Math 132 Midterm Examination 2 Solutions March 26 2012 6 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 of Z 1 x2 dx 1 f 3 4 2 Find the Simpson s Rule approximation using 4 subintervals of Z 1 x2 dx 1 e 2 3 3 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 17 19 13 13 4 Simpson s Rule applied to the integral Z e 1 k 1 Since 1 x 1 dx with n 20 will be closest to x 1 on 1 e and then the error bound of 5 Find the average value of sin x over the interval 0 d 2 1 e 1 5 204 is quite small 6 The decay of a certain radioactive isotope of the element rabbitonium is governed by the differential equation y 0 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 per minute Then k is f ln 3 45 Part II was long answer 1 Differential equations a 8 points Solve the differential equation y 0 x xy subject to the initial condition y 0 5 Separating the equation we have y0 x 1 y hence Z Z 1 dy x dx 1 y x2 ln 1 y C 2 2 1 y Aex 2 2 y Aex 2 1 The initial condition y 0 5 Ae0 1 gives that A 6 so y 6ex 2 2 1 b 8 points At time t 0 there is 1000 liters of water in a tank with 80 kg of salt dissolved in it Distilled water flows into the tank at 10 L min and water flows out of the tank at the same rate The tank is continually stirred and the salt is kept mixed evenly through the tank Set up a differential equation you needn t solve it for the mass of salt in the tank at time t Your answer should be of the form y 0 Inflow of salt 0 outflow of salt amount of salt in tank 1000 10 so if y amount of salt in tank then y0 The initial condition is y 0 80 y 10 1000 2 Arc lengths and approximate integration a 6 points Set up a definite integral representing the length of the curve y x3 between x 0 and x 4 Z 4q 1 3x2 2 dx 0 b 10 points The first several derivatives of f x x 1 x2 12x2 3 f 4 x 1 x2 7 2 f 0 x 1 x2 are as follows 1 3x 3 f x 1 x2 3 2 1 x2 5 2 45x 60x3 f 5 x 2 x 1 9 2 f 00 x Find with justification an n such that the Simpson s Rule approximation Sn for Z 4 1 x2 dx has error at most 0 001 1 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 12 x2 3 12x2 3 f 4 x 1 x2 7 2 1 x2 7 2 The top is 12 42 3 on 1 4 and the bottom is 1 everywhere hence 195 f 4 x 12 16 3 1 Approach 2 to bounding f 4 take another derivative The 5th derivative is continuous on 1 4 and has roots at 0 and 23 We approximate these points and the endpoints using the triangle inequality to simplify f 4 1 3 4 f 2 f 4 0 3 4 f 2 f 4 4 9 9 12 3 6 2 7 2 1 1 2 8 3 12 4 3 6 6 16 3 7 2 7 7 2 3 6 2 9 1 4 4 2 3 7 2 3 1 3 4 is the same as f 2 189 189 12 16 3 1 1 16 7 2 167 2 47 Since the max of f 4 x on 1 4 occurs at one of the above points as it is clearly zero at the points where it fails to be differentiable we have that f 4 x 3 Finding the bound Using the error bound for Simpson s rule and letting M be as found in Approach 1 or Approach 2 we want M 4 1 5 1 3 4 180n 10 so that s 103 M 55 180 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 n 4 Finding an integer value for n optional We can factor and round up to find an n that works We showed that it suffices to take s 4 n 103 M 55 180 s 4 23 58 M 22 32 5 s 4 2 57 M 32 If we followed approach 1 then it is convenient to notice that 195 200 as 200 has a very nice factorization s s s s 7 7 7 4 10 5 4 2 5 M 4 2 5 195 4 2 5 200 4 2 5 2 2 5 50 2 2 2 2 3 3 3 3 9 and we see that n 50 suffices Similarly for approach 2 3 Calculations 2 a 6 points Find an upper bound for 2e x 1 12 sin x 1 2 on the interval 3 3 Using the triangle inequality 2 2e x 1 12 sin x 1 2 2 2e x 1 12 sin x 1 2 2 2 e x 1 12 sin x 1 2 2 1 12 1 14 b 7 points Evaluate Z x2 cos x dx We apply integration by parts 2 times First take u1 x2 and dv1 cos x dx so that du1 2x dx and v1 sin x We get Z 2 2 x cos x dx x sin x Z 2x sin x dx Then take u2 2x and dv2 sin x dx so that du2 2 dx and v2 cos x We get the integral to be x2 sin x 2x cos x Z 2 cos x dx x2 sin x 2x cos x 2 sin x C c 6 points Evaluate Z 1 0 x dx 1 x2 We substitute u 1 x2 so that du 2x dx and the integral becomes Z 1 0 d 6 points Evaluate Z 2 x 1 du ln u dx 2 1 x 2 1 u 2 Z 1 2 1 ln 2 0 2 x tan 1 x dx 1 We apply integration by parts with u tan 1 x and dv x dx so that du 2 1 dx and v x2 We get 1 x2 Z 1 1 1 x dx tan x tan 1 We notice that 1 2 1 …
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