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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Homework 7 (February 23, 2013) \\
\end{center}
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\noindent 1. Continuation of Midterm 1, Problem~5: 

\noindent Consider $\alpha_1=\alpha_2=\alpha$ and find 
the angle $\alpha_0>0$, so that for $0<\alpha<\alpha_0$ 
\begin{eqnarray} \label{Tension}
  \left|\vec{T}^{\,1}\right| = \left|\vec{T}^{\,2}\right| > 
  \left|\vec{F}^{\,2}\right| 
\end{eqnarray} 
holds (6 points).
\medskip

Solution: Let $T^1=|\vec{T}^{\,1}|$, $T^2=|\vec{T}^{\,2}|$ and 
$F=|\vec{F}|$.  As $\alpha_1 =\alpha_2=\alpha$ we have (second 
solution of the midterm)
\begin{eqnarray} \nonumber
  T^1 &=& F\,\frac{\cos(\alpha)}{\sin(2\alpha)}\ =\ T^2
\end{eqnarray} 
and we are looking for
\begin{eqnarray} \nonumber
  1 &=& \frac{\cos(\alpha_0)}{\sin(2\alpha_0)}\ =\
  \frac{\cos(\alpha_0)}{2\,\sin(\alpha_0)\,\cos(\alpha_0)}\\ \nonumber
  &=& \frac{1}{2\,\sin(\alpha_0)}\ \Rightarrow\
  \sin(\alpha_0)\ =\ \frac{1}{2}\\ \nonumber
\end{eqnarray} 
\begin{eqnarray} 
  \alpha_0\ =\ \frac{\pi}{6}\,.
\end{eqnarray} 
\bigskip

\noindent 2. Use Stoke's Theorem to calculate the line integral
of the previous homework (4~points).
\medskip

Solution: The area of the triangle is
\begin{eqnarray} \nonumber
  A\ =\ \frac{a\,h}{2}\ =\ \frac{a^2\,\sqrt{3}}{4}\,,
  ~~~\vec{A}\ =\ A\,\hat{z} 
\end{eqnarray} 
and 
\begin{eqnarray} \nonumber
  \nabla\times\vec{F}\ =\ 2\,\hat{z}\,.
\end{eqnarray} 
Using Stoke's theorem,
\begin{eqnarray} 
  \oint_{\triangle}\vec{F}\cdot d\vec{s}\ =\ 
  2\,\hat{z}\cdot \vec{A}\ =\ \frac{a^2\,\sqrt{3}}{2}\,.
\end{eqnarray} 
Obviously, this is much shorter than the explicit calculation.

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