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8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll see how his sme heorem, when comined wih he clculus concep of limis, llows us o find he lenghs of coninuous plne curves defined s grphs of funcions or s grphs of prmeric equions. We ll do his y wy of he Riemnn inegrl you were inroduced o in prior clculus course. Consider piece of curve defined y funcion y f ( x) over n inervl x. To pproxime he lengh of his curve, one cn sudivide he inervl [, ] ino suinervls (o he righ we ve used four suinervls). This gives us he following poins on he curve: P ( x, y ), P ( x, y ). P x, y, P x, y, P x, y, 1 1 1 Here s usul, y f x i i. for ech of hese poins. The curve lengh is hen pproximely he sum of he lenghs of he line segmens connecing hese poins: + + + ( dis P, P dis P, P dis P, P dis P, P 1 1 ( ) ( y ) x x + y y + x x + y y + x x + y y + x x + y 1 1 1 1 Using summion noion his is shorer: ( x x ) + ( y y ) Lengh k+ 1 k k+ 1 k k Of course if we use more suinervls he pproximion will e even closer o he rue lengh of he curve. Going from suinervls o n suinervls is very esy using summion noion: ) k ( x x ) ( y y ) Lengh + k+ 1 k k+ 1 k

To mke hings more precise nd specific, we divide he inervl [, ] ino n equl suinervls, ech of lengh xn Δ x. Then xk + k Δ x for k, 1,,, n. Defined his wy, x nd n. Since ech suinervl is of lengh x k k+ 1 k Δ, we hve h xk 1 x Δ y y y, so our lengh pproximion formul ecomes: + k x Δ for ll k. Also, les define xk xk yk yk x yk k k ( + 1 ) ( + 1 ) Δ Lengh + Δ + If we now fcor ou he Δ x, we ge: Δy k Δyk Lengh Δ x +Δyk 1+ x 1 Δ + Δ k k Δx k Δx x Tking he limi s he numer of sudivisions n goes o infiniy, our summion ecomes Riemnn inegrl nd he pproximion ecomes he exc curve lengh. In fc, h s how we will define curve lengh: Δyk dy Lengh lim 1+ x 1 n Δ + k Δx dy Lengh 1+ 1 + ( f '( x) ) Exmple 1. Find he lengh of he curve defined y x x y for x. Since 1 1 x x dy y x x x, we hve 1 dy Lengh 1+ 1+ x 1 + x Mking he susiuion u 1 + x du, we ge

1 u Lengh udu u du 1 ( ) 1 ( 1) 1 1 1 1 Exmple. Find he lengh of he prolic rc defined y x y for x 1. Since 1 x dy x y x, we hve 1 1 dy Lengh 1+ 1+ x Mking he susiuion x n sec d θ θ θ, we ge Lengh 1+ n θ sec θ dθ secθ sec θ dθ sec θ dθ Invoking ulr inegrion: + sec x sec x sec x n x n x We ge: sec x sec xn x sec xn x sec xn x sec x sec x 1 sec x sec x + sec xn x sec x x x x+ x x x+ ln sec x + n x sec sec n sec sec n Thus: sec Lengh sec xn x+ ln sec x+ n x x + C sec θ dθ secθ nθ + ln secθ + nθ θ θ

1 Lengh sec n ln sec n sec n ln sec n + + + 1 + ln 1+ ( 1+ ln + 1 1 ln 1+ ) 1.177957696 x Exmple. Find he lengh of he piece of cenry curve defined y y cosh for x. Since x dy x 1 x y cosh sinh sinh, we hve x sinh dy x x Lengh 1 1 sinh cosh + + 1 sinh sinh ( sinh1 sinh ) ( sinh1 ) sinh1 ( e ) or 1 1 e e e e e e 1 1 e e e Exmple. Find he rc lengh of he curve defined y x + 8 y for x. x x x x + 8 x + 8 dy y x x dy x 16 8x Hence: ( x 16) ( x ( x + 8 ) ) x 8 ( x 16) x 8 8 ( x + ) x x 8 8 6x x x x + x 16 dy 1 6x x 6x dy x 16 x x + 56 6x + x x + 56 1+ 1+ 1 1 + + 8x 6x 6x 6x + + 56 + 56 x + 16 + 6 8x

And so: dy x + 16 x 16 1 x x 8x 8x 8x 8 Lengh 1+ + + 1 1 x x 1 1 1 + ( ) ( ) 8 1 7 1 7 1 17 + 6 ( 1) 1 1 Even finding he lengh of he grph of pr of simple nd degree polynomil wsn so esy. Unforunely mos rc lengh inegrls re no solvle, excep numericlly. Bu if you cn les se up he correc rc lengh inegrl, you cn find numericl pproximions o he exc nswer h re s ccure s you like, using he pproprie sofwre (or clculor). If you ke look ny clculus ex s coverge of he opic of rc lengh, you ll noice h here re no mny exmples; nor re here mny homework prolems. No only his, u you ll see he sme prolems! There jus ren mny rc lengh inegrls h cn e solved in he usul sense. We ll ge few more using prmeric equions! Le curve e defined y prmeric equions () x f y g, for. To pproxime he curve s lengh, we ll divide he inervl ino n equl suinervls, ech of lengh Δ. Then we define k + k Δ, for n k, 1,,, n. We hve he poins P ( x, y ), P ( x, y ). P x, y, P x, y, P x, y, 1 1 1

Here x f ( ) nd y g( ) for ech of hese poins. The curve lengh is hen pproximely he k k k k sum of he lenghs of he line segmens connecing hese poins: + + + ( dis P, P dis P, P dis P, P dis P, P 1 1 ( ) ( y ) x x + y y + x x + y y + x x + y y + x x + y 1 1 1 1 Using summion noion his is shorer: ( x x ) + ( y y ) k Lengh k k k k + 1 + 1 As efore, if we use more suinervls he pproximion will e even closer o he rue lengh of he curve. And s efore, going from suinervls o n suinervls is very esy using summion noion: k ( x x ) ( y y ) Lengh + k+ 1 k k+ 1 k ) Leing Δ ( nd y g( ) g( ) x f f k k k + 1 ) Δ for k, 1,,,, he ove formul k k+ 1 k ecomes: Lengh Δ x +Δy k k k Nex we muliply y Δ Δ : Δ Δ xk +Δyk Δxk Δyk k k k Δ k Δ k Δ Δ Lengh Δ x +Δy Δ + Δ Tking he limi s he numer of sudivisions n goes o infiniy, our summion ecomes Riemnn inegrl nd he pproximion ecomes he exc curve lengh. In fc, h s how we will define curve lengh for prmericlly defined curves: Δxk Δy k dy lim n k Δ Δ d d Lengh + Δ + d dy Lengh + d ( f '() ) + ( g '() ) d d d

Exmple 5. Find he rc lengh of he curve prmericlly defined y x y for 1. dy dy nd + ( ) + ( ) 9 + 9 + d d d d And so: 1 1 dy Lengh + 9 + d d d d Mking he susiuion u 9 + du 18 d du d, our inegrl ecomes: 18 1 1 1 1 1 1 1 1 1 u d 18 u du 18 u du 18 Lengh 9 + 1 1 1 1 1 8 1 1 1 1 18 7 7 7 Exmple 6. Find he lengh of he curve deermined y x rsin y rcos for. This prmerizion rces ou circle of rdius r once (clockwise nd sring 1:), so he lengh of he rc is simply he circumference of h circle. Since he circumference of circle wih rdius r is r, h s he lengh of he curve. Bu h ws oo esy! Le s do i gin, using he defining prmeric equions: dy dy rcos nd rsin + rcos + rsin d d d d r cos + sin r 1 r And so: dy Lengh + d r d r d r r ( ) r d d

Exmple 7. Find he lengh of he curve deermined y ( ) ( ) x e cos y e sin for 5. dy e cos e sin nd e sin + e cos d d ( ) ( ) ( ) ( ) dy + + + d d ( e cos( ) e sin ( ) ) e sin ( ) e cos( ) ( cos( ) sin ( )) sin ( ) + cos( ) e + ( ) ( ) e C S S C + + Noice in h ls line how he noion is simplified. This kind of susiuion mkes he lger much esier, nd eses he wrier s crmp! To coninue e C CS + S + S + SC + C e C + S + ( S + C ) e 1+ Thus, 5 5 5 dy Lengh + d e 1+ d 1+ e d 1+ e d d 5 ( e e ) ( e ) 5 5 1+ 1 1+ Exmple 8. Find he lengh of he curve deermined y x + 1 y n ln 1 for 1. 1 1 1 1 d ln + 1 d ln ( + 1) d ln ( + 1) d ln ( + 1) d d d d d + 1 + 1 1 + n + + dy d 1 dy 1 1 1 d d 1+ d d + 1 + 1 + 1 + 1 Hence, 1 1 dy d Lengh + d d d + 1

Mking he susiuion n d sec d θ θ θ, our inegrl ecomes: 1 d sec θ dθ sec θ dθ Lengh + 1 n θ + 1 secθ θ θ sec + sec n secθ dθ sec + n θ θ θ secθ dθ dθ ln sec + n secθ + nθ secθ + nθ θ θ ( + ) ln sec + n ln sec + n ln + 1 ln 1+ ln 1 Exmple 9. Find n inegrl h equls he lengh of he sine curve over one period. One period of he sine curve cn e prmerized y x y sin for. Thus: Hence, dy dy 1 nd cos + 1 + ( cos) 1+ cos d d d d dy Lengh + d 1+ cos d d d This inegrl cn e solved y ny elemenry inegrion echniques one will ever encouner in Clculus I. ( Bu in cse you re ineresed, h lengh is pproximely 7.6955785558. ) In he ls secion of his chper you ll lern some echniques for evluing inegrls numericlly. Numericl echniques re essenil for evluing inegrls like he one we jus encounered in our ls exmple. And such inegrls cully end o e he rule rher hn he excepion. Bu sophisiced numericl inegrion lgorihms now hrdwired ino mny inexpensive clculors, nd powerful sofwre like Mple re lso ville o esily del wih reclcirn inegrls!