Name: Discussion Section: ID: This exam consists of 16 questions: 14 Multiple Choice Questions 5 Points Each Free Response Questions 30 Points Total INSTRUCTIONS: Read each problem carefully and answer the question as written. You may use a non-graphing calculator and a standard sized (no larger than 4 6 ) index card worth of notes for the exam, but you may use no other aids. Record your answer to the multiple choice questions on the accompanying answer card. Show your work on the written problems and write clearly the ease with which your answer can be read will be a factor in your grade. Be sure to write your name on the first page of the exam as well as on all pages of the written problems. 1. A particle moves in a straight line and its position at time t seconds is given by the function s t =ln t 4 in meters. What is the average velocity of the particle from t= to t=4? (Pick the closest answer.) (a) 0 m/sec (b) 0.15 m/sec (c) 0.5 m/sec (d) 0.46 m/sec (e) 0.51 m/sec (f) 0.55 m/sec (g) 0.69 m/sec (h) 0.75 m/sec (i) 1.00 m/sec (j) 1.65 m/sec Solution: (d). Average velocity is given by s 4 s = 4 ln 0 ln 8 = ln.5 0.46 meters/sec. Page 1 of 16
. What is lim x 1 x 1 x 1? (a) (b) 1 (c) 1 (d) 0 (e) (f) 1 (g) 1 (h) (i) 1 (j) The limit does not exist. Solution: (e). lim x 1 x 1 x 1 =lim x 1 1 x 1 = 1. Page of 16
3. Which of the following best describes the graph of the function f x = x 1 sin x at the point x = 0? (a) increasing, concave up (b) decreasing, concave up (c) increasing, concave down (d) decreasing, concave down (e) relative minimum (f) relative maximum (g) increasing, inflection point (h) decreasing, inflection point (i) critical point, inflection point Solution: (g). We apply the first and second derivative tests: By the Product Rule, f ' x = x sin x x 1 cos x. Plugging in x = 0, we get f ' 0 =1. By the Product Rule again, f ' ' x = sin x x cos x x cos x x 1 sin x. So, f ' ' 0 = 0 0 1 0 1 1 0=0. We can check that concavity actually changes using, for example, a phase diagram. Page 3 of 16
4. We wish to enclose a 75 square foot rectangular garden. On three sides, we wish to use a brick wall that will cost $10 per foot and, on the remaining side, we will use a wooden fence which costs $5 per foot. Find the cost of the wall for the garden whose dimensions minimize the cost of the wall. (That is, find the cost of the cheapest possible wall given these parameters.) (a) $00 (b) $75 (c) $300 (d) $350 (e) $375 (f) $400 (g) $450 (h) $600 (i) $750 (j) $900 Solution: (c). We start with a rectangle of dimensions x y and let the wooden fence be on one of the x sides. Thus, the total cost of the fence will be C=15 x 0 y and is subject to the constraint equation x y=75. So, y= 75 x. Substituting back into the cost equation gives us C=15 x 1500. This is the equation we want to optimize. x Taking two derivatives gives us: C ' =15 1500 and C ' ' = 3000. x x 3 The second derivative will be positive for all x > 0, and so we know that any critical point will be a minimum. We now find the critical point by setting C ' =0 and getting 15= 1500 x or x =100 or x=10. Plugging that back into our constraint equation, we get y = 7.5. Plugging x = 10 and y = 7.5 back into the cost equation gives C=15 10 0 7.5 =300. Page 4 of 16
5. An analysis of the market for Tchotchkies tells us that as of October 1, 008 the demand for, and price of, Tchotchkies had the following characteristics: dx dx Demand: x = 150, =1.5 (x is in thousands of units, dt dp Price: p = $10, =0.1 (in $/month). dt dt is in thousands of units per month.) At what rate is the total revenue for Tchotchkies changing on October 1, 008? (Answers are in thousands of $/month.) (a) 1.6 (b) 10 (c) 11.5 (d).5 (e) 30 (f) 45 (g) 77.5 (h) 15 (i) 150 (j) 6 Solution: (e). Recall that Revenue is given by: R= x p. So, by the Product Rule: dr dt = dx dt p x dp dt =1.5 10 150 0.1 =15 15=30. So, overall revenues are increasing at a rate of $30,000/month as of October 1, 008. Page 5 of 16
6. On the last trading day of November 1994, the closing stock price of Berkshire Hathaway was $0,55 per share. On the last trading day of November 008, the closing stock price of Berkshire Hathaway was $104,000 per share. During this time there were no stock splits or cash dividends paid. Find the annual rate of return of holding Berkshire Hathaway stock over this 14 year period assuming continuous compounding. (Round to the nearest basis point. A basis point is one-one hundredth of a percentage point.) (a) 11.59% (b) 1.% (c) 13.64% (d) 15.91% (e) 16.1% (f) 17.9% (g) 0.3% (h) 7.17% (i) 33.33% (j) 36.19% Solution: (a). By the formula for continuous compounding, we know that A= P e r t. In this case, A = 104,000, P=0,55, and t = 14 and we are solving for r: So, 104,000=0,55 e 14 r So, e 14 r 5.067 So, r ln 5.067 0.1159 or about 11.59%. 14 Page 6 of 16
7. f x =ln[ 3 x 1 3 x 1]. What is f ' 5? (Select the nearest answer.) x 1 (a) -.7 (b) -1.6 (c) -1.00 (d) 0.00 (e) 0.14 (f) 0.38 (g) 1.00 (h) 1.65 (i).7 (j) 3.08 Solution: (e). By the rules of logs, f x = 1 [ ln 3 x 1 3 x 1 1 / x 1 ] = 3 ln 3 x 1 1 4 ln x 1 1 ln x 1. So, f ' x = 3 3 3 x 1 1 4 x 1 1 x x 1. So, f ' 5 = 9 3 1 18 5 6 0.14. Page 7 of 16
8. Suppose price and demand for J-Pods are related by the equation x= f p =300 e 0.05 p where p is in dollars and x is in thousands of J-Pods. What is the elasticity of demand when p = 150? (a) -0.05 (b) 0.05 (c) 0.17 (d) 0.5 (e) 0.75 (f) 1 (g) 1.5 (h) 4.75 (i) 7.5 (j) 15 Solution: (i). By definition, E p = p f ' p. Notice that f ' p = 0.05 f p so this formula f p becomes: E p = 150 0.05 f p =0.05 150 =7.5. f p Page 8 of 16
9. Suppose f(x) is a function so that f ' x = e x and f(0) =. What is f()? (Give the closest answer.) (a) 0.61 (b) 1.61 (c).14 (d).5 (e).39 (f) 3.14 (g) 3.7 (h) 5.14 (i) 6.39 (j) 11.39 Solution: (h). f x = e x dx= x e x C. We know that = f 0 =0 e 0 C and so C=1. Thus, f x =1 x e x and f =5 e 5.14. Page 9 of 16
10. Estimate the area under the curve y= x 1 from x=0 to x = using a Riemann sum with 4 rectangles and left endpoints. Choose the closest answer. (a) 1.1 (b) 1.4 (c) 1.41 (d) 1.79 (e).00 (f).16 (g).4 (h).41 (i).61 (j) 3.14 Solution: (i). Splitting the interval [0,] into four equal pieces gives us the intervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, ] each of which has length 0.5. The Riemann Sum we want uses the left endpoints and so it is calculated by: [ f 0 f 0.5 f 1 f 1.5 0.5 ]=[ 1 1.5.5 0.5 ].61. Page 10 of 16
11. What is the area below the graph of y = cos x between x = 0 and x=? (a) 0 (b) (c) (d) (e) 1 4 (f) 1 (g) (h) 3 (i) (j) Solution: (f). Since cos x 0 for x between 0 and, this area is given by: A= / / 0 cos x=sin x 0 =sin sin 0=1. Page 11 of 16
1. The demand curve for Whirligigs is given by p=40 0.16 x. Find the Consumer Surplus for Whirligigs at a consumption level of x = 100. (a) $100.30 (b) $133.33 (c) $145.00 (d) $3.67 (e) $66.67 (f) $35.5 (g) $360.00 (h) $400.00 (i) $45.5 (j) $493.33 Solution: (b). If x = 100, then the corresponding price is given by p=40 0.16 100 =36. Thus, the Consumer Surplus is given by: CS= 0 100 [ 40 0.16 x 36]dx= 0 100 4 0.4 x 1 / dx=4x 0.8 3 x3/ 0 100 400 66.67 133.33. Page 1 of 16
13. Find the volume of the solid obtained by rotating the area under the curve y=1 x from x = 0 to x = about the x- axis. (Select the nearest answer.) (a) 3.1 (b) 3.9 (c) 7.8 (d) 1. (e) 4.4 (f) 31.4 (g) 36. (h) 45.7 (i) 51.0 (j) 60.0 Solution: (e). We calculate: / 4 x3 V = 0 1 x dx= 0 1 x x dx= x 3 x 0= 4 8 3 4.4. Page 13 of 16
14. Find x ln 1 x dx. (a) ln ln 1 x C (b) ln x C (c) ln x C (d) (e) (f) (g) x ln 1 x C x ln x C x ln x x C ln x x C (h) x ln ln 1 x C Solution: (b). Observe that by the properties of logs, x ln 1 x dx= ln x x dx. Letting u=ln x, du= dx x ln x x dx= u du= u C = u C= ln x C. x ln 1 ln x =. Thus, x x we get: Page 14 of 16
Name: ID: Discussion Section: Written Problem 1 Instructions: Answer below. Show your work. Write clearly. You may receive partial credit for partially worked-out answers. 15. (15 Points) At time t=0 hours, there are 1,000 bacteria in a petri dish. The bacteria colony grows exponentially and 4 hours later there are 6,000 bacteria present. (a) (5 Points) Express the population in the petri dish as a function of t. Solution: We know that P t =P 0 e k t. We need to find k. So, we have: 6,000= P 4 =1,000 e 4 k. Thus, 4 k=ln 6 or k= ln 6 4 0.45. So, P t =1,000 e 0.45 t. (b) (5 Points) How long does it take the bacteria colony to double in size? Solution: We want to find t so that P(t) = P_0. So, we want t so that P t =,000=1,000 e 0.45 t. This means that e 0.45 t = or t= ln 1.54 hours. 0.45 (c) (5 Points) How fast is the population growing at time t = hours? Solution: P ' t =0.45 1,000 e 0.45 t =450 e 0.45 t. Thus, P ' =450 e 0.9 1,100 bacteria/hr. Page 15 of 16
Name: Discussion Section: ID: Written Problem Instructions: Answer below. Show your work. Write clearly. You may receive partial credit for partially worked-out answers. 16. (15 Points) Consider the area enclosed by the curves y=x and y=8 x. (a) (5 Points) Find all points where these two curves intersect. Solution: Set x =8 x. So either x=0 or x 3 / =8. So, either x = 0 or x = 4. (b) (5 Points) Express the enclosed area as a definite integral or a sum of definite integrals. Solution: Plot points or graph to find that 8 x x on [0,4 ]. Knowing that, A= 0 4 8 x x dx. (c) (5 Points) Find the enclosed area. / 4 Solution: A= 0 8 x x dx= 16 x3 x3 3 3 4 0= 18 3 64 3 0 0 = 64 3. Page 16 of 16