203 lines
5.1 KiB
Text
203 lines
5.1 KiB
Text
#import "@youwen/zen:0.1.0": *
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#import "@preview/cetz:0.3.1"
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#set math.equation(numbering: "(1)")
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#show math.equation: it => {
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if it.block and not it.has("label") [
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#counter(math.equation).update(v => v - 1)
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#math.equation(it.body, block: true, numbering: none)#label("")
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] else {
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it
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}
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}
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#show: zen.with(
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title: "Math 6A Course Notes",
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author: "Youwen Wu",
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date: "Winter 2025",
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subtitle: [Taught by Nathan Schley],
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)
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#outline()
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= Lecture #datetime(day: 7, month: 1, year: 2025).display()
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== Review of fundamental concepts
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You can parameterize curves.
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#example[Unit circle][
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$
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x = cos(t) \
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y = sin(t)
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$
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]
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For an implicit equation
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$ y = f(t) $
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Parameterize it by setting
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$ x = t \ y = f(t) $
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Parameterize a line passing through two points $arrow(p)_1$ and $arrow(p)_2$ by
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$ arrow(c)(t) = arrow(p)_1 + t (arrow(p)_2 - arrow(p)_1) $
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Take the derivative of each component to find the velocity vector. The
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magnitude of velocity is speed.
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#example[
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$
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arrow(c)(t) = <5t, sin(t)> \
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arrow(v)(t) = <5, cos(t)>
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$
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]
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== Polar coordinates
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Write a set of Cartesian coordinates in $RR^2$ as polar coordinates instead, by
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a distance from origin $r$ and angle about the origin $theta$.
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$ (x,y) -> (r, theta) $
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= Lecture #datetime(day: 9, month: 1, year: 2025).display()
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== Vectors
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A dot product of two vectors is a generalization of the sense of size for a
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point or vector.
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#example[
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How far is the point $x_1, x_2, x_3$ from the origin? \
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Answer: $x_1^2 + x_2^2 + x_3^2$
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]
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#definition[
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For vectors $u$ and $v$, where
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$ v = vec(v_1, v_2, dots.v, n), u = vec(u_1, u_2, dots.v, n) $
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The dot product is defined as
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$ sum_(i=1)^n v_i dot u_i $
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]
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#proposition[
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The dot product of two vectors is the product of their magnitudes and the cosine of the angle between.
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$ arrow(v) dot arrow(w) = ||arrow(v)|| dot ||arrow(w)|| cos theta $
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]
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= Lecture #datetime(day: 23, month: 1, year: 2025).display()
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Midterm is next Thursday in class!
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== Arclength and curvature
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Easy way of finding curvature: reparameterize curve with speed 1, then
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curvature is acceleration. If we can't do that then we need some other
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technique.
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Given $arrow(c)(t) = <2t^(-1), 6, 2t>$, find the curvature $kappa(t)$.
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$
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kappa (t) = (||arrow(c)'(t) times arrow(c)''(t)||) / (||arrow(c)'(t)||^3)
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$
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== Arclength parameterization
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Find an arc-length parameterization of $arrow(c)(t) = <e^t sin(t), e^t cos(t), 5e^t>$.
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Let $s = 0$ when $t = 0$ and let $s$ be the arc-length that has traveled along
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the curve after $t$ seconds, then we can find $s$ by integrating the curve's
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speed over $t$.
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$
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s(t) = integral^t_0 ||arrow(c)'(u)|| dif u
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$
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= Lecture #datetime(day: 12, year: 2025, month: 2).display()
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== Chain rule for multivariate functions
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We find motivation for the chain rule.
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Consider a hiker whose path is given by
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$
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arrow(c) (t) = <x(t), y(t)>
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$
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and
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$
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f(x,y) = x dot y
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$
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What does $x'(t)$ represent? Speed in $x$-direction. Likewise for $y'(t)$.
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Say $x'(t) = 3$, $y'(t) = 4$. Then how far did we travel in $t$ seconds?
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Suppose our slope in the $x$ direction is given by $m_x = 2$. Suppose the slope
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in $y$ is $m_y = -2$. In fact $m_x = f_x (x,y)$ and $m_y = f_y (x,y)$ (here
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$f_k$ is the partial derivative with respect to $k$).
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So each change in $t$ of 1 leads to a change in elevation up 6 meters in
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$x$-axis and down 8 meters in $y$-axis.
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So the total change $Delta z$ is given by
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$
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Delta z = m_x dot Delta x + m_y dot Delta y
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$
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and analogously in calculus land
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$
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(dif z) / (dif t) = (diff f) / (diff x) dot x'(t) + (diff f) / (diff y) dot y'(t)
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$<chain-rule>
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In fact @chain-rule is the chain rule.
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#fact[
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$
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(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)
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$
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]
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#example[
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Consider $f(x) = x^x$. What is $f'(x)$?
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We can do this with logarithmic differentiation but we can also do this with the multivariable chain rule.
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$
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f(x,y) =
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$
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]
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#example[
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Find the derivative $dif/(dif t) (f(x,y))$, where $f(x,y) = x^y$, $x(t) = t$,
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and $y(t) = 1$. Assume $t > 0$.
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]
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#example[
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Find the partial derivative $diff/(diff s) f(x,y,z)$ where $f(x,y,z) = x^2 y^2 + z^3$, and
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$
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x(s,t) = s t \
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y(s, t) = s^2 t \
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z(s,t) = s t^2
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$
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]
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== Implicit differentiation
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Review from single variable: given $f(x,y)$ we can differentiate each term with
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respect to $x$, then collect all $(dif y)/(dif x)$ terms together and solve for
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it as a variable to obtain $(dif y)/(dif x) = f'(x,y)$.
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We do something similar for more variables. Main idea: extraneous variables are
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held constant in practice.
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Example: consider the surface $3x^2 + 5y z + z^3 = 0$. We want $(diff y)/(diff
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z)$ at some point. Use implicit differentiation by viewing the surface as a
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level set of some larger function $F(x,y,z) = 3x^2 + 5y z + x^3$ (the level set
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part is when $F(x,y,z) = 0$).
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By applying the product rule (really the chain rule @chain-rule)
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$
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(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 \
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(diff y) / (diff z) = - (5y + 3z^2) / (5z)
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$
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