Where should the lifeguard enter the water?
Tonight's idea
A lifeguard runs fast and swims slowly. The straight line to a swimmer in trouble is the shortest route, but it spends a long time in slow water. Running further along the sand first is longer and quicker.
The quickest route bends at the water's edge. Light does the same: it is slower in water than in air, and a ray bends at the surface. Fermat's principle says why: light takes a route whose travel time is stationary, meaning its derivative is zero as you slide the route a little. Usually that is the quickest.
Measure both angles from the perpendicular to the boundary. With speeds \(v_1\) and \(v_2\) on the two sides, the time is stationary when \(\dfrac{\sin a}{v_1} = \dfrac{\sin b}{v_2}\). This is Snell's law.
Much of physics can be written this way. In mechanics, a thrown ball follows the path that makes a quantity called the action stationary. Laws like these are called variational principles.
Watch one: two routes, timed
A model: a straight shoreline, a steady 5 m/s on sand and 1 m/s in water, and straight legs. The lifeguard starts 20 m back from the water. The swimmer is 10 m out and 30 m along the beach. She enters the water \(x\) metres along. Time is distance over speed, and each distance is a hypotenuse:
\[T(x) = \frac{\sqrt{400 + x^2}}{5} + \frac{\sqrt{100 + (30 - x)^2}}{1}.\]
Straight line, crossing at \(x = 20\): \(T(20) = \sqrt{800}/5 + \sqrt{200} = 5.66 + 14.14 = 19.80\) s, over 42.43 m. Run until level with the swimmer, \(x = 30\): \(T(30) = \sqrt{1300}/5 + 10 = 17.21\) s, over 46.06 m. That is 3.6 m longer and about 2.6 s quicker. Can she do better still?
Your turn
- Compute \(T(x)\) to two decimal places for \(x = 10\), 25, 28 and 29. Together with \(T(20)\) and \(T(30)\) above, which of the six entry points is quickest?
- Differentiate \(T\). Show that \(T'(28) \lt 0\) and \(T'(29) \gt 0\), and say what that tells you. Then check that \(x = 28.34\) makes \(T'\) very nearly zero, and find the time for that route. How much quicker is it than the straight line?
- In the figure, \(a\) and \(b\) are the angles the sand leg and the water leg make with the perpendicular to the shore. Write \(\sin a\) and \(\sin b\) in terms of \(x\), and show that \(T'(x) = 0\) says \(\sin a / 5 = \sin b / 1\). Find \(a\) and \(b\) at \(x = 28.34\). What would the rule look like with angles measured from the shoreline?
- The speed of light in water is its speed in air divided by about 1.33. A ray in air meets a flat pond at \(45^\circ\) to the perpendicular. Find its angle to the perpendicular in the water. Does it bend towards the perpendicular or away, and by how many degrees? What about a ray arriving along the perpendicular?
Stretch: stationary is not always least
A circular mirror has radius 1 m and centre \(O\). Points \(A\) and \(B\) lie on a diameter, each 0.75 m from \(O\) on opposite sides. Light goes from \(A\) to a point \(P\) on the mirror and then to \(B\), where \(OP\) makes angle \(t\) with \(OB\). The cosine rule gives the route length
\[L(t) = \sqrt{1.5625 + 1.5\cos t} + \sqrt{1.5625 - 1.5\cos t}.\]
By symmetry, a ray really does reflect from \(A\) to \(B\) at \(t = 90^\circ\). Compute \(L\) at \(90^\circ\), \(80^\circ\) and \(60^\circ\), and find \(L'(90^\circ)\). Is the real ray the shortest of these routes? What does "stationary" mean here?
Answers
- \(T(10) = 4.47 + 22.36 = 26.83\) s. \(T(25) = 6.40 + 11.18 = 17.58\) s. \(T(28) = 6.88 + 10.20 = 17.08\) s. \(T(29) = 7.05 + 10.05 = 17.10\) s. With \(T(20) = 19.80\) and \(T(30) = 17.21\), the quickest of the six is \(x = 28\). She should run most of the way along before entering.
- \(T'(x) = \dfrac{x}{5\sqrt{400 + x^2}} - \dfrac{30 - x}{\sqrt{100 + (30 - x)^2}}\). Then \(T'(28) = 0.1627 - 0.1961 = -0.033\) and \(T'(29) = 0.1646 - 0.0995 = +0.065\). \(T\) is still falling at 28 and is rising at 29, so the minimum is between them. At \(x = 28.34\), \(T' = 0.16341 - 0.16376 \approx -0.0004\), and \(T = 6.94 + 10.14 = 17.07\) s, which is 2.72 s quicker than the straight line. (A computer gives \(x = 28.3436\). Each square root term curves upward, so \(T'\) increases all the way along and this is the only stationary point: the minimum.)
- Each leg is the hypotenuse of a right triangle whose side along the shore is opposite the angle: \(\sin a = \dfrac{x}{\sqrt{400 + x^2}}\) and \(\sin b = \dfrac{30 - x}{\sqrt{100 + (30 - x)^2}}\). So \(T'(x) = \dfrac{\sin a}{5} - \dfrac{\sin b}{1}\), and \(T' = 0\) is exactly \(\dfrac{\sin a}{5} = \dfrac{\sin b}{1}\). At \(x = 28.34\): \(\sin a = 0.817\), \(a = 54.8^\circ\), and \(\sin b = 0.164\), \(b = 9.4^\circ\); indeed \(0.817/5 = 0.163\). Angles from the shoreline are \(90^\circ - a\) and \(90^\circ - b\), so the same rule would use cosines. Snell's law with sines needs angles from the perpendicular.
- Air is the fast side: \(\dfrac{\sin 45^\circ}{1.33} = \dfrac{\sin b}{1}\), so \(\sin b = 0.7071/1.33 = 0.5317\) and \(b = 32.1^\circ\). The ray bends by \(45 - 32.1 = 12.9^\circ\), towards the perpendicular, just as the lifeguard's water leg is closer to the perpendicular than her sand leg. Along the perpendicular, \(\sin a = 0\) forces \(\sin b = 0\): the ray goes straight on, slower but not bent.
Stretch. \(L(90^\circ) = 1.25 + 1.25 = 2.5\) m, \(L(80^\circ) = 2.491\) m, \(L(60^\circ) = 1.521 + 0.901 = 2.422\) m. Differentiating, \(L'(t) = \dfrac{-0.75\sin t}{\sqrt{1.5625 + 1.5\cos t}} + \dfrac{0.75\sin t}{\sqrt{1.5625 - 1.5\cos t}}\), and at \(90^\circ\) both square roots are 1.25, so \(L'(90^\circ) = 0\). The real ray is the longest of the routes off the mirror, not the shortest: \(\cos^2 t\) is smallest at \(90^\circ\), and \(L^2 = 3.125 + 2\sqrt{1.5625^2 - 2.25\cos^2 t}\). Stationary means the graph of \(L\) is flat there, so nearby routes differ in length only by a tiny second-order amount. A flat point can be a minimum, a maximum or neither. Fermat's principle asks only for flat. (The shortest route, 2 m at \(t = 0\), is also a real ray: straight along the diameter and back.)