Do the same as in .1 but find the length L of a path on a

The shortest path between two points on a curved surface, such as the surface of a sphere, is called a geodesic. To find a geodesic, one has first to set up an integral that gives the length of a path on the surface in question. This will always be similar to the integral (6.2) but may be more complicated (depending on the nature of the surface) and may involve different coordinates than x and y. To illustrate this, use spherical polar coordinates (r, 0, 0) to show that the length of a path joining two points on a sphere of radius R is 02 L = R f \11+ sin2 0'(9)2 d0 (6.41) e, [Eq. (6.13)] if (01, 01) and (82, 02) specify the two points and we assume that the path is expressed as 0 = (0). (You will find how to minimize this length in Problem 6.16.)

Do the same as in Problem 6.1 but find the length L of a path on a cylinder of radius R, using cylindrical polar coordinates (p, (1), z). Assume that the path is specified in the form 0 = (z).

Consider a ray of light traveling in a vacuum from point P1 to P2 by way of the point Q on a plane mirror, as in Figure 6.8. Show that Fermat's principle implies that, on the actual path followed, Q lies in the same vertical plane as P1 and P2 and obeys the law of reflection, that 01 = 02. [Hints: Let the mirror lie in the xz plane, and let P1 lie on the y axis at (0, yi, 0) and P2 in the xy plane at (x2, y2, 0). Finally let Q = (x, 0, z). Calculate the time for the light to traverse the path P1Q P2 and show that it is minimum when Q has z = 0 and satisfies the law of reflection.]

A ray of light travels from point P1 in a medium of refractive index n1 to P2 in a medium of index n2, by way of the point Q on the plane interface between the two media, as in Figure 6.9. Show that Fermat's principle implies that, on the actual path followed, Q lies in the same vertical plane as P1 and P2 and obeys Snell's law, that n1 sin 01 = n2 sin 02. [Hints: Let the interface be the xz plane, and let P1 lie on the y axis at (0, h1, 0) and P2 in the x, y plane at (x2, h2, 0). Finally let Q = (x, 0, z). Calculate the time for the light to traverse the path P1 Q P2 and show that it is minimum when Q has z = 0 and satisfies Snell's law.]

Fermat's principle is often stated as "the travel time of a ray of light, moving from point A to B, is minimum along the actual path." Strictly speaking it should say that the time is stationary, not minimum. In fact one can construct situations for which the time is maximum along the actual path. Here is one: Consider the concave, hemispherical mirror shown in Figure 6.10, with A and B at opposite ends of a diameter. Consider a ray of light traveling in a vacuum from A to B with one reflection at P, in the same vertical plane as A and B. According to the law of reflection, the actual Figure 6.9 Problem 6.4 Figure 6.10 Problem 6.5 path goes via point Po at the bottom of the hemisphere (6 = 0). Find the time of travel along the path APB as a function of 6 and show that it is maximum at P = Po.

In many problems in the calculus of variations, you need to know the length ds of a short segment of a curve on a surface, as in the expression (6.1). Make a table giving the appropriate expressions for ds in the following eight situations: (a) A curve given by y = y(x) in a plane, (b) same but x = x(y), (c) same but r = r (0), (d) same but 0 = 0(r); (e) curve given by 0 = (z) on a cylinder of radius R, (f) same but z = z(0); (g) curve given by 9 = 0(0) on a sphere of radius R, (h) same but 0 = 0(0).

Consider a right circular cylinder of radius R centered on the z axis. Find the equation giving \(\phi\) as a function of z for the geodesic (shortest path) on the cylinder between two points with cylindrical polar coordinates \(\left(R, \phi_{1}, z_{1}\right)\) and \(\left(R, \phi_{2}, z_{2}\right)\) . Describe the geodesic. Is it unique? By imagining the surface of the cylinder unwrapped and laid out flat, explain why the geodesic has the form it does .

Verify that the speed of the roller coaster car in Example 6.2 (page 222) is ,I2gy. (Assume the wheels have negligible mass and neglect friction.)

Find the equation of the path joining the origin 0 to the point P (1, 1) in the xy plane that makes the integral g (y'2 yy' + y2) dx stationary.

In general the integrand f (y, y', x) whose integral we wish to minimize depends on y, y', and x. There is a considerable simplification if f happens to be independent of y, that is, f = f (y', x). (This happened in both Examples 6.1 and 6.2, though in the latter the roles of x and y were interchanged.) Prove that when this happens, the EulerLagrange equation (6.13) reduces to the statement that f = const. (6.42) Since this is a first-order differential equation for y(x), while the EulerLagrange equation is generally second order, this is an important simplification and the result (6.42) is sometimes called a first integral of the EulerLagrange equation. In Lagrangian mechanics we'll see that this simplification arises when a component of momentum is conserved.

Find and describe the path y = y(x) for which the integral .1 + y' 2 dx is stationary.

Show that the path y = y(x) for which the integral x0 y'2 dx is stationary is a sinh function.

In relativity theory, velocities can be represented by points in a certain "rapidity space" in which the distance between two neighboring points is ds = [2/(1 r2)]./dr2 r 2 d02, where r and 0 are polar coordinates, and we consider just a two-dimensional space. (An expression like this for the distance in a non-Euclidean space is often called the metric of the space.) Use the EulerLagrange equation to show that the shortest distance from the origin to any other point is a straight line.

(a) Prove that the brachistochrone curve (6.26) is indeed a cycloid, that is, the curve traced by a point on the circumference of a wheel of radius a rolling along the underside of the x axis. (b) Although the cycloid repeats itself indefinitely in a succession of loops, only one loop is relevant to the brachistochrone problem. Sketch a single loop for three different values of a (all with the same starting point 1) and convince yourself that for any point 2 (with positive coordinates x2, y2) there is exactly one value of a for which the loop goes through the point 2. (c) To find the value of a for a given point x2, y2 usually requires solution of a transcendental equation. Here are two cases where you can do it more simply: For x2 = n-b, y2 = 2b and again for x2 = 27rb, y2 = 0 find the value of a for which the cycloid goes through the point 2 and find the corresponding minimum times.

Consider again the brachistochrone problem of Example 6.2 (page 222) but suppose that the car is launched from point 1 with a fixed speed vo. Show that the path of minimum time to the fixed point 2 is still a cycloid, but with its cusp (the top point of the curve) a height v, I2g above point 1.

Use the result (6.41) of Problem 6.1 to prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f (0, 0', 6) in (6.41) is independent of 0, so the EulerLagrange equation reduces to afoot = c, a constant. This gives you 01 as a function of 6. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero, and describe the corresponding geodesics.]

Find the geodesics on the cone whose equation in cylindrical polar coordinates is z = Ap. [Let the required curve have the form 0 = 0(p).] Check your result for the case that 0.

Show that the shortest path between two given points in a plane is a straight line, using plane polar coordinates.

A surface of revolution is generated as follows: Two fixed points (x1, yi) and (x2, y2) in the x, y plane are joined by a curve y = y(x). [Actually you'll make life easier if you start out writing this as x = x(y).] The whole curve is now rotated about the x axis to generate a surface. Show that the curve for which the area of the surface is minimum has the form y = yo cosh[(x xo)I yo], where xo and yo are constants. (This is often called the soap-bubble problem, since the resulting surface is usually the shape of a soap bubble held by two coaxial rings of radii yl and y?.)

If you haven't done it, take a look at Problem 6.10. Here is a second situation in which you can find a "first integral" of the EulerLagrange equation: Argue that if it happens that the integrand f (y, y', x) does not depend explicitly on x, that is, f = f (y, y'), then df of , of dx ay Y ay'Y Use the EulerLagrange equation to replace of/ay on the right, and hence show that df d ,af ) dx dx ay' This gives you the first integral f af = const. ay' (6.43) This can simplify several calculations. (See Problems 6.21 and 6.22 for examples.) In Lagrangian me-chanics, where the independent variable is the time t, the corresponding result is that if the Lagrangian function is independent of t, then energy is conserved. (See Section 7.8.)

In Example 6.2 (page 222) we found the brachistochrone by exchanging the variables x and y. Here is a method that avoids that exchange: Write the time as in Equation (6.19) but using x as the variable of integration. Your integrand should have the form f (y, y', x) = -1(y12 + 1)/y. Since this is independent of x, you can invoke the "first integral" (6.43) of Problem 6.20. Show that this differential equation leads you to the same integral for x as in Equation (6.23) and hence to the same curve as before.

You are given a string of fixed length l with one end fastened at the origin 0, and you are to place the string in the xy plane with its other end on the x axis in such a way as to enclose the maximum area between the string and the x axis. Show that the required shape is a semicircle. The area enclosed is of course f y dx, but show that you can rewrite this in the form fo f ds, where s denotes the distance measured along the string from 0, where f = \ y'2, and y' denotes dy Ids. Since f does not involve the independent variable s explicitly, you can exploit the "first integral" (6.43) of Problem 6.20.

An aircraft whose airspeed is vo has to fly from town 0 (at the origin) to town P, which is a distance D due east. There is a steady gentle wind shear, such that vwind = V y X, where x and y are measured east and north respectively. Find the path, y = y(x), which the plane should follow to minimize its flight time, as follows: (a) Find the plane's ground speed in terms of vo, V, 0 (the angle by which the plane heads to the north of east), and the plane's position. (b) Write down the time of flight as an integral of the form fop f dx. Show that if we assume that y' and 0 both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand f takes the approximate form f = (1 + I y'2)/(1 ky) (times an uninteresting constant) where k = V (c) Write down the EulerLagrange equation that determines the best path. To solve it, make the intelligent guess that y(x) = Ax(D x), which clearly passes through the two towns. Show that it satisfies the Euler-Lagrange equation, provided X = (-V4 2k2D2 2)/ (k D2). How far north does this path take the plane, if D = 2000 miles, vo = 500 mph, and the wind shear is V = 0.5 mph/mi? How much time does the plane save by following this path? [You'll probably want to use a computer to do this integral.]

Consider a medium in which the refractive index n is inversely proportional to r2; that is, n = a I r2, where r is the distance from the origin. Use Fermat's principle, that the integral (6.3) is stationary, to find the path of a ray of light travelling in a plane containing the origin. [Hint: Use two-dimensional polar coordinates and write the path as 0 = 4> (r). The Fermat integral should have the form f f (0, 0', r) dr, , where f (0, 4>', r) is actually independent of 0. The EulerLagrange equation therefore reduces to afoot= const. You can solve this for 0' and then integrate to give 0 as a function of r. Rewrite this to give r as a function of 0 and show that the resulting path is a circle through the origin. Discuss the progress of the light around the circle.]

Consider a single loop of the cycloid (6.26) with a fixed value of a, as shown in Figure 6.11. A car is released from rest at a point Po anywhere on the track between 0 and the lowest point P (that Figure 6.11 Problem 6.25 is, Po has parameter 0 < 00 < 7). Show that the time for the cart to roll from Po to P is given by the integral time (Pc, P) = 1 cos 0 d6 cos 00 cos6 and prove that this time is equal to 7 N/a I g. Since this is independent of the position of Po, the cart takes the same time to roll from P, to P, whether Po is at 0, or anywhere between 0 and P, even infinitesimally close to P. Explain qualitatively how this surprising result can possibly be true. [Hint: To do the mathematics, you have to make some cunning changes of variables. One route is this: Write = Tr 2a and then use the relevant trig identities to replace the cosines of 0 by sines of a. Now substitute sin a = u and do the remaining integral.]

Give in detail the argument that leads from the stationary property of the integral (6.30) to the two EulerLagrange equations (6.34).

Prove that the shortest path between two points in three dimensions is a straight line. Write the path in the parametric form x = x(u), Y = Au), and z = z(u) and then use the three EulerLagrange equations corresponding to (6.34).