6. Arc length - length of curvesConsider an arbitrary curvey=f(x), recall that when we introduced differentiation we discussed thata sufficiently small portion of this curve can be approximated by a straight line. Then the derivativedydxof the functiony=f(x)at the midpoint of this curve portion can be approximated by the fractiontriangleytrianglex,wheretrianglexandtriangleyare respectively the differences inx-coordinates andy-coordinates of the endpointsof the curve portion. In this case,trianglex,triangley→0andtriangleytrianglexcan be interpreted as theslopeof the curve atthe midpoint of this small curve portion.As shown in the above diagram, let the curve extends fromx=atox=b.We divide the interval[a, b]intonsub-intervals so that each has a difference inx-coordinatestrianglex=b-an. Iftrianglex→0fornissufficiently large (n→ ∞), in thek-th sub-interval fork= 1, . . . , n, we may approximate the length ofthe curve portion astriangleLk=radicalbigtrianglex2+triangley2k, wheretriangleykis the difference iny-coordinates of the endpointsin thek-th sub-interval.If the functionf(x)isdifferentiablein the interval[a, b],triangleykcan be approximated bytrianglexdydx|x=¯xk=fprime(¯xk)trianglexfortrianglex→0as discussed before, where¯xkis thex-coordinate of a point within thek-thsub-interval fork= 1, . . . , n.
2.Taking the infinite Riemann sum of the length of all curve portions, we may obtain the total lengthLorusually calledarc lengthof the curve in the interval[a, b]asArc lengthL= limn→∞n∑k=1triangleLk= limn→∞n∑k=1radicalBig1 + [fprime(¯xk)]2trianglex=´baradicalBig1 + [fprime(x)]2dx.E.g. 1. Find the arc length of the curvey=12(ex+e-x)in thex-interval[-ln 2,ln 2].[ans: 3/2]13