This toolkit contains all of the definitions, properties, facts and theorems you need to memorize all on one webpage. These are your tools on quizzes and exams.
Table of Contents
A set is a collection of objects, along with a condition of membership. We write \[A = \{\text{objects} : \text{membership condition}\}\]
If $a$ is an element of the set $S$, we write $a \in S$.
$\mathbb{R}$ denotes the set of all real numbers. The familiar $xy$-plane is \[\mathbb{R}^2 = \{(x, y) : x, y \in \mathbb{R}\}\]
The 3D rectangular coordinate system is \[\mathbb{R}^3 = \{(x, y, z) : x, y, z \in \mathbb{R}\}\]
Movement in each coordinate $a, b, c$ of the point $(a, b, c)$ is parallel to the $x$-, $y$-, and $z$-axis, respectively.
There are two directions the positive $z$-axis could point. Mathematicians decided on the right hand rule: flatten your hand and point your four fingers along the $x$-axis. Curl your fingers in the shortest rotation from the $x$-axis to the $y$-axis. Then your thumb points in the direction of the positive $z$-axis.
The distance between the points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$, denoted $|P_1P_2|$, is \[|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]
A sphere is the set of all points $P(x, y, z)$ at distance $r$ away from the center $C(h, k, \ell)$. Its equation is \[(x - h)^2 + (y - k)^2 + (z - \ell)^2 = r^2\]
If the center is $O(0,0,0)$, then the equation is \[x^2 + y^2 + z^2 = r^2\]
A vector is a mathematical quantity with both direction and magnitude. Vectors carry no information about position.
A displacement vector $\ray{AB}$ is a vector that takes you from a point $A$ to a point $B$.
Two vectors are equivalent (we write $\ray{AB} = \ray{CD}$) if they have the same direction and magnitude. Their initial points may be different, yet the vectors are equivalent.
The zero vector, denoted $\vec{0}$, is the vector with magnitude $0$. It is the one exception to the definition above: it has no assigned direction.
Given two vectors $\vec{u}$ and $\vec{v}$, position the tail of $\vec{v}$ on the tip of $\vec{u}$. Then $\vec{u} + \vec{v}$ is the vector from the tail of $\vec{u}$ to the tip of $\vec{v}$.
Let $c$ be a scalar and $\vec{v}$ a vector. The scalar multiple $c\vec{v}$ is the vector whose magnitude is $|c|$ times the magnitude of $\vec{v}$ and whose direction is the same as $\vec{v}$ if $c > 0$ and opposite to $\vec{v}$ if $c < 0$.
If $c = 0$ or $\vec{v} = \vec{0}$, then $c\vec{v} = \vec{0}$.
Scaling changes magnitude and either preserves or reverses direction. Every scalar multiple of $\vec{v}$ lies along the same line. Collecting all of them at once, the set \[\left\{c\,\vec{v} : c \in \mathbb{R}\right\}\] is a line in the direction of $\vec{v}$.
The difference of two vectors is defined as \[\vec{u} - \vec{v} = \vec{u} + (-\vec{v})\]
Think Reverse the direction of $\vec{v}$, then add normally.
If the initial points of both vectors are the same, $\vec{u} - \vec{v}$ is simply the vector from the tip of $\vec{v}$ to the tip of $\vec{u}$.
If we let the initial point of a vector $\vec{a}$ be the origin, then the terminal point has coordinates $(a_1, a_2)$ or $(a_1, a_2, a_3)$, and we write \[\vec{a} = \vc{a_1, a_2} \qquad \text{or} \qquad \vec{a} = \vc{a_1, a_2, a_3}\]
Angled brackets distinguish a vector from a coordinate $(a_1, a_2)$. These vectors are called position vectors.
Given points $A(a_1, a_2, a_3)$ and $B(b_1, b_2, b_3)$, the (position) vector $\vec{v}$ with representation $\ray{AB}$ is \[\vec{v} = \vc{b_1 - a_1,\ b_2 - a_2,\ b_3 - a_3}\] A vector is a difference of points (destination minus start). This is why a vector has no information about position: the difference forgets where the vector started.
Let $\vec{a} = \vc{a_1, a_2, a_3}$, $\vec{b} = \vc{b_1, b_2, b_3}$ and $c \in \mathbb{R}$.
The set of all $n$-dimensional vectors is \[V_n = \left\{\vc{a_1, a_2, \dots, a_n} : a_i \in \mathbb{R}\right\}\]
$V_n$ is called a vector space: a set of vectors, together with addition and scalar multiplication, required to satisfy the 8 properties below (the vector space axioms, for those in Calculus IV).
Let $\vec{a}, \vec{b}, \vec{c} \in V_n$ and $c, d \in \mathbb{R}$.
A unit vector is a vector with magnitude 1. Given $\vec{a} \neq \vec{0}$, the corresponding unit vector is \[\vec{u} = \dfrac{\vec{a}}{|\vec{a}|}\]
Any vector $\vec{a} \neq \vec{0}$ can be decomposed into a product of its length and its unit direction: \[\vec{a} = \underbrace{\abs{\vec{a}}}_{\text{length}} \cdot \underbrace{\dfrac{\vec{a}}{|\vec{a}|}}_{\text{unit direction}}\]
An arbitrary vector $\vec{a} = \vc{a_1, a_2, a_3}$ can be broken up into vectors in each dimension: \begin{align} \vec{a} &= \vc{a_1, a_2, a_3} \\ &= \vc{a_1, 0, 0} + \vc{0, a_2, 0} + \vc{0, 0, a_3} \\ &= a_1\vc{1,0,0} + a_2\vc{0,1,0} + a_3\vc{0,0,1} \\ &= a_1\iv + a_2\jv + a_3\kv \end{align} where $\iv = \vc{1,0,0}$, $\jv = \vc{0,1,0}$, $\kv = \vc{0,0,1}$ are the standard basis vectors.
Given two vectors $\vec{a} = \vc{a_1, a_2, a_3}$ and $\vec{b} = \vc{b_1, b_2, b_3}$, the dot product $\vec{a}\cdot\vec{b}$ is defined \[\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]
Let $\vec{a}, \vec{b}, \vec{c} \in V_3$ and $c \in \mathbb{R}$. Then
If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then \[\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\]
Key Idea $\vec{a}\cdot\vec{b}$ is the Pythagorean defect: how far off $\vec{a}$ and $\vec{b}$ are from forming a right angle.
If $\vec{a}, \vec{b} \neq \vec{0}$, then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ satisfies \[\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\]
In $\mathbb{R}^3$, two nonzero vectors $\vec{a}$ and $\vec{b}$ span a plane. The angle $\theta$ between $\vec{a}$ and $\vec{b}$ means the same thing in $\mathbb{R}^2$ as it does in $\mathbb{R}^3$: place them tail to tail and measure the angle inside the plane, taking $0 \leq \theta \leq \pi$.
Two vectors are orthogonal (perpendicular) precisely when the angle between them is $\theta = 90^\circ$.
Fact $\vec{a}$ and $\vec{b}$ are orthogonal if and only if $\vec{a}\cdot\vec{b} = 0$.
The angle $\theta$, and in turn the dot product, tells us how two nonzero vectors are oriented:
Vectors are parallel when $\theta = 0^\circ$ or $180^\circ$, or equivalently when $\vec{b} = c\vec{a}$ for some scalar $c$.
$\vec{a}\cdot\vec{b}$ really measures "how much" $\vec{b}$ is pointing in $\vec{a}$'s direction, relative to $|\vec{a}|$.
When $|\vec{a}| = 1$, $\vec{a}\cdot\vec{b}$ answers the question "How much of $\vec{b}$ points along $\vec{a}$, measured in $\vec{a}$-lengths?"
In general, $\vec{a}$ is not a unit vector, but \[\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \qquad \text{implies} \qquad |\vec{b}|\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|} = \dfrac{\vec{a}}{|\vec{a}|}\cdot \vec{b}\] which is exactly the dot product of $\vec{b}$ with the unit vector in the direction of $\vec{a}$.
The scalar projection of $\vec{b}$ onto $\vec{a}$ (also called the component of $\vec{b}$ along $\vec{a}$) is the signed length of the shadow of $\vec{b}$ in the direction of $\vec{a}$: \[\text{comp}_{\vec{a}}\vec{b} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}\]
The vector projection of $\vec{b}$ onto $\vec{a}$ is that shadow as a vector (scalar projection times the unit vector in the direction of $\vec{a}$): \[\text{proj}_{\vec{a}}\vec{b} = \left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}\right)\dfrac{\vec{a}}{|\vec{a}|} = \left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\right)\vec{a}\]
If $A, B, C$ are distinct points on circle $O$ where $\overline{AB}$ is a diameter, then $\angle ACB$ is a right angle.
If $A(x_1, y_1)$ and $B(x_2, y_2)$ are the endpoints of a diameter, then the circle is the set of points $C(x, y)$ with $\ray{CA}\cdot\ray{CB} = 0$, giving \[(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\]
A determinant of order 2 is defined by \[\begin{vmatrix} a & b \\ c & d\end{vmatrix} = ad - bc\]
Insert the coordinates of $\vec{a}$ and $\vec{b}$ into the determinant column-wise, $\vec{a}$ first. The result is the signed area of the parallelogram determined by $\vec{a}$ and $\vec{b}$.
The sign comes from the right hand rule, with the left column vector rotated onto the right column vector:
Swapping the order of the columns (rotating $\vec{b}$ onto $\vec{a}$ instead) flips the sign.
Let $\vec{a} = \vc{a_1, a_2, a_3}$ and $\vec{b} = \vc{b_1, b_2, b_3}$. The cross product of $\vec{a}$ and $\vec{b}$ is the vector \[\vec{a}\times\vec{b} = \vc{a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1}\]
A determinant of order 3 is defined by expansion along the first row: \[\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix} = a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3\end{vmatrix} - a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3\end{vmatrix} + a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2\end{vmatrix}\]
Note the alternating signs $+, -, +$.Let $\vec{a} = \vc{a_1, a_2, a_3}$ and $\vec{b} = \vc{b_1, b_2, b_3}$. The cross product $\vec{a}\times\vec{b}$ is \[\vec{a}\times\vec{b} = \begin{vmatrix} \iv & \jv & \kv \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix} = \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\iv - \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3\end{vmatrix}\jv + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\kv\]
The vector $\vec{a}\times\vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$: \[(\vec{a}\times\vec{b})\cdot\vec{a} = 0 \qquad \text{and} \qquad (\vec{a}\times\vec{b})\cdot\vec{b} = 0\]
Among the two directions orthogonal to both, the direction of $\vec{a}\times\vec{b}$ is given by the right hand rule.
Careful Always curl your fingers from the vector left of $\times$ towards the vector on the right.
A vector $\vec{a}$ that is orthogonal to every vector in the plane.
If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ with $\theta \in [0, \pi]$, then \[|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta\]
The length of the cross product $\vec{a}\times\vec{b}$ is equal to the area of the parallelogram determined by $\vec{a}$ and $\vec{b}$.
Let $\vec{a}, \vec{b} \in V_3$ be nonzero. Then \[\vec{a}\times\vec{b} = \vc{A_{yz},\ A_{zx},\ A_{xy}}\] where $A_{yz}$ is the signed area of the parallelogram determined by $\vec{a}, \vec{b}$ after projecting onto the $yz$-plane, and likewise for $A_{zx}$ and $A_{xy}$.
Given $\vec{a}\times\vec{b}$, after you apply the right hand rule, look down the positive axis normal to that plane (toward the origin).
Projecting a surface onto a coordinate plane: set the coordinate along the plane's normal to 0.
Let $\vec{a}, \vec{b}, \vec{c} \in V_3$ and $c \in \mathbb{R}$. Then
In particular, $\vec{a}\times\vec{a} = \vec{0}$ for every $\vec{a} \in V_3$.
The product $\vec{a}\cdot(\vec{b}\times\vec{c})$ is called the scalar triple product.\[\vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}\]
The volume of the parallelepiped determined by the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is the absolute value of their scalar triple product: \[V = \left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|\]
In particular, $\vec{a}, \vec{b}, \vec{c}$ lie in the same plane (are coplanar) if and only if $\vec{a}\cdot(\vec{b}\times\vec{c}) = 0$.
A line $L$ in $\mathbb{R}^3$ is completely determined if we know two things: a point on the line and a direction vector for $L$, which is a vector $\vec{v}$ parallel to the line.
The vector equation of a line through a point $P_0(x_0, y_0, z_0)$ with position vector $\vec{r}_0$ and parallel to the vector $\vec{v}$ is \[\vec{r} = \vec{r}_0 + t\vec{v}, \qquad t \in \mathbb{R}\]
Letting $\vec{v} = \vc{a, b, c}$, $\vec{r} = \vc{x, y, z}$ and $\vec{r}_0 = \vc{x_0, y_0, z_0}$, this becomes \[\vc{x, y, z} = \vc{x_0 + ta,\ y_0 + tb,\ z_0 + tc}\]
Parametric equations for a line through the point $(x_0, y_0, z_0)$ and parallel to the vector $\vec{v} = \vc{a, b, c}$ are \[x = x_0 + at, \qquad y = y_0 + bt, \qquad z = z_0 + ct\]
The line segment from $\vec{r}_0$ to $\vec{r}_1$ is given by the vector equation \[\vec{r}(t) = (1 - t)\vec{r}_0 + t\vec{r}_1, \qquad 0 \leq t \leq 1\]
A plane in $\mathbb{R}^3$ is completely determined if we know two things:
The vector equation of a plane through a point $P_0(x_0, y_0, z_0)$ with position vector $\vec{r}_0$, normal vector $\vec{n}$, and arbitrary point $P(x, y, z)$ with position vector $\vec{r}$ is \[\vec{n}\cdot\left(\vec{r} - \vec{r}_0\right) = 0\] $\vec{r} - \vec{r}_0$ sweeps out every point in the plane.
A scalar equation of the plane through the point $P_0(x_0, y_0, z_0)$ with normal vector $\vec{n} = \vc{a, b, c}$ is \[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]
Think of parameter $t$ as time, and $(x(t), y(t))$ as the location of a moving point. This allows us to trace curves which do not pass the vertical line test.
Vector $\vc{x(t), y(t)}$ is a position vector with tip tracing out the curve.
A conic section is an intersection of a double cone and a plane in $\mathbb{R}^3$.
| Conic | Rectangular Equation | Parametric Equations |
|---|---|---|
| Circle | \( (x-h)^2 + (y-k)^2 = r^2 \) | \( \begin{aligned} x &= h + r\cos t \\ y &= k + r\sin t \end{aligned} \) \( t \in [0, 2\pi) \) |
| Ellipse | \( \dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1 \) | \( \begin{aligned} x &= h + a\cos t \\ y &= k + b\sin t \end{aligned} \) \( t \in [0, 2\pi) \) |
| Hyperbola | \( \dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1 \) | \( \begin{aligned} x &= h + a\sec t \\ y &= k + b\tan t \end{aligned} \) \( t \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) \cup \left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right) \) |
A vector-valued function is a function $\vec{r} : \mathbb{R} \to \mathbb{R}^3$. It eats real numbers and spits out vectors: \[\vec{r}(t) = \vc{x(t), y(t), z(t)} = x(t)\iv + y(t)\jv + z(t)\kv\]
The component functions $x(t), y(t), z(t)$ are parametrics in disguise. The space curve traced by $\vec{r}$ is the path swept out by the tip of the position vector $\vec{r}(t)$.
Think Input a time $t$, output a position.
The domain of $\vec{r}(t)$ is the set of all $t$ you can plug in and get a vector out. That is, it is the intersection of the domains of the component functions.
To parametrize the curve where two surfaces meet:
If $\vec{r}(t) = \vc{f(t), g(t), h(t)}$, then \[\lim_{t\to a}\vec{r}(t) = \vc{\lim_{t\to a} f(t),\ \lim_{t\to a} g(t),\ \lim_{t\to a} h(t)}\] provided the limits of the component functions exist.
Let $C$ be the curve traced by $\vec{r}(t)$. The derivative $\vec{r}\,'(t)$ is defined \[\vec{r}\,'(t) = \dfrac{d\vec{r}}{dt} = \lim_{h\to 0}\dfrac{\vec{r}(t + h) - \vec{r}(t)}{h}\] if this limit exists. $\vec{r}\,'(t)$ is called the tangent vector to $C$.
The tangent line to $C$ at $\vec{r}(t_0)$ is the line parallel to $\vec{r}\,'(t_0)$ at the point of tangency.
Think If $\vec{r}(t)$ is the track of a rollercoaster and you are sitting at the tip of $\vec{r}(t)$, then the direction of $\vec{r}\,'(t)$ is your line of sight.
If $\vec{r}(t) = \vc{f(t), g(t), h(t)} = f(t)\iv + g(t)\jv + h(t)\kv$ where $f, g$ and $h$ are differentiable functions, then \[\vec{r}\,'(t) = \vc{f'(t), g'(t), h'(t)} = f'(t)\iv + g'(t)\jv + h'(t)\kv\]
A unit vector that has the same direction as $\vec{r}\,'(t)$ is \[\vec{T}(t) = \dfrac{\vec{r}\,'(t)}{|\vec{r}\,'(t)|}\]
If $\vec{r}(t)$ describes the position vector of an object, then \[\vec{v}(t) = \vec{r}\,'(t)\] describes the rate of change of position of the object with respect to time.
The speed of the object is the magnitude $|\vec{v}(t)| = |\vec{r}\,'(t)|$. Speed is a scalar function; velocity is a vector function.
The acceleration of the object is \[\vec{a}(t) = \vec{v}\,'(t) = \vec{r}\,''(t)\]
$\vec{a}(t)$ records how the tip of $\vec{v}(t)$ is changing: its direction says which way the tip of $\vec{v}(t)$ is being dragged next, and its magnitude says how fast the tip of $\vec{v}(t)$ is changing.
Acceleration changes velocity in two different ways:
Suppose $\vec{u}$ and $\vec{v}$ are differentiable vector functions, $c \in \mathbb{R}$ and $f$ is a real-valued function.
Let $\vec{r}(t) = \vc{x(t), y(t), z(t)}$ be a continuous vector function. Then \[\int_a^b \vec{r}(t)\,dt = \vc{\int_a^b x(t)\,dt,\ \int_a^b y(t)\,dt,\ \int_a^b z(t)\,dt}\]
The indefinite integral is \[\int \vec{r}(t)\,dt = \vc{\int x(t)\,dt,\ \int y(t)\,dt,\ \int z(t)\,dt} + \vec{c}\] where $\vec{c} = \vc{c_1, c_2, c_3}$ is a constant vector.
Think Just like differentiation, integrate each component separately.
If a curve $C$ has vector equation $\vec{r}(t) = \vc{x(t), y(t), z(t)}$ where $a \leq t \leq b$ and the curve is traversed exactly once as $t$ increases from $a$ to $b$, then the arc length of $C$ is \[L = \int_a^b \sqrt{\left[x'(t)\right]^2 + \left[y'(t)\right]^2 + \left[z'(t)\right]^2}\,dt = \int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 + \left(\dfrac{dz}{dt}\right)^2}\,dt\]
Since $\vec{r}\,'(t) = \vc{x'(t), y'(t), z'(t)}$, we can simply write \[L = \int_a^b \left|\vec{r}\,'(t)\right|\,dt\]
In the staircase paradox, a staircase approximation to the diagonal of the unit square has length 2 at every stage, yet the diagonal has length $\sqrt{2}$.
The issue is that tangency to the curve is not preserved: the staircase's corners are not on the line you are trying to estimate. In a proper arc length calculation, all estimating points are on the curve, so tangency is preserved under the limit.
Suppose the curve $C$ is traversed exactly once by $\vec{r}(t) = \vc{x(t), y(t), z(t)}$ where $t \in [a,b]$ as $t$ increases from $a$ to $b$. The arc length function is \[s(t) = \int_a^t \left|\vec{r}\,'(u)\right|\,du\]
By the Fundamental Theorem of Calculus, \[\dfrac{ds}{dt} = \left|\vec{r}\,'(t)\right|\] which says "the rate of change of arc length with respect to $t$ is exactly the speed of $\vec{r}(t)$."
A parametrization $\vec{r}(s)$ is a unit speed parametrization (or a parametrization with respect to arc length) if \[\left|\vec{r}\,'(s)\right| = 1\] for every $s$ in the domain. Plugging in $s$ then gets you exactly $s$ units of arc length along the curve.
The curvature of a curve is \[\kappa = \left|\dfrac{d\vec{T}}{ds}\right|\] where $\vec{T}$ is the unit tangent vector.
Curvature describes how the unit tangent vector $\vec{T}$ changes with respect to arc length. Since $\vec{T}$ is a unit vector, only changes in direction contribute to the rate of change of $\vec{T}$.
Think $\kappa(t)$ is a scalar function that informs us how $\vec{T}$'s direction is changing.
There are two ways: \[\kappa(t) = \dfrac{\left|\vec{T}\,'(t)\right|}{\left|\vec{r}\,'(t)\right|} \qquad \text{ or } \qquad \kappa(t) = \dfrac{\left|\vec{r}\,'(t)\times\vec{r}\,''(t)\right|}{\left|\vec{r}\,'(t)\right|^3}\]
Fact A circle of radius $a$ has curvature $\kappa = \frac{1}{a}$.At unit speed, a particle's velocity vector turns at $\frac{1}{a}$ radians per unit time. The turning is slow for large circles but fast for small circles (since the speed of the particle is the same).
If $|\vec{r}(t)| = c$ is constant, then $\vec{r}\,'(t)$ is orthogonal to $\vec{r}(t)$ for all $t$: \[\vec{r}(t)\cdot\vec{r}\,'(t) = 0\]
In particular, this applies to the unit tangent vector: $\vec{T}\,'(t)$ is orthogonal to $\vec{T}(t)$.
At any point where $\kappa \neq 0$, the unit normal vector is \[\vec{N}(t) = \dfrac{\vec{T}\,'(t)}{\left|\vec{T}\,'(t)\right|}\]
The vector $\vec{B}(t) = \vec{T}(t)\times\vec{N}(t)$ is the binormal vector. It is a unit vector orthogonal to both $\vec{T}$ and $\vec{N}$.
The direction of $\vec{N}$ always points in the direction the curve is turning.
The direction of $\vec{a}_{\text{normal}}$ is $\vec{N}$, and the direction of $\vec{a}_{\text{tangential}}$ is $\vec{T}$.
Given vectors $\vec{v}_1$ and $\vec{v}_2$, a linear combination is any vector $c_1\vec{v}_1 + c_2\vec{v}_2$ where $c_1, c_2 \in \mathbb{R}$.
The coefficient $c_i$ tells you how many copies of $\vec{v}_i$ to walk along. If $\vec{v}_i$ is a unit vector, then $c_i$ is how far you walk in $\vec{v}_i$'s direction.
The acceleration vector $\vec{a}(t)$ is a linear combination \[\vec{a} = a_T\vec{T} + a_N\vec{N}\] where \[a_T = \text{comp}_{\vec{v}}\vec{a} = \dfrac{\vec{v} \cdot \vec{a}}{\left|\vec{v}\right|} = \dfrac{\vec{r}\,'(t)\cdot\vec{r}\,''(t)}{\left|\vec{r}\,'(t)\right|} \qquad \text{and} \qquad a_N = \kappa \left|\vec{v}\right|^2 = \dfrac{\left|\vec{v} \times \vec{a}\right|}{\left|\vec{v}\right|} = \dfrac{\left|\vec{r}\,'(t)\times\vec{r}\,''(t)\right|}{\left|\vec{r}\,'(t)\right|}\]
Imagine you are driving. Acceleration is felt in two ways:
Also, $a_N =\dfrac{\left|\vec{v} \times \vec{a}\right|}{\left|\vec{v}\right|} = \dfrac{\text{Area}}{\text{base}}$ of the parallelogram determined by $\vec{v}$ and $\vec{a}$.
A function of two variables is a function $f : \mathbb{R}^2 \to \mathbb{R}$, written $z = f(x, y)$. A function of three variables is a function $g : \mathbb{R}^3 \to \mathbb{R}$, written $w = g(x, y, z)$.
The domain $D$ of $f(x, y)$ is the set of all points you can plug in and get a real number out. $D = \{(x, y) : \text{conditions}\}$
The graph of $f(x, y)$ is the surface $\{(x, y, f(x, y)) : (x, y) \in D\}$. f(x, y)$ is the height of the surface above or below each point $(x, y, 0)$.
Traces are the curves of intersection of a surface with planes parallel to the coordinate planes ($x = k$, $y = k$ or $z = k$).
Technique To sketch a surface, sketch several traces in each direction and assemble them.
Level curves of $f(x, y)$ are the traces $z = k$ projected onto the $xy$-plane. A collection of level curves for several values of $k$ is a contour plot.
Think A level curve is the set of all points where the surface has the same height $k$.
For $w = f(x, y, z)$, the same technique gives level surfaces $f(x, y, z) = k$ which live in $\mathbb{R}^3$.
For a single variable limit $\displaystyle\lim_{x\to a} f(x)$, there are only two ways for $x$ to approach $a$: from the left and from the right.
For a multivariable limit $\displaystyle\lim_{(x,y)\to(a,b)} f(x, y)$, there are infinitely many paths for $(x, y)$ to approach $(a, b)$.
If $f(x, y) \to L_1$ as $(x, y) \to (a, b)$ along a path $C_1$, and $f(x, y) \to L_2$ as $(x, y) \to (a, b)$ along a path $C_2$, where $L_1 \neq L_2$, then $\displaystyle\lim_{(x,y)\to(a,b)} f(x, y)$ does not exist.
Think Approaching along a path turns the problem into a one-variable (Calculus I) limit. Always try first: $y = 0$ (the $x$-axis), $x = 0$ (the $y$-axis), and all lines $y = mx$ at once.
Careful Paths that agree prove nothing. Even if every line $y = mx$ gives the same limit, the limit may still not exist. Try curves such as $x = y^2, y = -x + x^3$, etc. Look at the graph for ideas, or try to engineer cancellations of indeterminate form $0/0$.
A function $f(x, y)$ is continuous at $(a, b)$ if $\lim_{(x,y)\to(a,b)} f(x, y) = f(a, b)$
Think Continuous means the limit is found by plugging in.
Elementary functions (polynomials, rational functions, $e^x$, $\ln x$, trig functions, and their compositions) are continuous on their domains. If $(a, b)$ is in the domain, the limit is just $f(a, b)$.
If $g(x, y) \leq f(x, y) \leq h(x, y)$ near $(a, b)$ and \[\lim_{(x,y)\to(a,b)} g(x, y) = \lim_{(x,y)\to(a,b)} h(x, y) = L\] then $\displaystyle\lim_{(x,y)\to(a,b)} f(x, y) = L$.
Technique Bound $|f(x, y)|$. If $|f(x, y)| \leq g(x, y)$, then $-g(x, y) \leq f(x, y) \leq g(x, y)$.
Given $\displaystyle\lim_{(x,y)\to(0,0)} f(x, y)$, substitute $x = r\cos\theta$, $y = r\sin\theta$, where $r = \sqrt{x^2 + y^2}$. Then $(x, y) \to (0, 0)$ exactly when $r \to 0^+$: \[\lim_{(x,y)\to(0,0)} f(x, y) = \lim_{r\to 0^+} f(r\cos\theta, r\sin\theta)\]
Think Polar substitution either creates an easier limit, or checks all lines $y = mx$ at once.
As you zoom in on a point on the curve of $f(x)$, the curve becomes indistinguishable from its tangent line. The tangent line is the best linear approximation of $f$ near each point.
At any point $(a, b)$, tracing $f(x, y)$ with the plane $y = b$ or $x = a$ gives a curve, on which we can find tangent lines.
We take these derivatives by holding one variable constant.
If $f(x, y)$ is a function of two variables, its partial derivatives are the functions $f_x$ and $f_y$ defined by \[f_x(x, y) = \lim_{h\to 0}\dfrac{f(x + h, y) - f(x, y)}{h} \qquad f_y(x, y) = \lim_{h\to 0}\dfrac{f(x, y + h) - f(x, y)}{h}\]
If $z = f(x, y)$, we also write \[f_x = \dfrac{\partial f}{\partial x} = \dfrac{\partial z}{\partial x} = \dfrac{\partial}{\partial x}f(x, y) \qquad f_y = \dfrac{\partial f}{\partial y} = \dfrac{\partial z}{\partial y} = \dfrac{\partial}{\partial y}f(x, y)\]
To find $f_x$, treat $y$ like a constant. To find $f_y$, treat $x$ like a constant.
Partial derivatives are slopes of tangent lines in a 2D trace, so the sum, product, quotient and chain rules all hold for partial differentiation.
For $w = f(x, y, z)$, finding $\dfrac{\partial w}{\partial x}$ means holding both $y$ and $z$ constant. Similarly for $w_y$ and $w_z$.
The partial derivatives $f_x$, $f_y$ are functions, so we can take their partial derivatives too: \[f_{xx} = \dfrac{\partial^2 f}{\partial x^2} \qquad f_{xy} = (f_x)_y = \dfrac{\partial^2 f}{\partial y\,\partial x} \qquad f_{yx} = (f_y)_x = \dfrac{\partial^2 f}{\partial x\,\partial y} \qquad f_{yy} = \dfrac{\partial^2 f}{\partial y^2}\]
Think In subscript notation, differentiate left to right. In $\partial$ notation, differentiate right to left (in the denominator).
Suppose $f$ is defined on a disk $D$ that contains the point $(a, b)$. If $f_{xy}$ and $f_{yx}$ are both continuous on $D$, then $f_{xy}(a, b) = f_{yx}(a, b)$.
A partial differential equation (PDE) is an equation involving partial derivatives. For example, the one dimensional wave equation is $u_{tt} = a^2 u_{xx}$ where $u(x, t)$ is the displacement of a vibrating string at distance $x$ from one end at time $t$.
Draw the dependent variable at the top, the variables it depends on below it, and so on. Do this for each function, down to the independent variables. To differentiate the top variable with respect to a bottom variable:
A function is in explicit form when the dependent variable is isolated: $y = f(x)$ or $z = f(x, y)$.
A function is in implicit form when variables are moved to one side: $F(x, y) = 0$ or $F(x, y, z) = 0$.
Think $y = f(x)$ lives on the level curve $F(x, y) = 0$, and $z = f(x, y)$ lives on the level surface $F(x, y, z) = 0$.
If $F(x, y) = 0$ defines $y$ implicitly as a function of $x$, then \[\dfrac{dy}{dx} = -\dfrac{F_x}{F_y} \qquad (F_y \neq 0)\]
If $F(x, y, z) = 0$ defines $z$ implicitly as a function of $x$ and $y$, then \[\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z} \qquad \dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z} \qquad (F_z \neq 0)\]
$f_x$ and $f_y$ only tell us the rate of change of $z$ in the $x$ or $y$ directions: they are slopes of the traces with the planes $y = y_0$ and $x = x_0$. Now allow the vertical plane to rotate around $(x_0, y_0)$. Each rotated plane creates a trace, and its tangent slope is the rate of change of $z$ in that direction. We use a unit vector $\vec{u}$ to describe the direction.
The directional derivative of $f$ at $(x_0, y_0)$ in the direction of a unit vector $\vec{u} = \vc{a, b}$ is \[D_{\vec{u}}f(x_0, y_0) = \lim_{h\to 0}\dfrac{f(x_0 + ah, y_0 + bh) - f(x_0, y_0)}{h}\] if this limit exists.
Note $f_x$ and $f_y$ are the special cases $\vec{u} = \iv$ and $\vec{u} = \jv$, respectively.
Algebraically The step from $(x_0, y_0)$ to $(x_0 + ah, y_0 + bh)$ has length $h\sqrt{a^2 + b^2}$. Since we divide by $h$ in $D_{\vec{u}}f$, we need $\sqrt{a^2 + b^2} = |\vec{u}| = 1$.
Geometrically $f_x(x_0, y_0)$ is a slope: for every one unit increase in $x$, $z$ changes by $f_x(x_0, y_0)$. A unit vector maintains this one unit step in the direction of $\vec{u}$, so $D_{\vec{u}}f$ is an actual number that represents slope.
If $f$ is a differentiable function of $x$ and $y$, then $f$ has a directional derivative in the direction of any unit vector $\vec{u} = \vc{a, b}$, given by \[D_{\vec{u}}f(x, y) = f_x(x, y)\,a + f_y(x, y)\,b\]
Given $f(x, y)$, the gradient of $f$ (read "del $f$") is the vector function $\nabla f : \mathbb{R}^2 \to \mathbb{R}^2$ defined by \[\nabla f(x, y) = \vc{f_x(x, y),\ f_y(x, y)}\]
If $f(x, y, z)$ is a function of three variables, the gradient of $f$ is the vector function $\nabla f : \mathbb{R}^3 \to \mathbb{R}^3$ defined by \[\nabla f(x, y, z) = \vc{f_x(x, y, z),\ f_y(x, y, z),\ f_z(x, y, z)}\]
Note For $z = f(x, y)$, $\nabla f$ lives in the $xy$-plane. $\nabla f$ has no $z$-component.
If $f : \mathbb{R}^n \to \mathbb{R}$ and $\vec{u} \in \mathbb{R}^n$ is a unit vector, then \[D_{\vec{u}}f = \nabla f \cdot \vec{u}\]
Careful $\vec{u}$ must be a unit vector. If it is not, normalize it first: $\vec{u} = \dfrac{\vec{v}}{|\vec{v}|}$.
Think $D_{\vec{u}}f = \text{comp}_{\vec{u}}\nabla f$ is the scalar projection of $\nabla f$ onto $\vec{u}$ (since $\vec{u}$ is a unit vector).
Let $f$ be a differentiable function of two or three variables. The maximum value of $D_{\vec{u}}f(\vec{x})$ is $|\nabla f(\vec{x})|$, and it is attained when $\vec{u}$ has the same direction as $\nabla f(\vec{x})$.
Key Idea Walking in the direction of $\nabla f$ is the fastest uphill direction. Walking in the direction of $-\nabla f$ is the fastest downhill direction.
Note $D_{\vec{u}}f(\vec{x})$ and $D_{\vec{u}}f(x, y)$ mean the same thing (overloaded notation).
Suppose $\vec{u}$ is orthogonal to $\nabla f(\vec{x})$. Then $D_{\vec{u}}f(\vec{x}) = 0$.
Geometrically, $\vec{u}$ is tangent to the level curves of $f$, and $\nabla f$ is orthogonal to the level curves of $f$.
Think In the plane, rotate $\nabla f = \vc{x, y}$ by $90^\circ$ clockwise to get a direction of no change: $\vc{y, -x}$.
If $F(x, y, z)$ is a function of three variables, then $\nabla F$ is orthogonal to the level surfaces $F(x, y, z) = k$.
Why Let $\vec{r}(t) = \vc{x(t), y(t), z(t)}$ be any curve on the surface with $\vec{r}(t_0) = \vc{x_0, y_0, z_0}$. Since $F(x(t), y(t), z(t)) = k$, the chain rule gives \[\dfrac{\partial F}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial F}{\partial y}\dfrac{dy}{dt} + \dfrac{\partial F}{\partial z}\dfrac{dz}{dt} = 0 \qquad \text{so} \qquad \nabla F(x_0, y_0, z_0) \cdot \vec{r}\,'(t_0) = 0\] So $\nabla F$ is orthogonal to the tangent vector of every curve on the surface through that point.
The tangent plane to the surface $F(x, y, z) = k$ at the point $P(x_0, y_0, z_0)$ is \[F_x(P)(x - x_0) + F_y(P)(y - y_0) + F_z(P)(z - z_0) = 0\] The normal vector of the plane is $\nabla F(P)$.
Note If you have $z = f(x, y)$, first convert to implicit form $F(x, y, z) = z - f(x, y) = 0$. Then you can use the equation.
The tangent plane at $P(x_0, y_0, z_0)$ is the plane containing the tangent lines to the $x$- and $y$-traces with slopes $f_x$ and $f_y$.
An equation of the tangent plane to the surface $z = f(x, y)$ at the point $P(x_0, y_0, z_0)$ is \[z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]
Assumes $f$ is differentiable at $(x_0, y_0)$.
The linear approximation $L : \mathbb{R}^2 \to \mathbb{R}$ of $z = f(x, y)$ at $P(x_0, y_0, z_0)$ is \[L(x, y) = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]
If the implication $P \implies Q$ is true, then $Q$ is a necessary condition for $P$ (if $P$ is true). On the other hand, $P$ is a sufficient condition for $Q$ (if $Q$ is true).
$f(x, y)$ is differentiable at $(x_0, y_0)$ if and only if the graph of $f$ becomes indistinguishable from a non-vertical plane with respect to a dilation at $(x_0, y_0, f(x_0, y_0))$ as the zoom factor $\lambda \to \infty$.
A homothety or dilation: an equal scaling expansion or contraction in all dimensions $x, y, z$ with respect to a point.
The secant cone is the surface created by sweeping out secant rays in all directions $\vec{u} = \vc{\cos\theta, \sin\theta}$ at $(x_0, y_0)$, as $\theta$ goes from $0$ to $2\pi$. It uses the one-sided directional derivative \[D^+_{\vec{u}}f(x_0, y_0) = \lim_{h\to 0^+}\dfrac{f(x_0 + h\cos\theta, y_0 + h\sin\theta) - f(x_0, y_0)}{h}\]
As $h \to 0^+$, the secant cone converges to the tangent cone. If $f$ is differentiable at $P$, then the tangent cone is the tangent plane.
Think Tangent lines in all directions $\vec{u}$ live in the tangent plane. If the secant cone doesn't converge to a plane, then $f$ is not differentiable.
\[\text{continuous partials} \implies \text{differentiable (locally linear)} \implies \text{secant cone converges to tangent plane}\]
Continuity of partials is a sufficient condition for differentiability. Convergence of the secant cone to a plane is a necessary condition for differentiability.
If $f_x$ and $f_y$ exist near $(a, b)$ and are continuous at $(a, b)$, then $f$ is differentiable at $(a, b)$.
Suppose $y = f(x)$ is a differentiable function. The differential of $x$, denoted $dx$, is any nonzero number. The differential of $y$, denoted $dy$, is \[dy = f'(x)\,dx\]
$\Delta y$ is the actual change in $f$ from $x$ to $x + \Delta x$.
Think $dy$ is the rise. $dy = $ slope $\cdot$ (run of $dx$).
Suppose $z = f(x, y)$ is differentiable. Define $dx$ and $dy$ to be independent variables. Then the total differential $dz$ is defined by \[dz = f_x(x, y)\,dx + f_y(x, y)\,dy\]
Key Ideas $dz$ is the sum of independent linear changes in $x$ and $y$.
$\Delta z \approx f_x(a, b)\,dx + f_y(a, b)\,dy$, where $\Delta z = f(a + \Delta x, b + \Delta y) - f(a, b)$.
Suppose $f : \mathbb{R}^n \to \mathbb{R}$ is differentiable and $\vec{u} \in \mathbb{R}^n$ is a unit vector. Define $ds$ to be an independent variable representing the distance moved in the direction of $\vec{u}$. Then the total differential $df$ in the direction of $\vec{u}$ is \[df = D_{\vec{u}}f\,ds = \left(\nabla f \cdot \vec{u}\right)ds\]
Let $f(x, y)$ be a function of two variables. A critical point $P$ of $f$ is a point $P$ in the domain of $f$ where $\nabla f(P) = \vec{0}$ or $\nabla f(P)$ is undefined.
Intuition Generally speaking, the tangent plane at critical points is either horizontal or does not exist.
The open ball of radius $r$ centered at $(a, b)$ is $B\left((a, b), r\right) = \left\{(x, y) : (x - a)^2 + (y - b)^2 < r^2\right\}$.
Let $f(x, y)$ be a function of two variables and $(a, b)$ a point in its domain.
If $f$ has a local maximum or minimum at $(a, b)$, then $(a, b)$ is a critical point of $f$.
A saddle point is a critical point where the concavity chanes depending on the direction.
Suppose the second partial derivatives of $f$ are continuous on a ball with center $(a, b)$, and suppose that $f_x(a, b) = 0$ and $f_y(a, b) = 0$ [so $(a, b)$ is a critical point of $f$]. Let \[D = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}_{(a, b)} = f_{xx}(a, b)\,f_{yy}(a, b) - \left[f_{xy}(a, b)\right]^2\]
Think $f_{xx}$ and $f_{yy}$ only measure concavity in the $\iv$ and $\jv$ directions. The mixed partial $f_{xy}$ accounts for every other direction $\vec{u}$.