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\begin{center}
History and Applications of Calculus 
Homework 4\\
James Tiberius Kirk\\
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\noindent

\vspace{2mm}
\noindent
1. A basic physics fact is that ``work is
force times distance.'' This means that to pull something along with constant force
$F$ Newtons for a total distance of $x$ meters takes $Fx$ Joules of energy...
In this question, you'll also need Newton's Law of Gravity: the
gravitational force
between mass $M$ and mass $m$ which are a distance $r$ apart is
$F = G M m / r^2$ where $G$ is the universal gravitational constant.
\begin{itemize}
\item[(a)] The constant $G$ is roughly $6.674\times 10^{-11} \operatorname{m}^2 / \operatorname{kg}.\operatorname{sec}^2$.
Also the radius of the earth is about 4000 miles. What is the mass of the earth?
\item[(b)] Suppose you have a mass $m$ distance $r$ from another mass $M$.
How much work do you need to do in order to move the mass $m$ very
very far away from the mass $M$ (``to infinity'')? 
\item[(c)] In the lectures this week, we showed that the eccentricity of a planetary orbit around the sun
  satisfies the formula
$e = \left(\frac{v_0}{v_{\operatorname{crit}}}\right)^2  - 1$ where $v_{\operatorname{crit}} = \sqrt{GM} /
\sqrt{r_0}$, $v_0$ is the initial velocity of a planet of
mass $m$ moving in the tangential direction, $r_0$ is its initial distance from
the sun, and $M$ is the mass of the sun.
How much kinetic energy does the planet have when $v_0 = \sqrt{2} v_{\operatorname{crit}}$?
\item[(d)] Explain the connection between parts (b) and (c).
\end{itemize}
\vspace{2mm}
\noindent
{\em Hints.}
(a) You'll also need to use $g  = 9.8 \operatorname{m}/ \operatorname{sec}^2$.
(b) You need to add up ``force times distance'' using 
an integral $\int_r^\infty F \, dr$
since the force is changing with the distance.
(c) Kinetic energy is $\frac{1}{2} m v^2$...
(d) You should find your answers to (b) and (c) are the same and have
to explain the coincidence!


\vspace{2mm}
\noindent
2. Recall that a function $f(x)$ with domain $\R$ is called {\em even} if $f(x) =
f(-x)$ and
{\em odd} if $f(x) = - f(-x)$ for all $x$.
\begin{itemize}
\item[(a)]
Suppose that $f(x)$ is {\em any} function with domain $\R$.
Prove that there exist unique functions $E(x)$ and $O(x)$ such that
$f(x) = E(x) + O(x)$ for every $x$, with $E(x)$ being even and $O(x)$
being odd. These functions $E(x)$ and $O(x)$ are called the {\em even} and {\em odd} parts of
$f(x)$, respectively.
\item[(b)]
Suppose that $f(x) = e^x$ is the exponential function.
Sketch the graphs of $f(x)$, and its even and odd parts $E(x)$ and
$O(x)$, all on the same axes.
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
For (a), start from the equations $f(x) = E(x) + O(x)$, 
$E(x) = E(-x)$ and $O(x) = -O(-x)$.
Try to ``solve'' to find formulae for $E(x)$ and $O(x)$ in terms of
$f(x)$ and $f(-x)$.

\vspace{2mm}
\noindent
3. The functions $E(x)$ and $O(x)$ in 2(b) are called the {\em hyperbolic
  functions} $\cosh x$ and $\sinh x$, respectively.
Explicitly: $$
\cosh x = \frac{1}{2}(e^x + e^{-x}), \qquad
\sinh x =
\frac{1}{2}(e^x - e^{-x}).
$$
\begin{itemize}
\item[(a)] Show that the derivative of $\sinh x$ is $\cosh x$ and the
  derivative of $\cosh x$ is $\sinh x$.
\item[(b)] Show that $\cosh^2 x - \sinh^2 x = 1$.
\item[(c)] Let $\tanh x := \frac{\sinh x}{\cosh x}$
and $\sech x := \frac{1}{\cosh x}$. Prove that
the derivative of $\tanh x$ is $\sech^2 x$.
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
For (a) and (b), just plug in the explicit definitions in terms of
$e^x$ and do some algebra!
For (c), use the quotient rule.

\vspace{2mm}
\noindent
4. Consider the differential equation
$$
f''(x) - f(x) = 0.
$$
\begin{itemize}
\item[(a)]
Prove that the functions $\cosh x$ and $\sinh x$ both give solutions
of this differential equation, hence, so does any linear combination
of the form $a \cosh x + b \sinh x$ for $a, b \in \R$.
In fact, the latter is the {\em general solution} of this differential
equation; you may assume this without proof in the remainder of the question.
\item[(b)] Prove that $\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh
  y$. In particular, $\cosh 2x = \cosh^2 x + \sinh^2 x = 2 \cosh^2 x -
  1 = 1 + 2 \sinh^2 x$.
\item[(c)] What is the analogous ``double angle formula'' for $\sinh 2x$?
\end{itemize}

\vspace{2mm}
\noindent
{\em Hints.}
For (b), one approach is to mimic what we did for question 4 on last
week's homework. Another valid approach, which is probably more work, would
be to replace $\cosh x$ by $\frac{1}{2}(e^x+e^{-x})$ everywhere and
use laws of exponents...


\vspace{2mm}
\noindent
5. In the lectures this week, 
we
used laws of motion to derive
the differential equation
$$
\frac{d^2 u}{d \theta^2} + u = \frac{f(r) r^2}{(v_0)^2 (r_0)^2 m}
$$
where $u = \frac{1}{r}$ and $f(r)$ was the central force, $m$ was the
mass of the planet, and $v_0$ and $r_0$ came from the initial conditions
we were assuming (which implied $u = \frac{1}{r_0}$ and
$\frac{du}{d\theta} = 0$ when $\theta = 0$). Then we
assumed the central force was given by Newton's inverse square law
$f(r) = \frac{G M m}{r^2}$,
and solved this equation to obtain the final equation of motion; it
was an ellipse, parabola or hyperbola according to the initial
velocity $v_0$. I suggest you review this before
attempting this question!

Taking the same general setup and initial conditions, suppose instead that 
we live in a parallel universe in which the
central force satifies the {\em inverse cube law}
$$
f(r) = \frac{G M m}{r^3}
$$
for some alien constant $G$ (it even has different units compared to our $G$!).
What can you say about orbits of planets in this universe?

\vspace{2mm}
 \noindent
{\em Hints.}
Take the same differential equation, convert the $r$'s on the right
hand side into $u$'s, and solve this differential equation! 
You should find things depend on $v_{\operatorname{crit}}$ which now should be
defined as $\sqrt{GM} / r_0$. The easiest case is when $v_0 = v_{\operatorname{crit}}$,
and you should find for this that the orbit is a circle. What happens
when $v_0 > v_{\operatorname{crit}}$? What about when $v_0 < v_{\operatorname{crit}}$? Perhaps questions
2, 3 and 4 were set for a reason!

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