Math 116 Second Midterm March 23, 2010

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1 Math 6 Second Midterm March 23, 2 Name: EXAM SOLUTIONS Instructor: Section:. Do not open this exam until you are told to do so. 2. This exam has pages including this cover. There are 9 problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck. 3. Do not separate the pages of this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical questions, so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation for each problem, so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad. However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3 5 note card. 7. If you use graphs or tables to find an answer, be sure to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers, and remove all headphones. Problem Points Score Total

2 Math 6 / Exam 2 (March 23, 2 page 2. [ points] There is a bucket, shaped like a cylinder, with a radius of 5 inches and a height of 2 inches. It has a circular hole in the bottom which has a radius of inch. The bucket begins full of water, but it flows out the hole in the bottom. Let t be the number of seconds since the water began dripping from the bucket, and let V (t denote the volume (in inches 3 of water remaining in the bucket at time t. Let h(t be the depth of the water in the bucket at time t. a. [2 points] Write a formula for the volume of water in the bucket, V (t, as a function of the depth of the water in the bucket, h(t. V (t = 25πh(t b. [8 points] The volume of water changes such that it satisfies the differential equation dv dt =.6π 9.6h. Solve for the depth of the water at time t =. Be sure to include units in your answer. We know that V = 25πh, so h = V 25π.6π =.6.784πV. Using separation of variables, we have 9.6V 25π dv V /2 =.6.784πdt.946dt 2V /2 =.6.784πt + C V /2 =.3.784πt + C. Substituting this we get that dv dt = V = (.3.784πt + C 2 (.478t + C 2 When t =, V = 5π, so V = 5π = C 2, giving C = 5π , so V = (.3.784πt + 5π 2 (.478t When t =, V = inches 3, so h inches.

3 Math 6 / Exam 2 (March 23, 2 page 3 2. [ points] Determine if each of the following integrals diverges or converges. If the integral converges, find the exact answer. If the integral diverges, write DIVERGES. Show ALL work and use proper notation. Calculator approximations will not receive credit. a. [5 points] 2 3 x /3 dx 2 The integral converges to 9 2 (22/ dx = lim dx x/3 a + a x/3 = lim a + 9 = lim a + = 9 2 (22/3. 2 x2/3 2 a ( 9 2 (22/3 9 2 a2/3 b. [5 points] 2 e /x dx x 2 Make the substitution u = x, so du = dx. Then we have the general x 2 antiderivative e /x dx = e u du = e u + C = e /x + C. That gives us x 2 2 e /x The integral converges to e /2. x 2 dx = lim a + 2 e /x a x 2 dx = lim a e /x 2 + a ( = lim e /2 e /a a + = e /2.

4 Math 6 / Exam 2 (March 23, 2 page 4 3. [6 points] with DEtools : dfieldplot diff y x, x = y$ Ky, y x, x =..2, y =K.2...2, arrows = line, dirgrid = 5, 5, color = BLACK, axes = framed, labels = x, y ; Warning, y is present as both a dependent variable a. [3 points] The following figure is the slope field for dy and a name. Inconsistent specification of the dependent variable dx = is ay(b y, where a and b are deprecated, and it is assumed that the name is being used in constants. place of the dependent variable y K x Which of the following statements is true regarding a and b? Circle only one answer. A. a >, b > B. a >, b < C. a <, b < D. a <, b > with DEtools : b. [3 points] The following figure is the slope field for dy dfieldplot diff y x, x =.5 x K.25, y x, x =..2, dx = ax + b, where a and b are y =K.2...2, arrows = line, dirgrid = 5, 5, color = BLACK, axes = framed, labels = x, y ; constants y K x Which of the following statements is true regarding a and b? Circle only one answer. A. a >, b > B. a >, b < C. a <, b < D. a <, b >

5 Math 6 / Exam 2 (March 23, 2 page 5 4. [2 points] A bank account earns 2.5% annual interest compounded continuously. Continuous payments are made out of the account at a rate of $5, per year for 8 years. a. [4 points] Write a differential equation describing the balance B = f(t, where t is in years satisfying t 8. db dt =.25B 5, =.25(B 6, b. [4 points] Solve the differential equation you found in part (a given an initial balance of B. db =.25(B 6, dt db =.25dt (B 6, ln B 6, =.25t + C B 6, = Ae.25t B = Ae.25t + 6, Given B = B when t =, we have B = A + 6,, so A = B 6,, giving B = (B 6, e.25t + 6,. c. [4 points] What was the initial balance if the account has $, remaining 8 years after the account was opened? Give your answer to the nearest penny. Solving for B given that B =, when t = 8, we have, = (B 6, e.25(8 + 6,, giving 59, = (B 6, e.45, which leads to B = 59, e , 223, The initial balance would be approximately $223,

6 Math 6 / Exam 2 (March 23, 2 page 6 5. [ points] Let t be the number of minutes a student waits for the Bursley-Baits bus. For constants a and b, the probability density function giving the distribution of t is { if t < p(t = ae bt if t <. According to this density function, the mean waiting time for the bus is 8 minutes. a. [6 points] Determine the exact values of the constants a and b. Answers supported only by calculator work will not receive full credit. Write your final answers on the spaces provided. Because this is a density function, ae bt dt =. We have lim c c ae bt dt = lim c a b e bt c = lim c [ a b a b e bc] = a b =. This gives us the condition a = b. We know the mean time is 8, so we also have 8 = tae bt dt. Then we have 8 = c lim c ate bt dt. Use integration by parts using u = at, du = adt, dv = e bt dt, v = b e bt. This gives us ( lim a c c b te bt c a + b e bt dt = lim ( a c b ce bc a b 2 e bt c ( = lim a c b ce bc a b 2 e bc + a b 2 = a b 2 = 8. Using a = b, we have 8 = b b 2 = b, so that b = 8 = a. a = /8 b = /8 b. [4 points] Using your answers from part (a, determine the exact value for median waiting time. Include units in your answer. Answers supported only by calculator work will not receive full credit. Integrating we get Let M be the median so that.5 = This gives us e M/8 =.5, so that M 8 M 8 e t/8 dt..5 = e t/8 M = e M/8 +. = ln(.5, and M = 8 ln(.5 = 8 ln(2 minutes.

7 Math 6 / Exam 2 (March 23, 2 page 7 6. [2 points] The position of a particle at time t is given by x = cos ( e t, and y = cos ( 3e t, where both x and y are measured in cm, and t is measured in seconds. a. [5 points] Find the exact speed of the particle at time t =. Show enough work to support your answer and include units. Calculator approximations will not receive full credit. We first find dx dt Speed = and dy dt. dx dt = et sin(e t and dy dt = 3et sin(3e t ( e t sin(e t 2 + ( 3e t sin(3e t 2 = Evaluating at t = gives us Speed = sin 2 ( + 9 sin 2 (3 cm/sec. e 2t sin 2 (e t + 9e 2t sin 2 (3e t b. [7 points] Use derivatives to determine the concavity of the particle s path at time t =. d 2 y dx 2 = d dt From part (a, dy dx = 3et sin(3e t e t sin(e t = 3 sin(3et sin(e t. d dt ( 3 sin(3e t sin(e t / dx ( dy / dx dx dt. dt = 9et cos(3e t sin(e t 3e t sin(3e t cos(e t sin 2 (e t /( e t sin(e t = 9et cos(3e t sin(e t 3e t sin(3e t cos(e t e t sin 3 (e t. Evaluating at t =, we have d2 y dx 2 path is concave up when t =. = 9 cos(3 sin( 3 sin(3 cos( sin 3 ( >. Therefore the

8 Math 6 / Exam 2 (March 23, 2 page 8 7. [2 points] Suppose that functions f(x, g(x, and h(x are continuous and differentiable for x and satisfy the condition that f(x g(x h(x for x. Furthermore, suppose that g(xdx converges. You do not need to show your work for this page. No partial credit will be given. a. [4 points] Consider the following group of statements: I. h(xdx diverges. II. h(xdx converges. III. 3 h(xdx converges. Which of the above statements must be true? Circle ONE of the following choices: A. I is true. B. II is true. C. III is true. D. I and III are true. E. II and III are true. b. [4 points] Consider the following group of statements: I. f(xdx diverges. II. f(xdx converges. III. 3 f(xdx converges. Which of the above statements must be true? Circle ONE of the following choices: A. I is true. B. II is true. C. III is true. D. I and III are true. E. II and III are true. c. [4 points] Consider the following group of statements: I. (f(x + g(xdx converges. II. (h(x + g(xdx converges. III. g(x x dx converges. Which of the above statements must be true? Circle ONE of the following choices: A. I is true. B. II is true. C. III is true. D. I and III are true. E. II and III are true.

9 Math 6 / Exam 2 (March 23, 2 page 9 8. [4 points] Consider the area contained above the line y =.8, and below the curve r = 2 sin(3θ. You may find the following figure helpful. y r = 2sin(3θ y =.8 x a. [4 points] Find the (x, y coordinates for the two points where y =.8 and r = 2 sin(3θ intersect as shown in the figure above. Show enough work to support your answer. We convert y =.8 into r =.8.8 sin θ, then solve for sin θ = 2 sin(3θ, which gives θ.4296, The corresponding coordinates are (.7464,.8 and (.6854,.8. b. [4 points] Write an expression for the area that is specified. You do not need to evaluate your expression. Area = sin 2 (3θdθ sin 2 (θ dθ c. [6 points] Calculate the perimeter that surrounds the specified area. You may round your final answer to two decimal places. The distance along the line y =.8, is =.6. Along the curve r = 2 sin(3θ, x = 2 sin(3θ cos(θ and y = 2 sin(3θ sin(θ. We use this to find dx dθ = 2 sin(3θ sin(θ + 6 cos(3θ cos(θ and dy dθ arc length = (dx 2 + dθ = 2 sin(3θ cos(θ + 6 cos(3θ sin(θ. ( dy dθ The total perimeter is approximately =

10 Math 6 / Exam 2 (March 23, 2 page 9. [4 points] An ice cube melts at a rate proportional to its surface area. Let V (t denote the volume (in cm 3 of the ice cube, and let x(t denote the length (in cm of a side of the ice cube t seconds after it begins to melt. a. [4 points] Write a differential equation for V (t, the ice cube s volume t seconds after it started melting. Your differential equation may contain V, t and an unknown constant k. We know that V = x 3, so x = V /3. The surface area of the cube is given by 6x 2. That gives us dv dt = 6kx2, and substituting x in terms of V, we have dv dt = 6kV 2/3. b. [4 points] The ice cube s initial volume is V >. Solve the differential equation you found in part (a, finding V in terms of t, k, and V. dv Using separation of variables, we have = 6kdt. This gives 3V /3 = V 2/3 6kt + C, and V /3 = 2kt + C, or V = (2kt + C 3. When t =, V = V, which gives us V = C 3, so that C = V /3. The solution is then V = (2kt + V /3 3. c. [6 points] Graph the volume of the ice cube versus time given V ( = V. Be sure to label your axes and any important features of your graph, including the time at which the ice cube has completely melted. The vertical intercept is V = V. The horizontal intercept is t = 2k V /3.

Math 116 Second Midterm March 20, 2013

Math 116 Second Midterm March 20, 2013 Math 6 Second Mierm March, 3 Name: EXAM SOLUTIONS Instructor: Section:. Do not open this exam until you are told to do so.. This exam has 3 pages including this cover. There are 8 problems. Note that the