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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Applications of the derivative}

\begin{example}
A man, walking at night, is walking directly towards a streetlight. The light
is 10 feet off the ground, and the man is 6 feet tall and walking at 4 feet
per second.  When the man is 8 feet from the streetlight, how fast is the  length
of his shadow changing?
\end{example}
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{\par\centering \includegraphics{streetlight.ps} \par}
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\subsubsection*{Solution}

This one is unusual, in that it does not involve the Pythagorean Theorem. It
is actually simpler than it appears, but confuses a lot of people because you
are now expecting to use the Pythagorean Theorem on every one of these problems. 

Reducing the question to its basic parts, let \( s \) be the length of the
shadow, and let \( x \) be the distance from the man to the streetlight. You
do \emph{not} want to know the length of that ray of light from the streetlight
to the ground; you also don't know anything about it, so leave it out of the
problem. The only other facts you have are the height of the man (6 feet) and
that of the streetlight (10 feet). That information gives the following drawing:

\vspace{0.3cm}
{\par\centering \includegraphics{light-measurements.ps} \par}
\vspace{0.3cm}

You \textbf{know} that \( dx/dt=-4 \) (walking towards the streetlight), and
that the height of the man is \( 6 \), the light is \( 10 \). You \textbf{want}
to know what \( ds/dt \) is \textbf{when} \( x=8 \). 

Both these triangles are right triangles (presuming that both the streetlight
and the man are standing upright, and that the ground is flat), and so both
triangles have a right angle. They also both share the angle made by the ground
and the ray of light at the tip of the shadow, so all three angles of the triangle
are the same; that is, the two triangles are similar. Similar triangles have
the same ratios of corresponding sides, so the ratio of base/altitude is the
same for the two triangles. For the larger triangle, the altitude is 10, and
the base is \( (x+s) \); for the smaller triangle, the base is 6 and the altitude
is \( s \), so\[
\frac{(x+s)}{10}=\frac{s}{6}.\]
Simplifying, \begin{eqnarray*}
\frac{x}{10} & = & \frac{s}{6}-\frac{s}{10}\\
 & = & \frac{5s-3s}{30}\\
 & = & \frac{s}{15}.
\end{eqnarray*}
Differentiating both sides, and substituting \( dx/dt=-4 \) (he is walking
towards the streetlight).\begin{eqnarray*}
\frac{1}{10}\frac{dx}{dt} & = & \frac{1}{15}\frac{ds}{dt},\\
\frac{1}{10}(-4) & = & \frac{1}{15}\frac{ds}{dt},\: \textrm{or}\\
\frac{ds}{dt} & = & -6,
\end{eqnarray*}
so the shadow is shrinking at a rate of 6 feet/sec. This is the value for all
values of \( x \), the part about \( x=8 \) is irrelevant. 
\end{document}
