Mathematical Methods for Physics and Engineering - Matematica.NET

CALCULUS OF VARIATIONSB(b, y(b))dsdxdyA(a, y(a))Figure 22.2An arbitrary path between two fixed points.◮Show that the shortest curve joining two points is a straight line.Let the two points be labelled A and B and have coordinates (a, y(a)) and (b, y(b))respectively (see figure 22.2). Whatever the shape of the curve joining A to B, the lengthof an element of path ds is given byds = [ (dx) 2 +(dy) 2] 1/2=(1+y ′2 ) 1/2 dx,and hence the total path length along the curve is given byL =∫ ba(1 + y ′2 ) 1/2 dx. (22.7)We must now apply the results of the previous section to determine that path which makesL stationary (clearly a minimum in this case). Since the integral does not contain y (orindeed x) explicitly, we may use (22.6) to obtaink = ∂F y ′=∂y ′ (1 + y ′2 ) . 1/2where k is a constant. This is easily rearranged and integrated to giveky = x + c,(1 − k 2 )1/2which, as expected, is the equation of a straight line in the form y = mx + c, withm = k/(1 − k 2 ) 1/2 . The value of m (or k) can be found by demanding that the straight linepasses through the points A and B and is given by m =[y(b) − y(a)]/(b − a). Substitutingthe equation of the straight line into (22.7) we find that, again as expected, the total pathlength is given byL 2 =[y(b) − y(a)] 2 +(b − a) 2 . ◭778