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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<meta
name="viewport"
content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no"
/>
<title>APMA E2000 - Chain Rule</title>
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rel="stylesheet"
href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/6.4.2/css/all.min.css"
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src="https://jsxgraph.org/distrib/jsxgraphcore.js"
></script>
<style>
.smaller {
font-size: smaller;
}
</style>
</head>
<body>
<div class="reveal">
<div class="slides">
<section>
<section>
<h3 class="framelabel">Lecture 10</h3>
<h1>Chain Rule</h1>
<h2>APMA E2000</h2>
<div class="r-stretch"></div>
<p style="text-align: right">
Drew Youngren <code>[email protected]</code>
</p>
</section>
<section>
<!-- Set up standard LaTeX macros -->
$\gdef\RR{\mathbb{R}}$ $\gdef\vec{\mathbf}$
$\gdef\bv#1{\begin{bmatrix} #1 \end{bmatrix}}$
$\gdef\proj{\operatorname{proj}}$ $\gdef\comp{\operatorname{comp}}$
<h2>Announcements</h2>
<ul>
<li class="fragment">Quiz 3 (on HW 4) this week</li>
<li class="fragment">HW5 due next Tuesday</li>
</ul>
</section>
</section>
<section>
<section>
<h1>1-minute review</h1>
</section>
<section>
<h6 class="framelabel">Linearization</h6>
<p>
The <strong>linearization</strong> of a differentiable function
$f(x,y)$ about a point $(a,b)$ is the linear function \[ L(x,y) =
f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) \] $ \approx f(x,y)$. Its
graph is tangent to the graph of $f$.
</p>
</section>
<section data-auto-animate>
<h6 class="framelabel">Error</h6>
<p>\[ f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) \]</p>
</section>
<section data-auto-animate>
<h6 class="framelabel">Error</h6>
<p>\[ f(x,y) - f(a,b) \approx f_x(a,b)(x-a) + f_y(a,b)(y-b) \]</p>
</section>
<section data-auto-animate>
<h6 class="framelabel">Error</h6>
<p>\[ \Delta f \approx f_x(a,b)\Delta x + f_y(a,b)\Delta y \]</p>
</section>
<section data-auto-animate>
<h6 class="framelabel">Error</h6>
<p>\[ df = f_x(a,b) dx + f_y(a,b)dy \]</p>
<p class="fragment">
This <strong>differential</strong> of $f$ estimates the error of
$f$. That is, it approximates how much the output will change for
given changes of the input.
</p>
</section>
<section>
<h3 class="framelabel">Theorem</h3>
<p>
If all partial derivatives of a function of several variables $f$
exist <em>and are continuous</em> in a neighborhood of a position
$\vec a$, then $f(\vec x)$ is differentiable at $\vec a$.
</p>
</section>
<section>
<h6 class="framelabel">Non-example</h6>
<p>\[ f(x,y) = \frac{xy}{\sqrt{x^2 + y^2}} \]</p>
<div class="r-stretch">
<iframe
src="https://3demos.ctl.columbia.edu/?currentChapter=How+To&showPanel=false&obj0_kind=graph&obj0_color=%230d0887&obj0_params_a=-2&obj0_params_b=2.1&obj0_params_c=-2&obj0_params_d=2&obj0_params_z=x+y+%2F+%28sqrt%28x%5E2+%2B+y%5E2%29%29+&obj0_params_t0=0&obj0_params_t1=1"
frameborder="0"
width="100%"
height="100%"
></iframe>
</div>
</section>
</section>
<section>
<section data-background-image="assets/PatchedCabinets.jpeg">
<h1>The Chain Rule</h1>
<p
class="fragment"
style="
position: absolute;
right: 2px;
bottom: 2px;
font-size: x-small;
"
>
Photo credit:
<a
href="http://paradiso.media.mit.edu/synth.html"
target="_blank"
rel="noopener noreferrer"
>Joe Paradiso</a
>
</p>
</section>
<section
data-background-video="assets/chain_rule_1d.mp4"
data-background-opacity="0.3 "
>
<h6 class="framelabel">1-D Chain Rule</h6>
<div class="r-stack">
<p class="fragment current-visible fade-out">
\[ \frac{d}{dx} (f\circ g)(x) = f'(g(x))g'(x) \]
</p>
<p class="fragment fade">
\[ \frac{d}{dx} f(g(x)) = \frac{df}{du} \frac{du}{dx} \] where
$u = g(x)$.
</p>
</div>
</section>
<section>
<h6 class="framelabel">Main problem</h6>
<p>
A smooth curve $\left\langle x(t), y(t) \right\rangle$ moves
through a domain of a differentiable function $f(x,y)$. What is
$\frac{df}{dt} \Big\vert_{t = t_0}$?
</p>
<p class="fragment">
<em>Hint:</em>
<a
href="https://3demos.ctl.columbia.edu/?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"
target="_blank"
rel="noopener noreferrer"
>Replace</a
>
$f$ with $L$.
</p>
</section>
<section>
<p>
Let $L(x,y)$ be the linearization of $f$ around $\left\langle a, b
\right\rangle = \left\langle x(t_0), y(t_0) \right\rangle$.
</p>
<p class="fragment">
\[ \frac{df}{dt} \Big\vert_{t = t_0} = \frac{dL }{dt} \Big\vert_{t
= t_0} L(x(t), y(t)) = \]
</p>
<p class="fragment">
\[ \frac{d}{dt} \Big\vert_{t_0}(f(a, b) + f_x(a,b)(x(t) - a) +
f_y(a,b)(y(t) - b)) \]
</p>
<p class="fragment">\[ = f_x(a,b) x'(t_0) + f_y(a,b) y'(t_0) \]</p>
</section>
<section>
<div class="r-frame">
<h3 class="framelabel">Multivariable Chain Rule</h3>
<p>
If $f(x_1, \ldots, x_n)$ is a differentiable function and each
$x_i$ depends smoothly on $t$, then \[ \frac{df}{dt} =
\sum_{i=1}^n \frac{\partial f}{\partial x_i} \frac{d x_i}{dt} \]
</p>
</div>
</section>
<section>
<h3 class="framelabel">Notation</h3>
<p>
We write \[\nabla f = \left\langle \frac{\partial f}{\partial
x_1}, \ldots, \frac{\partial f}{\partial x_n} \right\rangle \]
</p>
<p class="fragment">
Thus, the chain rule becomes \[ \frac{d}{dt} f ( \vec x(t)) =
\nabla f(\vec x(t)) \cdot \vec x'(t) \]
</p>
</section>
<section>
<h6 class="framelabel">Other Forms</h6>
<p>
Other cases can be handled by extrapolating from the case above.
\[\frac{\partial}{\partial t} f(x(s,t),y(s,t),z(s,t)) =\]\[
\frac{\partial f}{\partial x} \frac{\partial x}{\partial t} +
\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
+\frac{\partial f}{\partial z}\frac{\partial z}{\partial t} \]
</p>
</section>
</section>
<section>
<section>
<h1>Examples</h1>
</section>
<section>
<h6 class="framelabel">Elevation change</h6>
<p>
A particle circles the $z$-axis
<a
href="https://3demos.ctl.columbia.edu/?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"
target="_blank"
rel="noopener noreferrer"
>every second</a
>
at a distance of 1 unit. If it is confined to the surface $z=xy$,
find the maximum of $\frac{dz}{dt}$.
</p>
</section>
<section>
<strong>Solution</strong>.
<p class="fragment">
$z = xy$ and $\left\langle x(t), y(t) \right\rangle = \left\langle
\cos(2 \pi t), \sin (2 \pi t) \right\rangle $. \[\frac{dz}{dt} =
\frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial
z}{\partial y} \frac{dy}{dt} \]
</p>
<p class="fragment">
\[ = y (-2\pi\sin (2\pi t)) + x(2\pi \cos(2\pi t)) \] \[ =
-2\pi\sin^2 (2\pi t) + 2\pi \cos^2(2\pi t) \] \[ = 2\pi \cos(4\pi
t) \] which has a maximum of $2\pi$.
</p>
</section>
<section>
<h6 class="framelabel">Related Rates</h6>
<p class="smaller">
The base radius of a
<a
href="https://3demos.ctl.columbia.edu/?Y3VycmVudENoYXB0ZXI9SG93K1RvJnNjYWxlPTAuNjQmc2hvd1BhbmVsPWZhbHNlJm9iajBfa2luZD1zdXJmYWNlJm9iajBfY29sb3I9JTIzMmMyN2QzJm9iajBfYW5pbWF0aW9uPXRydWUmb2JqMF9wYXJhbXNfYT0wJm9iajBfcGFyYW1zX2I9MSZvYmowX3BhcmFtc19jPTAmb2JqMF9wYXJhbXNfZD0yK3BpJm9iajBfcGFyYW1zX3g9JTI4NCslMkIrMSt0JTI5K3UrY29zJTI4diUyOSZvYmowX3BhcmFtc195PSUyODQrJTJCKzErdCUyOSt1K3NpbiUyOHYlMjkrJm9iajBfcGFyYW1zX3o9JTI4OCstKzIrdCslMjkrJTI4MSstK3UlMjkmb2JqMF9wYXJhbXNfdDA9LTEmb2JqMF9wYXJhbXNfdDE9MQ=="
target="_blank"
rel="noopener noreferrer"
>right cone</a
>
is increasing at a rate of 1 cm/min, and its height is decreasing
at 2cm/min. When the base has radius 4 cm and the height is 8 cm,
is the volume increasing or decreasing?
</p>
<div class="r-stretch">
<iframe
src="https://3demos.ctl.columbia.edu/?Y3VycmVudENoYXB0ZXI9SG93K1RvJnNjYWxlPTAuNjQmc2hvd1BhbmVsPWZhbHNlJm9iajBfa2luZD1zdXJmYWNlJm9iajBfY29sb3I9JTIzMmMyN2QzJm9iajBfYW5pbWF0aW9uPXRydWUmb2JqMF9wYXJhbXNfYT0wJm9iajBfcGFyYW1zX2I9MSZvYmowX3BhcmFtc19jPTAmb2JqMF9wYXJhbXNfZD0yK3BpJm9iajBfcGFyYW1zX3g9JTI4NCslMkIrMSt0JTI5K3UrY29zJTI4diUyOSZvYmowX3BhcmFtc195PSUyODQrJTJCKzErdCUyOSt1K3NpbiUyOHYlMjkrJm9iajBfcGFyYW1zX3o9JTI4OCstKzIrdCslMjkrJTI4MSstK3UlMjkmb2JqMF9wYXJhbXNfdDA9LTEmb2JqMF9wYXJhbXNfdDE9MQ=="
frameborder="0"
width="100%"
height="100%"
></iframe>
</div>
</section>
<section>
<strong>Solution</strong>.
<p class="fragment">
$V = \frac\pi3 r^2 h$ and thus \[ \frac{dV}{dt} = \frac{\partial
V}{\partial r}\frac{dr}{dt} + \frac{\partial V}{\partial
h}\frac{dh}{dt} \] \[ = \frac23 \pi r h \frac{dr}{dt} + \frac13
\pi r^2 \frac{dh}{dt} \] \[ = \frac23 \pi (4) (8) (1) + \frac13
\pi 4^2 (-2) = \frac{32 \pi}{3} > 0 \]
</p>
</section>
<section>
<h6 class="framelabel">Polar Coordinates</h6>
<p>
Suppose $f(x,y)$ has continuous derivatives of all order. Express
$\frac{\partial^2 f}{\partial r\partial \theta}$ in terms of the
partial derivatives of $f$.
</p>
<p>$r$ and $\theta$ are the usual polar coordinates.</p>
</section>
<section>
<p><strong>Solution</strong>.</p>
<p class="fragment">
\[\frac{\partial f}{\partial r} = f_x \frac{\partial x}{\partial
r} + f_y \frac{\partial y}{\partial r} = f_x \cos \theta + f_y
\sin \theta \]
</p>
<p class="fragment r-stretch">
\[\frac{\partial^2 f}{\partial \theta \partial r } = \left(f_{xx}
\frac{\partial x}{\partial \theta} + f_{xy} \frac{\partial
y}{\partial \theta}\right)\cos \theta - f_x \sin \theta \]\[+
\left(f_{yx} \frac{\partial x}{\partial \theta} + f_{yy}
\frac{\partial y}{\partial \theta}\right)\sin \theta + f_y \cos
\theta \] \[ = (f_{yy} - f_{xx})r\sin\theta \cos \theta + f_{xy} r
\cos (2\theta) + f_y \cos \theta - f_x \sin\theta \]
</p>
</section>
<section>
<h6 class="framelabel">Chain chain chain</h6>
<p>
Suppose \[ \begin{align*} z &= z(x,y) \\ x &= x(u,v) \\ y &=
y(u,v) \\ u &= u(s,t) \\ v &= v(s,t) \\ \end{align*}\] are all
differentiable. Find an expression for $\frac{\partial z}{\partial
s}$.
</p>
</section>
<section>
<p><strong>Solution</strong>.</p>
<p class="fragment">
\[ \begin{align*} \frac{\partial z}{\partial s} =& \frac{\partial
z}{\partial x}\frac{\partial x}{\partial u}\frac{\partial
u}{\partial s} + \frac{\partial z}{\partial x}\frac{\partial
x}{\partial v}\frac{\partial v}{\partial s} \\ &+ \frac{\partial
z}{\partial y}\frac{\partial y}{\partial u}\frac{\partial
u}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial
y}{\partial v}\frac{\partial v}{\partial s} \\ \end{align*} \]
</p>
</section>
</section>
<section>
<section>
<h1>Implicit Differentiation</h1>
</section>
<section>
<h3>Implicit Relationships</h3>
<p>
Every implicit relationship among variables $x$ and $y$ can be
expressed as a level set \[ F(x,y) = c. \]
</p>
<div class="fragment">
<h6 class="framelabel">Examples</h6>
<ul>
<li class="fragment">
The circle
<span class="r-stack">
<span class="fragment current-visible"
>$y = \pm\sqrt{R^2 - x^2}$</span
>
<span class="fragment fade-in">$ x^2 + y^2 = R^2$</span>
</span>
</li>
<li class="fragment">
<a
href="https://3demos.ctl.columbia.edu/?Y3VycmVudENoYXB0ZXI9SG93K1RvJm9iajBfa2luZD1ncmFwaCZvYmowX2NvbG9yPSUyMzBkMDg4NyZvYmowX3BhcmFtc19hPS0yJm9iajBfcGFyYW1zX2I9MiZvYmowX3BhcmFtc19jPS0yJm9iajBfcGFyYW1zX2Q9MiZvYmowX3BhcmFtc196PWNvcyUyOHgreSUyOSslMkIreCstK2V4cCUyOC14JTI5K3krLSsxJm9iajBfcGFyYW1zX3QwPTAmb2JqMF9wYXJhbXNfdDE9MQ=="
target="_blank"
rel="noopener noreferrer"
>Something</a
>
else
<span class="r-stack">
<span class="fragment current-visible"
>$\cos(xy) + x - 1 = e^{-x}y$</span
>
<span class="fragment fade-in"
>$\cos(xy) + x - e^{-x}y = 1$</span
>
</span>
</li>
</ul>
</div>
</section>
<section>
<h6 class="framelabel">Main Problem</h6>
Given $y$ implicitly defined as a function of $x$ via $F(x,y) = c$,
find an expression for $\frac{dy}{dx}$ in terms of derivatives of
$F$.
</section>
<section>
\[ \frac{d}{dx}F(x,y) = 0\]
<div class="fragment">
<strong>WARNING</strong> Mind your $d$'s and $\partial$'s.
</div>
<div class="fragment">\[ \frac{d}{dx}F(x,y(x)) = 0 \]</div>
<div class="fragment">
\[ \frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial
F}{\partial y}\frac{dy}{dx} = 0\]
</div>
<div class="fragment">\[ \frac{dy}{dx} = -\frac{F_x}{F_y} \]</div>
</section>
<section
data-background-image="assets/implicit_example.png"
data-background-opacity="0.6"
>
<h6 class="framelabel">Example</h6>
<p>
Find the slope of the tangent line to \[ x \sin(y) - \frac12 =
\sqrt{2} - 2\cos(xy)\] at the point
$\left(\frac12,\frac\pi2\right)$.
</p>
</section>
<section>
<p><strong>Solution</strong>.</p>
<p class="fragment">
\[F(x,y) = x \sin y + 2\cos(xy) = \frac12 + \sqrt 2 \] \[
\frac{dy}{dx} = \left.-\frac{F_x}{F_y} \right\rvert_{(1/2,\pi/2)}
= \left.-\frac{\sin y -2\sin(xy)y}{x\cos y -2
\sin(xy)x}\right\rvert_{(1/2,\pi/2)} \] \[ = - \frac{1 -
\frac{\pi}{\sqrt2}}{-\frac{1}{\sqrt2}} = \sqrt2 - \pi \]
</p>
</section>
</section>
<section>
<section><h1>Learning Outcomes</h1></section>
<section id="learning-outcomes">
<h6 class="framelabel">You should be able to...</h6>
<ul>
<li>Sketch a dependence tree for compositions.</li>
<li>
Select the correct form of the chain rule for arbitrary
compositions of scalar fields.
</li>
<li>Compute higher order derivatives via the chain rule.</li>
<li>
Implicitly find rates of change by reexpressing relations
between quantities as level sets.
</li>
</ul>
</section>
</section>
</div>
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