Math 116 Practice for Exam 1

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1 Math 6 Practice for Exam Generated February 2, 25 Name: SOLUTIONS Instructor: Section Number:. This exam has 9 questions. 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. 2. Do not separate the pages of the exam. If any pages do become separated, write your name on them and point them out to your instructor when you hand in the exam. 3. Please read the instructions for each individual exercise carefully. One of the skills being tested on this exam is your ability to interpret questions, so instructors will not answer questions about exam problems during the exam. 4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that the graders can see not only the answer but also how you obtained it. Include units in your answers where appropriate. 5. 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. 6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 7. You must use the methods learned in this course to solve all problems. Semester Exam Problem Name Points Score Winter 2 5 Fall olive oil 3 Winter 23 8 tortoise+hare Fall 2 7 AC current 3 Winter 2 7 paint truck Fall 23 4 shawarma kafta 7 Winter wine glass 2 Fall 24 3 Winter 2 5 ramp 8 Total 8 Recommended time (based on points): 5 minutes

2 Math 6 / Exam (February 8, 2) page 6 5. [ points] For each statement below, circle TRUE if the statement is always true; otherwise, circle FALSE. There is no partial credit on this page. a. [2 points] The function sin x x has an anti-derivative. True False d x 2 b. [2 points] dx x et2 dt = 4x 3 e x4 2xe x2. True False c. [2 points] The average of the function f(x) = x from x = to x = 3 is ln( 3). True False b a d. [2 points] f(x)dx is greater than or equal to at least one of LEFT(n), RIGHT(n), TRAP(n), or MID(n) regardless of what f(x) or n is. True False e. [2 points] If b a f(x)dx > then f(b) > f(a). True False Winter, 2 Math 6 Exam Problem 5 Solution

3 Math 6 / Final (December 7, 23) page 3 9. [3 points] Olive oil have been poured into the Math Department s starfish aquarium! The shape of the aquarium is a solid of revolution, obtained by rotating the graph of y = x 4 for x around the y-axis. Here x and y are measured in meters. The aquarium contains water up to a level of y =.6 meters. There is a layer of oil of thickness.2 meters floating on top of the water. The water and olive oil have densities and 8 kg per m 3, respectively. Use the value of g = 9.8 m per s 2 for the acceleration due to gravity. a. [6 points] Give an expression involving definite integrals that computes the total mass of the water in the aquarium. Mass water =.6 π( 4 y) 2 ()dy =.6 π y()dy b. [7 points] Give an expression involving definite integrals that computes the work necessary to pump all the olive oil to the top of the aquarium. Work oil =.8.6 π( 4 y) 2 (8)(9.8)( y)dy =.8.6 π y(8)(9.8)( y)dy Fall, 23 Math 6 Exam 3 Problem 9 Solution

4 Math 6 / Exam (February, 23) page 3 8. [ points] A tortoise and a hare decide to race. They decide to race a straight 5 kilometer course. The race starts at 2pm. The hare is much faster than the tortoise, so he s confident that he ll win. The hare runs very fast for 3 minutes, getting to what it knows is the half-way point. The hare is tired (it had been studying for exams the night before), so it decides to take a nap. It falls asleep for 5 hours, wakes up, discovers that (now that it s 5:3) it s dark, and runs to the finish line, arriving at 6pm. When it gets there, it s surprised to see the tortoise is already there. I hope you enjoyed your nap! I ve been here for an hour, since 5 o clock! the tortoise says. Steady and slow is the way to go: I kept going the same speed the whole time. Let H(t) be the hare s velocity and T(t) be the tortoise s velocity, in km per hour, where t is measured in hours after 2pm. Let R(t) = t H(s)ds t T(s)ds. a. [ point] At times when R(t) >, who is winning the race? The hare b. [2 points] What is the practical interpretation of the function R(t)? Include units. R(t) is the distance in km between the tortoise and the hare t hours after 2pm. c. [3 points] For what values of t 6, does R(t) =? t =,t = 2.5,t = 6. d. [2 points] For what values of t 6 is the function dr dt <?.5 < t < 5. e. [3 points] Write down a definite integral that represents the hare s average velocity from 2 to 2:3. What is the value of the hare s average velocity during this time? /2.5 H(s)ds. We know that /2 H(s)ds = 2.5, because the Hare has gotten halfway by 2:3. Therefore, the average velocity is 5 km/hr. Winter, 23 Math 6 Exam Problem 8 Solution

5 Math 6 / Exam (October 2, 2) page 9 7. [3 points] Household electricity in the United States is supplied in the form of an alternating current that varies sinusoidally with a frequency of 6 cycles per second (Hz). The voltage is given by the equation E(t) = 7 sin(2πt), where t is given in seconds and E is in volts. a. [7 points] Using integration by parts, find sin 2 θdθ. Show all work to receive full credit. (Hint: sin 2 θ +cos 2 θ =.) We first note that sin 2 θdθ = sinθ(sinθ)dθ, so that we may take u = sinθ,dv = sinθdθ (and du = cosθdθ,v = cosθ). Then integration by parts gives sin 2 θdθ = sinθ( cosθ) cosθ(cosθ)dθ = sinθcosθ + cos 2 θdθ. Using the trig. identity given in the hint, we obtain ( sin sin 2 θdθ = sinθcosθ + 2 θ ) dθ = sinθcosθ + dθ sin 2 θ. The integral on the far right also appears on the left, so combining like terms, we get 2 sin 2 θdθ = sinθcosθ + dθ sin 2 θdθ = ( ) sinθcosθ + dθ 2 = 2 sinθcosθ + θ 2 +C. b. [6 points] Voltmeters read the root-mean-square (RMS) voltage, which is defined to be the square root of the average value of [E(t)] 2 over one cycle. Find the exact RMS voltage of household current. Fall, 2 Math 6 Exam Problem 7 (AC current) Solution

6 Math 6 / Exam (October 2, 2) page Since the frequency of the current is 6 cycles per second, one cycle is completed every 6 seconds. Thus RMS voltage = 6 E(t) 2 dt = Substituting w = 2πt,dw = 2πdt, we get w( RMS voltage = 6 = 7 2 2π w() 6 ) 2π 7 2 sin 2 (2πt)dt. 7 2 sin 2 (w) sin 2 (w)dw. 2π dw Using the antiderivative we found in part (a) with C =, the Fundamental Theorem of Calculus gives 7 RMS voltage = 2 ( 2π 2 sinθcosθ + θ ) 2π 2 = 7 2 2π (( 2 sin(2π)cos(2π)+ 2π2 ) ( 2 sin()cos()+ 2 ) ) = 7 2 Volts. Note that due to the periodicity of the sine function, the average value over one cycle could also have been computed over t (or any other number of periods). Fall, 2 Math 6 Exam Problem 7 (AC current) Solution

7 Math 6 / Exam (February 2) page 9 7. [ points] A truck carrying a large tank of paint leaves a garage at 9AM. The tank starts to leak in such a way that x miles from the garage, the density of paint on the road is e x2 /5 gallons per mile. At AM, a cleaning crew leaves from the same garage and follows the path of the truck, scrubbing the paint from the road as it travels until it catches up to the leaking truck. At t hours after AM, the leaking truck is 5ln(t+2) miles from the garage, and the cleanup crew is 35t miles from the garage. You may use your calculator to evaluate any definite integrals for this problem. a. [4 points] Calculate the total amount of paint that has leaked from the truck by AM. The truck at AM is at 5ln3 = miles from the garage. Total amount of paint leaked from the truck is 5ln3 e x2 /5 dx = gallons. b. [2 points] At time t hours after AM, what interval I of the road is still covered in paint? (you may assume that t represents a time before the trucks meet) The truck is at 5ln(t+2) miles from the garage and the crew is at 35t miles from the garage. I = [35t,5ln(t+2)]. c. [3 points] Let P(t) represent the amount of paint in gallons on the road t hours after AM. Find a formula (which may include a definite integral) for P(t). P(t) = 5ln(t+2) 35t e x2 /5 dx d. [2 points] Calculate P (). ( P (t) = e (5ln(t+2))2 /5 5 t+2 P () = e (5ln(3))2 /5 ( 5 3 ) 35 ) 35 (e ) (35t)2 /5 ( e (35)2 /5 ) = gal/hr Winter, 2 Math 6 Exam Problem 7 (paint truck) Solution

8 Math 6 / Exam (October 9, 23) page 5 4. [7 points] a. [8 points] The delicious chicken shawarma platter is served on an elliptical plate, described by the equation x 2 +4y 2 = 4. The mass density of the platter, including the food, is a function of y, given by δ(y) = +.5y grams per cm 2. In this problem, you do not need to evaluate any integrals. i) (4 points) Find an expression containing a definite integral that computes the mass of the chicken shawarma platter (including the food). m = 2 4 4y 2 (+.5y)dy. ii) (4 points) Find expressions for the coordinates x, ȳ of the center of mass of the platter. If your expression does not involve an integral, include a justification. x =, since both the shape and density function are symmetric about the y-axis. ȳ = y 2 4 4y 2 (+.5y)dy 2 4 4y 2 (+.5y)dy. Fall, 23 Math 6 Exam Problem 4 (shawarma kafta) Solution

9 Math 6 / Exam (October 9, 23) page 6 b. [9 points] The mouthwatering kafta kabob platter is served on a circular plate, with radius 2 cm. Including the food, the overall mass density of the platter is given by δ(r) = 5 2+r 2 grams per cm 2, where r is the distance from the center of the plate (in cm). i) (4 points) Write a definite integral that computes the mass of the kafta kabob platter (including food). You do not need to evaluate the integral. m = 2 2πr 5 2+r 2dr. ii) (3 points) Write an estimate for your expression in part i) of the mass of the platter using LEFT(3). Show all the terms in the sum. You do not need to evaluate the sum. LEFT(3) = 2 3 = 2 3 ( 5 2 2π +2π ( ) = ( 2 3 )2 +2π ( 4 3 )2 ) iii) (2 points) Where is the center of mass of this platter? Justify. At the center of the plate, since both the shape and density function are symmetric about the origin. Fall, 23 Math 6 Exam Problem 4 (shawarma kafta) Solution

10 Math 6 / Final (April 2) page 9 7. [2 points] a. [5 points] You rotate the region shown about the y-axis to create a drinking glass. Write an expression that represents the volume of material required to construct the drinking glass (your answer may contain f(y)). 3 x = ( ) y /5 3 x = f(y) -3 Volume = 3 ( π f(y) 2 ( ) y 5 dy + πf(y) 3)2 2 dy. 3 b. [7 points] Consider the vessel shown below. It is filled to a depth of foot of water. Write an integral in terms of y (the distance in ft from the bottom of the vessel) for the work required to pump all the water to the top of the vessel. Water weighs 62.4 lbs/ft 3. 5 ft. 3 ft. 2 ft. ft. y Using similar triangles: Volume of a slice=5 ( 3 y) y 2 Work= 5( 3 y) (62.4)(2 y)dy = 468y(2 y)dy. 2 Winter, 2 Math 6 Exam 3 Problem 7 (wine glass) Solution

11 Math 6 / Exam (October 8, 24) page 2. [3 points] Let g(x) be a differentiable, odd function and let G(x) be an anti-derivative of g(x) with G(2) =. A table of values for g(x) and G(x) is provided below. Be sure to show all of your work. x g(x) G(x) a. [2 points] Write down a formula for G(x) in terms of the function g(t). x G(x) = 2 g(t)dt b. [2 points] Compute g(x)dx. g(x)dx = G() G() = 4 ( 7) = 3 c. [3 points] Compute 2 4 g(x)dx. 2 4 g(x)dx = = g(x)dx g(x)dx = (G(4) G(2)) = G(2) G(4) = 9 d. [3 points] Compute 3 3 xg (x)dx. xg (x)dx = xg(x) 3 3 g(x)dx = 3g(3) g() (G(3) G()) = 2 2 (5 ( 4)) = e. [3 points] Compute g(3x)dx = /3 g(3x)dx. 3 g(x)dx = /3(G(3) G()) = /3(5 ( 7)) = 4 Fall, 24 Math 6 Exam Problem Solution

12 Math 6 / Exam (February 2) page 7 5. [8 points] A company wants to design a bicycle ramp using the shape of the graph of the function f(x) = 4 3 x3 2,where x is the length in meters of the base of the ramp. f x s Ramp L x Find the length s of a ramp with base of length L. Show all your work. s = = = L L L +(f (x)) 2 dx +(2 x) 2 dx +4xdx = 6 (+4x)3/2 L = 6 (+4L)3/2 6 Winter, 2 Math 6 Exam Problem 5 (ramp) Solution