Year: 2006
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Polar coordinates
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show that in polar coordinates the gradient of any curve
at the point $(r,\theta)$ is
\[
\frac{ \ \ \dfrac{\d r }{\d\theta} \tan\theta + r \ \ }
{ \dfrac{\d r }{\d\theta} -r\tan\theta}\,.
\]
\noindent \begin{center}
\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*}(-0.6,-3)(6.8,3) \psline(0,0)(6.54,0) \rput[tl](4.13,-0.22){$O$} \rput[tl](-0.47,0.07){$L$} \rput{-270}(5.75,0.08){\psplot[plotpoints=500]{-12}{12}{x^2/2/3}} \psline(2,1.5)(5.42,1.5) \psline(3.73,-0.74)(5.42,1.5) \psline[linewidth=0.4pt]{->}(3,1.5)(4,1.5) \psline[linewidth=0.4pt]{->}(5.42,1.5)(4.99,0.93) \psline(3.84,0.78)(6.62,2.05) \end{pspicture*}
\par\end{center}
A mirror is designed so that if an incident ray of light is parallel
to a fixed line $L$ the reflected ray passes through a fixed point $O$
on $L$. Prove that the mirror intersects any plane containing $L$ in
a parabola. You should assume that the angle between the incident
ray and
the normal to the mirror is the same as the
angle between the reflected ray and the normal.