2025-01-24 23:00:23 -08:00
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#import "@youwen/zen:0.1.0": *
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#import "@preview/cetz:0.3.1"
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2025-01-24 23:00:23 -08:00
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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 Scheley],
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)
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#outline()
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2025-01-07 16:58:41 -08:00
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= Lecture #datetime(day: 7, month: 1, year: 2025).display()
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2025-01-09 16:15:18 -08:00
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== Review of fundamental concepts
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You can parameterize curves.
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2025-01-09 16:15:18 -08:00
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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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2025-01-10 12:44:05 -08:00
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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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2025-01-24 23:00:23 -08:00
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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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