#import "@youwen/zen:0.1.0": *
#import "@preview/cetz:0.3.1"

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#show math.equation: it => {
  if it.block and not it.has("label") [
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#show: zen.with(
  title: "Math 6A Course Notes",
  author: "Youwen Wu",
  date: "Winter 2025",
  subtitle: [Taught by Nathan Schley],
)

#outline()

= Lecture #datetime(day: 7, month: 1, year: 2025).display()

== Review of fundamental concepts

You can parameterize curves.

#example[Unit circle][
  $
    x = cos(t) \
    y = sin(t)
  $
]

For an implicit equation
$ y = f(t) $
Parameterize it by setting
$ x = t \ y = f(t) $

Parameterize a line passing through two points $arrow(p)_1$ and $arrow(p)_2$ by
$ arrow(c)(t) = arrow(p)_1 + t (arrow(p)_2 - arrow(p)_1) $

Take the derivative of each component to find the velocity vector. The
magnitude of velocity is speed.

#example[
  $
    arrow(c)(t) = <5t, sin(t)> \
    arrow(v)(t) = <5, cos(t)>
  $
]

== Polar coordinates

Write a set of Cartesian coordinates in $RR^2$ as polar coordinates instead, by
a distance from origin $r$ and angle about the origin $theta$.

$ (x,y) -> (r, theta) $

= Lecture #datetime(day: 9, month: 1, year: 2025).display()

== Vectors

A dot product of two vectors is a generalization of the sense of size for a
point or vector.

#example[
  How far is the point $x_1, x_2, x_3$ from the origin? \
  Answer: $x_1^2 + x_2^2 + x_3^2$
]

#definition[
  For vectors $u$ and $v$, where
  $ v = vec(v_1, v_2, dots.v, n), u = vec(u_1, u_2, dots.v, n) $
  The dot product is defined as
  $ sum_(i=1)^n v_i dot u_i $
]

#proposition[
  The dot product of two vectors is the product of their magnitudes and the cosine of the angle between.

  $ arrow(v) dot arrow(w) = ||arrow(v)|| dot ||arrow(w)|| cos theta $
]

= Lecture #datetime(day: 23, month: 1, year: 2025).display()

Midterm is next Thursday in class!

== Arclength and curvature

Easy way of finding curvature: reparameterize curve with speed 1, then
curvature is acceleration. If we can't do that then we need some other
technique.

Given $arrow(c)(t) = <2t^(-1), 6, 2t>$, find the curvature $kappa(t)$.
$
  kappa (t) = (||arrow(c)'(t) times arrow(c)''(t)||) / (||arrow(c)'(t)||^3)
$

== Arclength parameterization

Find an arc-length parameterization of $arrow(c)(t) = <e^t sin(t), e^t cos(t), 5e^t>$.

Let $s = 0$ when $t = 0$ and let $s$ be the arc-length that has traveled along
the curve after $t$ seconds, then we can find $s$ by integrating the curve's
speed over $t$.

$
  s(t) = integral^t_0 ||arrow(c)'(u)|| dif u
$

= Lecture #datetime(day: 12, year: 2025, month: 2).display()

== Chain rule for multivariate functions

We find motivation for the chain rule.

Consider a hiker whose path is given by

$
  arrow(c) (t) = <x(t), y(t)>
$

and

$
  f(x,y) = x dot y
$

What does $x'(t)$ represent? Speed in $x$-direction. Likewise for $y'(t)$.

Say $x'(t) = 3$, $y'(t) = 4$. Then how far did we travel in $t$ seconds?

Suppose our slope in the $x$ direction is given by $m_x = 2$. Suppose the slope
in $y$ is $m_y = -2$. In fact $m_x = f_x (x,y)$ and $m_y = f_y (x,y)$ (here
$f_k$ is the partial derivative with respect to $k$).

So each change in $t$ of 1 leads to a change in elevation up 6 meters in
$x$-axis and down 8 meters in $y$-axis.

So the total change $Delta z$ is given by
$
  Delta z = m_x dot Delta x + m_y dot Delta y
$
and analogously in calculus land

$
  (dif z) / (dif t) = (diff f) / (diff x) dot x'(t) + (diff f) / (diff y) dot y'(t)
$<chain-rule>

In fact @chain-rule is the chain rule.

#fact[
  $
    (dif f) / (dif t) = (diff f) / (diff x) dot (diff x) / (diff t) + (diff f) / (diff y) dot (diff y) / (diff t) + (diff f) / (diff z) dot (diff z) / (diff t)
  $
]

#example[
  Consider $f(x) = x^x$. What is $f'(x)$?

  We can do this with logarithmic differentiation but we can also do this with the multivariable chain rule.

  $
    f(x,y) =
  $
]

#example[
  Find the derivative $dif/(dif t) (f(x,y))$, where $f(x,y) = x^y$, $x(t) = t$,
  and $y(t) = 1$. Assume $t > 0$.
]

#example[
  Find the partial derivative $diff/(diff s) f(x,y,z)$ where $f(x,y,z) = x^2 y^2 + z^3$, and

  $
    x(s,t) = s t \
    y(s, t) = s^2 t \
    z(s,t) = s t^2
  $
]

== Implicit differentiation

Review from single variable: given $f(x,y)$ we can differentiate each term with
respect to $x$, then collect all $(dif y)/(dif x)$ terms together and solve for
it as a variable to obtain $(dif y)/(dif x) = f'(x,y)$.

We do something similar for more variables. Main idea: extraneous variables are
held constant in practice.

Example: consider the surface $3x^2 + 5y z + z^3 = 0$. We want $(diff y)/(diff
z)$ at some point. Use implicit differentiation by viewing the surface as a
level set of some larger function $F(x,y,z) = 3x^2 + 5y z + x^3$ (the level set
part is when $F(x,y,z) = 0$).

By applying the product rule (really the chain rule @chain-rule)
$
  (diff F) / (diff x) = diff / (diff z) (3x^2 + 5 y z + z^3) = diff / (diff z) z^3 = 0 + (5 (diff y) / (diff x) z + 5y) + 3z^2 \
  (diff y) / (diff z) = - (5y + 3z^2) / (5z)
$