Application: Arc Length

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Bryan Bates
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Appliction: Arc Length 7 The Generl Problem The Riemnn integrl hs wide vriety of pplictions In this section, using the subdivide nd conquer strtegy we will show how it cn be used to determine the

1 Appliction: Arc Length 7 The Generl Problem The Riemnn integrl hs wide vriety of pplictions In this section, using the subdivide nd conquer strtegy we will show how it cn be used to determine the lengths of certin curves EXAMPLE 7 Verizon is hnging fiber optic cble round Genev It wnts to know how much cble it will need Since the cble must be hung with some slck (why?), the engineers cn t simply mesure the distnce between the poles to know how much cble to use The curve tht n idelized hnging chin or cble ssumes when supported t its ends nd cted on only by its own weight is clled ctenry The eqution for the grph of ctenry curve is hyperbolic cosine function ( x ) cosh (ex/ + e x/ ), where is constnt How cn Verizon use this eqution to find the length of the cble it needs? Let s try to ttck this problem in more generl fshion Suppose y f (x) is continuous function on the closed intervl [, b] Find the length of the grph of y f (x) on this intervl This length is clled the rc length of the curve Think bout how we pproched the re problem We used the only figures for which we hd n re formul (rectngles) nd used those figures to pproximte the re under curve Well, the only curve whose length we re certin of is stright line segment If the segment hs endpoints (x, y ) nd (x, y ), then the length of the segment is given by the distnce formul Segment Length (x x ) + (y y ) Figure 7: A hnging chin forms ctenry wiki/ctenry You might protest tht you know the circumference of circle: πr But why is tht formul true? So somehow we must use line segments (the only curves we know how to mesure) to obtin the length of more generl continuous curve Well, let s do the usul thing: subdivide nd conquer Suppose f is continuous on [, b] Given ruler, we might mrk off successive points Q, Q,, Q n long the curve nd tke the resulting polygonl rc s n pproximtion to the curve (see Figure 7) We could mesure the length of ech segment of the polygonl rc with our ruler (or the distnce formul) nd then summing these vlues together we would hve n pproximtion of the length of the curve This ide (nd the lck of precise definition of length for curves) motivtes the following process Assume tht f is continuous over the closed, bounded intervl [, b] Let P {x, x,, x n } be regulr prtition of [, b] into n equl width subintervls For i to n let Q i (x i, f (x i )) be the corresponding set of points on the grph of f Then the polygonl rc from Q to Q n is just the sequence of line segments

2 mth 3 ppliction: rc length Q Q f (x i ) f (x i ) Q i x Q i Q n Figure 7: A curve y f (x) nd polygonl pproximtion using regulr prtition x x x i x i x n b Q Q, Q Q,, Q n Q n The length of this polygonl rc is just the sum of the lengths of the individul segments Using the Pythgoren theorem the length of the ith segment is Length(Segment i) (x i x i ) + [ f (x i ) f (x i )] (7) We cn simplify (7) in two wys As usul, we let x x i x i Next, by the Men Vlue Theorem (yet once gin!), if we ssume tht f is lso differentible on [, b], then f (x i ) f (x i ) x i x i f (c i ) for some point c i between x i nd x i Consequently, f (x i ) f (x i ) f (c i ) [x i x i ] f (c i ) x Mking both of these chnges to (7) we get Length(Segment i) ( x) + [ f (c i ) x] + [ f (c i )] x (7) Adding the lengths of ll n line segments together we hve Length of Curve n n Length(Segment i) + [ f (c i )] x i i Notice tht we now hve Riemnn sum! To improve the pproximtion we do the stndrd thing: We let the number of polygonl pieces get lrge nd tke the limit We find n n i Length of Curve lim + [ f (c i )] x + [ f (x)] dx To be certin tht this limit of the Riemnn sums is, in fct, the definite integrl we need to know tht + ( f (x)) is continuous It will be, if we ssume tht f (x) is continuous With these conditions, we find THEOREM 7 (Arc Length Formul) If f is differentible nd f is continuous on the closed intervl [, b], then the rc length of f on [, b] is Arc Length of f + [ f (x)] dx EXAMPLE 7 Find the rc length of f over [, ] for f (x) 4 3 x3/

3 mth 3 ppliction: rc length 3 SOLUTION First determine the derivtive: f (x) x / on [, ] So by Theorem 7 Arc Length of f + [ f (x)] dx + (x / ) dx + 4x dx ( + 4x)3/ (5 7) 49 3 Notice tht we used mentl djustment in the integrtion: u + 4x EXAMPLE 73 Find the rc length of f over [, π/4] for f (x) ln(cos x) SOLUTION Determine the derivtive f (x) by Theorem 7 Arc Length of f + [ f (x)] dx cos x ( sin x) tn x on [, π/4] So π/4 + tn x dx π/4 sec x dx π/4 sec x dx ln sec x + tn x π/4 ln( + ) ln ln( + ) EXAMPLE 74 (Circumference of Circle) Find the rc length (circumference) of circle of rdius r SOLUTION To mke things esy we will ssume tht the circle is centered t the origin so tht its eqution is f (x) r x on the intervl [ r, r] Actully this is the eqution for the upper semi-circle of rdius r We d need to use the negtive root to get the lower hlf of the circle Ok, we cn del with tht by simply doubling the nswer The derivtive f (x) ( x) r x x r x 3 nd is not defined t r nd r where the denomintor is To void this, let s use one-twelfth of circle (3 ) by restricting the domin of f to [, r ] See Figure 73 r r Since we re using only one-twelfth of the circle, we need to multiply the nswer Figure 73: One-twelfth of circle (3 ) of rdius r

4 mth 3 ppliction: rc length 4 by, so r/ Circumfrence of Circle + x r x dx r/ (r x ) + x r x dx r/ r r x dx r/ r r x dx ( x ) r/ r rcsin r [ ( ) ] r rcsin rcsin() ( π ) r πr Awesome! We hve now proven tht the circumference of circle of rdius r is πr Erlier we showed tht the volume of sphere of rdius r is 4 3 πr3 Interestingly, we hve not yet shown tht the re of circle of rdius r is πr The reson for this is tht to determine the re we need to clculte r r x dx r but we do not yet know n ntiderivtive of r x (Look it up in tble! Find the re of circle using this) EXAMPLE 75 (Length of Line Segment) Our rc length formul better work for the length of n ordinry line segment Let s check Find the length of the segment between (3, 4) nd (8, ) SOLUTION Since the segment is prt of non-verticl line, its eqution hs the form y f (x) mx + b, so f (x) m The slope of the segment is Using Theorem 7 we find m 4 ( ) 3 8 Arc Length of Segment 5 + [ f (x)] dx dx 3 8 5x 3 5(8 3) 5 5 Using the distnce formul, the length of the segment is (8 3) + ( 4) , + () dx which is the sme nswer we got using the rc length formul Phew! EXAMPLE 7 (Simplifying) Let f (x) x5 + x 3 on [, ] Find the rc length

5 mth 3 ppliction: rc length 5 SOLUTION Ok, time to fess up There re not mny rc length integrls we cn do This is one of them, but it requires some lgebric mnipultion Once you see this you should recognize similr integrls The derivtive is f (x) x4 x 4 Notice tht the exponents re negtives of ech other This is importnt becuse when we squre f (x), [ f (x)] ( x4 ) x 4 4 x8 + 4 x 8, We get constnt of for the middle term Consequently, when we dd one to this, we get for the middle term, + [ f (x)] + 4 x8 + 4 x 8 4 x8 + + ( 4 x 8 x4 + ) x 4, which is gin perfect squre Now we re redy to clculte the rc length using Theorem 7 we find ( + [ f (x)] dx x4 + ) x 4 dx Arc Length x4 + x 4 dx x5 x 3 ( 3 ) ( 48 ) Messy, but doble! (Notice the similrity between the originl function f nd the ntiderivtive of + [ f (x)] which is chrcteristic of this type of problem) It s sd fct, rc lengths involve complicted integrnds so without dditionl integrtion methods (coming soon!) there re not mny rc lengths tht we cn compute Here re few more we d like to do EXAMPLE 77 Wht bout the rc length of prbol, sy f (x) x on [, ] Check tht Arc Length + 4x dx, but we do not yet hve n ntiderivtive for this problem Similrly If g(x) sin x on [, π], then π Arc Length + cos x dx Or if f (x) e x on [, ], then Arc Length + e x dx If you know how to use Wolfrm Alph, you might try these problems But for the moment, we re stuck! YOU TRY IT 7 Find the rc lengths of the following functions over the given intervls () f (x) 3 x3/ on [, 8] (Answer: 8 3 (3/ )) (b) f (x) x4 4 + on [, ] (Answer: 3 8x 3 ) (c) g(x) x 3 + x 49 on [, 3] (Answer: 8 ) (d) h(x) cosh x on [, ln ] Remember, cosh x ex + e x (Answer: 3 4 )

6 mth 3 ppliction: rc length YOU TRY IT 7 Here re few dditionl problems () Find the exct rc length of f (x) 5 x 3/ on the intervl [, ] (Answer: 74) (b) Find the rc length of y 8 x4 + 4 x on the intervl [, ] Simplify the integrnd! (Answer: 33 ) (c) Find the rc length of f (x) ln sec x on [, π/4] Use trig identity! (Answer: ln( + )) YOU TRY IT 73 (Extr Credit) Solve the problem in Exmple 7 on the intervl [, ]

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde