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\title[]{MA140-Engineering Calculus}
\author[ Lecture 28 ]{\large{ Lecture 28} 
\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}

{\textcolor{red}{Arc Length:}}
Let C be a curve given by the equation $y=f(x)$. It may be helpful to imagine the curve as
the path of a particle moving from point $A$ to point $B$. We subdivide the path (or arc) $AB$ into $n$ pieces at points $A=P_{0}, P_{1}, P_{2}, \cdots , P_{n}=B $. Join successive points of this subdivision by straight line segments 
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec700.pdf}
		\end{center}
\end{frame}
\begin{frame}
A representative line segment: 
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec701.pdf}
		\end{center}
		has length: $$L_{k}=\sqrt{(\Delta x_{k})^{2}+(\Delta y_{k})^{2}}$$
\end{frame}
\begin{frame}
An intuitive approximation to the length of the curve $AB$, $S$,  is the sum of all the lengths $L_{k}:$ 

$$S\cong \sum_{k=1}^{n} L_{k}=\sum_{k=1}^{n} \sqrt{(\Delta x_{k})^{2}+(\Delta y_{k})^{2}} $$ 
So $$S\cong \sum_{k=1}^{n} \sqrt{(\Delta x_{k})^{2}(1+(\frac{\Delta y_{k}^{2}}{\Delta x_{k}^{2}})) }$$ 
\end{frame}
\begin{frame}
So: $$S\cong \sum_{k=1}^{n} \sqrt{(1+(\frac{\Delta y_{k}^{2}}{\Delta x_{k}^{2}})) } (\Delta x_{k})$$
Using a Riemann sum approach.\\
Let $\Delta x\to 0$ or $n\to \infty$, we get: $$S=\lim_{n\to \infty} \sum_{k=1}^{n} \sqrt{(1+(\frac{\Delta y_{k}^{2}}{\Delta x_{k}^{2}})) } (\Delta x_{k})$$
So: $$S=\int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^{2}} dx$$
\end{frame}
\begin{frame}
\begin{theorem}
If $f$ is continuously differentiable on the closed interval $[a,b]$, the length of the curve (graph) $y=f(x)$ from $x=a$ to $x=b$ is: $$S=\int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^{2}} dx=\int_{a}^{b} \sqrt{1+(f'(x))^{2}} dx$$
\end{theorem}
\end{frame}
\begin{frame}
\begin{example}
Find the length of the curve $$y=\frac{4\sqrt{2}}{3}x^{3/2}-1, \quad 0\leq x \leq 1$$
\end{example}
We use the theorem with $a=0$, $b=1$, and 
$$y=\frac{4\sqrt{2}}{3}x^{3/2}-1$$ 
First we take the derivative of $y$: $$\frac{dy}{dx}=\frac{4\sqrt{2}}{3}\cdot \frac{3}{2}x^{1/2}=2\sqrt{2}x^{1/2}$$
So $$(\frac{dy}{dx})^{2}=(2\sqrt{2} x^{1/2})^{2}=8x$$
\end{frame}
\begin{frame}
The length of the curve from $x=0$ to $x=1$ is:
$$S=\int_{0}^{1} \sqrt{1+(\frac{dy}{dx})^{2}} dx=\int_{0}^{1} \sqrt{1+8x} dx$$
$$=\frac{2}{3}\cdot \frac{1}{8} (1+8x)^{3/2} \Big ]^{1}_{0}=\frac{13}{6}$$
\end{frame}
\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}

\end{document}