16—Calculus of Variations 40816.22 The equation (16.25) is an approximate solution to the path for light above a hot road. Is there
a functionn=f(y)representing the index of refraction above the road surface such that this equation
would be its exact solution?
16.23 On the first page of this chapter, you see the temperature dependence of length measurements.
(a) Take a metal disk of radiusaand place it centered on a block of ice. Assume that the metal
reaches an equilibrium temperature distributionT(r) =T 0 (r^2 /a^2 −1). The temperature at the edge
isT= 0, and the ruler is calibrated there. The disk itself remains flat. Measure the distance from the
origin straight out to the radial coordinater. Call this measured radiuss. Measure the circumference
of the circle at this radius and then express this circumference in terms of the measured radiuss.
(b) On a sphere of radiusR(constant temperature) start at the pole (θ= 0) and write the distance
along the arc at constantφdown to the angleθ. Now go around the circle at this constantθand write
its circumference. Express this circumference in terms of the distance you just wrote for the “radius”
of this circle.
(c) Show that the geometry you found in (a) is the same as that in (b) and find the radius of the
sphere that this “heat metric” expresses. Ans:R=a/ 2
√
αT 0 (1−αT 0 )≈a/ 2
√
αT 0
16.24 Using the same techniques as in section16.5, apply these methods to two concentric spheres.
Again, use a linear and then a quadratic approximation. Before you do this, go back to Eq. (16.30)
and see if you can arrive at that form directly,withoutgoing through all the manipulations of solving
forαandβ. That is, determine how you could have gotten to (16.30) easily. Check some numbers
against the exact answer.
16.25 For the variational problem Eq. (16.45) one solution isy=bx/a. Assume thatα, β > 0 and
determine if this is a minimum or maximum or neither. Do this also for the other solution, Eq. (16.47).
Ans: The first is min ifb/a >
√
β/ 6 α. The kinked solution is always a minimum.
16.26 If you can construct glass with a variable index of refraction, you can make
a flat lens with an index that varies with distance from the axis. What function