A RIGOROUS PROOF OF THE ARC LENGTH AND LINE INTEGRAL FORMULA USING THE RIEMANN INTEGRAL

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1 A RGOROUS ROOF OF THE ARC LENGTH AND LNE NTEGRAL FORMULA USNG THE REMANN NTEGRAL ZACHARY DESTEFANO Abstract. n this paper, provide the rigorous mathematical definiton of an arc in general Euclidean Space. then use that to provide an intuitive definition of its length. then find an analytic formula for the length and proof that it is equivalent to the intuitive definition. then apply the same idea to the Line ntegral over a Scaler Field. 1. ntroduction to Arcs and Curves n order to begin this study, we will have to rigorosly define an arc and a curve. We are not only considering curves moving through R 2 or R 3 but general Euclidean space R m. With this in mind, there are various definitions for a curve that we can think of. The following definiton is what has turned to be the most useful and precise definition of a curve. n order to prevent overcounting when finding the length, we are leaving out curves that cross themselves or have overlaps. Additionally, we will define as curve as having possibily infinite length whereas that does hold for an arc. Definition A set C R m is called a C p curve if and only if there exists nondegenerate interval and C p function φ : R m such that φ is 1-1 on 0 and C = φ(). f = [a, b] for some a, b R and a < b, then C is called an arc. Now, many arcs do not have lengths that are easy to compute. Many arcs are smooth curves whose length can only be approximated by lines. However, we can use the fact that as the number of lines approaches infinity, the approximation gets better. This will motivate the geometric description of arc length. Using analysis, we can precisely describe what happens as the number of lines goes to infinity. Roughly speaking, the length of one of the lines is approximately equal to the magnitude of the derivative vector. This will motivate the analytic definition of arc length. We will then prove that the two definitions are equivalent. 2. Geometric Description of Arc Length Definition The Approximation of the Arc Length by partition = {x 0, x 1,..., x n } n of [a, b] is L(C, ) = φ(x i ) φ(x i 1 ) Since we want the number of lines to approach infinity, the following geometric defintion makes sense as our arc length. Definition The Arc Length (geometric) of a curve C is C = lim L(C, ). 0 There are a few problems with this defintion. t is not easy to work with when you want to prove true the analytic formula for arc legnth. The following definition will be better to use. We will soon prove that the two geometric descriptions are equivalent. 1

2 2 ZACHARY DESTEFANO Definition L(C) = sup L(C, ) A curve is rectifiable if its length is finite. t is defined in terms of L(C). Definition C is rectifiable if L(C) is finite. Before we prove that L(C) = C for every rectifiable curve C, we will need a few lemmas. Lemma 2.1. The Triangle nequality can be extended to an arbitrary number of n points, so if {y 0,..., y n } are arbitrary points in R m, then y n y 0 y i y i 1 roof. By induction on n. For n = 1 it s obviously true. For n = 2, it s true by the triangle inequality. k 1 Assume that it s true for n = k 1, so that y k 1 y 0 y i y i 1. Let s now look at y k y 0. By the triangle inequality, By induction hypothesis, y k y 0 y k y k 1 + y k 1 y 0 k 1 k y k y 0 y k y k 1 + y i y i 1 = y i y i 1 Lemma 2.2. f Q is finer than, then L(A, Q) L(A, ). roof. = {x 0, x 1,..., x n } and let Q = {y 0, y 1,..., y N } be a refinement of so that for all 1 i n 1, 1 j N and k 1 such that x i = y j and x i+1 = y j+k. By 2.1, φ(x i+1 ) φ(x i ) j+k h=j+1 φ(y h ) φ(y h 1 ) Let g(i), k(i) be functions defined so that g(i) = j and k(i) = g(i + 1) g(i). Summing the above inequality over all i gets us n 1 L(C, ) = φ(x i+1 ) φ(x i ) i=0 n 1 g(i)+k(i) i=0 h=g(i)+1 φ(y h ) φ(y h 1 ) = We will now prove formally that the two geometric definitons given are equivalent. Most of the proof is in the following lemma. Since the lemma is more general, it will be used later for other purposes Lemma 2.3. f we have a function F defined on all partitions of and F () F (Q) if Q is finer than, then if sup F () exists and is finite, then lim F () 0 exists and is finite. Furthurmore, sup F () = lim F (). 0 N 1 j=0 φ(y j+1 ) φ(y j ) = L(C, Q)

3 ARC LENGTH AND LNE NTEGRAL FORMULA 3 roof. For shorthand, let F = sup F () and let F () = lim F (). Let E be a 0 sequence of partitions such that E = { n }, n+1 is finer than n and n 0 as n. (i): Definition of F () E : Take any sequence of partitions E then {F ( n )} is increasing and bounded above by F, hence by Monotone Convergence Theorem, {F ( n )} converges to some value F () E as n. (ii): F () E F : Assume that F () E > F, then let ɛ = F () F. By definition, partition such that Thus it holds that F () E ɛ < F () < F () E + ɛ F < F () which is a contradiction since F is the supremum over all partitions. (iii): F F () E : Assume that F > F () E, then let ɛ 1 = F F () E. By definition, partition such that thus it holds that F ɛ 1 < F () < F F () E < F () Let ɛ 2 = F () F () E and find a partition Q finer than such that thus it must hold that F () E ɛ 2 < F (Q) < F () E + ɛ 2 F (Q) < F () which is a contradiction by Lemma 2.2 since Q is a refinement of. (iv): F () E = F = F (): By (ii), F () E F and by (iii), F F () E hence it must hold that F = F () E. Since this holds for any sequence of partitions where the norm converges to zero, the limit as the norm goes to zero must be that value, hence F () E = F () so finally F () = F. Theorem 2.4. Assuming C is rectifiable, C = L(C) roof. L(C, ) is a function defined on all partitions. By Lemma 2.2, it holds that if Q is finer than, then L(C, Q) L(C, ). Then, by Lemma 2.3, it must hold that L(C) = sup L(C, ) = lim L(C, ) = C 0

4 4 ZACHARY DESTEFANO 3. The Analytic Description of Arc Length What was said previously tells us little about actually computing the arc length. n order to do that, we will need a good analytic formula. Definition The arc length (analytic) of C is defined to be s(c) = φ (t) dt t seems likely that the two definitions are equal and the idea behind the two definitions being equal is relatively simple. By the Mean Value Theorem, each term in C, φ(t j ) φ(t j 1 ) is approximately equal to φ (t j ) (t j t j 1 ), each term of s(c). For all of the following proofs, we will denote = {t 0, t 1,..., t n } as a partition of = [a, b]. We will let i = [t i 1, t i ] and denote t i = (t i, t i,..., t i ), u i = (u i, u i,..., u i ) where u i is any arbitrary value in i. Lemma 3.1. Let G(x 1, x 2,..., x m ) = (φ (x 1 ),..., φ (x m )). Then, ɛ > 0, fine enough so that for all 1 i n, it holds that s i, u i ( i ) m, G( s i ) G( u i ) < ɛ. roof. Since C is a C p arc, G is uniformly continuous on m, hence for all ɛ > 0, δ > 0 such that x y < δ G( x) G( y) < ɛ (3.1) Take any ɛ > 0. Choose a partition fine enough so that < δ. Then, m s i u i t i t i 1 = m (t i t i 1 ) 2 < hence by equation 3.1, G( s i ) G( u i ) < ɛ. m( δ m ) 2 = δ Lemma 3.2. Take the function G as defined in Lemma 3.1. For all partitions of, with every 1 i n, s i ( i ) m such that φ(t i ) φ(t i 1 ) = (t i t i 1 )G( s i ) roof. By the one-dimensional Mean Value Theorem, 1 i n, 1 j m, s ij [t i 1, t i ] such that φ j (t i ) φ j (t i 1 ) = φ j (s ij)(t i t i 1 ) hence m (φ j (t i ) φ j (t i 1 )) 2 = j=1 m (φ j(s ij )(t i t i 1 )) 2 j=1 m (φ j (t i ) φ j (t i 1 )) 2 m = (t i t i 1 ) φ j(s ij ) 2 j=1 Let s i = (s i1, s i2,..., s im ) then j=1 φ(t i ) φ(t i 1 ) = (t i t i 1 )G( s i ) Lemma 3.3. For all ɛ > 0, fine enough so that for all 1 i n and all u i i, it holds that φ (u i ) (t i t i 1 ) φ(t i ) φ(t i 1 ) < ɛ and φ (u i ) (t i t i 1 ) φ(t i ) φ(t i 1 ) < ɛ i

5 ARC LENGTH AND LNE NTEGRAL FORMULA 5 roof. By Lemma 3.1, we can choose fine enough so that thus of course G( s i ) G( u i ) < ɛ b a G( s i )(t i t i 1 ) G( u i )(t i t i 1 ) < ɛ(t i t i 1 ) ɛ (3.2) b a and by lemma 3.2 and definition of φ (u i ) ( φ(t i ) φ(t i 1 ) ) ( φ (u i ) (t i t i 1 )) < ɛ and of course by equation 3.2, it also holds that ( φ(t i ) φ(t i 1 ) ) ( φ (u i ) (t i t i 1 )) < ɛ(t i t i 1 ) b a Lemma 3.4. For all ɛ > 0, fine enough so that n φ (u i ) (t i t i 1 ) L(A, ) < ɛ roof. Take any ɛ > 0. We will use lemma 3.3 to say that ( φ(t i ) φ(t i 1 ) ) ( φ (u i ) (t i t i 1 )) < ɛ(t i t i 1 ) ɛ b a Summing over all the intervals get us n n φ(t i ) φ(t i 1 ) φ (u i ) (t i t i 1 ) < ɛ Rewriting again, we finally get L(C, ) n φ (u i ) (t i t i 1 ) < ɛ Lemma 3.5. For all ɛ > 0, fine enough so that φ (t) dt ɛ < L(C, ) < φ (t) dt + ɛ roof. By definition, for all ɛ > 0, fine enough so that φ (t) dt ɛ n 2 < φ (u i ) (t i t i 1 ) < φ (t) dt + ɛ 2 By Lemma 3.2, it holds that (3.3) n φ (u i ) (t i t i 1 ) ɛ n < L(C, ) < 2 φ (t i ) (t i t i 1 ) + ɛ 2 (3.4) Combining Equations 3.4 and 3.3, it holds that φ (t) dt ɛ < L(C, ) < φ (t) dt + ɛ (3.5)

6 6 ZACHARY DESTEFANO Lemma 3.6. Take a function g(t) integrable on and a function F () on all partitions of. f ɛ > 0, fine enough so that g(t) dt ɛ < F () < g(t) dt + ɛ then g(t) dt = sup F () roof. For shorthand, denote G() = g(t) dt and F = sup F () (i): Now, G() ɛ < F () F and since this holds for any ɛ > 0, it must be true that G() F (ii): Also, F () < G() + ɛ. Taking the supremum over all partitions, it holds that F G() + ɛ and since this must hold for any ɛ > 0, it s proven that F G() (iii): By (i), it must hold that F G() and by (ii), it must hold that G() F thus we ve proven that G() = F Theorem 3.7. s(c) = C roof. Let g(t) = φ (t) and let F () = L(C, ), then sup F () = C. Lemma 3.5, it satisifes the conditions of Lemma 3.6, so we can say that φ (t) dt = s(c) = C By 4. The Line ntegral f we take the arc C and a continuous function g : C R then we will want to integrate this function with respect to our arc. The line integral with then be the area covered by the function f as it travels along C and it will be denoted as L g (C). As with the arc length, there will be an analytic definition that will be introduced later. Definition The Approximation of the Line ntegral of g over C by partition L g (C, ) is defined to be n G i φ(t i ) φ(t i 1 ) where G i = inf g(φ(t)) t [t i 1,t i ] We will define L g (C) = lim L g(c, ) (4.1) 0 As with arc length, we will want a better geometric definiton of the line integral. Before we introduce it and show that it is equivalent to the one above, we will need a few lemmas. Lemma 4.1. f Q is finer than, then L g (C, Q) L g (C, ).

7 ARC LENGTH AND LNE NTEGRAL FORMULA 7 roof. = {x 0, x 1,..., x n } and let Q = {y 0, y 1,..., y N } be a refinement of so that for all 1 i n 1, x i = y j and x i+1 = y j+k for some 1 j N and k 1. Let G xi = inf g(φ(t)) and G y j = inf g(φ(t)) By Lemma 2.1, t [x i 1,x i ] t [y j 1,y j ] and by definition G xi G yh φ(x i+1 ) φ(x i ) G xi φ(x i+1 ) φ(x i ) j+k h=j+1 φ(y h ) φ(y h 1 ) for j + 1 h j + k, thus j+k h=j+1 G yj φ(y h ) φ(y h 1 ) Let g(i), k(i) be functions defined so that g(i) = j and k(i) = g(i + 1) g(i). Summing the above inequality over all i gets us n 1 G xi φ(x i+1 ) φ(x i ) i=0 n 1 g(i)+k(i) i=0 h=g(i)+1 G yh φ(y h ) φ(y h 1 ) = By definition of L g (C, ), L g (C, Q), we ve now proven that L g (C, ) L g (C, Q) N 1 j=0 G yj φ(y j+1 ) φ(y j ) With this lemma in mind, it now makes sense to define the line integral in a similiar way as we did previously by saying that L g (C) = sup L g (C, ) We are again only considering recitifable curves and continuous functions defined on them, hence L g (C) exists and is finite. Theorem 4.2. sup L g (C, ) = lim L g(c, ) 0 roof. Let F () = L g () then sup F (C, ) exists and is finite and if Q is finer than then F () F (Q), hence By Lemma 2.3, sup F (C, ) = lim F (C, ) 0 Theorem 4.3. g ds = g(φ(t)) φ (t) dt roof. Take any ɛ > 0. Let G = C fine enough so that φ(t i ) φ(t i 1 ) φ (t i ) (t i t i 1 ) < Thus, we have the equation inf g(φ(t)). By Lemma 3.2, we can assert that t [a,b] ɛ 2G (b a) (t i t i 1 ) G ti φ(t i ) φ(t i 1 ) G ti φ (t i ) (t i t i 1 ) < Summing this over all intervals, we end up with ɛ 2G ti (b a) (t i t i 1 ) ɛ 2(b a) (t i t i 1 )

8 8 ZACHARY DESTEFANO n G ti φ(t i ) φ(t i 1 ) n G ti φ (t i ) (t i t i 1 ) < ɛ 2 By definition of integrals, it holds that fine enough so that n g(φ(t)) φ (t) dt G ti φ (t i ) (t i t i 1 ) < ɛ 2 By combining equations 4.2 and 4.3, we get g(φ(t)) φ (t) dt ɛ < L g (C, ) < g(φ(t)) φ (t) dt + ɛ Finally, by Lemma 3.6, it must hold that g(φ(t)) φ (t) dt = sup L g (C, ) (4.2) (4.3)