Preclass Notes

MATH 126

Table of contents

Week 1 Monday (12.1)
Week 1 Wednesday (12.2)
Week 1 Friday (12.3)
Week 2 Monday (12.4, 12.5)
Week 2 Wednesday (12.5)
Week 2 Friday (12.6)
Week 3 Monday (13.1)
Week 3 Wednesday (13.2)
Week 3 Friday (13.3a)
Week 4 Monday (13.3b)
Week 4 Friday (13.4)
Week 5 Monday (14.1)
Week 5 Wednesday (14.7)
Week 6 Wednesday (15.1)
Week 6 Friday (15.2)
Week 7 Monday (10.3)
Week 7 Wednesday (15.3)
Week 7 Friday (15.4)
Week 8 Wednesday (Taylor Notes 1)
Week 8 Friday (Taylor Notes 2)
Week 9 Monday (Taylor Notes 3)
Week 9 Wednesday (Taylor Notes 4)
Week 9 Friday (Taylor Series 5)
Week 10 Wednesday

Week 1 Monday (12.1)

Three-dimensional spaces
We need an origin - a reference point. Begin with two axes that are perpendicular. The third axis must extend perpendicular to both of the existing axes.
Right hand rule: helps you determine in which direction a third axis points.
We can describe every point space with some set of coordinates $(a, b, c)$/$(x, y, z)$.
$\mathbb{R} = (-\infty, \infty)$ - the set of real unmbers.
Set notation. All points in 3D form:

\[\mathbb{R} \times \mathbb{R} \times \mathbb{R} = \mathbb{R}^3\] \[= {(x, y, z) | x, y, z \in \mathbb{R}\]

Set theory: {\text{bag of possibilities} \text{ rule(s)}}$$
Surfaces: $${(x, y, z) x, y, z \in \mathbb{R}, z = 3}$gives you the$x-y$plane hovering at level$z = 3$$.
Three important planes: at $x = 0, y = 0, z = 0$.
Just stating the rule $z = 3$ (the equations) is also acceptable.

Week 1 Wednesday (12.2)

Vector: depicted as an arrow, with a length as its magnitude, an initial point, and a terminal point
Adding two vectors. The sum of two vectors is the vector from the initial point of one vector to the terminal point of the other vector.
Scalar multiplication. Scalar - a real number. Preserves parallel properties.
A position vector is a vector extending from the origin to a specific point.
Components - number of dimensions of a vector.
Length/magnitude of a vector can be calculated using Pythagorean theorem
Standard basis vectors - unit vectors along the dimensions.
- You can decompose any vector into a linear sum of the standard basis vectors.

Week 1 Friday (12.3)

Dot product/scalar product/inner product: element-wise weighted sum. For a vector $\vec{a} = \langle a_1, a_2, ..., a_n \rangle$ and $\vec{b} = \langle b_1, b_2, ..., b_n \rangle$, the dot product is $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + ... + a_n b_n$.
The input is a set of vectors, but the output is a scalar. The dot product produces a different mathematical structure.
\[\vec{v} \cdot \vec{v} = \| \vec{v} \|^2\]
Theorem: if the angle between two vectors $\vec{a}$ and $\vec{b}$ is $\theta$, then $\vec{a} \cdot \vec{b} = \| \vec{a} \| \| \vec{b} \| \cos \theta$.
- Alternatively, $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{\| \vec{a} \| \| \vec{b} \| }$
Two vectors are orthogonal if and only if $\vec{v} \cdot \vec{w} = 0$.
- A zero-vector is orthogonal to every other vector.
Projection - what does some $\vec{b}$ do to $\vec{a}$?
- Scalar projection of $\vec{b}$ onto $\vec{a}$: $\text{comp}_{\vec{a}} \vec{b} = \frac{ \vec{a} \cdot \vec{b}}{\| \vec{a} \|}$. This component is a real number. Exploits a trigonometric relationship.
- To obtain the vector projection, scale the vector to a unit vector and multiply by the scalar projection.

Week 2 Monday (12.4, 12.5)

Section 12.4

The cross-product holds only for three-dimensional vectors.
Determinant method.
The resulting vector $\vec{v} = \vec{a} \times \vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$.
If you swap the order, you will get the negative product.
$\| \vec{a} \times \vec{b} \| = \| \vec{a} \| \| \vec{b} \| \sin \theta$, where $\theta$ is the smallest angle between $\vec{a}$ and $\vec{b}$. This is a good test for parallelity (equals 0).
Area of parallelogram: $\| \vec{a} \times \vec{b} \|$

Section 12.5 Part I

Vector equations: equations in which we want to equate a vector on the LHS to a vector on the RHS.
Map a vector to a point
You can multiply a vector by a scalar and it will remain on the same parallel line.
Multiply the direction vector - make it shorter, longer, switch the direction, etc.
If $t \in \mathbb{R}$, we get all points on that line.
\[\vec{r} = \vec{r_0} + t\vec{v}\]
- $\vec{r_0}$ - position vector
- $t \in \mathbb{R}$ - parameter
- $\vec{v}$ - direction vector
Sometimes, we write vectors vertically instead of horizontally.
Properties:
- Two lines are parallel if their direction vectors are parallel
- Two lines intersect if they have a point $(x, y, z)$ in common
- Two lines are skew if they are not parallel and do not intersect.

Week 2 Wednesday (12.5)

Vector form of a line: $\vec{r} = \vec{r}_0 + t\vec{v}$
- $\vec{v}$ is the direction vector
- $\vec{r}_0$ is the position vector.
Parameter form of a line: $x = x_0 + ta, y = y_0 + tb, z = z_0 + tc$
- You can obtain this simply by multiplying out the parameter form of the line, in which you find each component of the line $\langle x, y, z\rangle$.
- The slopes are the direction vectors, and the $y$-intercepts are the position vectors
Symmetric form of a line if $a, b, c \neq 0$:

\[\frac{x - x_a}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\]

Obtained by solving each equation for $t$. $t$ should be the same for points.
Two lines are parallel if their direction vectors are parallel
Two vectors intersect if they have a point in common
Two lines are skew if they are not parallel and do not intersect.
You can specify a certain $t$ range.
Planes - to find vector equations for all points on a plane.
Normal vector - any vector that sticks out in a perpendicular function.
Suppose we are given a normal vector (can be found using the cross product), a point $R$, and an unknown point $P(x, y, z)$.
- $\vec{r} - \vec{r}_0$ is in the plane.
- $\vec{r} - \vec{r}_0$ is orthogonal to $\vec{n}$, the normal vector.
- Vector form: $(\vec{r} - \vec{r}_0) \cdot \vec{n} = 0$.
If $\vec{n} = \langle a, b, c \rangle, \vec{r} = \langle x, y, z \rangle, \vec{r}_0 = \langle x_0, y_0, z_0 \rangle$, then we have

\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]

This is the standard form/scalar equation of the plane.

If two planes are not parallel, they intersect along a line; the acute angle of intersection is the acute angle between their normal vectors.

\[\vec{n}_1 \cdot \vec{n}_2 = \| \vec{n}_1 \| \| \vec{n_2} \| \cos \theta\]

Week 2 Friday (12.6)

Goal: to become familiarized with 7 three-dimensional shapes and names: cylinders, cones, ellipsoids, paraboloids (2 types), hyperboloids (2 types).
Two-dimensional shape review
- Line
- Parabola: $ax^2 + by = c$ or $ax + by^2 = c$.
- Ellipse: $ax^2 + by^2 = c$, $a, b, c > 0$.
- Hyperbola: $ax^2 - by^2 = c$ or $-ax^2 + by^2 = c$.
Cylinders
- If one variable is absent, then the graph is a two-dimensional curve extended into the third dimension.
- Circular cylinder: $x^2 + y^2 = 1$
- Cosine cylinder: $z = \cos x$
Quadratic surfaces
- Any surface given by an equation involving sum of first and second powers of $x, y, z$.
- $ax^2 + bx + cy^2 + dy + ez^2 + fz = g$.
- To visualize, use traces. Fix one variable at a time and then find the resulting 2D shape. Fix the other two to find the cross-section and find the resulting plane.
Elliptical/Circular Paraboloid
- \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z}{c}\]
Hyperbolic Paraboloid
- \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{z}{c}\]
- Traces build together a surface.
Ellipsoid/Sphere
- \[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]
Circular/Elliptical Cone
- \[\frac{x^2}{a^2} + \frac{y^2}{eb2} = \frac{z^2}{c^2}\]
Hyperboloid of one sheet
- \[\frac{x^2}{a^2} + \frac{b^2}{e^2} - \frac{z^2}{f^2}= 1\]
Hyperboloid of two sheets
- \[\frac{x^2}{a^2} + \frac{b^2}{e^2} - \frac{z^2}{f^2}= 1\]
In all of the above, replace $\text{var}$ with $(\text{var} - r$ to shift the future.

Week 3 Monday (13.1)

Goals:
1. Given curve equations, describe what the curve looks like
2. Given surfaces, find curve equations for the intersections
We allow for vector components to be expressions of an independent variable $t$. You can read out the parametric equations component-wise.
Space curve. A line is indeed a space curve with a specific term in which the $t$ can be factored out.
How does the space curve look like in the $x, y, z$ coordinate systems? There is no $t$ in the coordinate system - $t$ is just a parameter, a time.
Eliminate $t$ from the equation and find the curve in the $x, y, z$ coordinate system.
Surface of motion - the surface upon which the particle moves. We must have one equation.
When trigonometry is involved, invoke the Pythagorean theorem. In general algebra, try to use substitution to obtain only one equation involving $x$ and/or $y$ and/or $z$.
Intersection between a curve and a surface: plug in and solve for times, then plug back into the curve or the surface to obtain points.
Intersection between two curves: do they intersect? Do they collide?
Intersecting two surfaces. Begin by finding a parametrization for two surfaces.
- Let one of $x, y, z$ be equal to $t$ and solve the others in terms of $t$.
  - Set the most complicated independent variable to $t$.
- For circles/ellipses, try $x = a\cos t$ and $y = b\sin t$.

Week 3 Wednesday (13.2)

Derivatives and integrals of vector functions.
$t$ can be interpreted as a time.
Limit of a vector: take the limits of the individual components of a vector function.
Derivative of the vector: take component-wise derivative.
Continuous vector functions
The unit tangent vector $\vec{T}(t) = \frac{\vec{r}'t}{\| \vec{r}'(t) \|}$.
The tangent vector cannot be a zero-vector.
The tangent line is parallel to $\vec{r}'(t)$ and passes through the point $\vec{r}(t)$.
The integral of a vector is done component-wise.
The distance traveled on a curve from time $a$ to $b$ is:

\[\int_a^b \| \vec{r}'(t) \| dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} dt\]

This is arc length if the arc is not traversed multiple times.

Week 3 Friday (13.3a)

Given a curve, we will assume that its derivative is always continuous.
Continuity is considered component-wise.
We can compute the distance along a curve by integrating the magnitude of the derivative.
Arclength function: feed arc-length with a variable and make the arc length dependent on a variable - the endpoint $t$.

\[s(t) = \int_a^b \| \vec{r}'(u) \| du\]

Fundamental theorem of analysis: $f(x) = \frac{d}{dx} \int_a^x f(u) du$.
$\frac{s(t)}{dt}$ is the speed function - distance over time.
Re-parametrization: describe the same curve with different equations.
Parametrization in terms of $s$: where is the vector location after it has traveled $s$ units along the curve?
$\vec{T}'$ measures change in the direction of the curve, represented via $\vec{T}$>
Curvature $K$ or $\kappa$ - measure of how quickly a curve changes direction at some point.

\[\kappa = \| \frac{d\vec{T}}{ds} \| \approx \| \frac{\text{change in direction}}{\text{change in distance}} \|\] \[K(t) = \frac{\| \vec{r}'(t) \times \vec{r}''(t) \|}{ \| \vec{r}'(t) \|^3} = \frac{\| \vec{T}'(t)}{\| \vec{r}'(t) \|}\]

For two dimensional vectors, expand dimensions to the third dimension and simplify.
$\frac{1}{\kappa(t)}44 is the radius of the curvature: it is the radius of the circle that best matches the curve, dependent on$t$$.

Week 4 Monday (13.3b)

Unit tangent vector for a smooth curve - $\vec{T}(t)$.
Principal unit normal vector/normal vector - $\vec{N}(t)$.

\[\vec{N}(t) = \frac{\vec{T}'(t)}{\| \vec{T} '(t) \| }\]

$\vec{T}(t)$ and $\vec{T}'(t)$ are orthogonal.
Binormal vector - $\vec{B}(t)$$

\[\vec{B}(t) = \vec{T}(t) \times \vec{N}(t)\]

We can create a time-dependent coordinate system with three orthogonal axes using these vectors. This is a TNB frame.

\[\| \vec{B}(t) \| = \| \vec{T}(t) \| \cdot \| \vec{N} (t) \| \cdot \sin \theta \implies 1 = 1 \cdot 1 \cdot 1\]

$\vec{B}(t)$ is a unit vector.
Normal plane - plane spanned by $\vec{N}(t)$ and $\vec{B}(t)$.
The normal plane is orthogonal to the unit tangent vector; thus, some point on a curve $P$ and the derivative of the curve at $P$ can be plugged into the plane equation to define the normal plane.

Week 4 Friday (13.4)

Velocity vector $\vec{v}$. Speed $s = \| \vec{v} \|$.
Derivative of velocity - acceleration.

\[\vec{F} = m \cdot \vec{a}\]

Problem. An object of mass 10kg is being acted on by a force $\vec{F}(t) = \langle 130t, 10e^t, 10e^{-t} \rangle$. You are given $\vec{v}(0) = \langle 0, 0, 1 \rangle$ and $\vec{r}(0) = \langle 0, 1, 1 \rangle$. Find the position function $\vec{r}(t)$.

Solution. If we integrate both sides, we have that $\vec{a} = \vec{r}'' \implies \langle \frac{130}{2}t^2, 10e^t - 10, -10e^-t + 20 \rangle = \int 10 \cdot \vec{r}'(t)$. Integrating again gives $\langle \frac{65}{3}t^3, 10e^t - 10t, 10e^{-t} + 20t \rangle = 10 \vec{r}(t)$. Dividing everything by 10 yields the curve.

Reference to acceleration and velocity in that very moment.
Two components: a tangent and normal component of acceleration.

\[\nu = \text{speed}\] \[\vec{T}(t) = \frac{\vec{v}(t)}{\nu} \implies \vec{T} \nu = \vec{v}(t)\]

Differentiating both sides:

\[\vec{T}' \nu + \vec{T} \nu' = \vec{v}'(t)\]

Using the formula for curvature, we obtain

\[\kappa = \frac{\| \vec{T}' \|}{v} \implies \kappa \nu = \| \vec{T}' \|\] \[\vec{N} = \frac{\vec{T}'}{\| \vec{T}' \| \implies \vec{T}' = \kappa \nu \vec{N}\] \[\kappa \nu^2 \vec{N} + \nu' \vec{T} = \vec{a}(t)\]

We have written acceleration as a sum of $\vec{T}$ and $\vec{N}$.
\[a_\vec{T} = \nu', a_\vec{N} = kappa \nu^2\]

Week 5 Monday (14.1)

Multivariable functions
A real function $f$ of two variables assigns each pair
Domain: the set of all values that are allowed.
Range: the set of all values the output takes on.
Look for domain restrictions - square roots, logarithms, rationals, absolute values, etc.
Intercepts are helpful (esp. when planes are involved).
Use traces to generate level curves - contour map/elevation map as a projection of level curves.

Week 5 Wednesday (14.7)

Critical numbers - places where the critical number is zero or not defined in the domain.
Secpmd derivative test - check concavity
A local maximum occurs at $9a, b)$ if $f(a, b) \ge f(x, y)$4 for all points$(x, y)$near$(a, b)$$. This is the local minimum value.
Some extrema can happen at special locations. A critcial point is a point at which $f_x(a, b) = 0$ and $f_y(a, b)=0$, or $f_x(a, b)44 or$f_y(a, b)$$ do not exist.
Saddle point - point which is not a minimum or maximum
Second derivative test - directional derivatives.
Compute $D = f_{xx} (a, b) f_{yy}(a, b) - f_{xy}^2(a, b)$.
- If $D > 044, concavity is the same in all directions. If$f_{xx} > 0$, it is a local minimum (concave up in all directions). If$f_{xx} < 0$$, it is concave down in all directions, a local maximum.
- If $D < 0$, saddle point.
- If $D = 0$, inconclusive.

Week 6 Wednesday (15.1)

Double integral - signed volume between a function and the $xy$ plane.
Integrate across both dimensions.

Week 6 Friday (15.2)

Given $x$ in the range $a \le x \le b$, we have $g_1(x) \le y \le g_2(x)$. Top/bottom. Compute integral as $\int_a^b \left( \int_{g_1(x)}^{g_2(x)} f(x, y) dy \right)$

Week 7 Monday (10.3)

Polar coordinates - stand at the origin facing the positive $x$-axis.
Rotate by $\theta$
Conversion: $x = r \cos \theta$, $y = r \sin theta$, $\tan \theta = \frac{y}{x}$, $x^2 + y^2 = r^2$

Week 7 Wednesday (15.3)

In polar coordinates, $x = r \cos \theta, y = r \sin \theta$
To set up a double integral, describe the region with polar coordinates, replace $x$ and $y$, and replace $dA$ with $r dr d \theta$.
Why replace $dA$ with $r dr d \theta$? The arc length is $r d \theta$, the ‘length’ is $dr$.

Week 7 Friday (15.4)

Mass and center of mass
The density $\rho(x, y)$ of some lamina varies with $x, y$; $\rho(x, y)$ is measured in units of mass p unit area.
Mass of whole lamina: $\int \int_D \rho (x, y) dA$
Inverse/proportionality: adapt a proportionality constant $k$
Moment of the lamina about the $x$-axis: $M_x = \int \int_D y \rho(x, y) dA$
The center of mass can lie outside of the lamina.

Week 8 Wednesday (Taylor Notes 1)

Taylor polynomial of degree 1: $T_1(x) = f(b) + f'(b)(x - b)$
Approximation error
Bounds - given a function $f(x)$ - a number is an upper bound if all values are less than or equal to that value.
You don’t need a ‘best’ bound, just a bound that works.
Error boudns: given an interval $I$ aroudn $b$, $b - \epsilon \le x \le b + \epsilon$: if $\| f''(x) \| \le M$ for all $x$ in $I$, then

\[\| f(x) - T_1(x) \| \le \frac{M}{2} \| x - b \|^2\]

Week 8 Friday (Taylor Notes 2)

1st Taylor Polynomial: the tangent line. The error is bounded by $\frac{M}{2} \| x - b \|^2$ for $\| f''(x) \| M$. You can make this independent of $x$ by finding the maximum possible RHS expression on the domain.
The second Taylor polynomial of a function $f(x)$ based at $b$ is

\[T_2(x) = T_1(x) + \frac{1}{2} f''(b)(x-b)^2\]

This error will be less than or equal to $\frac{M}{6} \| x - b \|^3$ for $\| f'''(x) \| le M$.

\[T_3(x) = T_2(x) + \frac{1}{6} f'''(b)(x-b)^3\] \[T_n(x) = T_{n-1}(x) + \frac{1}{n!} f^n(b)(x-b)^n\]

The error for the $n$th Taylor polynomial is $\frac{M}{(n+1)!} \| x-b \|^{n+1}$
We are matching the derivative on multiple different levels.

Week 9 Monday (Taylor Notes 3)

Sigma notation - allows us to express complex patterns, like exponentials of $-1$ to change sign
$n$th Taylor polynomial - as $n$ increases, the error bound decreases.
We let $n \to \infty$

Taylor Series for $f(x)$ at base $b$:

\[\lim_{n\to\infty} T_n(x) = \sum_{k=0}^\infty \frac{1}{k!} f^k (b)(x-b)^k\]

If the limit exists for an $x$, it converges at $x$. Otherwise, it diverges at $x$.
Interval of convergence - largest open interval over which the series converges.
Convergence only happens within an interval.
Important series:
- $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$, converges for all $x$
- $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \pm ...$, converges for all $x$
- $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \pm ...$, converges for all $x$

Week 9 Wednesday (Taylor Notes 4)

$n$th Taylor polynomial of $f$ on $I$: $T_n(x) = \sum_{k=0}^n \frac{1}{k!} f^k (b) (x-b)^k$
Error bound: error $\lke \frac{M}{(n+1)!} \|x-b\|^{n+1}$
Taylor series - take $n\to\infty$, we see error $\to 0$.
Word of caution - sometimes a function is differentiable and $T_n$ doesn’t diverge, but it does not work out.
Interval of convergence: when do we see convergence? Which $x$ leads to convergence when $n\to\infty$?
Consider $(1 - x) T_n(x) = ...$, slve for $T_n (x)$.

Week 9 Friday (Taylor Series 5)

Developing additional ability to write Taylor series
On the interval of convergence, we can pull in constants and merge Taylor series.
Whenever we have an integral/derivative and a sum, we can distribute across it term-wise.

\[\frac{d}{dx} A(x) = \sum_{k=0}^\infty \frac{d}{dx} \left( \frac{f^k (b)}{k!} (x - b)^k \right) = \sum_{k = 0}^\infty \frac{f^k (b)}{(k-1)!} (x - b)^{k-1}\]

Same logic can be applied to the integral
The differentiation/integration will have the same interval of convergence as the original

Week 10 Wednesday

Take derivatives repeatedly and find patterns in evaluation at certain points.
For $f(x) = \sum_{k=0}^\infty _k (x - b)^k$, $f^n b = c_n \cdot n!$, $c_n = \frac{f^n (b)}{n!}$