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VECTOR CALCULUS Solved Exercises Line Integrals VECTOR CALCULUS Solved exercises Equacions Diferencials Grau en Enginyeria en Tecnologies Industrials (GETI) ETSEIB { UPC Departament de Matem`atiques ETS d'Enginyeria Industrial de Barcelona (ETSEIB) Universitat Polit`ecnicade Catalunya (UPC) Version: May 2021 https://mat-web.upc.edu/etseib/ed/ Note: All the underlined text provides links to websites which may help you better understand each problem, or obtain further information. Line integrals 1. Length of the arc of the cycloid σ(t) = (R(t − sin t);R(1 − cos t)), 0 ≤ t ≤ 2π. Resolution: A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slipping. To get the parameterization, we first parameterize the rim by (R cos(3π=2 − t);R sin(3π=2 − t)), 0 ≤ t ≤ 2π, since the wheel goes clockwise and we measure the angle from the first contact point whose angle is 3π=2. As the center of the wheel travels like (Rt; R), the parameterization of the cycloid turns out to be σ(t) = (R(cos(3π=2 − t) + Rt; sin(3π=2 − t) + R) = R(t − sin t; 1 − cos t) ; 0 ≤ t ≤ 2π: 0 p t Therefore σ0(t) = R(1 − cos t; sin t), so that σ (t) = R 2(1 − cos t) = 2R sin and 2 2π 2π t=2π t t t L(cycloid) = 2R sin dt = 2R sin dt = 2R −2 cos = 8R: ˆ0 2 ˆ0 2 2 t=0 2. Length of the epicycloid with k cusps parameterized by σ(t) = (R(k + 1) cos t − R cos((k + 1)t);R(k + 1) sin t − R sin((k + 1)t)), 0 ≤ t ≤ 2π. Resolution: An epicycloid or hypercycloid is a plane curve produced by tracing the path of a chosen point on the circumference of a circlecalled an epicyclewhich rolls without slipping around a fixed circle. It is a particular kind of roulette. To get the length, as we have the parameterization, we first have to calculate its first derivative which is σ0(t) = (−R(k + 1) sin(t) + R sin((k + 1)t)(k + 1);R(k + 1) cos(t) − R cos((k + 1)t)(k + 1)) Now let's simplify this expression saying that a = k + 1 : σ0(t) = R(−a sin(t) + sin(at)a; a cos(t) − cos(at)a) The norm is 0 p σ (t) =R (−a sin(t) + sin(at) a)2 + (a cos(t) − cos(at) a)2 = q =R a2(sin2(t) − 2 sin(t) sin(at) + sin2(at) + cos2(t) − 2 cos(t) cos(at) + cos2(at)) = q =aR sin2(t) − 2 sin(t) sin(at) + sin2(at) + cos2(t) − 2 cos(t) cos(at) + cos2(at) = =aRp2 − 2 sin(t) sin(at) − 2 cos(t) cos(at) = v u =aRu2 − 2 (sin(t) sin(at) + cos(t) cos(at)) = t | {z } | {z } 1 1 2 (cos(t−at)−cos(t+at)) 2 (cos(t−at)+cos(t+at)) p =aRp2 − 2 cos(t − at) = aR 2p1 − cos(t − at) = p p =aR 2p1 − cos((1 − a)t) = (cos x is an even function so) = aR 2p1 − cos((a − 1)t) = p (k + 1)R 2p1 − cos(kt) (where we have changed again k = a − 1) Finally let us compute the length 2π p L(epicycloid) = (k + 1)R 2p1 − cos(kt) dt ˆ0 p 2π = (k + 1)R 2 p1 − cos(kt) dt ˆ 0 | {z }p q cos(kt) 1−cos(kt) sin kt = 1 − = p j ( 2 )j 2 2 2 2π 2π p p kt kt = (k + 1)R 2 2 sin dt = 2(k + 1)R sin dt: ˆ0 2 ˆ0 2 kt k 2 Now we apply the change u = 2 ; du = 2 dt () dt = k du and use that jsin uj is π-periodic and that jsin uj = sin u ≥ 0 for 0 ≤ u ≤ π, πk π k + 1 u=π L(epicycloid) = 4 R jsin uj du = 4(k + 1)R jsin uj du = 4(k + 1)R [− cos u]u=0 k ˆ0 ˆ0 = 8(k + 1)R: 8(k+1)R An epicycloid with k cusps has k archs, each one with length k . You can see a figure for some k just clicking on it: k = 1, k = 2, k = 3, k = 4... 3. Compute the theoretical time of navigation between Lisbon and New York if we follow a trajectory of constant direction and we sail at speed of 30 knots. Resolution: In this exercise we need to compute a great-circle distance, which is the shortest distance between two points on the surface of a sphere, measured along the surface of the sphere. We are going to establish that Lisbon is our Point 1 and New York is the Point 2. Let θ1;'1 and θ2;'2, be the geographical longitude and latitude in radians of the two points, and ∆θ; ∆' be their absolute differences. 2 Illustration of the central angle α, between two points P1 and P2. The actual arc length d on a sphere of radius r can be trivially computed as d = Rα, where α is the angle between its two endpoints P1 and P2, measured from the center O of the sphere. If −π ≤ θj ≤ π and −π=2 ≤ 'j ≤ π=2, then: hOP~ 1; OP~ 2i cos α = = cos '1 cos '2 (cos θ1 cos θ2 + sin θ1 sin θ2) + sin '1 sin '2 ~ ~ OP1 OP2 = sin '1 sin '2 + cos '1 cos '2 cos(∆θ): [Obs.: This formula can be directly applying the spherical law of cosines to the spherical triangle whose vertices are P1, P2 and one of the poles an auxiliary third point; if the North Pole is chosen, the angles of this triangle are π=2 − '1, π=2 − '2 and α.] First let's collect some data that we need to solve the problem, first of all we need the mean radius of the Earth which is approximately R ≈ 6371 km, then we need the coordinates of Lisbon which o 0 o 0 are (θ1;'1) = (9 8 W; 38 43 N) ≈ (0:1594067; 0:6757333) rad, and also the coordinates of New York o 0 o 0 which are (θ2;'2) = (74 0 W; 40 43 N) ≈ (1:2915436; 0:7106399) rad. Using such data and the formula obtained above, we can calculate the distance, which is d = Rα = 0:8510653 · 6371 = 5422:14 km. Now that we know the distance, we only need to finish the problem by calculating the time we need to go from Lisbon to New York at a constant speed of 30 knots, knowing that a knot is 1:852 km=h the theoretical time of navigation t will be: 5422:14 t = = 97:59 h; 30 · 1:852 which is a little bit more than 4 days. 4. A cassette tape with thickness of 16µm (micrometers) is rolled in a spool with internal radius r0 = 1:11cm and external radius r1 = 2:46cm. How much does the tape of the spool measure? [Hint: Approximate the form of the spool by the curve described in polar coordinates through the equation r = g(θ) = cθ,θ0 ≤ θ ≤ θ1, being c > 0 a constant to be determined, so that rj = cθj .] Resolution: To solve this problem we shall look at it as if we were to find the length of the curve (C) as an Archimedean spiral. See more information about Archimedean spiral in wikipedia. 3 Archimedean spiral −2 −2 −6 Our data are r0 = 1:11 · 10 m, r1 = 2:46 · 10 m and the increment per revolution is 16 · 10 m. We are given C in polar coordinates as r = r(θ) for θ0 ≤ θ ≤ θ1, so in cartesian coordinates C can be parameterized as C = σ([θ0; θ1]): (x; y) = σ(θ) = (r(θ) cos θ; r(θ) sin θ); θ0 ≤ θ ≤ θ1: To find θ0 and θ1 first we have to determine c. We will do this imposing that the radius increases 16 · 10−6m for each rotation. So, with any rotation 16 · 10−6 = g(θ + 2π) − g(θ) = 2πc; so that 8 · 10−6 c = : π Now we can calculate: r 0:0111 · π θ = 0 = ; 0 c 8 · 10−6 r 0:0246 · π θ = 1 = : 1 c 8 · 10−6 We can define both r(θ) and r0(θ) because it will be useful for us to work with them during the definition of the parameterization: r (θ) = c · θ and r0 (θ) = c Finally, we know that the arc length in polar coordinates is: s dr 2 d` = r2 + dθ; dθ and so, the total arc length of the tape is: θ1 θ1 q θ1 p 2 p d` = r(θ)2 + r0(θ)2 dθ = (c · θ) + c2 dθ = c 1 + θ2 dθ ˆC ˆθ0 ˆθ0 ˆθ0 1 p 1 p θ=θ1 = c · · θ · 1 + θ2 + ln θ + 1 + θ2 ≈ 94:6m; 2 2 θ=θ0 t a2 p 2 2 p 2 2 p 2 2 where we have used the primitive t + a dt = t + a + ln t + t + a + ctant, which ˆ 2 2 p can be found either by the change of variables t = a sinh(x) or the change by parts u = t2 + a2, dv = dt. 4 5. Mass of a spiral of the helix of radius R and height h parameterized by ht σ(t) = R cos t; R sin t; ; 0 ≤ t ≤ 2π; 2π if the density at each point is proportional to the distance from the origin, being k > 0 the constant.
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