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\ctrline {\bf A small sample of mathematical typing}
\vskip 10 pt
\ctrline (taken from ThomasIs \sl Calculus,\rm 1956, pages 200 and 401)
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The period of vibration of a pendulum of mass $m$ is given by
$$T =2pi\sqrt{I\over mgl},$$
in which $I$ is the moment of inertia of the pendulum about an axis through the
pivot, $g$ is the acceleration due to gravity, and $l$ is the distance from the
pivot to the center of gravity (see Fig. 5--11).

\noindent {\bf Example 1}quad (Fig.5--12) Using Eq. (5), find the period of vibrat
ion of a pendulum which is simply a long slender bar of length $L$ pivoted at
one end.
\vskip 3pt
\noindent {\it Solution.}quad Suppose the rod coincides with the $x$-axis
from $x$ =0 to $x$ = $L$. Divide the interval $0≤x≤L$ into subintervals of
width $\Delta x$.Let $\Delta m$denote the mass between $x$ and $x+\Delta x$.
Then the moment of inertia of $\Delta m$ differs from $x↑2\Delta m$
by an infinitesimal of higher  order than $\Delta x$ as $\Delta x →0$ (for
a homogeneous rod of constant cross section), and aalso $\Delta  m = \rho \Delta x$,
where $\rho =M/L$ is the mass pper unit length.
Note that if $L$ is in feet and $g$  in ft/sec↑2  then $T$ will be the time in 
seconds required for one complete swing of the pendulum.

{\bf ll--5 Tangential vectors.}\quad  As the point $P$  moves along a given curve
in the $zy$-plane, we may imagine its position as being specified by the length
of arc $s$ from some arbitrarily chosen reference point $P↓0$ on the curve. The
vector $$R=ix+jy$$ from $O$ to $P(xy)$ corresponding to the value $s$, Then
$${\Delta R\over\Delta s}= i{\Delta x\over\Delta s}+j{\Delta y\over\Delta s}=
{PQ\over\Delta s