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\bigskip

\begin{center}
History and Applications of Calculus Homework 3\\
Johannes Kepler\\
\end{center}


\bigskip

Some sample lines of math written in LaTeX for you to play around
with:
\begin{align*}
x &= 2y+3\\
x-3 &= 2y\\
(x-3) / 2 &= y.
\end{align*}
If you remove the $*$ then it numbers all of the lines so you can
refer to them later:
\begin{align}
x &= 2y+3\\
x-3 &= 2y\\
(x-3) / 2 &= y. \label{myname}
\end{align}
If you only wanted to number the final equation (\ref{myname}) you can use {\tt
  notag}:
\begin{align}
x &= 2y+3\notag\\
x-3 &= 2y\notag\\
(x-3) / 2 &= y.
\end{align}
 Here is a very long equation split over several lines:
\begin{multline*}
f(x) = x + \frac{x^2-3}{x+1} + \det
\left[\begin{array}{rr}2&4\\1&-2\end{array}\right] + f'(2x+4) + e^{x^2}\\+ 248 \cdot 3 + 248.3 + \int_0^{100}
\cosh x \,dx.
\end{multline*}
\medskip

\hrule

\medskip

\noindent

1.
The cycloid is defined parametrically by the equations
$$
x = a(\theta - \sin \theta),\qquad
y = a(1 -\cos \theta).
$$
Calculate the area under one ``cycle'' of this cycloid and above the
$x$-axis.

\vspace{2mm}
\noindent
{\em
Hints.}
See the lectures last week for pictures of the cycloid and
cardioid. You'll want to use the area under curve formula $\int_a^b
y \,dx$. If you do the substitution $x = x(\theta)$, so $\,dx =
x'(\theta) \, d \theta$, this will transform into an integral in terms
of $\theta$ which you can calculate.

\vspace{2mm}
\noindent
2. The cardioid is defined in polar coordinates by the equation 
$$
r =
2a(1-\cos \theta).
$$
Calculate the area of the cardioid.

\vspace{2mm}
\noindent
{\em
Hints.}
Remember the formula derived in the lectures for the area of a sector of a curve in polar coordinates:
$\frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \,d\theta$.

\vspace{2mm}
\noindent

\vspace{2mm}
\noindent
3. Suppose $F_1$ and $F_2$ are two given foci 
in the plane which are a distance $2c$ apart,
and $0 < a < c$.
The {\em hyperbola} is the locus of points
$P$ such that $\big||PF_2| - |P F_1|\big| = 2a$.
\begin{itemize}
\item[(a)] Assuming
  that the foci are at coordinates $(\pm c, 0)$, show that the Cartesian equation for this hyperbola
  has the form
$$
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
$$
for some $b > 0$.
Write $b$ in terms of $a$ and $c$ and show that $c = ae$ where $e =
\sqrt{1+\frac{b^2}{a^2}}$ is the {\em eccentricity}.
\item[(b)] Find the polar equation for this ``left half'' of this
  hyperbola, that is, the curve 
defined by the equation
$|PF_2| - |P F_1| = 2a$ (I've removed the absolute values!). Assume for
this that $F_1$ is
the origin and $F_2$ is the point $(2c,0)$.
\end{itemize}

\vspace{2mm}
\noindent
{\em
Hints.}
For (a), look back at the question about the ellipse on last week's
homework. It is a similar method, I think.
For (b), your answer should take the form
$r = \frac{r_0(1+e)}{1 + e \cos \theta}$
for some $r_0$ and $e > 1$ (both of which you should determine in terms of
$a$ and $c$). Look how we did it for the ellipse in the lectures!

\vspace{2mm}
\noindent
4.
In this question you can assume without proof that the general
solution of the differential equation
$$
f''(x) + f(x) = 0
$$
is $f(x) = a \cos x + b \sin x$
where $a = f(0)$ and $b = f'(0)$, i.e., they are constants you can
determine from the initial conditions.
Use this to prove the multiple angle formulae
\begin{align*}
\sin(x+y) &= \sin x \cos y + \cos x \sin y,\\
\cos(x+y) &= \cos x \cos y - \sin x \sin y,\\
\tan(x+y) &= \frac{\tan x + \tan y}{1 - \tan x \tan y}.
\end{align*}


\vspace{2mm}
\noindent
{\em
Hints.}
To prove the first identity involving sine, consider the function $f(x) =
\sin (x+y)$ where $y$ is viewed as a constant. Show that this function
satisfies the differential equation, then compute $f(0)$ and $f'(0)$
and apply the result you were given. The proof for cosine is
similar. To get the one for tangent, use
$\tan = \frac{\sin}{\cos}$.

\vspace{2mm}
\noindent
5. If you have a graph $r = f(\theta)$ in polar coordinates, you can
also consider the {\em inversion} of this graph, which is $r =
\frac{1}{f(\theta)}$.
Note this transformation switches points inside the unit circle with
points outside...
\begin{itemize}
\item[(a)]
The graphs $r = 1 -\cos \theta$ and $r = \frac{1}{1-\cos \theta}$ are the
inversions of each other. Sketch these two graphs on the same
axes. What are these two graphs called?
\item[(b)] Consider the straight line whose Cartesian equation is $x = c$
for some constant $c > 0$. Show that its inversion is a circle
passing through the origin.
\item[(c)] Show that the inversion of a circle which does {\em not} pass through the origin is another such circle.
\end{itemize}

\vspace{2mm}
\noindent
{\em
  Hints.}
For (a), note that $$
\cos(\pi + \theta) = \cos \pi \cos \theta - \sin
\pi \sin \theta = - \cos \theta.
$$ 
This means that the graph of $r = 1 / (1-\cos \theta)$ is the graph of
$r = 1 / (1+\cos \theta)$ rotated through angle $\pi$ around the
origin. You saw the latter graph in class...
For (b), the first step is to write the equation $x = a$ in polar
coordinates as $r = f(\theta)$ for some function $f$.
You're on your own for (c)!

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