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\title[]{MA140-Engineering Calculus}
\author[ Lecture 29]{\large{ Lecture 29} 
\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}
\begin{frame}
\titlepage
\end{frame}


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


\begin{frame}
\begin{example}
The cable of a suspension bridge takes the shape of the curve:
$$y=\frac{h}{l^{2}}x^{2}-\frac{2h}{l}x+h$$
Where $0\leq x\leq 2l$, $h>0$. Find the length of the cable.
\end{example}

\end{frame}
\begin{frame}
$$\frac{dy}{dx}=\frac{2h}{l^{2}}x-\frac{2h}{l}=\frac{2h}{l}(\frac{x}{l}-1)$$
So: $$(\frac{dy}{dx})^{2}=\Big [ \frac{2h}{l}(\frac{x}{l}-1) \Big ]^{2}$$
The length of the curve equals: $$S=\int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^{2}} dx=\int_{0}^{2l} \sqrt{1+\Big [ \frac{2h}{l}(\frac{x}{l}-1) \Big ]^{2}} dx$$
First we find the following indefinite integral:
$$\int \sqrt{1+\Big [ \frac{2h}{l}(\frac{x}{l}-1) \Big ]^{2}} dx, \quad (\textcolor{red} {\star})$$ 
\end{frame}
\begin{frame}
Let $u=\frac{2h}{l}(\frac{x}{l}-1)$, then $du=\frac{2h}{l^{2}}dx$ or $\frac{l^{2}}{2h}du=dx$\\
So: $$(\textcolor{red} {\star})=\frac{l^{2}}{2h}\int \sqrt{1+u^{2}} du$$
{\textcolor{red}{quick reminder:}}\\
\begin{itemize}
\item $\sinh (x)=\frac{e^{x}-e^{-x}}{2}$
\item $\cosh (x)=\frac{e^{x}+e^{-x}}{2}$
\item $\frac{d}{dx} \sinh (x)=\cosh (x)$
\item $\frac{d}{dx} \cosh (x)=\sinh (x)$
\item $\cosh^{2} (x)-\sinh^{2} (x)=1$
\end{itemize}

\end{frame}
\begin{frame}
Now let $u=\sinh (v)$, so $du=\cosh (v) dv$, so:
$$(\textcolor{red} {\star})=\frac{l^{2}}{2h} \int \sqrt{1+u^{2}} du=\frac{l^{2}}{2h} \int \sqrt{1+\sinh ^{2} (v)} \cosh (v) dv$$ $$=\frac{l^{2}}{2h} \int \cosh (v) \cosh (v) dv=\frac{l^{2}}{2h} \int \cosh^{2} (v) dv$$
We have: $$\cosh^{2} (v)=\frac{[e^{2v}+e^{-2v}+2]}{4}=\frac{1}{2} [\frac{e^{2v}+e^{-2v}+2}{2}]$$ $$=\frac{1}{2}[\frac{e^{2v}+e^{-2v}}{2}+1]=\frac{1}{2}[\cosh(2v)+1]$$
So the above integral is:
$$(\textcolor{red} {\star})=\frac{l^{2}}{2h} \int \frac{1}{2}[\cosh(2v)+1] dv=\frac{l^{2}}{4h} \Big [ \frac{1}{2} \sinh (2v)+v \Big ]+c$$
\end{frame}
\begin{frame}
So:
$$(\textcolor{red} {\star})=\frac{l^{2}}{4h}\Big [ \sinh(v)\sqrt{1+\sinh^{2} (v)}+v \Big ]+c=\frac{l^{2}}{4h}\Big [ u\sqrt{1+u^{2}}+\sinh^{-1} (u) \Big ]+c$$ $$=\frac{l^{2}}{4h}\Big [ \frac{2h}{l}(\frac{x}{l}-1)\sqrt{1+(\frac{2h}{l}(\frac{x}{l}-1))^{2}}+\sinh^{-1} (\frac{2h}{l}(\frac{x}{l}-1)) \Big ]+c$$
So the length of the cable is:
$$\frac{l^{2}}{4h}\Big [ \frac{2h}{l}(\frac{x}{l}-1)\sqrt{1+(\frac{2h}{l}(\frac{x}{l}-1))^{2}}+\sinh^{-1} (\frac{2h}{l}(\frac{x}{l}-1)) \Big ]+c \Big ]^{2l}_{0}$$
$$=\Big ( \frac{l^{2}}{4h} (\frac{2h}{l} \sqrt{1+(\frac{2h}{l})^{2}}+\sinh^{-1} (\frac{2h}{l})\Big )$$$$-\Big ( \frac{l^{2}}{4h} (\frac{-2h}{l} \sqrt{1+(\frac{2h}{l})^{2}}+\sinh^{-1} (\frac{-2h}{l})\Big )$$
\end{frame}
\begin{frame}
Using the fact that $\sinh^{-1}(-x)=-\sinh^{-1}(x)$, we can simplify the above answer:
$$=\sqrt{l^{2}+4h^{2}}+\frac{l^{2}}{2h}\sinh^{-1}(\frac{2h}{l}) $$
\end{frame}

\end{document}