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WUSTL MATH 132 - woodroofe_exam2_solns

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Math 132Midterm Examination 2 Solutions – March 26, 20126 multiple choice, 4 long answer. 100 points.Part I was multiple choice. Only the correct answers are listed here.1. Find the Trapezoid Rule approximation using 4 subintervals ofZ1−1x2dx.(f) 3/42. Find the Simpson’s Rule approximation using 4 subintervals ofZ1−1x2dx.(e) 2/33. Consider the system consisting of 3 point masses:10 kg at (3, −1)20 kg at (2, 10)100 kg at (1, 0)The center of mass is:(g) (1713,1913)4. Simpson’s Rule applied to the integralZe11xdx with n = 20 will be closest to:(k) 1(Since1x≤ 1 on [1, e], and then the error bound of1·(e−1)5204is quite small.)5. Find the average value of sin x over the interval [0, π].(d) 2/π6. The decay of a certain radioactive isotope of the element rabbitonium is governed by thedifferential equation y0= −ky. At t = 0 you have 300 mg of radioactive rabbitonium.At t = 45 minutes, you are left with only 100 mg of radioactive rabbitonium.Then k is per minute.(f) ln 3/45.Part II was long answer.1. Differential equations(a) (8 points) Solve the differential equation y0= x+xy subject to the initial conditiony(0) = 5.Separating the equation, we havey01 + y= xhenceZ11 + ydy =Zx dxln ||1 + y| =x22+ C1 + y = Aex2/2y = Aex2/2− 1.The initial condition y(0) = 5 = Ae0− 1 gives that A = 6, soy = 6ex2/2− 1.(b) (8 points) At time t = 0, there is 1000 liters of water in a tank, with 80 kg of saltdissolved in it. Distilled water flows into the tank at 10 L/min, and water flowsout of the tank at the same rate. The tank is continually stirred, and the salt iskept mixed evenly through the tank.Set up a differential equation (you needn’t solve it) for the mass of salt in thetank at time t. (Your answer should be of the form y0= .)Inflow of salt = 0,outflow of salt = (amount of salt in tank/1000) · 10,so if y = amount of salt in tank, theny0= −y · 101000.The initial condition is y(0) = 80.2. Arc lengths and approximate integration(a) (6 points) Set up a definite integral representing the length of the curve y = x3between x = 0 and x = 4.Z40q1 + (3x2)2dx.(b) (10 points) The first several derivatives of f(x) =√1 + x2are as follows:f0(x) =x√1 + x2, f00(x) =1(1 + x2)3/2, f(3)(x) =−3x(1 + x2)5/2,f(4)(x) =12x2− 3(1 + x2)7/2, f(5)(x) =45x − 60x3(x2+ 1)9/2.Find (with justification) an n such that the Simpson’s Rule approximation SnforZ4−1√1 + x2dx has error at most 0.001.The main step in this problem is finding an upper bound for f(4).Approach 1 to bounding f(4): (triangle inequality)We have that|f(4)(x)| =|12x2− 3||1 + x2|7/2≤12 |x2| + 3|1 + x2|7/2.The top is ≤ 12 · 42+ 3 on [−1, 4], and the bottom is ≥ 1 everywhere, hencef(4)(x)≤12·16+31= 195.Approach 2 to bounding f(4): (take another derivative)The 5th derivative is continuous on [1, 4], and has roots at 0 and ±√32. We approx-imate these points and the endpoints, using the triangle inequality to simplify:f(4)(−1) =12 − 3(1 + 1)7/2≤926/2=98f(4)(−√32)=12 ·34− 3(1 +34)7/2=6(74)7/2≤6(32)6/2=169f(4)(0) =317/2= 3f(4)(√32)is the same asf(4)(−√32)f(4)(4)| =12 · 16 −3(1 + 16)7/2≤189167/2=18947≤ 1Since the max of |f(4)(x)| on [−1, 4] occurs at one of the above points (as it isclearly zero at the points where it fails to be differentiable), we have thatf(4)(x)≤ 3Finding the bound: Using the error bound for Simpson’s rule, and letting Mbe as found in Approach 1 or Approach 2, we wantM · (4 − (−1))5180n4≤1103,so thatn ≥4s103· M · 55180.Writing that you take n to be the least even integer greater than this value(plugging in M to be 195, or 3, or whatever bound you found) gets full credit.Finding an integer value for n (optional): We can factor and round up tofind an n that “works”. We showed that it suffices to taken ≥4s103· M · 55180=4s23· 58· M22· 32· 5=4s2 · 57· M32.If we followed approach 1, then it is convenient to notice that 195 ≤ 200 (as 200has a very nice factorization).4s2 · 57· M32=4s2 · 57· 19532≤4s2 · 57· 20032=4s24· 51032= 2 · 52·√59≤ 50,and we see that n = 50 suffices. (Similarly for approach 2.)3. Calculations(a) (6 points) Find an upper bound for2e−(x+1)2+ 12 sin(x + 1)2 on the interval[−3, 3].Using the triangle inequality,2e−(x+1)2+ 12 sin(x + 1)2 ≤2e−(x+1)2+12 sin(x + 1)2= 2|e−(x+1)2| + 12|sin(x + 1)2|≤ 2 · 1 + 12 ·1 = 14.(b) (7 points) EvaluateZx2cos x dx.We apply integration by parts 2 times. First, take u1= x2and dv1= cos x dx, sothat du1= 2x dx and v1= sin x. We getZx2· cos x dx = x2· sin x −Z2x · sin x dx.Then take u2= 2x and dv2= sin x dx, so that du2= 2 dx and v2= −cos x. Weget the integral to be= x2sin x + 2x cos x −Z2 cos x dx = x2sin x + 2x cos x −2 sin x + C.(c) (6 points) EvaluateZ10x1 + x2dx.We substitute u = 1 + x2, so that du = 2x dx, and the integral becomesZ10x1 + x2dx =Z211udu2="ln |u|2#21=ln 22− 0.(d) (6 points) EvaluateZ1−1x tan−1x dx.We apply integration by parts with u = tan−1x and dv = x dx, so that du =11+x2dx and v =x22. We getZ1−1x tan−1x dx ="tan−1x ·x22#1−1−Z1−1x22·11 + x2dx.We notice that12x21+x2=121 −11+x2, hence the integral is="tan−1x ·x22#1−1+12Z1−111 + x2− 1 dx ="tan−1x ·x22+12tan−1x −x2#1−1=π4·12+12·π4−12−−π4·12−12·π4+12=π2− 1.4. Volumes and centroidsIn both problems on this page, we consider the region between the x-axis and the graphof y = exfor 0 ≤ x ≤ 2.(a) (11 points) Find the volume of the solid formed by rotating the given regionaround the y-axis.Solution 1: (easier) We use cylindrical shells:V = 2π ·Z20x · exdx = 2π [xex]20− 2πZ20ex= 2π [xex− ex]20= 2π(e2+ 1).Solution 2: (harder, sketched only) We use discs. The shape is between x = ln yand x = 2 for 1 ≤ y ≤ e2, and between x = 0 and x = 2 for 0 ≤ y ≤ 1. Thus, wegetV = πZ1022dy + πZe21(ln y)2dy.The integral of (ln y)2may be computed by two applications of integration byparts.(b) (8 points) Find the center of mass x with respect to x of the solid formed byrotating the given region around the x-axis.Half credit will be received for instead finding the center of mass x of the given(unrotated) region.Full credit: Assume uniform density 1. The


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WUSTL MATH 132 - woodroofe_exam2_solns

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