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\usepackage{amsmath,amsthm}
\usepackage{amssymb}
\usepackage{euscript}
\usepackage{graphicx}
\usepackage{tikz,pgfplots}
\pgfplotsset{compat=1.8}

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\newcommand{\R}{\mathbb{R}}  % Short for the real numbers
\newcommand{\cosec}{\operatorname{cosec}}

\begin{document}

\bigskip

\begin{center}
History and Applications of Calculus 
Homework 1\\
Charlie Brown
\end{center}

\bigskip

Some sample lines of math written in LaTeX for you to play around with (then delete when satisfied with your skills):
\begin{enumerate}
\item The quick brown fox jumps over the lazy dog.

\item {\it Italicized text}

\item {\bf Bold-face text}

\item Some normal text, {\it some italicized text}, and some {\bf
bold-face text}.

\item In LaTeX, all math must be inside dollar signs.  Notice the difference between
``Let x be an element of $\R$'' and ``Let $x$ be an element of $\R$''.

\item $\displaystyle\int_{0}^1 x^3\, dx= \Big [\frac{x^4}{4} \Big ]^1_0 =
\frac{1}{4}-\frac{0}{4}=\frac{1}{4}$.  

\item $f'(\theta)=\lim_{h\rightarrow 0} \left[
\frac{f(\theta+h)-f(\theta)}{h} \right]$  and
 $\lim\limits_{x\rightarrow\infty} \frac{\sin x}{x} =0$.

\item Let $T\colon \R^2 \rightarrow \R^2$ be given by $T(x,y)=(y,-x)$.  

\item $5<6$ and $6>2$.

\item For every $x\in (0,1)$ it is true that $x^2 < x$.

\item $A\subseteq B \iff (\forall x)[x\in A\Rightarrow x\in B]$

\item You all know the formula $\sum_{i=1}^n i = \frac{1}{2} n(n+1).$
I just wrote it in ``in-line'' math mode.  Here is the same formula in
``displayed'' math mode (double dollars):
$$
\sum_{i=1}^n i = \frac{1}{2} n(n+1).
$$
\end{enumerate}

\medskip

\hrule

\medskip

\vspace{2mm}
\noindent
1.  Some differentiation practice!
\begin{itemize}
\item[(a)] Starting from $\sin' = \cos$ and $\cos' = -\sin$, find the derivatives of $\tan x, \cot x, \sec x$ and $\cosec x$.
\item[(b)] Sketch the graphs of $y=\arcsin x$
and $y=\arccos x$ on the same axes. How does this demonstrate that $\arccos' = - \arcsin'$?
\item[(c)] You know that $\arcsin' x = \frac{1}{\sqrt{1-x^2}}$. So from (b) you get that
$\arccos' x = - \frac{1}{\sqrt{1-x^2}}$. Use implicit differentiation to prove this formula directly.
\end{itemize}

\vspace{2mm}
\noindent
{\em
Hints (which you should replace with your actual solution when you've
worked it out).}
(a)
Use the quotient rule and the chain rule! A useful LaTeX command for typesetting multiple equations with equals signs nicely lined up is
\begin{align*}
    \tan' x &= ,\\
    \cot' x &= ,\\
    \sec' x &= ,\\
    \cosec' x &=.
\end{align*}

\noindent
(b) Rather than trying to use LaTeX to draw these graphs, you could just draw them by hand, scan them in as a PDF, then include the PDF into your document using the {\tt includegraphics} command. You can google this command to find out how to use it, or look it up in the documentation included in Overleaf. It is also not so hard to work out how to draw the graphs using LaTeX. See question 3 below for an example. Again, you can usually find out how to do these things by doing a google search.

\noindent
(c) The issue here is that you have to take a square root... and so you have to make up your mind whether to take the positive or negative one. You have to know what slope you are expecting to resolve this ambiguity.

\vspace{2mm}
\noindent
2. At what angle to the ground should you launch a projectile in order
to get it to travel the furthest? (Ignore air resistance, assume the
ground is flat, etc...)

\vspace{2mm}
\noindent
{\em
Hints.}
Let $\theta$ be the launch angle and $v$ be the launch
speed. Let $x(t)$ be the horizontal distance traveled after $t$
seconds (so $x(0) = 0$). Let $y(t)$ be the vertical distance traveled
after $t$ seconds (so $y(0) = 0)$.
Recall that $\dot x(t)$, the derivative of $x(t)$ with respect to
time, is the horizontal velocity, $\ddot x(t)$, the second derivative,
is the horizontal acceleration. Of course that is zero since there are
no external horizontal forces acting on the projectile. So:
$\dot x(t) = v \cos \theta$.
Similarly $\dot y(t)$ is the vertical velocity and $\ddot y(t)$ is the
vertical acceleration. Since you are on the planet earth, you have
that $\ddot y(t) = -g$ where $g$ is a well-known constant, around $9.8 m/sec^2$.
Use this to work out formulae for $\dot y(t)$ and then $y(t)$. Then
work out the time when the projectile hits the ground again... how far
has it gone horizontally? What $\theta$ maximizes this distance?

\vspace{2mm}
\noindent
3. One of Archimedes' famous accomplishments was the quadrature of the
parabola. 
Here is the result. Take the usual parabola with graph $y = x^2$. Pick
two points on the parabola $A$ and $B$ with coordinates $(a,a^2)$ and
$(b,b^2)$, respectively, assuming for definiteness that $a < b$.
Join these two points up with a straight line $AB$.
Let $C$ be the point on the parabola in between $A$ and $B$ such that
the slope of the tangent line at $C$ is parallel to the line $AB$.

\begin{tikzpicture}
  \begin{axis}[samples=50,axis lines=middle]
  \addplot[blue,domain=-5:5](x,x*x);
  \end{axis}
  \draw[blue] (0,1) to (6.85,4);
	\draw[blue] (1.55,1.69) to (4,.16) to (6.18,3.7);
   \node at (1.2,1.9) {$\scriptstyle A$};
   \node at (6,3.9) {$\scriptstyle B$};
   \node at (3.9,.5) {$\scriptstyle C$};
\node at (1.55,1.69) {$\bullet$};
\node at (6.18,3.7) {$\bullet$};
\node at (4,.16) {$\bullet$};
\end{tikzpicture}

\vspace{2mm}
\noindent
{\bf Theorem.} {\em The area of the region below the line $AB$ and above the
parabola is equal to $\frac{4}{3}$ times the area of
the triangle $ABC$.}

\vspace{2mm}
\noindent
Give a proof of this theorem using all the tools from
calculus at your disposal.

\vspace{2mm}
\noindent
{\em
Hints.}
What are the coordinates of $C$? What is the area of $ABC$? What
integral could you compute to find the area under the parabola, above
the $x$-axis, to the left of $y=b$ and to the right of $y=a$?

\vspace{2mm}
\noindent
4. Use calculus to prove Snell's Law of Light Refraction: {\em If you
  shine a ray of
light
across a horizontal boundary between two media such that the speed of light in the top
medium is $u$ and the speed of light in the bottom medium is $v$, then
the light refracts in such a way that 
$$\frac{\sin \theta}{\sin \phi} = \frac{u}{v},$$ 
where $\theta$ is the
angle of incidence to the vertical above and $\phi$ is the angle
of incidence to the vertical below the boundary.}

\begin{tikzpicture}
	\draw[thick] (-3,0) to (3,0);
	\draw[<-] (-3,-3) to (-1,0) to (3,3);
	\draw[dotted,] (-1,-3) to (-1,3);
%\node at (-3,-3) {$\bullet$};
\node at (-3,-3.4) {$(0,-1)$};
%\node at (3,3) {$\bullet$};
\node at (3,3.4) {$(a,1)$};
\node at (-1,0) {$\bullet$};
\node at (-1.5,.4) {$(x,0)$};
\node at (-.6,.75) {$\theta$};
\node at (-1.2,-.75) {$\phi$};
\node at (4,1) {$\text{Speed of light in top medium} = u$};
\node at (4,-1) {$\text{Speed of light in bottom medium} = v$};
\end{tikzpicture}

\vspace{2mm}
\noindent
{\em Hints.} 
Light travels so as to minimize the total time taken (by the physics
of waves).
As in my picture, suppose the light starts at the coordinate $(a,1)$,
the boundary between the two media is the $x$-axis with the ray of
light crossing the $x$-axis at coordinate $(x,0)$, and the light ends
at the coordinate $(0,-1)$.
Find a formula for the total time $T(x)$ taken by light traveling
along the path shown (remember: time equals distance over speed).
Now you have to find the value of $x$ that minimizes this function.
This means you need to find the critical point where $T'(x) = 0$.
So differentiate $T(x)$, set the derivative to zero, and use this
equation to show that $v \sin \theta = u \sin \phi$.
It will help to write $\sin \theta$ and $\sin \phi$ in terms of $x$
before you start!


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