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\title[]{MA140-Engineering Calculus}
\author[ Lecture 16 ]{\large{ Lecture 16 } 
\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}{Tangents}

\begin{example}
Determine the (equation of the) tangent (line) to the function
$$
f(x) = x^2
$$
at the point $x=5$.
\end{example}

\vspace{-1cm}
\begin{equation}
\begin{split}
f(x) = x^2 \implies f'(x) = 2x\\
\mbox{At the point } x=5, y = f(x) = f(5) = 5^2 = 25.\\
\mbox{At the point } x=5, f'(x) = f'(5) = (2)(5) = 10.\\
\end{split}
\end{equation}
This means, the tangent line has slope 10 and goes 
through the point  $(5,25)$. The equation of a line with slope 
$m$ is $ y = mx + c $. So here we have $y=10x+c$. Since the line must
go through $(5,25)$, we insert this point in to the equation to get
$25 = (10)(5) + c \implies c = -25$. Finally our answer is 
$y=10x-25$.
\end{frame}


\begin{frame}

\begin{example}
Find the points on the curve $y=4x^3 - 6x$ where the tangent (line) is 
parallel to $2x + y + 5=0$.
\end{example}

%\vspace{-1cm}
To say two lines are ``parallel'' means they have the same slope:
\begin{enumerate}
\item
What is the slope of the line? Rewrite it as $y=-2x-5$, so the slope is 
$-2$. (The slope of a line $y=mx+c$ is $m$.)
\item
What is the slope at any point on the curve? 
$y=4x^3-6x \implies y' = 12x^2-6$.
\end{enumerate}
So we just set these two slopes equal to one another: $-2 = 12x^2-6
\implies 4 = 12x^2 \implies x=1/\sqrt{3}$ or $x = -1/\sqrt{3}$.
\begin{itemize}
\item
$x = 1/\sqrt{3} \implies y = 4(1/\sqrt{3})^3 - 6/\sqrt{3} = -14/(3\sqrt{3})$
\item
$x = -1/\sqrt{3} \implies y = -4(1/\sqrt{3})^3 + 6/\sqrt{3} = 14/(3\sqrt{3})$
\end{itemize}
\end{frame}

\begin{frame}
\begin{example}
Suppose a mug has the shape of a cylinder, and is being filled with water. At 
a particular point in time, the height of the water is rising at a rate of 
1cm per second. If the circumference of the mug is $6\pi$ cm,
what is the rate of increase of (the volume of) water in the mug?
\end{example}
$V = \pi r^2 h.$ Circumference = $2\pi r = 6\pi \implies r = 3 \implies V=9\pi h$.
$$
\frac{dV}{dt} = 9\pi\frac{dh}{dt} = 9\pi (1) = 9\pi \mbox{ (cubic cemtimetres 
per second)}
$$
\end{frame}



\end{document}