Math 118 Assignment 6 - Math 118 Honours Calculus II Winter 2014 Assignment 6 March 23 due April 3 1 Determine which of the following improper integrals

Math 118 Assignment 6 - Math 118 Honours Calculus II Winter...

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Unformatted text preview: Math 118: Honours Calculus II Winter, 2014 Assignment 6 March 23 due April 3 1. Determine which of the following improper integrals converge and which diverge. In the cases where they converge, except for question (e), compute their values. (a) ∞ 0 dx . a2 + x2 If a = 0, the integral diverges (due to the contributions near x = 0). For a = 0 we find T dx T π 1 lim = lim arctan = . 2 + x2 T →∞ 0 a T →∞ a a 2 |a| (b) 1− √ 0 δ lim δ→1− 0 dx π √ = lim arcsin δ = . 2 1 − x2 δ→1− (c) √ 1+ x √ dx. x ∞ 1 Since and √ ∞ 1 dx/ x dx . 1 − x2 √ 1 1+ x √ >√ x x diverges, we know from the Comparison Theorem that √ ∞ 1+ x √ dx x 1 also diverges. (d) 1 log x dx. 0+ 1 lim t→0+ t 1 · log x dx = lim t→0+ x log x − x dx x 1 t = lim [x log x − x]1 t t→0+ = lim (−1 − t log t + t) = −1 + lim t→0+ t→0+ Hence the integral converges to the value −1. 1 1/t = −1. −1/t2 (e) sin2 x dx. 1 + x3/2 ∞ 0 Since 1 sin2 x ≤ 3/2 3/2 1+x x 0≤ and ∞ 1 x3/2 1 dx converges, we know by the Comparison Theorem that ∞ 1 sin2 x dx 1 + x3/2 converges. Hence ∞ 0 sin2 x dx = 1 + x3/2 1 0 sin2 x dx + 1 + x3/2 ∞ 1 sin2 x dx 1 + x3/2 also converges. (f) 2 −2 1 dx. (x − 1)4/5 We need to split the integral into two pieces: 2 t t + lim 5(x − 1)1/5 −2 t→1+ t→1− t→1+ t t→1− −2 √ 5 = [0 − 5(−3)1/5 ] + [5 − 0] = 5 + 5 3. √ Thus this improper integral is convergent and has the value 5 + 5 5 3. (x − 1)−4/5 dx + lim lim (x − 1)−4/5 dx = lim 5(x − 1)1/5 2 t (g) 2 −2 1 dx. (x − 1)7/3 We need to split the integral into two pieces: t t→1− 2 (x−1)−7/3 dx+ lim lim −2 t→1+ (x−1)−7/3 dx = lim t→1− t 3 − (x − 1)−4/3 4 t + lim −2 t→1+ 3 − (x − 1)−4/3 4 Since these individual limits do not exist, the integral is divergent. 2. Let p and q be positive real numbers such that 1/p + 1/q = 1. (a) Prove that the function f (x) = xp p + x−q q ≥ 1 on (0, ∞). We note that f ′ (x) = xp−1 − x−q−1 = 0 when xp+q = xpq = 1; in other words, when x = 1. Since this is the only critical point and lim f (x) = lim f (x) = ∞, we see x→0+ x→∞ that f has a global minimum at x = 1, where f (1) = 1/p + 1/q = 1. 2 2 . t (b) Apply part (a) to a suitable choice for x to prove Young’s inequality: for all real numbers A ≥ 0 and B ≥ 0, AB ≤ Ap B q + . p q If either A or B is zero the result is trivial. Otherwise, on letting x = A/(AB)1/p , we see that 1/x = (AB)1/p /A = (AB)1−1/q /A = B/(AB)1/q , so that Young’s inequality follows immediately from part (a). (c) Given two positive integrable functions f and g on [a, b], show that the choices g(x) f (x) and B= A= 1/p 1/q b p b q f (x) dx g (x) dx a a leads to H¨lder’s inequality: o b a 1/p b fg ≤ f g a a fg ≤ F 1/p 1/q G b fp a F p + b gq a G =F q q . a b p a f (x) dx On integrating part (b) we see, denoting F = b 1/q b p 1/p 1/q G 1 1 + p q b q a g (x) dx, and 1/p b f = that a 1/q b p g q . a 3. A piece of wire is bent into a quarter unit circle centered about the origin with endpoints at (1, 0) and (0, 1). (a) Compute the coordinates (x, y) of the centroid of the wire using integration. √ A quarter circle can be parametrized as y(x) = 1 − x2 for x ∈ [0, 1], for which √ √ y ′ (x) = −x/ 1 − x2 and ds = 1/ 1 − x2 = 1/y. In terms of the length L = π/2 of the wire, we find 1 x = L−1 x ds = L−1 0 √ x dx = L−1 − 2 1−x 1 − x2 1 0 = L−1 = 2 π and, as expected from the symmetry principle, 1 y = L−1 y ds = L−1 1 dx = L−1 = 0 2 . π (b) Use Pappus’ first theorem to check your answers in part (a). We check the x centroid by computing the surface area 2πxL = 2π generated by revolving about the y axis, which is indeed the surface area of a hemisphere. One can also check the y centroid by revolving around the x axis. 3 (c) A quadratic Bezier approximation to a quarter unit circle is described by the curve x(t) = 1 − t2 and y(t) = 2t − t2 , with t ∈ [0, 1]. Show that the arc √ √ length of this curve is 1 + log(1 + 2)/ 2. 1 1 √ =2 2 1 t2 0 1 2 √ =4 2 √ √ 1 − t + dt = 2 2 2 τ2 0 sinh−1 (1) 2 cosh u du = 2 = 1 t− 2 0 τ2 0 √ 1√ sinh−1 (1) 2 1+1+ 2 2 1 1 2 √ 1 + dτ = 4 2 4 0 √ [−2t]2 + [2 − 2t]2 dt = 2 0 0 = 1 [x′ (t)]2 + [y ′ (t)]2 dt = L= 2 1 2 √ 1 + dt = 2 2 4 √ 1 + dτ = 2 2 4 1 u 2 sinh(2u) + 4 2 2t2 − 2t + 1 dt 0 1 −2 sinh−1 (1) 0 sinh−1 (1) τ2 + sinh2 u + 1 √ 1 = 2 sinh u 2 0 √ −1 log(1 + 2) sinh (1) √ √ =1+ , =1+ 2 2 where we have used the substitution t − 1 2 = τ followed by τ = 1 2 1 dτ 4 1 cosh u du 2 u 1 + sinh u + 2 sinh u. Note: The resulting arc length, 1.623, is about 3% greater than the arc length of a true quarter unit circle. However, if one uses a cubic Bezier approximation, the arc length is then accurate to 0.01%. 4. Calculate the surface area generated by revolving the curve y = 2x3/2 /3 from x = 0 to x = 1 about the y axis. The surface area is given by 1 1+ x 2π 0 dy dx 2 1 dx = 2π √ x 1 + x dx. 0 Now use the substitution u = 1 + x to obtain 2 2π 1 2 5/2 2 3/2 2 u − u 5 3 1 1 10 √ 3 1 5/2 1 3/2 1 1 12 √ 5 2− 2− = 4π 2 − 2 − + + = 4π 5 3 5 3 15 15 15 15 √ 8 π 2+1 . = 15 √ (u − 1) u du = 2π 2 (u3/2 − u1/2 ) du = 2π 5. Consider the curve γ described in polar coordinates as r(θ) = (θ − π)2 for θ ∈ [0, 2π]. (a) Sketch γ in the x–y plane. 4 sinh−1 (1) 2 0 y x (b) Determine the portion of the area enclosed by γ that lies outside the circle r(θ) = π 2 /4. Since γ intersects the circle of radius π 2 /4 at 3π/2 and π/2 and the area of the semicircle is π 5 /32, the desired area is π/2 π5 32 0 π/2 π5 (θ − π)4 dθ − = 32 0 1 A= 2 2 r 2 (θ) dθ − (θ − π)5 5 13 5 = π . 80 π/2 = (c) Use the fact that to γ at θ = π/2. dy dx = dy dx / dθ dθ 0 − π5 32 to compute the equation of the tangent line Since x(θ) = r(θ) cos θ and y(θ) = r(θ) sin θ we find at θ = π/2 that dy 2(θ − π) sin θ + (θ − π)2 cos θ 2(−π/2) 4 = = = . 2 sin θ 2 dx 2(θ − π) cos θ − (θ − π) −(π/2) π So the equation is y = π2 4 x+ . π 4 6. Let y = f (x) = x−3/2 − 1 and consider the curve γ = {(x, f (x)) : x ∈ [δ, 1]} in the limit δ → 0+ . (a) Does the area bounded by the curve and the lines y = 0, x = 0, and x = 1 approach a finite value as δ → 0+ ? Why or why not? If it does, compute the limiting area. No. The corresponding improper integral is divergent: 1 lim δ→0+ δ x−3/2 − 1 dx = lim −2x−1/2 − x δ→0+ 5 1 δ = ∞. (b) If the area in part (a) is fully rotated around the y axis, use the method of cross-sections to determine whether or not the resulting volume approaches a finite value as δ → 0+ . If it does, compute the limiting volume. On slicing the y axis, we find that the limiting volume is finite: ∞ π 0 ∞ 1 dy = π (y + 1)4/3 ∞ (y + 1)−4/3 dy = π −3(y + 1)−1/3 0 = 3π. 0 (c) If the area in part (a) is fully rotated around the y axis, use the method of cylindrical shells to determine whether or not the resulting volume approaches a finite value as δ → 0+ . If it does, compute the limiting volume. We now slice the x axis to obtain shells of radius x, height f (x), and thickness dx. The limiting volume is then 1 1 2π 0+ 1 (x−1/2 − x) dx = 2π 2x1/2 − x2 2 0+ x(x−3/2 − 1) dx = 2π 1 = 3π, 0+ in agreement with part (b). 7. The Euler-Mascheroni constant γ is defined as the limiting difference between the harmonic series and the logarithm: n . γ = lim n→∞ k=1 1 − log n . k (a) Prove for every n ∈ N that n k=1 n k =1 1 log 1 + k n log = k=1 n+1 = k=2 k+1 k 1 k = log(n + 1). n log 1 + n = k=1 log(k + 1) − log(k) k=1 n log(k) − k=1 log(k) = log(n + 1) − log(1) = log(n + 1). (b) Evaluate lim log n→∞ = log lim n→∞ n n+1 n n+1 = log 1 = 0, noting that the natural logarithm is continuous. 6 (c) Use parts (a) and (b) to prove that ∞ 1 − log k k=1 k+1 k = γ. In view of part (a) we find that ∞ k =1 1 − log k ∞ k+1 k n 1 1 − log 1 + k k = k=1 n = lim n→∞ k=1 = lim n→∞ k=1 n n+1 1 − log n + log k 1 − log(n + 1) k = γ + 0 = γ. (d) Compute ∞ 1 1 . − k k+1 k=1 n = lim n→∞ k=1 n 1 − k = lim n→∞ k=1 n k=1 n+1 1 − k n 1 k+1 k=2 1 k = lim n→∞ = lim n→∞ k=1 1− 1 − k 1 n+1 n+1 k=2 1 k = 1. (e) In the figure below, use parts (c) and (d) to determine the sum of the brown+green shaded areas: 1 and the sum of the green areas alone: γ. y y= (1,1) 1 2 ... k 7 1 x k+1 ... x ...
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