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2006 Paper 3 Q6
D: 1700.0
B: 1516.0
polar coordinates
gradient
curve sketching
parabola
reflection
differential equation
optics
geometric proof
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
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\par
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.
1989 Paper 2 Q5
D: 1600.0
B: 1561.1
polar coordinates
gradient
curve sketching
parabola
reflection
differential equation
optics
geometric proof
- Show that in polar coordinates, the gradient of any curve at the point
\((r,\theta)\) is
\[
\left.\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\tan\theta+r\right)\right/\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}-r\tan\theta\right).
\]
- A mirror is designed so that any ray of light which hits one side of the mirror and which is parallel to a certain fixed line \(L\) is reflected through a fixed point \(O\) on \(L\). For any ray hitting the mirror, the normal to the mirror at the point of reflection bisects the angle between the incident ray and the reflected ray, as shown in the figure. Prove that the mirror intersects any plane containing \(L\) in a parabola.
Solution:
- Suppose our curve is \(r(\theta)\), then \(y = r \sin \theta, x = r \cos \theta\) and \begin{align*} && \frac{\d y}{\d \theta} &= \frac{\d r}{\d \theta} \sin \theta + r \cos \theta \\ && \frac{\d x}{\d \theta} &= \frac{\d r}{\d \theta} \cos \theta - r \sin \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d \theta} \Bigg / \frac{\d x}{\d \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r \cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r \sin \theta} \\ &&&= \frac{\frac{\d r}{\d \theta} \tan\theta + r }{\frac{\d r}{\d \theta} - r \tan\theta} \end{align*} as required.
- Set up a system of polar coordinates such that the origin is at \(O\) and all points in the plane containing \(L\) are represented by \((r, \theta)\). The constraint we have is that the angle of the normal, is \(\frac12 \theta\). Let \(\tan \tfrac12 \theta = t\), then \(\tan \theta = \frac{2t}{1-t^2}\) \begin{align*} && \tan \frac12 \theta &= -\frac{\frac{\d r}{\d \theta} - r \tan\theta}{\frac{\d r}{\d \theta} \tan\theta + r } \\ \Rightarrow && t &= -\frac{r'-r\frac{2t}{1-t^2}}{r' \frac{2t}{1-t^2}+r} \\ &&&= \frac{2tr-(1-t^2)r'}{2tr'+(1-t^2)r} \\ \Rightarrow && (2t^2+1-t^2)r' &= (2t-t+t^3)r \\ && (1+t^2)r' &= t(t^2+1) r \\ \Rightarrow && r' &= t r \\ \Rightarrow && \frac{\d r}{\d \theta} &= \tan \tfrac12 \theta r \\ \Rightarrow && \int \frac1r \d r &= \int \tan \frac12 \theta \d \theta \\ && \ln r &= -2\ln \cos \tfrac12 \theta+C \\ \Rightarrow && r\cos^2 \frac12 \theta &= C \\ \Rightarrow && r + r\cos \theta &= D \\ \Rightarrow && r &= D-x \\ \Rightarrow && x^2 + y^2 &= D^2 - 2Dx + x^2 \\ \Rightarrow && y^2 &= D^2-2Dx \end{align*} Therefore it is a parabola