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\title[]{MA140-Engineering Calculus}
\author[ Lecture 35 ]{\large{ Lecture 35 } 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

\begin{example}
A robot ``jumps'' (vertically)
on the moon, with initial velocity $3ms^{-1}$. Assuming zero 
(air) resistance and that gravity is one sixth of gravity on the earth, solve the differential equation $d^2x/dt^2 = -g/6$ to calculate
\begin{enumerate}
\item
the height of the jump
\item
how long the jump lasts for (the time difference between jumping and landing
back in the same position)
\end{enumerate}
($x(t)$ is the vertical height as a function of time.)
\end{example}
This is a second order linear differential equation of the form $\ddot{x} = -g/6$.
We can integrate directly:
\begin{equation*}
\int \ddot{x} dt = \int \left(\frac{-g}{6}\right) dt
\implies \dot{x} = \frac{-gt}{6} + C_1
\end{equation*}
\end{frame}
\begin{frame}
We integrate a second time:
\begin{equation*}
\int \dot{x} dt = \int \left(\frac{-gt}{6} + C_1\right) dt
\implies x = \frac{-gt^2}{12} + C_1t + C_2
\end{equation*}
The information given in the question can be used to fix $C_1$ and $C_2$: At time
$t=0$, the initial position is $x(0)=0$ and the initial velocity is 
$\dot{x}(0) = 3$:
\begin{itemize}
\item
$x(0) = 0 \implies C_2 = 0$
\item
$\dot{x}(0) = 3 \implies C_1 = 3$
\end{itemize}
so $x(t) = -gt^2/12 + 3t$. When $x(t) = 0$, the robot is on the ground, so
\begin{equation*}
-gt^2/12 + 3t = 0 = t(-gt/12 + 3) \implies t = 0 \mbox{ or } 3-gt/12 = 0
\end{equation*}
which gives $t = 36/g$, or just under 4 seconds.
\end{frame}
\begin{frame}
At maximum height, the velocity is zero:
\begin{equation*}
\frac{-gt}{6} + 3 = 0 \implies t = \frac{18}{g}
\end{equation*}
and at this time the position is
\begin{equation*}
x(t) = x(18/g) = \left(\frac{18}{g}\right)\left(3 - \frac{18g}{12g}\right)
= \left(\frac{18}{g}\right)(3 - \frac{3}{2})
= 27/g
\end{equation*}
(just a little under 3 metres).

\end{frame}
\begin{frame}
\frametitle{Second Order Homogeneous Linear Differential Equations With Constant Coefficients}
General linear differential equation (of order $n$):
\begin{equation*}
\begin{split}
a_n(x)y^{(n)} 
+
&
a_{n-1}(x)y^{(n-1)} 
+
a_{n-2}(x)y^{(n-2)} 
+
\dots\\
&
\dots
+
a_2(x)y^{(2)} 
+
a_1(x)y^{(1)} 
+
a_0(x)y
=
b(x)
\end{split}
\end{equation*}
where
\begin{equation*}
y^{(n)} = \frac{d^ny}{dx^n}
\end{equation*}
(differentiate $n$ times).
\begin{enumerate}
\item
{\bf \color{red} Second Order} means $n=2$.
\item
{\bf \color{red} Homogeneous} means $b(x) = 0$.
\item
{\bf \color{red} Constant Coefficient} means all the ``functions'' $a_n(x), 
a_{n-1}(x), \dots , a_2(x), a_1(x), a_0(x)$ are just numbers (constants).
\end{enumerate}
\end{frame}
\begin{frame}
\begin{equation*}
ay^{\prime\prime} + by^\prime + c = 0
\end{equation*}
The \emph{auxilliary equation} (in variable $m$) is $am^2 + bm +c=0$ which
has solutions
\begin{equation*}
m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
\end{equation*}
Depending on the solutions, we have three cases:
\begin{enumerate}
\item
{\bf \color{red} Two identical roots} $b^2-4ac = 0$.\\
Solution $y = (A + Bx)e^{mx}$

\item
{\bf \color{red} Two different real roots} $b^2-4ac > 0$.\\
Solution $y = Ae^{m_1x} + Be^{m_2x}$

\item
{\bf \color{red} Two different complex roots} $b^2-4ac < 0$.\\
Solution $y = e^{px}(A \cos (qx) + B\sin(qx))$ (where the roots are $p\pm i q$)
\end{enumerate}
\end{frame}
\begin{frame}
\frametitle{Newtons Laws (as differential equations)}
\begin{enumerate}
\item
{\bf \color{red} First Law}
Every body continues in a state of rest or of uniform motion in a straight line
unless acted on by an external force.
\item
{\bf \color{red} Second Law}
Force equals mass times acceleration.
\end{enumerate}
In fact:
\begin{enumerate}
\item
The Second Law tells us which differential equation to solve.
\item
The First Law tells us a solution.
\end{enumerate}
\begin{itemize}
\item {\huge $ \ddot{x} = 0$\ } Uniform motion in a straight line
\item {\huge $ \ddot{x} = g$\ } Acceleration due to (constant) gravity
\item {\huge $ \ddot{x} = -kx$\ } Motion of a Spring (Force varies (linearly) with
position)
\end{itemize}


\end{frame}


\end{document}