrt{a1^2 + a2^2 + a_3^2} = \sqrt{a · a}$ , denoted as $|a|$ or $a$ .
Theorem: $a · b = |a| |b| cos θ$ , where $0 ≤ θ ≤ π$ is the angle between $a$ and $b$ .
- Proof:
  - Introduce $c := a - b$ .
  - $c · c = (a - b) · (a - b) = a · a - 2a · b + b · b$
  - $|c|^2 = |a|^2 - 2a · b + |b|^2$
  - By the law of cosines: $|c|^2 = |a|^2 + |b|^2 - 2|a| |b| cos θ$
  - Equating: $a · b = |a| |b| cos θ$
Corollary: Two vectors $a$ and $b$ are orthogonal if and only if $a · b = 0$ .
Example: $\hat{x} · \hat{y} = (1, 0, 0) · (0, 1, 0) = 1 × 0 + 0 × 0 + 0 × 0 = 0$ . $\hat{x}$ and $\hat{y}$ are orthogonal.
$\hat{x} · \hat{x} = 1$ etc.
The vectors $\hat{x}, \hat{y},$ and $\hat{z}$ are called an orthonormal triad.

The Vector or Cross Product

Definition: Given vectors $a$ and $b$ , the cross product is defined as
$a × b = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \ a1 & a2 & a3 \ b1 & b2 & b3 \end{vmatrix} = \hat{x}(a2b3 − a3b2) + \hat{y}(a3b1 − a1b3) + \hat{z}(a1b2 − a2b1)$ .
Properties:
- Skew-symmetry: $a × b = −b × a$
- Linearity: $(λa) × b = a × (λb) = λ (a × b)$ , for $λ ∈ R$ .
- Distributive: $a × (b + c) = a × b + a × c$
Note: $a × a = −a × a \implies 2a × a = 0 \implies a × a = 0$ .
Example: If $a = \hat{x} + 3\hat{y} + \hat{z}$ , $b = 2\hat{x} − \hat{y} + 2\hat{z}$ , then $a × b = 7\hat{x} - 7\hat{z}$ .
Example: The orthonormal triad $\hat{x}, \hat{y},$ and $\hat{z}$ satisfy $\hat{x} × \hat{y} = \hat{z}, \hat{y} × \hat{z} = \hat{x}, \hat{z} × \hat{x} = \hat{y}$ .

Geometrical Treatment of Cross Product

The definition of the cross product is independent of the choice of Cartesian axes.
Length of $a × b$ :
- $|a × b|^2 + (a · b)^2 = |a|^2|b|^2$
- $|a × b| = |a| |b|sin θ$ , where $0 ≤ θ ≤ π$ .
Direction of $a × b$ :
- $a · (a × b) = 0$ and $b · (a × b) = 0$ . $a × b$ is perpendicular to both $a$ and $b$ .
- The sense of $a × b$ is determined by the right-hand rule.
The cross product as an area:
- The area of a parallelogram with adjacent sides $a$ and $b$ is $A = |a × b| = |a| |b|sin θ$ .

The Scalar Triple Product and Volume

Scalar triple product: $a · (b × c)$ .
Theorem: The scalar triple product $a · (b × c)$ is equal to the determinant of the matrix with rows $a, b, c$ .
Volume of parallelepiped = $(|a| cos ϕ) × (|b| |c|sin θ) = a · (b × c)$ .
Corollary: Three nonzero vectors $a, b$ , and $c$ are coplanar if and only if $a · (b × c) = 0$ .

The Vector Triple Product

Vector triple product: $a × (b × c)$ .
The cross product is not associative.
Theorem: $a × (b × c) = b (a · c) − c (a · b)$ , mnemonic: ‘BAC minus CAB’
Proof involves brute-force calculation in a frame where the x- and y-axes lie in the plane generated by $b$ and $c$ .

The Geometry of Lines and Planes

Overview: Using vector operations to describe lines and planes in three-dimensional space.
Important for approximating general (smooth) curves and surfaces.

The Equation of a Line

Problem: Find the equation of a straight line passing through two given points $A$ and $B$ with position vectors $a$ and $b$ .
Solution:
- Let $r$ be the position vector of any point $P$ on the line.
- $\vec{OA} + \vec{AP} = \vec{OP} \implies a + \vec{AP} = r \implies \vec{AP} = r − a$
- $\vec{OA} + \vec{AB} = \vec{OB} \implies a + \vec{AB} = b \implies \vec{AB} = b − a$
- $\vec{AP}$ and $\vec{AB}$ are colinear, hence $\vec{AP} = t \vec{AB} = t(b − a)$ , where $t$ is some real number.
- $r = a + t(b − a)$
A line is specified by a vector $r0 := a$ and a vector $e = b − a$ that lies along the line: $r = r0 + te$ .
A straight line is a one-parameter curve.
Cartesian form: with $r0 = (x0, y0, z0), e = (ex, ey, e_z)$
- $x = x0 + tex, y = y0 + tey, z = z0 + tez$
- Eliminating $t$ : $\frac{x − x0}{ex} = \frac{y − y0}{ey} = \frac{z − z0}{ez}$
In the x-y plane ( $z = z0$ and $ez = 0$ ):
- $\frac{x − x0}{ex} = \frac{y − y0}{ey} \implies y = y0 + \frac{ey}{ex} (x − x0)$
- Standard equation: $y = y0 + m(x − x0)$ , with slope $m = \frac{ey}{ex}$ and y intercept $b = y0 − mx0$ .

The Perpendicular Distance Between a Point and a Line

Problem: Given a line $L$ as $r(t) = r_0 + te$ and a point $P(a)$ , find the shortest distance between the line and the point.
Solution:
- Shortest distance is the perpendicular distance.
- Suppose the perpendicular from $P$ to $L$ intersects $L$ at $r1$ , where $r1 = r0 + t1e$ .
- $e · (a − r1) = 0 \implies e · a = e · r1 = e · (r0 + t1e) = e · r0 + t1|e|^2$
- Solving for $t1$ : $t1 = \frac{e · (a − r_0)}{|e|^2}$
- $r1 = r0 + e \frac{e · (a − r0)}{|e|^2} := r0 + \hat{e} [\hat{e} · (a − r_0)]$ , where $\hat{e} = \frac{e}{|e|}$ is a unit vector along $L$ .
- Perpendicular distance: $d\perp = |a − r1| = |(a − r0) − [\hat{e} · (a − r0)] \hat{e}| = |(r0 − a) − [\hat{e} · (r0 − a)] \hat{e}|$
- $d^2\perp = (r0 − a)^2 − |\hat{e} · (r0 − a)|^2 = (r0 − a)^2 (1 − cos^2 θ) = (r0 − a)^2 sin^2 θ = |(r0 − a) × \hat{e}|^2$
- $d\perp = |(r0 − a) × \hat{e}|$
In the x-y plane:
- Line equation: $αx + βy + γ = 0$ , with slope $m = −\frac{α}{β}$ .
- $e = (ex, ey) = (1, m)$ , $\hat{e} = \frac{(β, −α)}{\sqrt{α^2 + β^2}}$
- Assigning $r0 = (x0, y0, 0), a = (a1, a2, 0)$ , then $\perp^2 = \frac{|a1α + a2β + γ|^2}{α^2 + β^2}=\frac{|ma1 − a_2 + b|^2}{1 + m^2}$ , where $m = −\frac{α}{β}$ and $b = −\frac{γ}{β}$ from $y = mx + b$

The Equation of a Plane

Problem: Find the equation of a plane passing through three given points $A, B, C$ with position vectors $a, b, c$ .
Solution:
- Vectors $v1 := \vec{BA} = a − b, v2 := \vec{BC} = c − b$ lie in the plane $\Pi$ .
- Normal to the plane: $n = v2 × v1 = (c − b) × (a − b) = a × b + b × c + c × a$ .
- Equation of a plane $\Pi$ is $r ∈ R^3|(r − b) · (a × b + b × c + c × a) = 0$
- Simplifying, $r · (a × b + b × c + c × a) = b · (c × a)$ .
General equation of a plane in three dimensions:
- $\Pi (n, r0) = r ∈ R^3|(r − r0) · n = 0$
- Parametrized by the normal vector $n$ and a reference vector $r_0$
- If $nz \neq 0$ , the Cartesian expression is $z = z0 + \frac{(y0 − y)ny}{nz} + \frac{(x0 − x)nx}{nz}$ .
- A point $z = z(x, y)$ on a surface is labeled by two parameters, $x$ and $y$ .
- A plane is a two-parameter object, while a line a one-parameter curve.

Skew Lines and Intersecting Lines in Three Dimensions

In two dimensions, two non-parallel lines intersect.
In three dimensions, lines need not intersect; they can be skew.
Intersecting lines: assuming intersection, computing the point of intersection in terms of the parameters of the two lines.
- Lines: $rL(t) = r0 + te, rM(u) = s0 + uf$
- Point of intersection: $(t0, u0)$ is solution to
 $\begin{pmatrix} (r0 − s0) · e \ (r0 − s0) · f \end{pmatrix} = \begin{pmatrix} −|e|^2 & e · f \ −e · f & |f|^2 \end{pmatrix} \begin{pmatrix} t0 \ u0 \end{pmatrix}$ .
A candidate point for the intersection is the solution of the previous equation, provided the solution exists.
Determinant of required matrix: $−|e|^2|f|^2 + (e · f)^2 = −|e × f|^2$ .
If $e × f \neq 0$ , the pair $(t0, u0)^T$ is candidate solution. Need $r0 − s0$ lies entirely in the plane generated by $e$ and $f$ .
Now, $r0 − s0 = αe + βf + γe × f$ . Require $γ = 0$ , or $(r0 − s0) · (e × f) = 0$ .
Sufficient conditions for the lines to intersect is:
- $e × f \neq 0$ AND
- $(r0 − s0) · (e × f) = 0$ .
Condition $(r0 − s0) · (e × f) = 0$ states that $r0 − s0$ lies entirely in the plane generated by $e$ and $f$ .
Lines that satisfy the first condition but not the second are called skew lines.

Ordinary Derivatives of Vectors

Overview: Vector in $R^3$ that varies continuously as a single parameter is varied (e.g., position $x$ of a particle as a function of time).
Curve: map $γ : R → R^3 , t → xγ (t) = (xγ (t), yγ (t), zγ (t))$ .
Notation: Dropping curve label $\gamma$ . Abuse of notation.
Definition (Derivative of a curve): Let $x(t)$ be a curve parametrized by $t$ . The derivative of the curve w.r.t. $t$ is defined as
$\frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) − x(t)}{\Delta t}$ , provided the limit exists.
Since $x(t) = (x(t), y(t), z(t)) = \hat{x}x(t) + \hat{y}y(t) + \hat{z}z(t)$ ,
$\frac{dx}{dt} = \hat{x} \frac{dx}{dt} + \hat{y}\frac{dy}{dt} + \hat{z}\frac{dz}{dt}$ .
Curves inherit all the properties of real-valued functions.
Theorem: The following properties are satisfied, for arbitrary differentiable curves $A(t), B(t),$ and $C(t)$ :
- $\frac{d}{dt} [A(t) + B(t)] = \frac{dA}{dt} + \frac{dB}{dt}$
- $\frac{d}{dt} [A(t) · B(t)] = A(t) · \frac{dB}{dt} + B(t) · \frac{dA}{dt}$
- $\frac{d}{dt} [A(t) × B(t)] = A(t) × \frac{dB}{dt} + \frac{dA}{dt} × B(t)$
- For a scalar function $f(t)$ , $\frac{d}{dt} [f(t)A(t)] = f(t)\frac{dA}{dt} + A\frac{df}{dt}$
- $\frac{d}{dt} [A · (B × C)] = A · \begin{bmatrix}B × \frac{dC}{dt} \end{bmatrix} + A · \begin{bmatrix}\frac{dB}{dt} × C \end{bmatrix} + \frac{dA}{dt} · (B × C)$
- $\frac{d}{dt} [A × (B × C)] = A × \begin{bmatrix}B × \frac{dC}{dt} \end{bmatrix} + A × \begin{bmatrix}\frac{dB}{dt} × C \end{bmatrix} + \frac{dA}{dt} × (B × C)$
Theorem: Let $x(t)$ be a curve in $R^3$ . Then $\frac{dx(t)}{dt}$ is everywhere tangent to the curve. Thus, $\frac{dx}{dt}$ is often called the tangent vector or the velocity vector.

Frenet–Serret Frame

Arc length: arc length along the curve, measured from a reference value $x0 = x(t = 0)$ is $s(t) = \int{0}^{s(t)} ds = \int{0}^{s(t)} \sqrt{dx^2 + dy^2 + dz^2} = \int{0}^{t} \sqrt{\begin{bmatrix}\frac{dx}{dt}\end{bmatrix}^2 + \begin{bmatrix}\frac{dy}{dt}\end{bmatrix}^2 + \begin{bmatrix}\frac{dz}{dt}\end{bmatrix}^2} dt = \int_{0}^{t} \begin{Vmatrix}\frac{dx}{dt}\end{Vmatrix} dt$ .
$\frac{ds}{dt} = \begin{Vmatrix}\frac{dx}{dt}\end{Vmatrix} ≥ 0$ , and the arclength is an increasing function of time.
Thus, inverse function $t = t(s)$ , enabling a reparametrization of the curve according to arclength (denoted with a tilde):
$\tilde{x}(s) = x(t(s)).$
$\frac{d\tilde{x}}{ds} = \frac{dx}{dt} \frac{dt}{ds} = \frac{dx}{dt} \frac{1}{\begin{Vmatrix}\frac{dx}{dt}\end{Vmatrix}}$ , (Chain Rule) and $\frac{d\tilde{x}}{ds}$ is a unit vector tangent to the curve.
Tangent = $T = \frac{d\tilde{x}}{ds}$ .
Now $T · T = 1$ , hence $0 = T · \frac{dT}{ds} + \frac{dT}{ds} · T \implies T · \frac{dT}{ds} = 0\implies \frac{dT}{ds}$ is perpendicular to the tangent vector $T$ .
A new unit vector $N ∝ \frac{dT}{ds}$ that is normal to the tangent:
$\frac{dT}{ds} = κ(s)N$ , and $N$ is the principal normal to the curve
$κ$ is the curvature.
Goal: Deriving a triple of axes that move with the curve. $T$ defines an axis everywhere parallel to the curve; $N$ defines an axis that is everywhere perpendicular to the curve.
In three dimensions, three axes are necessary: vector $B := T × N$ .
The triple $(T , N, B)$ of axes along the curve $\tilde{x}(s)$ parametrized by the arclength $s$ is called the Frenet–Serret frame.
$B · B = 1$ , hence $B · \begin{bmatrix}\frac{dB}{ds} \end{bmatrix} = 0$ .
$dB/ds$ is perpendicular to $T$ and $B$ , and must therefore lie along $N$ :
$\frac{dB}{ds} ∝ N$ .
$\frac{dB}{ds} = −τ (s)N$ , where $τ$ is the torsion.
Since $(T , N, B)$ form a right-handed system (by construction), and since $B = T × N$ , cyclic permutation gives $N = B × T$
Applying $\frac{d}{ds}$ on this gives:
$\frac{dN}{ds} = B × \frac{dT}{ds} + \frac{dB}{ds} × T = B × (κN) − τ (N × T ) = −κT + τB$ .
Assemble the results:
Theorem: Curvature $κ$ and torsion $τ$ are defined via the following equations:
- $\frac{dT}{ds} = κ(s)N$
- $\frac{dB}{ds} = −τ (s)N$
- $\frac{dN}{ds} = τB − κT$

Frenet-Serret - Worked examples

Solution: As we know from school, a curve in two dimensions can always be written in the form $y = f(x)$ .
In other words, $x = (x, f(x))$ .
Now here, $x$ is simply a label, which indicates that the first variable in the bracket pair $(x, f(x))$ ranges over the whole real line (or some interval thereof).
Thus, can re-write the curve as : $x = (t, f(t))$ .
The unit tangent vector is available immediately as $T = \frac{\dot{x}}{|\dot{x}|}$ , where $\dot{x} := \frac{dx}{dt} = (1, f'(t))$ , $|\dot{x} | = \sqrt{1+f'(t)^2}$ .
Henceforth, write $f$ instead of $f(t)$ & $f'(t)$ , the functional dependence of $f$ on $t$ being understood.
Hence, $T = \frac{(1, f')}{\sqrt{1+f'^2} }$ .
To find the principal normal vector, going to have to differentiate Eq. (5.2):
$\frac{dT}{dt} = ^…= \frac{f''}{(1+f'^2)^{(3/2)}} (-f',1).$
Also, $\frac{dT}{ds} = \frac{dT}{dt} \frac{1}{\begin{vmatrix}\frac{dx}{dt}\end{vmatrix}} = \frac{f''}{(1+f'^2)^{(3/2)}} (-f',1) \sqrt{1+f'^2} = κN$
What is unambigous is the obsolute value of $κ$ , which is sometimes refered to as the unsigned curvature is $κ{us} := |κ|$ , such that $κsNs = sign(κ)|κ|N = sign(κ)κ{us}N$
This gives an unsigned normal vector, $N{us} = sign(κ)N$ , such that $\frac{dT}{ds} = κ{us}N{us}$ , $κ{us} ≥ 0$
Because $(T, N)$ live in the $xy$ plane for all time, it follows that $B$ is in the $z$ direction : $B = \hat{z}$ .
Now, $τ \propto \frac{dB}{dt}$ , hence $τ = 0$ .
This makes sense: the torsion is actually a measure of how much the curve ``twists'' out of the plane generated by $(T, N)$ .
Since the curve lies in this plane for all time, it is impossible for it to ``twist'' out of this plane, hence $τ = 0$
Theorem: $τ = 0$ for a curve that lives entirely in the $xy$ plane.
# Ex:Find the curvature and torsion of the a right-handed helix, given by the following parametric equations
- $x(t) = r \cdot cost$
- $y(t) = r \cdot sint$
- $z(t) = v \cdot t$ t ∈ [0, ∞), r, v > 0.
Note: Equations \begin(5.3)\end, correspond to a right-handed helix.
Solution: we first compute the tangent vector:
- $\frac{dx}{dt} = \frac{d}{dt} (rcost, rsint, vt) = (-rsint, rcost, v)$
- $|\frac{dx}{dt}| = \sqrt{r^2+v^2}$
- $\frac{dx}{dt}| {\frac{dx}{dt}} = \frac{(-rsint, rcost, v)}{\sqrt{r^2+v^2}}$
Hence $T = \frac{(-rsint, rcost, v)}{\sqrt{r^2+v^2}}$
Also,
- $\frac{dT}{dt} = \frac{(-rcost, -rsint, 0)}{\sqrt{r^2+v^2}}$ ,and
- $\frac{dT}{dt} \frac{1}{\begin{vmatrix}\frac{dx}{dt}\end{vmatrix}} = \frac{(-rcost, -rsint, 0)}{(r^2+v^2)} = \frac{r}{r^2+v^2}(- cost,sint,0)$
- so, $kappa = \frac{\begin{vmatrix}r\end{vmatrix}}{r^2 +v^2}$
Here, by taking the postivie sign , the unsigned and the [singend] curvatures agree: $kappa _{us} = κ = \frac{r}{r2+v^2} := κ$ ;

Chapter 6 Overview

Here in this chapter is on the elementary theory of partial derivatives.
Although elementary partial differentiation is covered elsewhere .
- Definition 6:1 (Scalar field) A function φ (x1,x2, ∙ ∙ ∙ ,xn) of n variables is a map from a subset of Rn to R: φ : (Ω ⊂ Rn) → R (x1, x2, ∙ ∙ ∙ , xn) → φ (x1, x2, ∙ ∙ ∙ , xn).
  - The function φ assigns to each point (x1, x2, ∙ ∙ ∙ xn) ∈ Ω a real number (scalar), and is therefore called a scalar field.

The elevation above sea level at any point in Ireland is a function of latitude and longitude

    *   ### The pressure of an ideal gas is a function of temperature and density (Boyle's Law)
    *   ### The quantity theory of money says that the GDP of an economy is a function of the velocity of money and the quantity of (broad) money in circulation.
    In this section, we shall consider functions of two variables (x, y); the generalization to three .

Definitions

Definition 6.1. (Scalar field)
A function (\varphi (x1, x2, \dots , xn)) of n variables is a map from a subset of (R^n) to (R):. (\varphi : (\Omega \subset R^n) \rightarrow R (x1, x2, \dots , xn) \rightarrow \varphi (x1, x2, \dots , x_n)).
The function (\varphi) assigns to each point ((x1, x2, \dots x_n) \in \Omega) a real number (scalar), and is therefore called a scalar field.

Examples:

The elevation above sea level at any point in Ireland is a function of latitude and longitude.
The pressure of an ideal gas is a function of temperature and density (Boyle’s Law).
The quantity theory of money says that the GDP of an economy is a function of the velocity of money and the quantity of (broad) money in circulation.

Summary for Chapter 7

Taylor Series
- The taylor Series of a function $f(x)$ centered at $a$ is given by
  $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$
Maclaurin Series
- The Maclaurin series is a special case of the taylor series where $a=0$ and the formula becomes
  $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$
Radius of Convergence
- the radius of convergence $R$ of a power series determines the interval in which the series converses.
- it can be found using the ratio test
 $L = \lim{n\to \infty} |\frac{a{n+1}}{a_n}|$
- The series converges if $L<1$ , diverges if $L>1$ , and inconclusive if $L=1$
Common Maclaurin Series
- $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
- $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$
- $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$
- \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n , |x|<1
- Use these known series and manipulations to find the taylor or maclaurin series of related functions more efficiently.

Examples provided on how to derive and calculate the Maclaurin series using specific tests such as $e^x$ , $sin(x)$ , and $\frac{1}{(1-x)}$

Another example shows a short cut used to avoid computing too many complicated derivatives, such as
$f(x) = \frac{x^3}{x+8}$

Summary for Chapter 8

Taylor's Theorem
- Provides a way to approximate a smooth function by a polynomial. Under certain the error in approximation becomes exact when the degree of the polynomial becomes infinite.
- The general expression is
  $$f(x) = f(a) +