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Test 2 – Math 2300 – Fall 2012 On my honor as a University of Colorado at Boulder student I have neither given nor received unauthorized assistance on this exam. Name: Please select your section: 001 Scherer . . . . . . . . . (9am) 002 Davison . . . . . . . . (10am) 003 Gorokhovsky . . (11am) 004 Gorokhovsky . . (12pm) 005 Moore . . . . . . . . . . (2pm) 006 Rosenbaum . . . . . . (3pm) In order to receive full credit your answer must be complete, legible and correct. Show all of your work, and give clear explanations. Problem Max. points Points 1 15 2 10 3 20 4 10 5 10 6 20 7 15 Total 100 1 2 1. (15 pt) Find the area of the region in the polar coordinates bounded by the curve r = and rays θ = 0 and θ = π/2. Z β α Z π/2 √ Z π/2 2 1 2 1 1 r dθ = 2θ + 1 dθ = (2θ + 1)dθ 2 2 2 0 0 π/2 Z π/2 1 θ2 θ (π/2)2 π/2 = θ + dθ = + = + 2 2 2 2 2 0 0 √ 2θ + 1 3 2. (10 pt) Find the equation (in terms of x and y) of the tangent line to the curve r = θ at θ = π/2. x = r cos(θ) = θ cos(θ) and y = r sin(θ) = θ sin(θ) So dy dx = cos(θ) − θ sin(θ) and = sin(θ) + θ cos(θ) dθ dθ When θ = π/2, dx π dy = − and =1 dθ 2 dθ and (x, y) = (0, π/2). The tangent line is therefore 1 −2 y= x + π/2 = x + π/2 −π/2 π 4 3. A metal plate, with constant density 1 g/cm2 , has a shape bounded by the curve y = x2 and the line y = 1; and x, y are in cm. (a) (10 pt) Find the total mass of the plate (include units). The area is equal to the area of a rectangle with dimensions 1 × 2 minus the area under the curve y = x2 from x = −1 to x = 1, i.e., 1 Z 1 3 1 x −1 2 4 2 − x dx = 2 − = 2 − =2− = 2− 3 3 3 3 3 −1 −1 Therefore, since the density is 1 g/cm2 the mass is 4 3 g. (b) (10 pt) Find the coordinates of the center of mass of the plate. Since the plate has constant density and is symmetric about the y-axis, x = 0. √ A slice perpendicular to the y-axis has length 2 y and such slices can be taken from 0 to 1, so R1 R1 Z √ δA ydy 1 · 2( y)ydy 3 1 3/2 y 0 0 y= = = 2y dy Total Mass 4/3 4 0 1 12 3 2 5/2 = 2 y = 4 5 20 0 5 4. (10 pt) Find the length of the curve given by parametric equations x(t) = 2t, y(t) = t2 − ln t, 1 ≤ t ≤ 2. 2 x0 (t) = 2 and y 0 (t) = t − 1 t So 2 1 1 1 2 (x (t)) + (y (t)) = 4 + t − 2 + 2 = t + 2 + 2 = t + t t t and therefore for 1 ≤ t ≤ 2, s 2 p 1 1 0 2 0 2 =t+ (x (t)) + (y (t)) = t+ t t 0 2 0 2 2 Thus, the arclength is Z 2p Z (x0 (t))2 + (y 0 (t))2 dt = 1 1 2 2 1 1 t2 3 t + dt = + ln |t| = 2 + ln |2| − = + ln |2| t 2 2 2 1 6 5. (10 pt) Determine if the following improper integral Z∞ √ x dx x3 − 1 2 converges. Carefully explain your reasoning. When x ≈ ∞, 1 is insignificant compared to x3 so x x 1 x √ ≈ √ = 3/2 = 1/2 x x x3 − 1 x3 R ∞ dx So we predict this integral will diverge because 1 x1/2 diverges. Therefore, we wish to find some function smaller than √xx3 −1 whose integral diverges on [2, ∞). But x x 1 √ ≥ √ = 1/2 3 x3 R x R ∞ dx R ∞ dxx − 1 ∞ Since 1 x1/2 diverges so does 2 x1/2 and therefore 2 xdx 1/2 diverges. 7 6. For each of the series below determine if it is absolutely convergent, conditionally convergent or divergent. Carefully explain your reasoning. (a) (10 pt) ∞ X 1 √ 2 n +1 n=1 Note that √ 1/n n2 + 1 p √ = 1 + 1/n2 = n 1/ n2 + 1 p P 1 So limn→∞ 1 + 1/n = 1. Therefore, since ∞ n=1 n diverges so does the original sum. (b) (10 pt) ∞ X (−1)n n ln(n) n=2 1 1 The sequence n ln(n) is decreasing and limn→∞ n ln(n) = 0. So by the alternating series test, the series converges. On the other hand by using substitution with w = ln(x) we get Z ∞ dx = lim ln | ln(b)| − ln | ln(2)| = ∞ x ln(x) b→∞ 2 P 1 so that ∞ n=2 n ln(n) diverges. So the series is conditionally convergent. 8 7. Consider the region bounded by the graph of y = ex , line y = e and the y-axis. (a) (10 pt) Find the volume of the solid obtained by rotating this region about the x-axis Cross sections are washers with outer radius e and inner radius ex and thickness dx. Thus, cross sections have area (πe2 − π (ex )2 )dx = π(e2 − e2x )dx. Since these cross sections are realized for 0 ≤ x ≤ 1, the volume is 1 Z 1 1 2 1 1 2 1 1 2 2x 2x 2 2 π e − e dx = π e x − e e − =π e + =π e − 2 2 2 2 2 0 0 (b) (5 pt) Find the volume of the solid obtained by rotating this region about the y-axis A cross section is a thin cylinder with radius ln(y) and height dy. These cross sections have volume π(ln(y))2 dy and are taken from y = 1 to y = e. Thus, the volume is Z e π(ln(y))2 dy 1 We can evaluate this by the substitution u = ln(y) so that y = eu and dy = eu du. So Z Z 2 (ln(y)) dy = u2 eu du Z By using integration by parts we get Z Z 2 u 2 u u 2 u u u e = u e − 2ue du = u e − (2ue − 2eu ) = u2 eu − 2ueu + 2eu = (ln(y))2 y − 2 ln(y)y + 2y So Z 1 e e 2 2 π(ln(y)) dy = π((ln(y)) y − 2 ln(y)y + 2y) = π(e − 2e + 2e − 2) = π(e − 2) Other solutions are possible. 1