MATH 151, FALL SEMESTER 2011 COMMON EXAMINATION 3 - VERSION B - SOLUTIONS

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Name (print): Signature: MATH 5, FALL SEMESTER 0 COMMON EXAMINATION - VERSION B - SOLUTIONS Instructor s name: Section No: Part Multiple Choice ( questions, points each, No Calculators) Write your name, section number, and version letter (B) of the exam on the ScanTron form. Mark your responses on the ScanTron form and on the exam itself. The electric field between parallel plates produces the acceleration at 5 cost cm/sec on a charged particle, where t is in seconds. If the particle has initial position x0 cm and initial velocity v0 cm/sec, where is the particle at time t sec? a. x cm b. x cm c. x cm Correct Choice d. x 8 cm e. x 8 cm vt 5 sint C v0 C vt 5 sint xt 5 cost t K x0 5 K K 8 xt 5 cost t 8 x 5 cos 8. Find all solutions of lnx lnx ln5. a. x 5, only b. x 5, 0 only c. x only Correct Choice d. x 5 only e. x 0, only lnxx ln5 x x 5 0 x 5x 0 x 5 : ln5 is not defined, so not a solution. x : ln ln ln5. Compute arcsin cos 5. a. Correct Choice b. c. d. e. 5 5 cos 5 because 5 is in quadrant III. arcsin cos 5 arcsin arcsin.

. Compute ln x ln5 x a. 0 b. ln 5 c. ln d. ln ln 5 e. ln ln Correct Choice ln x ln5 x ln x 5 x ln x 5 x ln 5. A shrimp farm starts off with 500 shrimp. The number of shrimp quadruples after 8 weeks (to 000 shrimp). Assuming an exponential growth, how many shrimp are there after weeks? a. 000 b. 000 Correct Choice c. 500e 6 d. 500e /6 e. 500eln P P 0 t/ P 500 /8 500 / 500 000 6. If fx ln x / arctanx, then f x a. x x b. x c. x x d. x x Correct Choice e. x fx ln x arctanx 7. Find the line tangent to y ln x a. b. c. Correct Choice d. e e. e df dx at x e. Its y-intercept is fx ln x fe ln x ln e f x ln x x f e xln x eln e y fe f ex e e x e e x e x x x x x

8. Compute the derivative of fx x x. a. xx x b. xx x x c. x x lnx d. x x lnx x e. x x lnx x x Correct Choice fx x x e ln x x e xln x f x e xln x d dx x lnx e xln x lnx x x x x lnx x x 9. Find the locations of the absolute maximum and absolute minimum of fx x 5x 8x on the interval 0, 5. a. Abs Min at x 5 Abs Max at x 0 b. Abs Min at x Abs Max at x 0 Correct Choice c. Abs Min at x 0 Abs Max at x 5 d. Abs Min at x Abs Max at x e. Abs Min at x Abs Max at x 5 f x x 0x 8 x x Critical point at x. (x is not in the interval.) Endpoints at x 0, 5. f0 0 f 6 80 8 f5 5 5 0 0 0. For fx sinx x x which of the following is TRUE? a. The First Derivative Test says x 0 is a Local Maximum b. The Second Derivative Test says x 0 is a Local Minimum c. The Second Derivative Test says x 0 is a Local Maximum d. The Second Derivative Test says x 0 is an Inflection Point e. The Second Derivative Test Fails at x 0. Correct Choice f x cosx x f 0 0 f x sin x 6x f 0 0 The Second Derivative Test Fails.

. The plot at the right shows the graph of the FIRST DERIVATIVE f x of a function fx. Which of the following is FALSE? f ' (x) 0 0 a. fx has a local maximum at x 0 Correct Choice b. fx is increasing on 0, c. fx is increasing on, d. fx has an inflection point at x e. fx is concave up on, - 5-0 x (a) is false because fx has a local minimum at x 0 because to the left of 0, f is decreasing: f x 0 and to the right of 0 f is increasing: f x 0. (b) is true because f x 0 on 0, (c) is true because f x 0 on, (d) is true because to the left of f is increasing: f x 0.and to the right of f is decreasing: f x 0. So the concavity changes. (e) is true because f is increasing: f x 0. Compute x0 e x x x a. b. c. 0 d. Correct Choice e. e x x x0 x Apply l Hospital s Rule twice: l H e x l H e x x0 x x0 5. Compute i. i a. 56 b. 58 c. 60 d. 6 Correct Choice e. 6 5 i 5 6 i

Part Work Out Problems (5 questions. Points indicated. No Calculators) Solve each problem in the space provided. Show all your work neatly and concisely, and indicate your final answer clearly. You will be graded, not merely on the final answer, but also on the quality and correctness of the work leading up to it.. (6 points) Find the values of a and b so that fx ax b lnx will have an inflection point at, 5. f a 5 fx 5x b lnx f x 0x b x f x 0 b x f 0 b 0 b 0 Note: To be an inflection point, not only must f 0, but the concavity must actually change. In fact when a 5 and b 0, we have f x 0 0 0 x which is negative x x for 0 x and positive for x. 5. (8 points) Compute L x x x L x x x e ln x x x e x ln x x e P where P x ln x x ln x x x We can apply l Hospital s Rule because the numerator and denominator go to zero. We then multiply top and bottom by x. x 6 x P l H x x x 6 x x x So L e P e ALTERNATIVE SOLUTION FOR P : P x ln x x t0 ln t t t l H t0 6t t t 5

6. (0 points) A paint can needs to hold a liter of paint (000 cm ). The shape will be a cylinder of radius r and height h. The sides and bottom will be made from aluminum which costs $0. 0 per cm. The top will be made from plastic which costs $0. 05 per cm. What are the DIMENSIONS and COST of the cheapest such can? (Keep answers in terms of.) V r h 000 h 000 r C. 0rh r. 05r. rh. 5r 00 r. 5r C 00. 5r 0. 5r 00 r 00 r r r C 00 r. 5r 00 OR C. rh. 5r 5 h 000 000. 5 5. 5 5 To see that C is the absolute minimum, we note there is only one critical point and C 00 r. 5r approaches as r approaches 0 or. 7. (8 points) Derive the formula d dx arcsinx x. Note: The inverse function of sinx is denoted by either arcsinx or by sin x. SOLUTION : If y arcsinx then x siny. Use implicit differentaition to differentiate both sides with respect to x: cosy dy dy So dx dx But since y arcsinx, this says: cosy d dx arcsinx cosy cosarcsinx To get this in the correct form, if siny x then cosy sin y x. Notice we use the positive square root because y arcsinx is between and cosy is positive. So dx d arcsinx cosy x so that SOLUTION : If b arcsina then a sinb. The formula for the derivative of the inverse function says: dg d arcsinx So dx xa df dx xa d sinx cosx xb cosarcsina dx xb dx xb 6

8. (6 points) For the function fx xe x find a. all intervals where f is increasing or decreasing. f x e x xe x xe x. critical point: x For x, f x 0 and f is decreasing. For x, f x 0 and f is increasing. b. all intervals where f is concave up or concave down. f x e x e x xe x xe x. inflection point: x For x, f x 0 and f is concave down. For x, f x 0 and f is concave up. c. all local minima, local maxima, absolute minima and absolute maxima (give location and value). Since f is decreasing on the left of and f is increasing on the right of, there is a local and absolute minimum at x and f e x. Since there are no other critical points and no endpoints, there are no local or absolute maxima. d. all horizontal asymptotes (be sure to compute the its). x e x e no horizontal asymptote as x. x xex 0" indeterminate form. Apply l Hospital s Rule: x xex lh x e x x e x x ex 0 So y 0 is the horizontal asymptote as x. 7