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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}
Use differentials to find the relative error (the percentage of error) in the
computation of the volume of a cube, if the measurements of the sides might
be off by as much as 1\%.
\end{example}

\subsubsection*{Solution}

Let's say that the side of the cube is measured as \( x \). The percentage
error is of course (expressed as a percent) \( 100 \) times the fraction of
error \[
\frac{error}{value},\]
 so if the error in measurement is \( dx \), the fraction of the measurement
error is \[
.01=\frac{dx}{x},\]
 if the error is 1\%. Now, since the volume of the cube is \( V=x^{3} \), then
the differential of the volume is \( dV=3x^{2}dx \). The exact error in the
measurement of the volume is \( \Delta V \), which is approximated by \( dV \),
The fraction of the volume that is the error is \begin{eqnarray*}
\frac{\Delta V}{V} & \approx  & \frac{dV}{V}\\
 & = & \frac{3x^{2}dx}{x^{3}}\\
 & = & 3\frac{dx}{x}\\
 & = & .03,
\end{eqnarray*}
so the percentage of volume error is about 3\%, three times the measurement
error percentage of the sides.
\end{document}
