\noindent
\psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-0.57,-0.63)(8.51,6.23) \psline(0,0)(7,5) \psline(7,5)(7.75,1.98) \psline(7.75,1.98)(0,0) \parametricplot{-0.6740818217636368}{0.22208190190547994}{1*5.52*cos(t)+0*5.52*sin(t)+1.48|0*5.52*cos(t)+1*5.52*sin(t)+4.9} \psline(7,5)(5.79,1.45) \rput[tl](-0.4,-0.02){\(O\)} \rput[tl](5.76,1.29){\(P\)} \rput[tl](8.1,2.01){\(R\)} \rput[tl](7.2,5.26){\(Q\)} \psline(7.67,2.29)(7.37,2.22) \psline(7.37,2.22)(7.45,1.91) \end{pspicture*}
\par
In the diagram, \(O\) is the origin, \(P\) is a point of a curve \(r=r(\theta)\)
with coordinates \((r,\theta)\) and \(Q\) is another point of the curve,
close to \(P\), with coordinates \((r+\delta r,\theta+\delta\theta).\)
The angle \(\angle PRQ\) is a right angle. By calculating \(\tan\angle QPR,\)
show that the angle at which the curve cuts \(OP\) is
\[
\tan^{-1}\left({\displaystyle r\dfrac{\mathrm{d}\theta}{\mathrm{d}r}}\right).
\]
Let \(\alpha\) be a constant angle, \(0<\alpha<\frac{1}{2}\pi\). The
curve with the equation
\[
r=\mathrm{e}^{\theta\cot\alpha}
\]
in polar coordinates is called an
equiangular spiral. Show
that it cuts every radius line at an angle \(\alpha.\) Sketch the spiral.
Find the length of the complete turn of the spiral beginning at \(r=1\)
and going outwards. What is the total length of the part of the spiral
for which \(r\leqslant1\)?
{[}You may assume that the arc length \(s\) of the curve satisfies
\[
{\displaystyle \left(\frac{\mathrm{d}s}{\mathrm{d}\theta}\right)^{2}=r^{2}+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}.}]
\]