Arc Length. P i 1 P i (1) L = lim. i=1

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1 Arc Length Suppose tht curve C is defined by the eqution y = f(x), where f is continuous nd x b. We obtin polygonl pproximtion to C by dividing the intervl [, b] into n subintervls with endpoints x, x,...,x n nd equl width x. If y i = f(x i ), then the point P i (x i, y i ) lies on C nd the polygon with vertices P, P,...,P n is n pproximtion to C. The length L of C is pproximtely the length of this polygon nd the pproximtion gets better s we let n increse. Therefore, we define the length L of the curve C with eqution y = f(x), x b, s the limit of the lengths of the inscribed polygons with vertices P, P,... (if the limit exists): lim n n P i P i () i= Notice tht the procedure for defining rc length is very similr to the procedure we used for defining re nd volume: We divided the curve into lrge number of smll prts. We then found the pproximte lengths of the smll prts nd dded them. Finlly, we took the limit s n. The definition of rc length given by Eqution is not very convenient for computtionl purposes, but we cn derive n integrl formul for L in the cse where f hs continuous derivtive. [Such function f is clled smooth becuse smll chnge in x produces smll chnge in f (x).] If we let y i = y i y i, then P i P i = (x i x i ) 2 + (y i y i ) 2 = ( x) 2 + ( y i ) 2 By pplying the Men Vlue Theorem to f on the intervl [x i, x i ], we find tht there is number x i between x i nd x i such tht f(x i ) f(x i ) = f (x i)(x i x i ) tht is, Thus we hve y i = f (x i ) x P i P i = ( x) 2 + ( y i ) 2 = ( x) 2 + [f (x i ) x]2 = ( x) 2 + [f (x i )]2 ( x) 2 = ( + [f (x i )]2 )( x) 2 = + [f (x i )]2 ( x) 2 = + [f (x i )]2 x

2 Therefore, by the Definition bove, lim n n i= We recognize this expression s being equl to P i P i = lim n n + [f (x i )]2 x i= + [f (x)] 2 by the definition of definite integrl. This integrl exists becuse the function g(x) = + [f (x)] 2 is continuous. Thus we hve proved the following theorem: THEOREM: If f is continuous on [, b], then the length of the curve y = f(x), x b, is + [f (x)] 2 (2) If we use Leibniz nottion for derivtives, we cn write the rc length formul s follows: + ( ) 2 (3) EXAMPLE: Find the length of the segment of the horizontl line y = between the points (, ) nd (b, ). Solution: Since f(x) =, we hve f (x) =, nd (2) gives + [f (x)] 2 = b + = = b EXAMPLE: Find the length of the segment of the line y = x between the points (, ) nd (b, b). Solution: Since f(x) = x, we hve f (x) =, nd (2) gives + [f (x)] 2 = + = 2 = 2(b ) EXAMPLE: Find the length of the segment of the line y = mx + n between the points (, f()) nd (b, f(b)). Solution: Since f(x) = mx + n, we hve f (x) = m, nd (2) gives + [f (x)] 2 = + m2 = + m 2 = + m 2 (b ) EXAMPLE: Find the length of the rc of the semicubicl prbol y 2 = x 3 between the points (, ) nd (4, 8). 2

3 EXAMPLE: Find the length of the rc of the semicubicl prbol y 2 = x 3 between the points (, ) nd (4, 8). Solution: For the top hlf of the curve we hve nd so the rc length formul gives y = x 3/2 4 + ( ) 2 4 = = 3 2 x/ x If we substitute u = x, then du = 9 3. When x =, u = ; when x = 4, u =. Therefore udu = 9 3/4 9 2 ] [ 3 u3/2 = 8 ( ) ] 3/2 3 3/2 = 3/ (8 3 3) If curve hs the eqution x = g(y), c y d, nd g (y) is continuous, then by interchnging the roles of x nd y in (2) nd (3), we obtin the following formul for its length: d c d + [g (y)] 2 = + EXAMPLE: Find the length of the rc of the prbol y 2 = x from (, ) to (, ). c ( ) 2 (4) 3

4 EXAMPLE: Find the length of the rc of the prbol y 2 = x from (, ) to (, ). Solution: Since x = y 2, we hve / = 2y, nd (4) gives ( ) 2 + = + 4y2 We mke the trigonometric substitution y = 2 tnθ, which gives = 2 sec2 θdθ nd + 4y 2 = + tn 2 θ = sec θ. When y =, tnθ =, so θ = ; when y =, tnθ = 2, so θ = tn 2 = α, sy. Thus α sec θ 2 sec2 θdθ = 2 α sec 3 θdθ = 2 [ sec θ tn θ + ln sec θ + tn θ 2 = (sec α tn α + ln sec α + tnα ) 4 (We could hve used Formul 2 in the Tble of Integrls.) Since tn α = 2, we hve sec 2 α = +tn 2 α = 5, so sec α = 5 nd ln( 5 + 2) 4 Becuse of the presence of the squre root sign in Formuls 2 nd 4, the clcultion of n rc length often leds to n integrl tht is very difficult or even impossible to evlute explicitly. Thus we sometimes hve to be content with finding n pproximtion to the length of curve, s in the following exmple. EXAMPLE: (() Set up n integrl for the length of the rc of the hyperbol xy = from the point (, ) to the point 2, ). 2 (b) Use Simpson s Rule with n = to estimte the rc length. Solution: () We hve nd so the rc length is 2 y = x + ( ) 2 = = x 2 2 ] α + x 4 (b) Using Simpson s Rule (see Section 7.7) with =, b = 2, n =, x =., nd f(x) = + /x 4, we hve 2 + x x4 [f() + 4f(.) + 2f(.2) + 4f(.3) f(.8) + 4f(.9) + f(2)]

5 The Arc Length Function We will find it useful to hve function tht mesures the rc length of curve from prticulr strting point to ny other point on the curve. Thus if smooth curve C hs the eqution y = f(x), x b, let s(x) be the distnce long C from the initil point P (, f()) to the point Q(x, f(x)). Then s is function, clled the rc length function, nd, by (2), s(x) = x + [f (t)] 2 dt (5) We cn use Prt of the Fundmentl Theorem of Clculus to differentite (5) (since the integrnd is continuous): ds = ( ) 2 + [f (x)] 2 = + (6) The lst Eqution shows tht the rte of chnge of s with respect to x is lwys t lest nd is equl to when f (x), the slope of the curve, is. The differentil of rc length is ds = + nd this eqution is sometimes written in the symmetric form The geometric interprettion of Eqution 8 is shown in the Figure on the right. If we write ds, then from (8) either we cn solve to get (7), which gives (3), or we cn solve to get ( ) 2 ds = + which gives (4). ( ) 2 (7) (ds) 2 = () 2 + () 2 (8) EXAMPLE: Find the rc length function for the curve y = x 2 8 ln x tking P (, ) s the strting point. 5

6 EXAMPLE: Find the rc length function for the curve y = x 2 8 ln x tking P (, ) s the strting point. Solution: If f(x) = x 2 ln x, then 8 f (x) = 2x 8x + [f (x)] 2 = + + [f (x)] 2 = 2x + 8x ( 2x 8x Thus the rc length function is given by s(x) = x + [f (t)] 2 dt = x ) 2 = + 4x x = 2 4x ( 64x = 2x + ) 2 2 8x For instnce, the rc length long the curve from (, ) to (3, f(3)) is s(3) = ( 2t + ) dt = t 2 + ] x 8t 8 ln t = x ln x ln3 = 8 + ln

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