Instructor’s Lecture Notes

Introduction

Typographic Conventions

Generic variable names will be italicized, whereas names of constants or operators will be roman (upright, normal) type. For example in the equation \(f(x,y) = \sin\bigl(5\ln(x)+3y^2\bigr)\) the symbol \(f\) for the defined function and the symbols \(x\) and \(y\) for its parameters are italicized, whereas the symbols for the specific functions \(\sin\) and \(\ln\) and the constant \(3\) are roman. There is no analogue to this convention for handwritten mathematics except when writing certain ubiquitous sets. For example the real numbers will be denoted in type as \(\mathbf{R}\) in type but is often handwritten in “black-board bold” as \(\mathbb{R}.\)

Scalars will be set with the same weight as the surrounding type whereas vectors will be given a stronger weight. For example \(x\) must be a scalar and \(\bm{v}\) must be a vector. Similarly scalar-valued functions will be set with a standard weight like \(f\) whereas vector-valued functions will be set in a stronger weight like \(\bm{r}.\) It’s not worth bothering trying to draft bold symbols when handwriting mathematics, so instead vectors and vector-valued functions will be indicated with a small over-arrow like \(\vec v\) and \(\vec r.\)

Terminology & Definitions

The word space will be used to refer to our universe, the largest context we are working in, which will usually be three-dimensional real space \(\mathbf{R}^3.\) A point is a single, zero-dimensional element of space. A curve \(C\) is a locally one-dimensional collection of points in space, an simple example of which being a line. A surface \(S\) is a locally two-dimensional collection of points in space, an simple example of which being a plane. A region \(R\) is a general term referring to some connected collection of points — which could be in \(\mathbf{R}^1\) or \(\mathbf{R}^2\) or \(\mathbf{R}^3\) depending on the context — that typically serves as the domain of some function or integral. We sometime refer specifically to a region in \(\mathbf{R}^1\) as an interval \(I\) and a region in \(\mathbf{R}^3\) as an expanse \(E.\) We refer to the difference between the dimension of a collection of points and the dimension of the space it’s in as its codimension: e.g. a one-dimensional curve embedded in three-dimensional space has codimension two. To any of these collections of points, we may add the qualifier “with boundary” to indicate it might have, well, a boundary. The boundary of any collection of points has local dimension one fewer than that collection: e.g. the boundary of a surface-with-boundary is a curve, and the boundary of a curve-with-boundary are two points.

All of these previous terms have geometric connotations. If we wish to strip away that connotation and think of a space or a curve or a surface or a region as a collection of points we’ll refer to it as a set. One such of these sets may contain another — for a example a curve \(C\) may live entirely in a region \(R\) — and in this case we may say that the curve is a subset of the region, which we may denote symbolically \(C \subset R.\) If however a single point \(x\) is inside a set \(S\) we’ll refer to it as an element of \(S\) and denote this symbolically as \(x \in S.\)

Given a subset \(S\) of space we’ll let \(S^c\) denote its complement, the set of all points not in \(S.\) The notation \(\partial S\) will denote the boundary of \(S,\) all points in the space such that every ball centered on the point, no matter how small, intersects both \(S\) and \(S^c.\)

A set \(S\) is closed if \(\partial S \subset S\) and is open if its complement is closed. A set \(S\) is bounded if there exists some upper-bound \(M\) on the set of all distances from a point in \(S\) to the origin.

The set of all designated inputs of a function is its domain and the designated target set for the function’s potential outputs is its codomain; a function’s range is the subset of all outputs in the codomain that actually correspond to an input. Notationally, we’ll denote a function \(f\) having domain \(D\) and codomain \(R\) as \(f \colon D \to R.\) The symbol \(\times\) denotes the product of two spaces. For the function \(f \colon D \to R,\) the graph of \(f\) is the subset of \(D \times R\) consisting of all points \((x,y)\) for which \(x \in D\) and \(y \in R\) and \(f(x) = y.\)

An operator is a function whose inputs and outputs are functions themselves. For example, \(\frac{\mathrm{d}}{\mathrm{d}x}\) is the familiar differential operator; its input is a function \(f\) and its output \(\frac{\mathrm{d}}{\mathrm{d}x}(f)\) is the derivative of the function \(f.\) For a generic operator \(\mathcal{T}\) applied to a function \(f\) we may wish to evaluate the output function \(\mathcal{T}(f)\) at some value \(x.\) Doing so we would have to write \(\mathcal{T}(f)(x),\) or even \(\big(\mathcal{T}(f)\big)(x)\) to be precise. To avoid this cumbersome notation we will employ one of two notational tricks: (1) if there is only a single input function \(f\) in our context we omit the \(f\) and comfortably write \(\mathcal{T}(x)\) instead of \(\mathcal{T}(f)(x),\) trusting a reader perceives no ambiguity, or (2) the output function will be written in a “curried” form, \(\mathcal{T}_{f}(x)\) instead of \(\mathcal{T}(f)(x).\)

Analytic & Vector Geometry in Three-Dimensional Space

How does calculus extend to higher dimensional space? Or, how does calculus handle multiple independent or dependent variables?

In their previous experience studying calculus students learned the facts as they apply to functions \(f\colon \mathbf{R} \to \mathbf{R}\) that have a one-dimensional domain and one-dimensional codomain. These are called single-variable, scalar-valued functions, terms that allude to the one-dimensional-ness of their domain and codomain respectively. And in studying the geometry of these functions, they visualized their graphs in two-dimensional space \(\mathbf{R}^2.\) The goal now is to study how facts of calculus must be generalized in the case of functions whose domain or codomain have dimension greater than one. Functions that, via the graph or via an embedding of the domain into the codomain, can only be usefully visualized in a space of dimension greater than one. So we will start thinking about three-dimensional space \(\mathbf{R}^3.\)

First we’ll try to discuss space, surfaces, curves, etc, first from a purely analytic perspective, just to get used to thinking in higher dimensions. Then we’ll introduce vectors, and use vectors as a convenient means extend what we know about calculus of a single variable to multi-variable equations that correspond to surfaces and curves, and to extend common concepts from physics to three-dimensional space.

Three-Dimensional Space

In two-dimensional space \(\mathbf{R}^2\) we imposed a rectangular (Cartesian) coordinate system with two axes, the \(x\)-axis and \(y\)-axis, which divided the space into four quadrants, and we referred to each point by its coordinates \((x,y).\) In three-dimensional space \(\mathbf{R}^3\) we add a \(z\)-axis. The axes now divide space into eight octants, and we refer to each point by its coordinates \((x,y,z).\)

Distance and Midpoint Given points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) in three-dimensional space, their midpoint has coordinates \((\overline{x}, \overline{y}, \overline{z})\) given by the formulas \( \overline{x} = \frac{1}{2}(x_1\!+\!x_2) \) and \( \overline{y} = \frac{1}{2}(y_1\!+\!y_2) \) and \( \overline{z} = \frac{1}{2}(z_1\!+\!z_2),\) and the distance between them is calculated by the formula. \[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\] This formula results from applying the Pythagorean Theorem twice, and generalizes as expected to higher dimensional space. Fun fact, Pythagoras was mildly a cult leader.

Lines, Planes, and Surfaces

Codimension If you have a locally \(N\)-dimensional set sitting inside of \(M\)-dimensional space, we’ll refer to the difference \(M-N\) as the codimension of that set. The point of introducing the word “codimension” is to concisely state this fact: generically, a set with codimension \(n\) requires \(n\) equations to express.

Planes The template equation for a plane in space passing through the point \((x_0, y_0, z_0)\) is \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \quad\text{or}\quad Ax + By + Cz = D\,, \] where the coefficients \(A\) and \(B\) and \(C\) correspond to the “slopes” of the plane in various directions.

Lines In three dimensional space, a line is the intersection of two planes, but it’s more convenient to describe parametrically. A line through point \((x_0, y_0, z_0)\) is given by the equations

where the coefficients \(A\) and \(B\) and \(C\) describe the relative “direction” of the line; every increase of \(A\) in the \(x\)-direction will correspond to an increase of \(B\) in the \(y\)-direction and \(C\) in the \(z\)-direction. If necessary the line may be re-parameterized, and expressed succinctly with two planar equations.

Cylinders and Quadric Surfaces A quadric surface is a surface with corresponding equation containing terms with at most degree two, e.g. \(x^2\) or \(z^2\) or \(yz.\) A cylinder in general is a surface that results from translating a planar curve along a line orthogonal to the curve. The locus of a single point from the curve along the line is referred to as a ruling, and a single “slice” along a ruling will be a cross-section, more generally called a trace, congruent to the curve. A “cylinder” as you knew it previously is a specific example we’ll now refer to as a circular cylinder.

One last exercise if there is time.

Before next class, review polar coordinates in the plane, and read about the atan2 function.

Cylindrical and Spherical Coordinates

In two-dimensional space we’d usually refer to a point’s location with its rectangular coordinates, but there was a distinct other coordinate system, polar coordinates, by which we could refer to a point’s location using its distance from the origin and angle of inclination from the positive \(x\)-axis. Given a point with rectangular coordinates \((x,y),\) its polar coordinates would be \(\big(\sqrt{x^2+y^2}, \operatorname{atan2}(y,x)\big).\) Conversely given a point with polar coordinates \((r,\theta),\) its rectangular coordinates would be \((r\cos(\theta), r\sin(\theta)).\)

In three-dimensional space, there are three distinct coordinate systems we may use the describe the location of a point. The one we’ve been using, where a point is referred to by the triple of numbers \((x,y,z)\) that are it’s distance from the origin in the \(x\)-, \(y\)-, and \(z\)-directions respectively by is called rectangular (Cartesian) coordinates. The other two are based on polar coordinates in the \((x,y)\)-plane, but use a different method to describe where a point is above the plane.

In cylindrical coordinates, a point is referred to by a triple \((r, \theta, z)\) where \(r\) and \(\theta\) are the polar coordinates of the point in the \(xy\)-plane and \(z\) is the height of the point above the \(xy\)-plane. Similar to polar coordinates in two-dimensions, there is not a unique triple that describes a point, but conventionally, when given the choice, we use the triple for which \(r \geq 0\) and \(\theta\) is between \(-\pi\) and \(\pi.\) Explicitly, to convert between rectilinear and cylindrical coordinates: \[ \begin{align*} (x,y,z) &\longrightarrow \Big(\sqrt{x^2+y^2}, \operatorname{atan2}(y,x), z\Big) \\ \big(r\cos(\theta), r\sin(\theta),z\big) &\longleftarrow (r,\theta,z) \end{align*} \]

In spherical coordinates, a point is referred to by a triple \((\rho, \theta, \phi)\) where \(\rho\) is the point’s distance from the origin, \(\theta,\) the azimuth, is the angle in the \(xy\)-plane above which it lies, and \(\phi,\) the zenith, is the angle at which it’s declined from the positive \(z\)-axis. Again there is not a unique triple that describes a point, but conventionally we use the triple for which \(\rho \geq 0,\) \(\theta\) is between \(-\pi\) and \(\pi,\) and \(\phi\) is between \(0\) and \(\pi.\) Explicitly, to convert between rectilinear and spherical coordinates: \[ (x,y,z) \longrightarrow \Bigg(\sqrt{x^2+y^2+z^2}, \operatorname{atan2}(y,x), \operatorname{arccos}\bigg(\frac{z}{\sqrt{x^2+y^2+z^2}}\bigg)\Bigg) \\ \Big(\rho\sin(\phi)\cos(\theta), \rho\sin(\phi)\sin(\theta), \rho\cos(\phi)\Big) \longleftarrow (\rho,\theta,\phi) \]

A major reason for working in cylinder or spherical coordinates is that certain surfaces or regions are much easier to describe analytically in these coordinate systems. In cylindrical coordinates a cylinder of radius \(5\) is just \(r=5.\) In cylindrical coordinates \(r=z\) corresponds to a cone. In spherical coordinates a sphere of radius \(7\) is just \(\rho=7.\)

Before next class, if you’ve seen vectors in a previous class, review them.

Vectors in Three-Dimensional Space

A vector in three-dimensional space is a triple of real numbers \(\langle x,y,z \rangle.\) In terms of data, a vector \(\langle x,y,z \rangle\) is no different than a point \((x,y,z),\) However, it comes with a different connotation. A point is static, just a location in space, whereas a vector implies direction: the vector \(\langle x,y,z \rangle\) denotes movement, usually from the origin towards the point \((x,y,z),\) or in other contexts movement from some initial point in the \(\langle x,y,z \rangle\)-direction. For these reasons a point is illustrated with a dot, whereas a vector is illustrated with an arrow.

These examples have been of vectors in three-dimensional space, but certainly we have have two-dimensional vectors, or vectors of any higher dimension too.

If the initial and terminal point of a vector have been named, say \(A\) and \(B\) respectively, that vector from \(A\) to \(B\) will be denoted \(\overrightarrow{AB}.\) More often though we just assign the vector a variable name: a bold character like \(\bm{v} = \langle x,y,z\rangle\) when typeset, or decorated with an arrow like \(\vec{v} = \langle x,y,z\rangle\) when handwritten. Conventionally, like vector variables themselves, functions that return vectors are also typeset as bold characters

The magnitude of a vector \(\bm{v}\) (sometimes called its modulus) is its length, and is denoted as \(|\bm{v}|\) or sometimes as \(\|\bm{v}\|\). Explicitly for \(\bm{v} = \langle x,y,z \rangle,\) \[|\bm{v}| = \sqrt{x^2+y^2+z^2}\,.\] We’ve previously used those bars \(||\) to denote the absolute value of the number they contain, but this new use is not new, only a generalization, when we remember that a number’s absolute value can be interpreted as its distance from zero. The magnitude of the vector \(\langle x,y,z\rangle\) is the distance of the point \((x,y,z)\) from zero.

Speaking of zero, there is certainly a zero vector denoted \(\bm{0}\) that has a magnitude of zero.

Sometimes we wish to scale a vector, multiplying each of its components through by a constant to lengthen the vector, but maintain its direction.

This number out front of the vector is called a scalar, the act of distributing it, multiplying each component of the vector by it, is called scalar multiplication. Scaling a vector by a negative number will result in a vector going “backwards” along the same direction.

A vector consists of two pieces of information: a direction and magnitude. Sometimes it’s convenient to separate these two pieces of information. A unit vector is any vector of length one. Given a vector \(\bm{v},\) we’ll let \(\bm{\hat{v}}\) denote the unit vector. Writing \(\bm{v} = |\bm{v}|\bm{\hat{v}}\) effectively separates \(\bm{v}\) into its direction \(\bm{\hat{v}}\) and its magnitude \(|\bm{v}|.\)

Given two vectors \(\bm{u} = \langle u_1, u_2, u_3\rangle\) and \(\bm{v} = \langle v_1, v_2, v_3\rangle\) we can calculate their sum as \[ \bm{u} + \bm{v} = \langle u_1+v_1, u_2+v_2, u_3+v_3\rangle\,. \] This sum \(\bm{u} + \bm{v}\) is the vector that would result from moving along vector \(\bm{u}\) and then moving along vector \(\bm{v},\) or vice versa.

Sometimes it is convenient to break down a vector into a sum of its components in its coordinate directions. Let’s call the three vectors

The prototypical model for a vector is as a force. We will discuss this in more detail later.

Before next class, review what the cosine of an angle is, and review the law of cosines.

The Dot Product and Projections

There is no single “product” of vectors.

Given two vectors \(\bm{u} = \langle u_1, u_2, u_3 \rangle\) and \(\bm{v} = \langle v_1, v_2, v_3 \rangle,\) their dot product is the scalar value \[\bm{u}\cdot\bm{v} = u_1v_1 + u_2v_2 + u_3v_3\,. \] Sometimes this is called the scalar product, or the inner product of vectors. Some properties of this operations that should be explicitly pointed out.

An alternative characterization of the dot product, sometimes taken to be an alternative definition, involves the angle \(\theta\) between the vectors involved. \[\bm{u}\cdot\bm{v} = |\bm{u}||\bm{v}| \cos(\theta) \]

One interpretation of this characterization is that the dot product is a length-corrected measure of the angle between two vectors. Some specific facts following from this characterization:

• If two vectors are parallel, their dot product will simply be the product of their lengths.
• Two vectors are orthogonal — perpendicular, at a right angle to each other — if and only if their dot product is zero.
• The angle between two vectors is acute if and only if their dot product is positive, and the angle between two vectors is obtuse if and only if their dot product is negative.

But there is another interpretation of the dot product that is useful to always keep in mind. The projection of \(\bm{u}\) onto \(\bm{v},\) denoted \( \operatorname{proj}_{\bm v}(\bm u),\) is the vector parallel to \(\bm{v}\) that lies “orthogonally underneath” \(\bm{u},\) as if \(\bm{u}\) were casting a shadow onto \(\bm{v}.\) The projection is calculated as the formula \[ \operatorname{proj}_{\bm v}(\bm u) = \bigg(\frac{\bm{u}\cdot\bm{v}}{|\bm{v}|^2}\bigg) \bm{v}\,. \]

Now taking that last formula and taking the dot product of both sides by \(\bm{v}\) we get that \({\bm{u}\cdot\bm{v} = \operatorname{proj}_{\bm v}(\bm u) \cdot \bm{v}\,.} \) And since \(\operatorname{proj}_{\bm v}(\bm u)\) and \(\bm{v}\) are parallel, their dot product is simply the product of their lengths: \({\bm{u}\cdot\bm{v} = |\operatorname{proj}_{\bm v}(\bm u)||\bm{v}|\,.} \) Similarly, due to symmetry between the vectors \(\bm{u}\) and \(\bm{v},\) we also have \({\bm{u}\cdot\bm{v} = |\operatorname{proj}_{\bm u}(\bm v)||\bm{u}|\,.} \) The interpretation of the dot product we get from this is as a generalization of the usual product of numbers: the dot product of two vectors is the product of their lengths once one is projected onto the same line as the other.

In fact, this can be made a little stronger: given any third vector \(\bm{w},\) such that the angle between \(\bm{u}\) and \(\bm{w}\) is \(\theta\) and the angle between \(\bm{v}\) and \(\bm{w}\) is \(\phi,\) then projecting both \(\bm{u}\) and \(\bm{v}\) onto \(\bm{w}\) we get \[|\bm{u}||\bm{v}|\cos(\theta)\cos(\phi) = |\operatorname{proj}_{\bm w}(\bm u)||\operatorname{proj}_{\bm w}(\bm v)|\,.\] This left-hand-side is nearly the dot product of those vectors, especially in light of the identity \[ \cos(\theta)\cos(\phi) = \frac{1}{2}\Big(\cos(\theta+\phi)+\cos(\theta-\phi)\Big) \] and that right-hand-side is an average of the cosines of the sum and difference which seems to address the whole “sign” issue some folks seems to have, but I’m not sure how to cleanly consolidate this.

Physics intuition: work is the dot product of the force vector with the displacement vector.

Before next class, review the law of sines, and matrix determinants (if you’ve seen them), and that \(\frac{1}{2}AB\sin(\theta)\) area formula.

The Cross Product and Areas

Given two non-parallel vectors in three-dimensional space, there is a unique direction that is orthogonal to both of them. But how do we find that direction? I.e. how do we compute a vector pointed in that direction? The cross product is the answer to this question.

What is the cross product in relation to the determinant? Is it just a “partially applied” determinant? And how does it relate to the law of sines?

Given two vectors \(\bm{u} = \langle u_1, u_2, u_3 \rangle\) and \(\bm{v} = \langle v_1, v_2, v_3 \rangle,\) their cross product is the vector \[\bm{u}\times\bm{v} = \langle u_2v_3 - u_3v_2, u_1v_3 - u_3v_1, u_1v_2 - u_2v_1\rangle\,. \] But remembering this formula can be a bit tough, so we usually phrase it in terms of matrix determinants instead. \[\begin{align*} \bm{u}\times\bm{v} &= \det\begin{pmatrix} u_2 & u_3 \\ v_2 & v_3\end{pmatrix}\mathbf{i} + \det\begin{pmatrix} u_1 & u_3 \\ v_1 & v_3\end{pmatrix}\mathbf{j} + \det\begin{pmatrix} u_1 & u_2 \\ v_1 & v_2\end{pmatrix}\mathbf{k} \\\\ &= \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{pmatrix} \end{align*}\]

Always remember that this cross product returns a vector that, by design, is orthogonal to its factors. You can determine the direction of the cross product vector via the right-hand rule. On a related note, the cross product is not quite a commutative operation, but instead \(\bm{u} \times \bm{v} = -\bm{v} \times \bm{u}.\) Additionally, if two vectors are parallel, there is not a unique direction that is orthogonal to each of them, and so the cross product of parallel vectors will be the zero vector. Here are other properties of the cross product as a vector operation.

Just like how the dot product of two vectors has an alternative definition involving the trigonometry of the angle between them, there is a similar alternative definition of the cross-product \(\bm{u} \times \bm{v}\) as the unique vector orthogonal to \(\bm{u}\) and \(\bm{v}\) with magnitude \[|\bm{u}\times\bm{v}| = |\bm{u}||\bm{v}| \sin(\theta)\,.\]

The relationship between these two characterizations of the cross product is hidden slightly afield within the study of linear algebra. However the familiar right-hand side of the latter equation should give us a hint; it looks like the area formula for a triangle, and the determinant is a measure how much area scales. Without going into more detail, here’s the fact to know: the magnitude of the cross product \(\bm{u} \times \bm{v}\) is the area of the parallelogram determined by \(\bm{u}\) and \(\bm{v}.\) This previous fact generalizes to higher dimensions. The scalar triple product \[ \bm{u}\cdot(\bm{v}\times\bm{w}) = \det\begin{pmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{pmatrix} \] is the (signed) volume of the parallelepiped framed by the vectors \(\bm{u}\) and \(\bm{v}\) and \(\bm{w}.\)

Physics intuition: torque is the cross product of the radial position vector and the force vector.

Before next class, review lines and surfaces from last week.

Distances Between Points, Lines, and Planes

Previously we talked about lines and surfaces in three-dimensions space from a purely analytic perspective. Since introducing vectors, we should talk about how to denote the equations of lines and surfaces using vector-centric notation.

Recall that, generically, the equation for the plane passing through the point \((x_0, y_0, z_0)\) is \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\,, \] where the coefficients \(a\) and \(b\) and \(c\) correspond to the “slopes” in various directions. But these coefficients have another interpretation. A vector is normal to a surface at a point if it is orthogonal to any tangent vector to the surface at that point. I.e. it “sticks straight out” from the surface. The vector \(\langle a,b,c \rangle\) is normal to that plane.

Any two vectors and a point determine a unique plane in three dimensional space. Considering this last fact, we now know how to determine this plane.

Given any plane in space and a point not on that plane, we can calculate the (shortest) distance between the point and plane by taking any vector from a point on the plane to our point and projecting onto the normal vector to the plane.

normal/orthogonal/perpendicular are all the same-ish

Parametrically a line through point \((x_0, y_0, z_0)\) is given by the equations \[ x = x_0 + at \qquad y = y_0 + bt \qquad z = z_0 + ct\,, \] where the coefficients \(a\) and \(b\) and \(c\) describe the “direction” in which the line is headed. In the language of vectors, the line is headed in the direction \(\bm{v} = \langle a,b,c \rangle\) after starting at the initial point \(\bm{r}_0 = \langle x_0,y_0,z_0 \rangle.\) which means the line can be expressed in terms of vectors as \(\bm{r} = \bm{r}_0 + t\bm{v}.\)

Among points, lines, and planes in space, between any two of those things, how do we compute the distance?

Before next class, if you’ve seen them in a physics class, review the concepts of force, work, power, torque, etc.

Physics: Force, Work, Torque, etc

Blow one of these days on a pop quizzzz

Today we consolidate all this talk of vectors with concepts you may have seen in physics.

A force, having a direction and magnitude (measured in Newtons), can be modelled by a vector. You may already be familiar with this if you’ve been drawing free-body diagrams in your physics classes. The resultant (net) force on a mass is the sum of all the individual component forces acting on that mass.

Work done by a force acting on an object is the amount of force accumulated as the object is displaced; work it is a scalar quantity. The basic formula, \(W = F d,\) applies if the object is being displaced in the same direction as the force is being applied. But if the force and displacement are in different directions, we need to generalize this formula. Let \(\bm{F}\) be the (constant) force vector with magnitude \(F\) and let \(\bm{d}\) be the displacement vector with magnitude \(d.\) Calculating work, it’s no longer the magnitude of the force itself that matters, but only the magnitude of the component of the force parallel to the displacement: \(|\bm{F}|\cos(\theta)\) where \(\theta\) is the angle between the vectors. Specifically \[ W = \Big(|\bm{F}|\cos(\theta)\Big)|\bm{d}| = \bm{F}\cdot\bm{d}\,.\] I.e. work is the dot product of the force and displacement vectors.

Later, remembering that if the force being applied is variable then we need to compute work as an integral, we’ll generalize this further to calculating work as the line integral \[ W = \int\limits_C \bm{F} \cdot \mathrm{d}\bm{d}\,. \]

Power \(P\) is the derivative of work over time, also a scalar, that measures the rate at which work is being done. It can be computed as the dot product of the force and velocity vectors: \[ P = \dot{W} = \frac{\mathrm{d}W}{\mathrm{d}t} = \bm{F}\cdot\bm{v}\,. \]

The last formula is valid even in the more general situation when the force is nonconstant and applied along a curve \(C\) due to the fundamental theorem of calculus \[ P = \frac{\mathrm{d}W}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \int\limits_C \bm{F}\cdot\mathrm{d}\bm{d} = \frac{\mathrm{d}}{\mathrm{d}t}\int\limits_{\Delta t} \bm{F}\cdot\bm{v}\,\mathrm{d}t = \bm{F}\cdot\bm{v}\,. \]

Torque, a vector, is the rotational analogue of linear force, and measures the tendency of a body to rotate about the origin. Imagine you’re using a wrench to tighten a bolt. The bolt-head is the origin, the direction of the wrench is described by a (radial) position vector \(\bm{r}.\) If we apply a force \(\bm{F}\) to the tip of the wrench handle, the torque \(\bm{\tau}\) is computed as \[\bm{\tau} = \bm{r} \times \bm{F}\,.\] Note the order of the components of this cross product — the direction of the vector \(\bm{\tau}\) will be determined mathematically by the right-hand-rule, but be determined physically the direction of the threads on the bolt: righty-tighty lefty-loosey.

Torque is the first moment of force: \(\bm{\tau} = \bm{r} \times \bm{F}.\) Torque is the rate of the change of angular momentum: \(\bm{\tau} = \frac{\mathrm{d}\bm{L}}{\mathrm{d}t}.\) The derivative of torque over time is called rotatum.

Momentum is a vector quantity, is the product of an object’s mass with its velocity: \(\bm{p} = m\bm{v}.\) Force is the derivative of momentum: \(F = dp/dt,\) and impulse \(J\) is the change in momentum between t_1 and t_2 : \[ \Delta p = J = \int\limits_{t_1}^{t_2} F(t)\,\mathrm{d}t \] Angular momentum is first moment of momentum: \(\bm{R} = \bm{R}\times \bm{p}.\)

The moment of inertia (rotational inertia) is the second moment of mass, the torque needed for a desired angular acceleration. Mass|Force→Acceleration as RotationalInertia|Torque→AngularAcceleration.

Before next class, TK Question/Review for next class

Calculus of Vector-Valued Functions: Curves

TK

TK

Parametrically-Defined Curves and Surfaces

Calculus with Vector-Valued Functions

TK

Start class playing line-rider.

Have running example pulled up in Desmos

So far we haven’t talked about functions or calculus; this changes now.

The functions you’ve seen in previous calculus classes have been “real-valued functions of a real input,” \(f \colon \mathbf{R} \to \mathbf{R},\) the domain being that first \(\mathbf{R}\) and the codomain being that second \(\mathbf{R}.\) Today we generalize this to functions whose output is no longer one-dimensional.

A function is vector-valued (instead of scalar-valued) if the values of its outputs are vectors (instead of scalars). I.e. the codomain is \(\mathbf{R}^n\) instead of just \(\mathbf{R}.\) In particular we’ll study the vector-valued functions \(\bm{r}\colon \mathbf{R} \to \mathbf{R}^3.\)

The symbol \(\bm{r}\) is conventionally used because the curve is traced out \(\bm{r}\)adially.

Sines and cosines suggest a helix.

When sketching manually, imagine curves two components at a time.

Those individual functions that define each coordinate are called the component functions.

TK Limits continuity

space curves: which can be visualized as curves in three-dimensional space parameterized by the components of their outputs.

The derivative of a vector-valued function \(\bm{r}\) is \[ \frac{\mathrm{d}\bm{r}}{\mathrm{d}t} = \bm{r}'(t) = \lim_{h \to 0} \frac{\bm{r}(t+h) - \bm{r}(t)}{h}\,. \] You can take the derivatives of individual components.

We need a smooth parameterization — where \(\bm{r}'(t) \neq \bm{0}\) so it “never stops”.

Evaluating the derivative at a parameter \(t\) givens you a tangent vector. Sometimes we like to “normalize” this, make sure this vector has magnitude one, so we talk about the unit tangent vector \[ \mathbf{T}_{\bm{r}}(t) = \frac{\bm{r}'(t)}{|\bm{r}'(t)|} \]

There are two related definitions for the word smooth floating around: one for curves and one for parameterizations of curves. We’ll say a parameterization \(\bm{r}\) of a curve \(C\) is smooth if \(\bm{r}'\) exists and is nonzero at each point along the curve. Now, \(\bm{r}'\) being nonzero is a feature of the parameterization, whereas \(\bm{r}'\) existing is a feature of the curve \(C,\) so we’ll say a curve \(C\) is smooth of there exists a smooth parameterization \(\bm{r}\) of \(C.\) And all of this conflicts with the more general notion of the smoothness of a function anyways.

All the rules of differentiation we know and love from single-variable calc have analogs in higher dimensions.

This last one is the chain rule, and the three before are analoges of the product rule.

The definite integral of a vector-valued functions can also be defined component-wise, … but what does this even represent?

Before next class, TK Question/Review for next class

Arclength, Curvature, and Torsion

Recall the formula for the arclength of a parameterized curve in \(\mathbf{R}^2.\) It’s hardly any different in higher dimensions. Given \(\bm{r}(t) = \langle x(t), y(t), z(t)\rangle,\) the arclength of the curve \((x,y,z) = \bm{r}(t)\) between \(t=a\) and \(t=b\) is \[ \int\limits_a^b \sqrt{\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2 + \bigg(\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)^2 + \bigg(\frac{\mathrm{d}z}{\mathrm{d}t}\bigg)^2} \,\mathrm{d}t \quad = \quad \int\limits_a^b \big|\bm{r}'(t)\big| \,\mathrm{d}t \]

We can re-parameterize a curve in terms of it's arclength to get a “canonical” parameterization. For a curve \(\bm{r}\) define an arclength function \(s\) as \[s(t) = \int\limits_0^t \big|\bm{r}'(u)\big| \,\mathrm{d}u\] and notice that \(s\) is a strictly increasing function, and so it must be invertible on its entire domain. The parameterization \(\bm{r}\circ s^{-1}\) is the canonical one we’re talking about: up to a choice of “anchor point” where \(t=0\), it has the property that the point corresponding to a specific \(t = \ell\) is at a distance to \(\ell\) from that anchor point.

The curvature of a smooth curve \((x,y,z) = \bm{r}(t)\) is a measure of how eccentrically it’s turning at a point — how tightly the curve bends — and is defined as \[ \kappa = \bigg|\frac{\mathrm{d}{\mathbf{T}}}{\mathrm{d}s}\bigg| = \bigg|\frac{\mathrm{d}{\mathbf{T}}/\mathrm{d}t}{\mathrm{d}s/\mathrm{d}t}\bigg| = \bigg|\frac{\mathbf{T}'(t)}{\bm{r}'(t)}\bigg| = \frac{\big|\bm{r}'(t) \times \bm{r}''(t)\big|}{\big|\bm{r}'(t)\big|^3} \] where \(s\) is the arclength function. That last equality deserves a proof (TK theorem 10 page 945). The thing to keep in mind: the curvature of a circle of radius \(R\) is \(\frac{1}{R}.\) (related to osculating circle later)

Velocity and Acceleration along Curves

Instead of \(\bm{r}(t)\) tracing out a curve in space, we can think of it as being the location of something — a particle — at a time \(t;\) the curve is the path of the particle. Its velocity is the vector \(\bm{r}'(t);\) it's magnitude is the “speed”. The second derivative vector \(\bm{r}'(t)\) to a curve is its acceleration vector. Etc.

For a projectile fired from the origin \((0,0,0)\) within the \(xy\)-plane under in the influence of gravitational acceleration \(g\) in the \(z\) direction, the parametric equations that describe its trajectory are

The acceleration vector \(\bm{\alpha}\) always lies in the osculating plane, and can be written as \( \bm{\alpha} = |\bm{v}|'\mathbf{T}+\kappa|\bm{v}|^2\mathbf{N}. \) As proof \[ \bm{v} = |\bm{v}|\mathbf{T} \\\implies \bm{\alpha} = |\bm{v}|'\mathbf{T} + |\bm{v}|\mathbf{T}' \\\implies \bm{\alpha} = |\bm{v}|'\mathbf{T} + |\bm{v}|\bigl(\mathbf{N}|\mathbf{T}'|\bigr) \\\implies \bm{\alpha} = |\bm{v}|'\mathbf{T} + |\bm{v}|\Bigl(\mathbf{N}\bigl(\kappa |\bm{v}|\bigr)\Bigr) \\\implies \bm{\alpha} = |\bm{v}|'\mathbf{T} + \kappa|\bm{v}|^2\mathbf{N}. \]

TK my notes are honed in on Example #3 and the IVP for some reason here.

Before next class, TK Question/Review for next class

The Frenet-Serret and Darboux Frame

Actually have a pop quiz today, and talk about this tomorrow.

Normal and binormal vectors. Since \(\mathbf{T}_{\bm{r}}(t)\) is a unit vector, it will be orthogonal to its derivative: thinking of the vector \(\mathbf{T}_{\bm{r}}(t)\) as anchored at the origin, this is true because \(\mathbf{T}_{\bm{r}}\) is constant, the terminal point of its outputs all lying on the unit sphere. We can define the unit normal vector as \[\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\big|\mathbf{T}'(t)\big|}\] which as a vector indicates the “direction the curve is turning”. Then we define the unit binormal vector as \[\mathbf{B}(t) = \mathbf{T}(t)\times\mathbf{N}(t)\,.\] (Note that the order is canonical, and important.)

TNB frame is Frenet-Serret Frame and normal plane is spanned by normal and binormal vectors osculating plane (latin for osculum for to kiss) is spanned by the tangent and normal vectors circle of curvature osculating circle

Just like the tangent vector can be used to measure how eccentrically the curve bends, its curvature, the normal and binormal vectors can be used similarly to measure how eccentrically a curve “twists”, its torsion, which we explicitly calculate as \[ \tau = -\frac{\mathrm{d}\mathbf{B}}{\mathrm{d}s} \cdot \mathbf{N} = - \frac{\mathbf{B}' \cdot \mathbf{N}}{\big|\bm{r}'\big|} = \frac{\Big(\bm{r}' \times \bm{r}''\Big) \cdot \bm{r}'''(t)}{\big|\bm{r}'\times\bm{r}''\big|^2} \]

That last equality deserves a proof (TK exercise #72). Torsion is how much the curve is pulling out of its osculating plane. Since \(\mathbf{B}'\) is normal to this plane, it makes sense that the formula for \(\tau\) is based on it. It’s necessary in that calculation to see that \(\mathbf{B}'\) is parallel to \(\mathbf{N}\) (exercise), Because \(\mathbf{N}\) is a unit vector though, \(|\mathbf{B}'| = \tau.\) so the torsion will be how proportional their lengths are. The negative sign is just a convention, but corresponds to a curve coming UP out of the plane as having positive torsion.

curvature is the failure of a curve to be linear and torsion is the failure of a curve to be planar.

TK Frenet-Serret Apparatus and the relations T'=(kappa)N and B'=-tN and N'=-(kappa)T+tB.

Before next class, TK Question/Review for next class

Kepler’s Laws of Planetary Motion

TK

Calculus of Multivariable Functions: Surfaces

How does the calculus of single-variable functions extend to multivariable functions?

Now we discuss the “dual” to vector-valued functions of a scalar input, scalar-valued functions of multiple variables, \(f \colon \mathbf{R}^n \to \mathbf{R}.\) We’ll talk about what continuity means in this context, generalize the notion of the first and second derivatives to gradient and Hessian respectively, and discuss the problems of linearization and optimization in this higher-dimensional context.

Multivariable Functions and their Graphs

How do we “visualize” multivariable functions?

Consider functions \(f \colon \mathbf{R}^n \to \mathbf{R},\) where the input is a vector and the output is a scalar. These are called multivariable, scalar-valued functions. In particular we’ll study the case when \(n=2,\) for functions \(f\colon \langle x,y \rangle \mapsto z,\) where we can think about the graph \(z = f(x,y)\) as a surface in three-dimensional space living over the \((x,y)\)-plane. But we’ll also consider functions for \(n=3.\)

Any specific output value \(z_0\) corresponds to a horizontal cross section of the graph \({z_0 = f(x,y).}\) Sketching these cross sections can help us visualize the graph. In general we call these cross sections level sets, but for a function \(f \colon \mathbf{R}^2 \to \mathbf{R}\) they will be curves, so we call them level curves. Altogether the plot of a healthy collection of a graph’s leve sets is called its contour plot. It’s like the topography map of terrain on the earth, where each level curve corresponds to an elevation.

For a function \(f \colon \mathbf{R}^3 \to \mathbf{R}\) the level sets will be surfaces, so we call them level surfaces. In fact since the graphs of these functions would need four dimensions to visualize, we really only have their contour plots to visualize them.

Before next class, review the notation of limits and the definition of continuity.

Limits & Continuity

Write the definition of the limit and continuity on the board to be updated for this class.

Recall the definite of a limit from single-variable calculus: we’ll say \(\lim_{x \to a} f(x) = L\) if for every neighborhood of \(L\) there exists a neighborhood of \(a\) such that for all \(x\) in that neighborhood of \(a\) \(f(x)\) will be in that neighborhood of \(L.\) a function \(f\) is continuous at a point \(a\) \(\lim_{x \to a} f(x) = f(a),\) and is continuous on its domain if its continuous at every point in its domain.

For multivariable functions we only need to update the fact that points in our domain are not numbers \(a\) but are now points \((a,b):\) we’ll say \(\lim_{(x,y) \to (a,b)} f(x,y) = L\) if for every neighborhood of \(L\) there exists a neighborhood of \((a,b)\) such that for all \((x,y))\) in that neighborhood of \((a,b)\) \(f(x,y)\) will be in that neighborhood of \(L.\) Then the definition of continuity in this context is roughly the same: a function \(f\) is continuous at a point \((a,b)\) \(\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b),\) and is continuous on its domain if its continuous at every point in its domain.

For single-variable, when evaluating limits, there are only two directions from which to approach a point, two directions from which to enter the neighborhood of \(a,\) and so only two values that must agree for a limit to exist. For multivariable functions things become more complicated. A limit only exists when the limit along any path approaching the point \((a,b)\) exists and that value is the same for any path. To show that a limit doesn’t exist we must exhibit two paths in the domain towards \((a,b)\) along which the values of the limit differ.

Broad advice for evaluating limits: (1) if you know in your heart a function is continuous at the point in question based on its formula — e.g. it’s a polynomial function — then the limit is the value at that point, (2) a visualization, either the graph or the contour plot, can help give you intuition on whether or not the limit exists, and (3) if you suspect a limit doesn’t exist its fairly quick to check the “straight line” paths \(x=0\) and \(y=0\) and \(y=x\) and \(y=-x.\) If the limit agrees along all those paths but you’re still convinced the limit doesn’t exist you’ll need to come up with a path more cleverly based on the formula for the function.

But how do we prove a limit does exist? How do we prove the value of \(\lim_{(x,y) \to (a,b)} f(x,y)\) is consistent along all paths terminating in \((a,b)?\) One clean way to do so is to shift our plane to center on \((a,b),\) convert the domain and function’s formula into polar coordinates, and show that the limit as \(r\) approaches \(0\) doesn’t depend at all on the value of \(\theta.\)

Before next class, review all your rules of single-variable differentiation, and review the geometric interpretations of derivatives in terms of rates, slopes, concavity, etc.

Partial Derivatives

What’s a “derivative” when you have more than one independent variable?

The derivative of a single-variable function gives you information about the rate at which the function is increasing or decreasing corresponding to the slope of a line tangent to the graph of the function. For a multivariable function there are infinitely many directions to measure the rate of change; i.e. there is no single tangent line to a point on a surface, but instead a tangent plane. The derivative of a multivariable function cannot simply be another function, but is actually a vector of functions. Today we talk about the components of that vector, the rates-of-change of a function in the \(x\)- and \(y\)-direction. These are called partial derivatives. \[ f_x(x,y) = \lim\limits_{h \to 0} \frac{f(x+h, y) - f(x,y)}{h} \qquad f_y(x,y) = \lim\limits_{h \to 0} \frac{f(x, y+h) - f(x,y)}{h} \] The partial derivative of a function \(f\) with respect to \(x\) may be written any one of these various ways:

The “\(\mathrm{D}\)” is a short-hand for the differential operator, and the “\(1\)” is a reference to \(x\) being the first parameter of \(f\) without requiring it to have a name. They’re only “partial” derivatives because the “full” derivative is be a vector pieced together from these partial derivatives. More on that next week. To compute \(f_x,\) take the derivative with respect to \(x\) and pretend that the other variables are just constants.

Second partial derivatives, [TK] but how do you think of them?

If \(f\) is defined on some disk containing the point \((a,b)\) and \(f_{xy}\) and \(f_{yx}\) are both continuous on that disk, then \(f_{xy}(a,b) = f_{yx}(a,b).\)

Before next class, review how to find the equation of a line tangent to the graph of a function at a point. Additionally, practice determining limits and taking partial derivatives; these are two concrete skills to get proficient with ASAP so that they don’t slow you down later.

Tangent Planes

What’s an equation for the plane tangent to a surface at a point?

I put it in that latter form so it looks like the general equation of a plane \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0,\) because we can answer the same question about a plane tangent to the graph of \(z = f(x,y)\) at a point.

To answer the question, we need the vector \(\langle A, B, C\rangle\) normal to the graph at that point. Recall that \(f_x\) and \(f_y\) represent slopes in the \(x\) and \(y\) direction respectively, from which we can infer that the vectors \(\bigl\langle 1,0,f_x \bigr\rangle\) and \(\bigl\langle 0,1,f_y \bigr\rangle\) span the tangent plane at that point. If those span the plane, their cross product will be normal to the plane: \[ \bigl\langle 0,1,f_y \bigr\rangle \times \bigl\langle 1,0,f_x \bigr\rangle = \bigl\langle f_x, f_y, -1 \bigr\rangle \] So in general, for a graph \(z = f(x,y),\) letting \(z_0 = f(x_0, y_0),\) the plane \[ f_x(x_0, y_0)\bigl(x-x_0\bigr) + f_y(x_0, y_0)\bigl(y-y_0\bigr) - \bigl(z-z_0\bigr) = 0 \] will be tangent to the graph at the point at \(\bigl(x_0, y_0, z_0\bigr).\)

Before next class, briefly review the chain rule and implicit differentiation. Additionally, practice determining limits and taking partial derivatives; these are two concrete skills to get proficient with ASAP so that they don’t slow you down later.

The Chain Rule & Implicit Differentiation

Pop Quiz today instead?

TK Why do we need this? Stare at the book. It's just when you have crazy composite functions, where like z = f(x,y) where x and y are dependent on some other parameters s and t. What a soup of variables.

Given \(z = f(x,y)\) such that \(x\) and \(y\) are functions of some parameter \(t,\) \[ \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{\partial f}{\partial x} \frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}t} \,. \]

The Gradient Vector & Directional Derivatives

If partial derivatives are only “partial” derivatives what’s the “full” derivative?

Given a plane in space, what’s its steepest slope, and what’s is slope in a particular direction?

The gradient vector of a function \(f,\) denoted \(\nabla f\) or sometimes as \(\operatorname{grad}f,\) is the vector of partial derivatives of \(f\). This is “full” derivative of a multivariable function. Explicitly, in two variables, \[ \operatorname{grad}f(x,y) = \nabla f(x,y) = \Big\langle f_x(x,y), f_y(x,y) \Big\rangle = \frac{\partial f}{\partial x}\bm{i} + \frac{\partial f}{\partial y}\bm{j} \,. \]

Two key facts to thinking about what this vector represents geometrically:

• In the domain of \(f\) on its the contour plot, the vector \(\nabla f(a,b)\) will be perpendicular to the level curve of \(f\) containing \((a,b).\)
• On the surface \(z = f(x,y)\) at the point \(\bigl(a,b, f(a,b)\bigr),\) the vector \(\nabla f(a,b)\) will point in the direction in which \(f\) is increasing the fastest — the direction in which the tangent plane at the point is steepest — and this maximal rate of increase will be \(\bigl|\nabla f(a,b)\bigr|.\)

Can I prove/justify those facts?

The gradient is the direction of steepest ascent (maximal increase), and the magnitude of the gradient is how steep it is, but how can we calculate the rate of increase in another direction? Just take the dot product of the gradient with that direction. For a differentiable function \(f\) and a unit vector \(\bm{v} = \langle v_1, v_2\rangle,\) the directional derivative of \(f\) in the direction \(\bm{v},\) denoted \(\operatorname{D}_{\bm{v}} f,\) is the function \(\nabla f \cdot \bm{v}.\) Written out explicitly, \[\operatorname{D}_{\bm{v}} f(x,y) = \nabla f(x,y) \cdot \bm{v} = f_x(x,y) v_1 + f_y(x,y) v_2\,.\] Then \(\operatorname{D}_{\bm{v}} f(a,b)\) will give the rate at which \(f\) is increasing (or decreasing) at the point \((a,b)\) in the direction \(\bm{v}.\) In effect, because \(\bm{v}\) is a unit vector, \(\operatorname{D}_{\bm{v}} f(a,b)\) is just \(\operatorname{proj}_{\bm{v}}\bigl(\nabla f(a,b)\bigr).\) If \(\theta\) is the angle between \(\nabla f(a,b)\) and \(\bm{v}\) then \(\operatorname{D}_{\bm{v}} f(a,b) = \bigl|\nabla f(a,b)\bigr|\cos(\theta).\)

Before next class, review optimization problems from single-variable calculus, the “first- and second-derivative tests” and the concavity of the graph of a function. Recall that a function \(f\) has local extrema only where the derivative of \(f\) is zero, where the tangent line is horizontal. You can imagine now that if a multivariable function \(f\) has a local max or min at a point \((a,b),\) then \(\nabla f(x,y) = \bm{0}\) — the tangent plane will be flat.

Local and Global Extrema

Define local minimum/maximum/extrema, and absolute versions.

[analog of Fermats theorem?] If \(f\) has a local extrema at \((a,b)\) and \(f_x(a,b)\) and \(f_y(a,b)\) exist, then \(f_x(a,b) = 0\) and \(f_y(a,b) = 0.\)

critical point saddle point

Hessian matrix \[ \mathbf{H}_f = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix} \] The determinant of the Hessian matrix, denoted \(\operatorname{det}\mathbf{H}_f,\) and sometimes called the discriminant of \(f\) and denoted \(D_f,\) can be explicitly expressed as \[ \det\mathbf{H}_f = f_{xx}f_{yy} - \big(f_{xy}\big)^2\,. \]

[TK] eigenvalues?

Second Derivative Test

If \((a,b)\) is a critical point of \(f\) and the second partial derivatives of \(f\) are continues in some neighborhood of \((a,b),\) then

• if \(\mathrm{D}_f(a,b) \gt 0\) and \(f_{xx}(a,b) \gt 0,\) then \(f(a,b)\) is a local minimum;
• if \(\mathrm{D}_f(a,b) \gt 0\) and \(f_{xx}(a,b) \lt 0,\) then \(f(a,b)\) is a local maximum;
• if \(\mathrm{D}_f(a,b) \lt 0\) then \((a,b)\) is a saddle points.

If \(f\) is continuous on a closed, bounded subset of its domain, then \(f\) attains an absolute minimum and maximum value on that subset.

So to find the absolute extrema on the closed bounded region you’ve gotta check the critical points and any local extrema on the boundary of the region.

Before next class, TK Question/Review for next class

Optimization with Lagrange Multipliers

To find the extreme values of \(f(x,y,z)\) subject to \(g(x,y,z) = k,\) assuming these values exist and \(\nabla g \neq 0\) on the surface \(g(x,y,z) = k,\) the critical points will be any \((a,b,c)\) such that \[\nabla f(a,b,c) = \lambda \nabla f(a,b,c)\] for some real number \(\lambda.\)

The number \(\lambda\) is called the Lagrange multiplier for that point.

Do two constraints? Optimization problems?

Do hella examples.

Before next class, TK Question/Review for next class

TK

pop quiz

Integration in Higher Dimensions

TK

TK

Double Integrals in Rectangular Coordinates

Review of one-dimensional integration. For a sample of \(n\) points \(x_i^\ast\) in the interval \([a,b]\) you can form the Riemann sum \[ \sum_{i=1}^n f\big(x_i^\ast\big) \Delta x \] and take the limit to define the definite integral: \[ \int\limits_a^b f(x) \,\mathrm{d}x = \lim_{n \to \infty} \sum_{i=1}^n f\big(x_i^\ast\big) \Delta x \]

Over higher-dimensional domains we define integration similarly. For a function \(f\colon \mathbf{R}^2 \to \mathbf{R}\) and for a rectangular region \(R = [a,b] \times [c,d]\) we define the double integral as an iterated integral \[ \iint\limits_R f(x,y) \,\mathrm{d}A \;\;=\;\; \int\limits_a^b\int\limits_c^d f(x,y) \,\mathrm{d}y\,\mathrm{d}x \] where \(\mathrm{d}A = \mathrm{d}x\,\mathrm{d}y.\) This integral computes the signed area of the region bound between the graph \(z = f(x,y)\) and the \(xy\)-plane. Mechanically speaking, to evaluate such integrals, just take it one-variable-at-a-time.

There is an analogue to Clairaut’s Theorem for integration.

If \(f\) is continuous on the rectangle \(R = [a,b] \times [c,d]\) then \[ \int\limits_a^b\int\limits_c^d f(x,y) \,\mathrm{d}y\,\mathrm{d}x = \int\limits_c^d\int\limits_a^b f(x,y) \,\mathrm{d}x\,\mathrm{d}y \]

Conceptualizing integration of a function \(f,\) it may help to think of \(f\) as a density function on the region, and think of the iterated integral \(\iint_R f(x,y)\,\mathrm{d}A\) as computing the mass of the region given that density. On that note, we can compute the area of the region \(R\) as \(\iint_R 1 \,\mathrm{d}A\) — the area of a region is the same as its mass if its density is uniformly \(1.\) And we can compute the average value of a function \(f\) on a region \(R\) by dividing the integral of \(f\) over that region by that region’s area. I.e. the average value of \(f\) over \(R\) is \[ \frac{1}{\operatorname{Area}(R)}\iint\limits_R f(x,y)\,\mathrm{d}A. \]

Before next class, TK Question/Review for next class

It’s a little bit trickier if the region we’d like to integrate over is not a rectangle, but only a little bit. We’ve just gotta describe the boundary of the region analytically.

Before next class, review integration in polar coordinates.

Double Integrals in Polar Coordinates

Recall that a point \((x,y)\) in rectangular (Cartesian) coordinates can be expressed in polar coordinates as \(\Big(\sqrt{x^2+y^2}, \operatorname{atan2}(y,x)\Big).\)

polar “rectangle”

For a polar rectangle \(R\) we have \[ \iint\limits_{R} f(x,y) \,\mathrm{d}A = \int\limits_\alpha^\beta \int\limits_a^b f\big(r\cos(\theta),r\sin(\theta)\big) r \,\mathrm{d}r\,\mathrm{d}\theta \] And we must similarly extrapolate if \(R\) is not a polar rectangle, but is instead has inner and outer radii defined as the graphs of functions.

Examples

Before next class, TK Question/Review for next class

Applications of Double Integrals

Density mass electric charge first- second-moments center of mass inertia probability?

The surface area of the graph \(z = f(x,y)\) above a region \(R\) is given by the integral \[ \iint\limits_{R} \sqrt{ \Big(f_x(x,y)\Big)^2 +\Big(f_y(x,y)\Big)^2 +1 } \,\mathrm{d}A \]

Before next class, TK Question/Review for next class

Change of Coordinates and the Jacobian

Triple Integrals in Rectangular Coordinates

Mechanically triple integrals can be evaluated just the same as double integrals: by considering them as three iterated integrals and evaluating them one-at-a-time. It’s easiest to think of them as a calculation of mass given a density function.

Fubini’s theorem still holds.

Examples

Centroid and moment examples are the same.

Before next class, TK Question/Review for next class

Triple Integrals in Cylindrical and Spherical Coordinates

pop quiz on the second day of this

Cylindrical coordinates \[ x = r\cos(\theta) \qquad y = r\sin(\theta) \qquad z = z \] and integrals look like \[ \iiint\limits_R f(x,y,z) \,\mathrm{d}V = \int\limits_{\alpha}^{\beta} \int\limits_{TK}^{TK} \int\limits_{TK}^{TK} f\big(r\cos(\theta),r\sin(\theta),z\big) r \,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta \]

Before next class, TK Question/Review for next class

Spherical coordinates \[ x = \rho\sin(\phi)\cos(\theta) \qquad y = \rho\sin(\phi)\sin(\theta) \qquad z = \rho\cos(\phi) \] and integrals look like \[ \iiint\limits_R f(x,y,z) \,\mathrm{d}V = \int\limits_{c}^{d} \int\limits_{\alpha}^{\beta} \int\limits_{\gamma}^{\delta} f\big(r\cos(\theta),r\sin(\theta),z\big) \rho^2\sin(\phi) \,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\phi \] with conditions on the wedge.

Pop quiz on day 2 of this topic.

Before next class, TK Question/Review for next class

Change of Coordinates and the Jacobian Redux

Suppose a transformation from … TK. Then the Jacobian of this transformation is given by \[ \frac{ \partial(x,y) }{ \partial(u,v) } = \operatorname{det}\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \] in two-dimensions, or in three-dimensional space \[ \frac{ \partial(x,y,z) }{ \partial(u,v,w) } = \operatorname{det}\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{pmatrix} \]

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Application of Triple Integrals

pop quiz

TK

Before next class, TK Question/Review for next class

Introduction to Vector Calculus

TK

TK

Vector Fields

A vector field is a function that assigns a vector to each point in your space. … \(\bm{F}\colon \mathbf{R}^n \to \mathrm{R}^n\) E.g. in three-dimensional space \(\mathbf{R}^3\) both the domain and the codomain of a

MANY Examples and visualizations.

For a function \(f,\) the gradient operator \(\nabla\) gives us a vector field \(\nabla f\) called the gradient vector field of \(f.\) A vector field is called conservative if it is the gradient vector field of a scalar function. In this situation that \(\bm{F} = \nabla f\) is conservative, we call \(f\) a potential function for \(\bm{F}.\)

What IS a conservative vector field?

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Line Integrals in Space

If \(f\) is defined on a smooth curve \(C\) parametrically-defined as \(\big(x(t), y(t)\big)\) for \(a \leq t \leq b,\) then the line integral — or more specifically the line integral with respect to arclength — of \(f\) along \(C\) is defined to be \[ \int\limits_C f(x,y) \,\mathrm{d}\ell = \int\limits_a^b f(x,y) \sqrt{\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2 + \bigg(\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)^2} \,\mathrm{d}t \] where \(\mathrm{d}\ell\) is a tiny change in arclength. If C is not smooth, but is piecewise smooth, then the integral of \(f\) over \(C\) is the sum of the integrals of \(f\) over each of the smooth components.

This generalized to curves in three-dimensions as one would expect.

If the curve \(C\) can be expressed as the graph of a function \(y = f(x),\) then we may express the same line integral with respect to \(x\) instead: \[ \int\limits_C f(x,y) \,\mathrm{d}x = \int\limits_a^b f(x,y) \,x'(t) \,\mathrm{d}t \] and similarly if the curve is \(x = f(y).\) When line integrals with respect to \(x\) and with respect to \(y\) occur together, it’s conventional to write \[ \int\limits_C P(x,y) \,\mathrm{d}x + \int\limits_C Q(x,y) \,\mathrm{d}y = \int\limits_C P(x,y) \,\mathrm{d}x + Q(x,y) \,\mathrm{d}y \]

It’s helpful to recall that a line segment with initial point \(\bm{v}_0\) and terminal point \(\bm{v}_1\) can be parameterized as \(\bm{r}(t) = (1-t)\bm{v}_0 + t\bm{v}_1\) for \(0 \leq t \leq 1.\)

A curve’s parameterization determines an orientation; i.e. the direction along which the curve is traversed. Given a curve \(C\) with implied orientation, we’ll let \(-C\) denote the same curve traversed in the opposite direction.

Line Integrals in Vector Fields

For a continuous vector field \(\bm{F}\) defined on a smooth curve \(C\) given by the vector-valued function \(\bm{r}(t)\) for \(a \leq t \leq b,\) the line integral of \(\bm{F}\) along \(C\) is \[ \int\limits_C \bm{F}\cdot\mathrm{d}\bm{r} = \int\limits_a^b \bm{F}\big(\bm{r}(t)\big)\cdot\bm{r}'(t)\,\mathrm{d}t = \int\limits_C \bm{F}\cdot\bm{T}\,\mathrm{d}\ell \] where \(\bm{T}\) is the unit tangent vector at a point on the curve.

OH WE"RE NOT DONE! ANOTHER CHAPTER

Given a smooth curve \(C\) given by the vector-valued function \(\bm{r}\) parameterized for \(a \leq t \leq b.\) If \(f\) is a differentiable function whose gradient vector \(\nabla f\) is continuous on \(C,\) then \[ \int\limits_C \nabla f \cdot \mathrm{d}\bm{r} = f\big(\bm{r}(b)\big) - f\big(\bm{r}(a)\big) \]

Independence of path, and simple closed curves.

And a BUNCH of theorems about conservative vector fields.

Before next class, TK Question/Review for next class

Surface Integrals in Vector Fields

Just like we can define a curve in two-dimensional space parametrically where each coordinate is a function of a single variable, we can parametrically define a surface (manifold) in three-dimensional space where each of the three coordinates is a function of two variables. E.g. a surface \(\mathcal{S}\) can be defined as \[ \mathcal{S}(u,v) = x(u,v)\mathbf{i} +y(u,v)\mathbf{j} +z(u,v)\mathbf{k} \]

Show examples, with grid curves along which \(u\) and \(v\) are constant. Surfaces of revolution and tangent planes and graphs as examples.

Given a smooth surface \(S\) over a domain \(R\) define as \[ \mathcal{S}(u,v) = x(u,v)\mathbf{i} +y(u,v)\mathbf{j} +z(u,v)\mathbf{k} \] such that \(S\) is "covered just once" as \(u,v\) vary (rectifiable?) the surface area of \(S\) can be calculated as \[ \iint\limits_R \big| S_u \times S_v\big| \,\mathrm{d}A \] where \[ S_u = \frac{\partial x}{\partial u}\mathbf{i} + \frac{\partial y}{\partial u}\mathbf{j} + \frac{\partial z}{\partial u}\mathbf{k} \qquad S_v = \frac{\partial x}{\partial v}\mathbf{i} + \frac{\partial y}{\partial v}\mathbf{j} + \frac{\partial z}{\partial v}\mathbf{k} \]

Before next class, TK Question/Review for next class

Surface Integrals : Surface Area :: Line Integrals : Arclength.

Given a parametrically-defined surface \(S\) over domain \(R\) defined as \[ \mathcal{S}(u,v) = x(u,v)\mathbf{i} +y(u,v)\mathbf{j} +z(u,v)\mathbf{k} \] and a function \(f\) defined on \(S,\) the surface integral of \(f\) over \(S\) is \[ \iint\limits_S f(x,y,z) \,\mathrm{d}S = \iint\limits_R f\big(S(u,v)\big)\,\big| S_u \times S_v\big| \,\mathrm{d}A \]

Oriented surfaces need to be cleared up before talking about flux — surface integrals over vector fields.

If \(\bm{F}\) is a continuous vector field defined on an oriented surface \(S\) with normal vector \(\mathbf{n}_S,\) then the surface integral of \(\bm{F}\) over \(S\) can be calculated as \[ \iint\limits_S \bm{F}\cdot\mathrm{d}\mathbf{S} = \iint\limits_S \bm{F}\cdot \mathbf{n}_S \,\mathrm{d}S \] this is also called the flux of \(\bm{F}\) across \(S.\)

Electric flux example

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Green’s Theorem

This relates the line integral over a boundary of a region with the double integral over the interior of the region.

A closed curve is positively oriented if it is being traversed counter-clockwise.

For a positively oriented, piecewise-smooth, simple closed planar curve \(C\) with interior region \(R\) if \(P\) and \(Q\) have continuous partial derivatives on some open neighborhood containing \(R,\) then \[ \int\limits_C P\,\mathrm{d}x + Q\,\mathrm{d}y = \iint\limits_R \bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\bigg)\,\mathrm{d}A \]

A short-hand notation for the boundary of a region \(R\) is \(\partial R\) which is a brilliant overload of the partial-differential operator in light of Stokes’ Theorem.

When the orientation of the curve \(C\) needs to be acknowledged we use the notation \[\oint\limits_C P\,\mathrm{d}x + Q\,\mathrm{d}y\] and sometimes even put a little arrow on that circle in \(\oint\) to specify the orientation.

We can extend Green’s theorem to non-simple closed regions.

Bio on George Green.

Before next class, TK Question/Review for next class

Curl and Divergence

For a vector field \(\bm{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\) the curl of \(\bm{F},\) denoted \(\operatorname{curl}\bm{F}\) is defined to be \[ \operatorname{curl}\bm{F} = \nabla \times \bm{F} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{pmatrix} = \bigg(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\bigg)\mathbf{i} + \bigg(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\bigg)\mathbf{j} + \bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\bigg)\mathbf{k} \]

If \(f\) has continuous second-order partial derivatives, then \(\operatorname{curl} \nabla f = \mathbf{0}.\)

If \(\bm{F}\) is defined on all of \(\mathbf{R}^3\) and if the components functions of \(\bm{F}\) have continuous second-order partial derivatives, and \(\operatorname{curl} \bm{F} = \mathbf{0},\) then \(\bm{F}\) is a conservative vector field.

For a vector field \(\bm{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\) the divergence of \(\bm{F},\) denoted \(\operatorname{div}\bm{F}\) is defined to be \[ \operatorname{div}\bm{F} = \nabla \cdot \bm{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \]

If \(\bm{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\) is defined on all of \(\mathbf{R}^3\) and if the components functions of \(\bm{F}\) have continuous second-order partial derivatives, then \(\operatorname{div}\operatorname{curl} \bm{F} = 0.\)

Laplace operator \(\nabla^2 = \nabla \cdot \nabla\)

Green’s theorem in vector form \[ \oint\limits_C \bm{F}\cdot\mathrm{d}r = \oint\limits_C \bm{F}\cdot \mathbf{T}_r \,\mathrm{d}\ell = \iint\limits_R \big(\operatorname{curl}\bm{F}\big) \cdot \mathbf{k} \,\mathrm{d}A \] or alternatively \[ = \oint\limits_C \bm{F}\cdot \mathbf{n}_r \,\mathrm{d}\ell = \iint\limits_R \operatorname{div}\bm{F}(x,y)\,\mathrm{d}A \]

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Stokes’ Theorem

Like Green’s theorem but in a higher dimension. I.e. you can evaluate flux just by looking at the boundary.

Given an oriented piecewise-smooth surface \(S\) that is bounded by a simple, closed, piecewise-smooth, positively oriented boundary curve \(\partial S,\) and a vector field \(\bm{F}\) whose components have continuous partial derivatives on a open region in space containing \(S,\) \[ \int\limits_{\partial S} \bm{F}\cdot\mathrm{d}\bm{r} = \iint\limits_S \operatorname{curl}\bm{F} \cdot \mathrm{d}\bm{S} \,. \]

Examples

Bio on George Stokes.

Before next class, TK Question/Review for next class

The Divergence Theorem

This is now the same theorem, but for regions whose boundary are surfaces

Given a simple solid region \(E\) where \(\partial E\) denotes the boundary surface of \(E\) taken to have positive (outward) orientation, and given a vector field \(\bm{F}\) whose components have continuous partial derivatives on a open region in space containing \(E,\) \[ \iint\limits_{\partial E} \bm{F}\cdot\mathrm{d}\bm{S} = \iiint\limits_E \operatorname{div}\bm{F} \,\mathrm{d}V \,. \]

Examples

Sometimes called Gauss’s Theorem, or Ostrogradsky’s Theorem

Before next class, TK Question/Review for next class

Maxwell’s Equations

pop quiz

The Helmholtz Decomposition Theorem