aa r X i v : . [ m a t h . HO ] A p r ARC LENGTH OF FUNCTION GRAPHS VIATAYLOR’S FORMULA
PATRIK NYSTEDT
University West, Department of Engineering Science, SE-461 86Trollh¨attan, Sweden
Abstract.
We use Taylor’s formula with Lagrange remainder toprove that functions with bounded second derivative are rectifi-able in the case when polygonal paths are defined by interval sub-divisions which are equally spaced. We discuss potential benefitsfor such an approach in introductory calculus courses. Introduction
One of the first experiences of measurements that we encounter inour lives is that of length . Even young children are involved in many ev-eryday activities that concern length measurements. Questions such as”How tall am I?” or ”How long can you jump?” or ”How far is it to myfriends house?” arise naturally from them. In the early years of school-ing we are taught how to measure lengths of straight lines using a rulerand express our findings in appropriate units. In middle school, we arepresented with the problem of measurement of the circumference a thecircle and how to relate this to the length of its diameter. For manystudents the transition from understanding straight line measurementsto comprehending length measurement of non-linear curves is not soeasily accomplished. Indeed, it is only natural for them to pose ques-tions such as ”How can we measure something curved using a straightruler?” or ”What do we really mean when we speak of the length ofa curve?”. As teachers, we have to treat these questions seriously, be-cause when pondering over this, the students are placed in very goodcompany. Indeed, over the millennia, many of our greatest thinkersfailed to provide satisfying answers to such questions. For instance,the Greek philosopher Aristotle (384-322 BC) stated the following con-cerning comparisons of motions along straight lines and along circles:
E-mail address : [email protected] . ”But, once more, if the motions are comparable, we aremet by the difficulty aforesaid, namely that we shallhave a straight line equal to a circle. But these are notcomparable.” [10, p. 141]With some exceptions (for instance Archimedes rectification of the cir-cle using a spiral, see e.g. [15]), Aristotle’s view on these matters per-sisted amongst scholars even up to the time of Descartes (1596-1650)who wrote the following in his work La G´eom´etrie from 1637:”...the ratios between straight and curved lines are notknown, and I believe cannot be discovered by humanminds, and therefore no conclusion based on such ratioscan be accepted as rigorous and exact.” [16, p. 91]Descartes would only 20 years later be proved wrong on this pointby Neil who showed how to rectify the semi-cubical parabola y = ax . Independently, both van Heuraet and Fermat came to the sameconclusion within a few years after Neil’s discovery [19]. After that,of course, Newton and Leibniz fully developed the calculus machineryincluding formulas for arc length using integrals [4, p. 217, p. 242].2. Arc length in calculus teaching
The first time students are exposed to arc length calculations of gen-eral functions is in introductory calculus courses. In popular calculusbooks (see e.g. [1, 9, 17]) the concept of curve length is typically definedin the following way.
Definition 1.
Let A and B be two points in the plane and let | AB | denote the distance between A and B . Let C be a curve in the planejoining A and B . Suppose that we choose points A = P , P , P , . . . , P n − and P n = B in order along the curve. The polygonal line P , P , P , . . . , P n constructed by joining adjacent pairs of these pointswith straight lines forms a polygonal approximation to C , having length L n = P ni =1 | P i − P | . The curve C is said to be rectifiable if the limit L of L n , as n → ∞ and the maximum segment length | P i − P i | → L is called the length of C .An obvious pedagogical difficulty for teachers using such a definitionis that then we are not calculating a limit of a sequence , in the usualsense that the students are used to, but rather the limit of a net [14].Not only is such a definition unsuitable for concrete calculations, forinstance using computer simulations, but also highly abstract. Disre-garding this difficulty, the typical calculus book (see loc. cit.) will then RC LENGTH OF FUNCTION GRAPHS 3 state some variant of the following result which is then used in exercisesto calculate lengths of function graphs in particular cases.
Theorem 2. If f is a real-valued function defined on [ a, b ] with theproperty that its derivative exists and is continuous on [ a, b ] , then f is rectifiable on [ a, b ] and its length L equals R ba p f ′ ( x ) dx . Inthat case, if G is a primitive function of p f ′ ) on [ a, b ] , then L = G ( b ) − G ( a ) . The typical ”proof” of this result runs as follows. For the partition { a = x < x < x < · · · < x n = b } , let P i be the point ( x i , f ( x i )),0 ≤ i ≤ n . By the mean-value theorem there exists c i ∈ ( x i − , x i ) suchthat f ( x i ) − f ( x i − ) = f ′ ( c i )( x i − x i − ). A few lines of calculation nowyield that L n = P ni =1 p f ′ ( c i ) ∆ x i which can be recognised as aRiemann sum for R ba p f ′ ( x ) dx which ends the proof by invokingthe fundamental theorem of calculus (FTC).The problem with this ”proof” is that it is, in fact, not a proof atall. Why? Well, because it relies on the FTC which is not provedin full detail in any of the popular calculus texts in use today. Sure,parts of it is proved, but the hardest part concerning the convergenceof Riemann sums is left out. The reason for skipping this is that apresentation including all details will be long and complicated. Forinstance, in Tao’s book [18] the definition of general Riemann sumsand proofs of properties these, including the FTC, takes more than30 pages, excluding an argument for the crucial fact that continuousfunctions on compact intervals are uniformly continuous, which wouldmake the presentation even longer.We sympathise with the method of ”cheating” with the theory incalculus courses. To be honest, we can, of course, not prove everystatement made in the course. However, we feel that leaving out avalid argument concerning such a central fact as the convergence ofRiemann sums should be regarded as cheating at the wrong place.In a recent article [13], we argue that the integral therefore shouldbe defined using equally spaced subdivisions of the interval using onlyleft (or right endpoints). We call the corresponding sums Euler sums ,inspired by the fact that Euler [5, Part I, Section I, Chapter 7] proposedsuch sums for the approximative calculations of integrals. In loc. cit.,we show, using an idea of Poisson (see [2] or [6]), utilizing Taylor’sformula with Lagrange remainder, that the following version of theFTC easily can be proved in just a few lines of calculation.
Theorem 3. If F is a real-valued function defined on [ a, b ] such thatits first derivative exists and is continuous on [ a, b ] , and its second ARC LENGTH OF FUNCTION GRAPHS derivative exists and is bounded on ( a, b ) , then f = F ′ is integrable on [ a, b ] and R ba f ( x ) dx = F ( b ) − F ( a ) . Simplified arc length
In this article, we parallel our investigations in [13] and use Euler-like sums to define length of function graphs (see Definition 4). Weprove (see Theorem 7), using our version of the FTC, assuming someregularity conditions, that length of function graphs can be calculatedvia integrals using the classical formula given in Theorem 2.
Definition 4.
Suppose that f is a real-valued function defined on aninterval [ a, b ]. For all n ∈ N we put ∆ x = ( b − a ) /n , and for all k ∈ { , , . . . , n − } , we put x k = a + k ∆ x and ∆ y k = f ( x k +1 ) − f ( x k ).We say that L n = P n − k =0 p (∆ x ) + (∆ y k ) is the n th polygonal length of f on [ a, b ] and we say that f is rectifiable on [ a, b ] if the limit L =lim n →∞ L n exists. In that case, we call L the arc length of f on [ a, b ].The above definition is mathematically crystal clear and the poly-gonal lengths of this form are easy for students to calculate in particularcases (see Section 5). To prove the main result of the article, we needTaylor’s formula with Lagrange remainder, a result which we now state,for the convenience of the reader. Theorem 5.
Let n be a non-negative integer. If f is a real-valuedfunction defined on [ a, b ] such that its n th derivative exists, is continu-ous on [ a, b ] , and is differentiable on ( a, b ) , then there exists c ∈ ( a, b ) such that f ( b ) = n X j =0 f ( j ) ( a ) j ! ( b − a ) i + f ( n +1) ( c )( n + 1)! ( b − a ) n +1 . Proof.
For a short proof, see e.g. [8, 13, 14]. (cid:3)
In the proof of our main result, we also need the following lemma.
Lemma 6. If A , B and C are real numbers, with A > , then there isa real number D , between and C , such that p A + ( B + C ) = √ A + B + ( B + D ) C p A + ( B + D ) . Proof.
Define the function g : R → R by g ( x ) = p A + ( B + x ) , for x ∈ R . Since A >
0, the function g is differentiable at all x ∈ R withderivative g ′ ( x ) = ( B + x ) √ A +( B + x ) . The claim now follows from Theorem 5with n = 0, a = 0 and b = C (that is, the mean value theorem). (cid:3) RC LENGTH OF FUNCTION GRAPHS 5
Theorem 7. If f is a real-valued function defined on [ a, b ] such that itsfirst derivative exists and is continuous on [ a, b ] , its second derivativeexists and is bounded on ( a, b ) , then f is rectifiable on [ a, b ] if and onlyif the function p f ′ ) is integrable on [ a, b ] . In that case, the length L of f on [ a, b ] equals R ba p f ′ ( x ) dx . If, in addition, p f ′ ) has an antiderivative G on [ a, b ] , then L = G ( b ) − G ( a ) .Proof. We use the notation introduced earlier. From Theorem 5 with n = 1, we get that ∆ y k / ∆ x = f ′ ( x k ) + f ′′ ( c )∆ x/ c ∈ ( x k , x k +1 ), depending on k and ∆ x , for k ∈ { , . . . , n − } .Thus, from Lemma 6, it follows that p y k / ∆ x ) = p f ′ ( x k ) + f ′′ ( c )∆ x/ = p f ′ ( x k ) + ( f ′ ( x k ) + D ) f ′′ ( c )∆ x/ p f ′ ( x k ) + D ) for some real number D between 0 and f ′′ ( c )∆ x/
2. Hence L n = n − X k =0 p (∆ x ) + (∆ y k ) = n − X k =0 p y k / ∆ x ) ∆ x = n − X k =0 p f ′ ( x k ) ∆ x + n − X k =0 ( f ′ ( x k ) + D ) f ′′ ( c )(∆ x ) / p f ′ ( x k ) + D ) which proves the claim, since (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − X k =0 ( f ′ ( x k ) + D ) f ′′ ( c )(∆ x ) / p f ′ ( x k ) + D ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (∆ x ) n − X k =0 | f ′′ ( c ) | ≤ M ( b − a ) n → , as n → ∞ , for any M satisfying | f ′′ ( x ) | ≤ M when a < x < b . Thelast part follows from Theorem 3. (cid:3) Remark 8.
From the above proof, we immediately get the error bound | L − L n | ≤ M ( b − a ) n , for all n ∈ N , where M = sup { | f ′′ ( x ) | ; a < x < b } , for the n th polygonal length. ARC LENGTH OF FUNCTION GRAPHS Primitives of p f ′ ) It seems to be a common opinion among mathematics teachers thatthere are few examples of functions f for which p f ′ ) has a primi-tive function. In this section, we show that this is far from true byrecalling two large classes of such functions.4.1. The examples of Neil, van Heuraet and Fermat.
All of thepersons mentioned above considered rectification of curves of the type f ( x ) n = ax n +1 , for positive integers n and positive real numbers a .Here, we will not follow their original approaches, but instead usemodern tools from a typical calculus class to investigate this problem.First of all, by taking n th roots we can always rewrite the equation as f ( x ) = bx /n for a positive real number b (we assume that x and y are non-negative). Therefore, p f ′ ( x ) = √ cx /n for somepositive real number c . Next, we make the substitution s = √ cx /n sothat p cx /n dx = es n − √ s ds for some positive real number e . It is well known that it is alwayspossible to find a primitive function to an expression which is rationalin s and √ s by making the substitution t = s + √ s . Indeed,from the equality ( t − s ) = 1 + s we get that s = ( t − / t and thus √ s = t − s = t − ( t − / t = ( t + 1) / t. From the equality s = ( t − / t we get that ds/dt = (2 t · t − ( t − / (2 t ) = ( t + 1) / t . Therefore Z s n − √ s ds = Z ( t − n − (2 t ) n − · t + 12 t · t + 12 t dt = 2 − n − Z ( t − n − ( t + 2 t + 1) t − n − dt. If we expand the product in the last integral we can write the integrandas a sum of powers of t which, of course, is easily integrated. To illus-trate the above procedure, we will carry out this analysis, in completedetail, in a few cases. The case when n = 1 and a = 1 / . This is the problem of the rectifi-cation of the parabola f ( x ) = x /
2. In this case c = 1 and x = s andthe integral that we seek therefore equals Z √ x dx = 2 − Z ( t + 2 t + 1) t − dt RC LENGTH OF FUNCTION GRAPHS 7 = 14 Z t + 2 t − + t − dt = t / t ) / − t − / C. To simplify this result, we note that t = 2 x + 1 + 2 x √ x and ( x + √ x + 1)( x − √ x + 1) = − t − = √ x + 1 − x which in turn implies that t − = 2 x + 1 − x √ x . All of this finally implies that Z √ x dx = x √ x / x + √ x ) / C. The case when n = 2 and a = 2 / . This is the problem of the rectifi-cation of the semicubical parabola f ( x ) = 4 x /
9. In this case we get f ( x ) = 2 x / / p f ′ ( x ) = √ x . Here we could, in the-ory, follow the general procedure suggested previously. However, thatwould lead to an unnecessarily long calculation since we immediatelysee that the sought after integral equals Z √ x dx = 2(1 + x ) / / C. The case when n = 3 and a = 3 / . This is the problem of the rectifi-cation of the curve f ( x ) = 3 x / /
4. In this case c = 1 and x / = s sothat e = 3 and the integral that we seek therefore equals Z p x / dx = 3 · − Z ( t − ( t + 2 t + 1) t − dt = 316 Z t − t − + t − dt = 3 t / − t ) / − t − /
64 + C. From the first example, we get that t = 8 s + 8 s + 1 + 4 s (2 s + 1) √ s and t − = 8 s + 8 s + 1 − s (2 s + 1) √ s so that Z p x / dx = 3 s (2 s + 1) √ s / − s + √ s ) / C = 3 x / (2 x / + 1) p x / / − x / + p x / ) / C ARC LENGTH OF FUNCTION GRAPHS
Pythagorean triples.
Suppose that we seek two functions p and q such that f ′ = p/q and 1 + ( f ′ ) = g where g is some func-tion to which we can find a primitive function G . This implies that1 + p /q = g or equivalently that ( p + q ) /q = g . One way to ac-complish this is if p + q = r for some function r of reasonably simpletype. This means that ( p, q, r ) is a Pythagorean triple of functions. Itis a classical result in number theory that such triples, consisting ofintegers, can be parametrized by p = k ( m − n ), q = k (2 mn ) and r = k ( m + n ), where k , m and n are positive integers with m > n ,and with m and n coprime and not both odd (see e.g. [12]). In [11]Kubota has shown that the same kind of result holds in any unique fac-torization domain (UFD). In particular, it holds for polynomial rings R [ X ], since they are Euclidean domains and hence UFD’s. The bottomline is that we can use this kind of parametrization to yield examplesof rectifiable curves in the following way. Choose any functions m and n and put p = m − n and q = 2 mn . Take a function f such that f ′ = p/q = m/ n − n/ m . Then p f ′ ) = p m/ n − n/ m ) = p m/ n ) − / n/ m ) = p ( m/ n + n/ m ) = m/ n + n/ m so that G ( x ) = Z p f ′ ( x ) dx = Z m/ n + n/ m dx. Let us illustrate the above algorithm in three examples.
Example 9.
A problem which often comes up in calculus textbooks isto calculate the length of a portion of the hyperbolic cosine function.Based on our calculations above, it is easy too see why. Indeed, if weput f ( x ) = cosh( x ), then f ′ ( x ) = sinh( x ) = m/ (2 n ) − n/ (2 m ) if weput m = e x and n = 1. Therefore, we get that G ( x ) = Z m/ (2 n ) + n/ (2 m ) dx = Z cosh( x ) dx = sinh( x ) + C. The corresponding task for the students could therefore be:
Problem 10.
Show that the length of f ( x ) = cosh( x )over the interval [0 ,
1] equals e/ − / (2 e ) . Example 11.
Take m = 4 x and n = x + 1. Then we need to find f so that f ′ ( x ) = m/ (2 n ) − n/ (2 m ) = 4 x/ (2 x + 2) − x/ − / (8 x ). We RC LENGTH OF FUNCTION GRAPHS 9 choose f ( x ) = log(2 x + 2) − x / − log( x ) /
8. Then, from the above,we get that G ( x ) = Z p f ′ ( x ) dx = Z m/ n + n/ m dx = Z x/ (2 x +2)+ x/ / (8 x ) dx = log(2 x +2)+ x / x ) / C. Now we can construct a challenging task for the students:
Problem 12.
Show that the length of f ( x ) = log(2 x + 2) − x / − log( x ) / ,
2] equals3 /
16 + log(5) − / . Example 13.
Take m = ( x + 2) and n = ( x + 1)( x + 1). Then weneed to find f so that f ′ ( x ) = m/ (2 n ) − n/ (2 m ) = ( x + 2) x + 1)( x + 1) − ( x + 1)( x + 1)2( x + 2) . Since ( x + 2) x + 1)( x + 1) = x x + 1) + 74( x + 1)and ( x + 1)( x + 1)2( x + 2) = x/ − / − x + 2) + 92( x + 2)we can choose f ( x ) = log( x + 1)8 + 7tan − ( x )4 − x x − x + 2) − x + 2)2 + C. Now we can construct a really challenging task for the students:
Problem 14.
Show that the length of f ( x ) = log( x + 1)8 + 7tan − ( x )4 − x x − x + 2) − x + 2)2over the interval [0 ,
1] equals7 π −
53 + 9log(3)2 − . Discussion
In this article, we have presented a simplified definition of arc lengthas a limit of polygonal sums where the subdivision of the interval isuniform. We feel that such an approach would support the students’learning of calculus for many reasons.First of all, we have provided a complete proof that the polygonallengths converge precisely when the associated integral Z ba p f ′ ( x ) dx exists. In many popular calculus books the proof of this fact is incom-plete since convergence of the nets associated to general Riemann sumsis not proved.Secondly and perhaps more importantly, the students can, using asimple computer program, easily calculate approximations of our sim-plified polygonal lengths, before using the formula L = Z ba p f ′ ( x ) dx. For instance, suppose the students are given the task of calculating thearc length of f ( x ) = 2 x / / , n ∈ N we havethat ∆ x = 5 /n and thus L n = n − X k =0 vuut n + (cid:18) k + 5 n (cid:19) / − (cid:18) kn (cid:19) / ! . Using a computer program, rounding off to four decimal places, we get L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L = 38 /
3. After this the students cantry to make the exact calculation, which, as we saw before, is therectification of the semicubical parabola. Namely, since f ′ ( x ) = x , weget, using theorem 2, that L = Z √ x dx = (cid:20) x ) / (cid:21) = 2 · / − · / f ( x ) = x / RC LENGTH OF FUNCTION GRAPHS 11 over the interval [0 , x = 1 /n and thus L n = n − X k =0 vuut n + 14 (cid:18) k + 1 n (cid:19) − (cid:18) kn (cid:19) ! . Using a computer program, rounding off to four decimal places, we get L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . L ≈ . . After this, the students could try to calculate the exact value of theintegral. From the discussion in the previous section this is the lengthof the parabola which equals Z √ x dx = √ / √ / . Finally, the students could try to calculate the length of f ( x ) = x / , L =1 . Z √ x dx which involves elliptic integrals (see e.g. [7]) and is impossible to cal-culate exactly using the elementary functions. It is our firm belief thatstudents should be subjected to the calculation of such integrals in atypical calculus course, in order for them to appreciate the numeri-cal calculations, which, after all, are crucially important for them in afuture work-life as e.g. engineers. References [1] R. A. Adams,
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