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\begin{document}
%MODULE 1: Intro to SAGE


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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{Related Rates}


\subtitle{The derivative expresses the \emph{instantaneous rate of change}.
We can use this fact to solve problems involving rates of change that
are related to each other.}

\makebeamertitle

\lyxframeend{}\lyxframe{{Defining the problem}}

\begin{question} 
\begin{itemize}
\item The radius of a ripple in a pond grows at a rate of 5~cm/s. How fast
is the circumference changing at 3~seconds? 
\item The radius of a ripple in a pond grows at a rate of 5~cm/s. How fast
is the area changing at 3~seconds? 
\item A camera is directed at the launch of a rocket. The camera is 3~km
from the rocket. Once the rocket rises 1~km, it has a speed of 1500~km/hr.
How fast must the angle of the camera change to remain fixed on the
rocket? 
\end{itemize}
\end{question}


\lyxframeend{}\lyxframe{{Leibniz notation}}

\noindent In most modules we have favored Newton's notation $\ec f'$
for the derivative of a function $\ec f$. In this module we will
use Leibniz notation $\ec df/dt$ instead. \emph{Why?} Clarity of
expression. All the problems can be solved using Newton's notation,
but Newton's notation does not indicate the independent variable.
Problems in related rates usually have several variables that depend
on time, and the derivatives are always \emph{with respect to time},
which the Leibniz notation $\ec df/dt$ emphasizes.


\lyxframeend{}\lyxframe{{First example: Experimentation}}

\begin{question} The radius of a ripple in a pond increasing at a
rate of 5~cm/s. How fast is the circumference changing at 3 seconds?\end{question}

\noindent The problem asks us to find $\ec dC/dt$, given that $\ec dr/dt=5$.


\pause{}

You can visualize this using the SAGElet {}``The Ripple Effect: Circumference''.
It animates the ripple, and approximates $\ec dC/dt$ using the secant
lines. Experiment with different values of $\ec dr/dt$ ({}``change
in radius per second'') and see if you can determine a pattern.


\pause{}

Interestingly, any fixed choice of $\ec dr/dt$ gives a constant value
for $\ec dC/dt$, \emph{regardless of the value of $\ec r$}. Here
are some values that we found:

\begin{center}
{\ec%
\begin{tabular}{c|cccc}
$\frac{dr}{dt}$  & 0  & 1  & 5  & 10 \tabularnewline
\hline 
$\frac{dC}{dt}$  & 0  & $2\pi$  & $10\pi$  & $20\pi$ \tabularnewline
\end{tabular}}
\par\end{center}

This suggests the relationship $\ec dC/dt=2\pi(dr/dt)$. Since $\ec dr/dt$
is constant, so is $\ec dC/dt$. But can we \emph{prove} it?


\lyxframeend{}\lyxframe{{First Example: Concept}}

\begin{question} The radius of a ripple in a pond increasing at a
rate of 5~cm/s. How fast is the circumference changing at 3 seconds?\end{question}

\noindent The problem asks us to find $\ec dC/dt$, given that $\ec dr/dt=5$.

If we had a relationship between $\ec r$ and $\ec C$, we could differentiate
it with respect to $\ec t$. Since the radius $\ec r$ and the circumference
$\ec C$ depend on time $\ec t$, the Chain Rule would give us quantities
$\ec dr/dt$ and $\ec dC/dt$. We would end up with a relationship
between the rates!


\pause{}

This is the general procedure for related rates:\begin{method} 
\begin{itemize}
\item find an equation that relates the quantities; 
\item differentiate with respect to time; and 
\item solve for the unknown rate. 
\end{itemize}
\end{method}


\lyxframeend{}\lyxframe{{First Example: Solution}}
\begin{overlayarea}
{\textwidth}{6in}

\begin{question} The radius of a ripple in a pond increasing at a
rate of 5~cm/s. How fast is the circumference changing at 3 seconds?\end{question}
\begin{onlyenv}%{}
<1-2>

\noindent The problem asks us to find $dC/dt$, given that $dr/dt=5$.


This is the general procedure for related rates:\begin{method} 
\begin{itemize}
\item find an equation that relates the quantities; 
\item differentiate with respect to time; and 
\item solve for the unknown rate. 
\end{itemize}

\end{method}
\begin{onlyenv}%{}
<1>An equation that relates $\ec r$ and $\ec C$ is 
\begin{align*}
\ec C & {\ec=2\pi r}.
\end{align*}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<2>Differentiating with respect to time gives 
\begin{align*}
\dims C & {\dims=2\pi r}\\
\ec\frac{dC}{dt} & {\ec=2\pi\frac{dr}{dt}}.
\end{align*}

\end{onlyenv}%{}
\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

Substituting $\ec dr/dt=5$, we have 
\begin{align*}
\dims\frac{dC}{dt} & {\dims=2\pi\frac{dr}{dt}}\\
\ec\frac{dC}{dt} & {\ec=10\pi}.
\end{align*}
 Since $\ec t$ does not appear in the equation, $\ec dC/dt=10\pi$
for \emph{all} values of $\ec t$. At 3~seconds, therefore, the circumference
is increasing at $\ec10\pi$~cm/s, approximately. This agrees with
the answer found by experimentation.

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{\noindent {Second Example: Experimentation}}

\begin{question} The radius of a ripple in a pond increasing at a
rate of 5~cm/s. How fast is the area changing at 3 seconds?\end{question}

\noindent The problem asks us to find $\ec dA/dt$, given that $\ec dr/dt=5$.


\pause{}
\begin{onlyenv}%{}
<2>

You can visualize this using the SAGElet {}``The Ripple Effect: Area''.
It animates the ripple, and approximates $\ec dA/dt$ using the secant
lines of $\ec A$. Experiment with different values of $\ec dr/dt$
({}``change in radius per second'') and see if you can determine
a pattern.

\end{onlyenv}%{}

\pause{}

Unlike the circumference problems, fixed values of $\ec dr/dt$ do
\emph{not} result in a constant $\ec dA/dt$, depending on the value
of $\ec r$. Here are some values that we found with $\ec r'=5$:

\begin{center}
{\ec%
\begin{tabular}{c|ccccc}
$t$  & 1  & 2  & 3  & 4  & 5 \tabularnewline
\hline 
$r$  & 5  & 10  & 15  & 20  & 25 \tabularnewline
\hline 
$dA/dt$  & $25\pi$  & $75\pi$  & $125\pi$  & $175\pi$  & $225\pi$ \tabularnewline
\end{tabular}}
\par\end{center}

Notice that $\ec dA/dt\approx125\pi$ when $\ec t=3$. It seems to
obey the pattern 
\[
{\ec\frac{dA}{dt}=2\pi r\cdot\frac{dr}{dt}-\left(\frac{dr}{dt}\right)^{2}\pi}.
\]
 (This pattern is not obvious! Don't worry if you couldn't guess it.)
But can we \emph{prove} it?


\lyxframeend{}\lyxframe{{Second Example: Concept}}

\begin{question} The radius of a ripple in a pond increasing at a
rate of 5~cm/s. How fast is the area changing at 3 seconds?\end{question}

\noindent The problem asks us to find $\ec dA/dt$, given that $\ec dr/dt=5$.

As in the previous example, if we had a relationship between $\ec r$
and $\ec A$, then we could differentiate it with respect to $\ec t$
and obtain a relationship between the rates $\ec dr/dt$ and $\ec dC/dt$.


\pause{}

Recall the general procedure for related rates:\begin{method} 
\begin{itemize}
\item find an equation that relates the quantities; 
\item differentiate with respect to time; and 
\item solve for the unknown rate. 
\end{itemize}
\end{method}


\lyxframeend{}\lyxframe{{Second Example: Solution}}
\begin{overlayarea}
{\textwidth}{6in}

\begin{question} The radius of a ripple in a pond increasing at a
rate of 5~cm/s. How fast is the area changing at 3 seconds?\end{question}


\noindent The problem asks us to find $dA/dt$, given that $dr/dt=5$.
\begin{onlyenv}%{}
<1-3>

Recall the general procedure for related rates:\begin{method} 
\begin{itemize}
\item find an equation that relates the quantities; 
\item differentiate with respect to time; and 
\item solve for the unknown rate. 
\end{itemize}

\end{method}
\begin{onlyenv}%{}
<1>

An equation that relates $\ec r$ and $\ec A$ is 
\begin{align*}
\ec A & {\ec=\pi r^{2}}.\\
\phantom{\frac{dA}{dt}} & \phantom{=2\pi r\cdot\frac{dr}{dt}.}
\end{align*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<2>

Differentiating with respect to time gives 
\begin{align*}
\dims A & {\dims=\pi r^{2}}\\
\ec\frac{dA}{dt} & {\ec=2\pi r\cdot\frac{dr}{dt}}.
\end{align*}


\end{onlyenv}%{}
\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3->

Substituting $\ec dr/dt=5$, we have 
\begin{align*}
\dims\frac{dA}{dt} & {\dims=2\pi r\cdot\frac{dr}{dt}}\\
\ec\frac{dA}{dt} & {\ec=10\pi r}.
\end{align*}
 But what is $\ec r$?

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

It can happen when solving a related-rate problem that you need to
determine the value of a quantity that was not given. In this case,
we know that the radius increases at a rate of 5~cm/s. This is a
linear relationship, so $\ec r=5t$. At this particular moment (3~seconds)
$\ec t=3$, so $\ec r=15$.

\end{onlyenv}%{}

\pause{}


\begin{center}
Hence ${\displaystyle \ec\frac{dA}{dt}=150\pi}$.
\par\end{center}

\end{overlayarea}

\lyxframeend{}\lyxframe{Second Example: Comparison to approximation}

\noindent Unlike the example with circumference, the approximation
here was somewhat off. Using secant lines, the SAGElet approximated
$\ec dA/dt\approx125\pi$. The correct answer was instead $\ec dA/dt=150\pi$.
The error is due to a term in the pattern we identified with the approximation,
$\ec dA/dt\approx2\pi r(dr/dt)-{\hilite(dr/dt)^{2}\pi}$. Increasing
the value of {}``units per second'' in the SAGElet obtains better
approximations, because the secant lines are computed closer to $\ec t=3$.
\emph{(Try it!)} For example, we obtained the following results with
different values of {}``units per second'': 

\begin{center}
\begin{figure}[h!]
{\ec

\begin{centering}
\begin{tabular}{c|ccccccccc}
{\tiny $\ec t$ } & {\tiny 2 } & {\tiny 2.25 } & {\tiny 2.5 } & {\tiny 2.75 } & {\tiny \hilite 3 } & {\tiny 3.25 } & {\tiny 3.5 } & {\tiny 3.75 } & {\tiny 4 }\tabularnewline
\hline 
{\tiny $n=1$ } & {\tiny $75\pi$ } &  &  &  & {\tiny $\hilite125\pi$ } &  &  &  & {\tiny $175\pi$ }\tabularnewline
{\tiny $n=2$ } & {\tiny $87.5\pi$ } &  & {\tiny $112.5\pi$ } &  & {\tiny $\hilite137.5\pi$ } &  & {\tiny $162.5\pi$ } &  & {\tiny $187.5\pi$ }\tabularnewline
{\tiny $n=4$ } & {\tiny $93.75\pi$ } & {\tiny $106.25\pi$ } & {\tiny $118.75\pi$ } & {\tiny $131.25\pi$ } & {\tiny $\hilite143.75\pi$ } & {\tiny $156.25\pi$ } & {\tiny $168.75\pi$ } & {\tiny $181.25\pi$ } & {\tiny $193.75\pi$ }\tabularnewline
\end{tabular}
\par\end{centering}

}\caption{ Values of $\ec dA/dt$ for different values $\ec n$ of {}``units
per second''.}
\end{figure}

\par\end{center}

\vspace{-2ex}
As $\ec n$ increases, the accuracy of the approximation of $\ec dA/dt$
increases.


\lyxframeend{}\lyxframe{{Third Example}}
\begin{overlayarea}
{\textwidth}{6in}

\begin{question} A camera is directed at the launch of a rocket.
The camera is 3~km from the rocket. Once the rocket rises 1~km,
it has a speed of 1500~km/hr. How fast must the angle of the camera
change to remain fixed on the rocket?\end{question}


\noindent The problem asks us to find $\ec d\alpha/dt$, given that
$\ec dh/dt=1500$.


\pause{}
\begin{onlyenv}%{}
<2>

How can we relate the two rates? The quantities $\ec\alpha$ and $\ec h$
are related by a right triangle: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.1]{figures/rocket_diagram} 
\end{figure}



\noindent Notice $\ec h$ lies on the side opposite $\ec\alpha$,
and the side adjacent $\ec\alpha$ is constant at 3~km. We can relate
the quantities using trigonometry:
\[
{\ec\tan\alpha=\frac{h}{3}}.
\]


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

This allows us to differentiate and relate the rates:
\begin{align*}
{\dims\tan\alpha} & {\dims=\frac{h}{3}}\\
\ec\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\ec=\frac{dh/dt}{3}}.
\end{align*}
 \begin{explanation}Remember from the Chain Rule that, since we differentiate
with respect to $t$,\\
 the derivatives must be multiplied by $dh/dt$ and $d\alpha/dt$
as appropriate.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4-5>

We can substitute known values at this instant:\vspace{-2ex}
\begin{align*}
{\dims\tan\alpha} & {\dims=\frac{h}{3}}\\
\dims\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\dims=\frac{dh/dt}{3}}\\
\ec\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\ec=\frac{1500}{3}=500}.
\end{align*}
 


\begin{center}
But what is the value of $\ec\alpha$?
\par\end{center}


\pause{}
\begin{onlyenv}%{}
<5>

\vspace{-2ex}
Since $\ec\tan\alpha=h/3$, then $\ec\alpha=\arctan(h/3)$. Thus\vspace{-2ex}
\[
{\ec\alpha=\arctan\frac{1}{3}}.
\]


\end{onlyenv}%{}
\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<6>

Thus:
\begin{align*}
{\dims\tan\alpha} & {\dims=\frac{h}{3}}\\
\dims\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\dims=\frac{dh/dt}{3}}\\
\dims\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\dims=\frac{1500}{3}=500}\\
\ec\sec^{2}\left(\arctan\frac{1}{3}\right)\cdot\frac{d\alpha}{dt} & {\ec=500}.
\end{align*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<7>

Thus:
\begin{align*}
{\dims\tan\alpha} & {\dims=\frac{h}{3}}\\
\dims\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\dims=\frac{dh/dt}{3}}\\
\dims\sec^{2}\alpha\cdot\frac{d\alpha}{dt} & {\dims=\frac{1500}{3}=500}\\
\dims\sec^{2}\left(\arctan500\right)\cdot\frac{d\alpha}{dt} & {\dims=500}\\
\ec\frac{d\alpha}{dt} & {\ec=\frac{500}{\sec^{2}\left(\arctan\frac{1}{3}\right)}}.
\end{align*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<8>

Thus:\vspace{-2ex}
\begin{align*}
{\dims\tan\alpha} & {\dims=\frac{h}{3}}\\
 & \dims\vdots\\
\ec\frac{d\alpha}{dt} & {\ec=\frac{500}{\sec^{2}\left(\arctan\frac{1}{3}\right)}}.
\end{align*}
 From trigonometry, $\ec\sec^{2}\left(\arctan\frac{1}{3}\right)=\frac{\sqrt{10}}{3}$.
Hence
\[
{\ec\frac{d\alpha}{dt}=\frac{500}{\frac{\sqrt{10}}{3}}=\frac{1500}{\sqrt{3}}\approx866}\mathrm{\ radians\ per\ hour}.
\]


\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item Problems that involve {\hilite related rates} can be solved by
taking derivatives. 
\item The method of solution is to 

\begin{itemize}
\item find an equation that relates the quantities; 
\item differentiate with respect to time; and 
\item solve for the unknown rate. 
\end{itemize}
\item Quantities that are constant should not be substituted in before taking
the derivative. 
\item Sometimes you will have to go back and re-examine an equation in order
to determine the value of a quantity. 
\end{itemize}

\lyxframeend{}
\end{document}
