Mathematics 105 Calculus I. Exam 1. February 13, Solution Guide

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1 Matematics 05 Calculus I Exam February, 009 Your Name: Solution Guide Tere are 6 total problems in tis exam. On eac problem, you must sow all your work, or oterwise torougly explain your conclusions. Tere is always at least one step preceding a final answer. Units may be requested for your final answer; a point deduction will apply if tey are omitted. On te portion of te exam marked NO CALCULATOR, you will be allowed 0 minutes during wic your calculator must be closed and put away. If you finis tis tion early, you may and in your work early. However, only after you and in te no calculators tion will you be permitted to use a calculator. You may not return to te no calculator portion after anding it in. Before beginning, ensure your calculator is set to Radians mode. You will ave 80 minutes to complete tis exam. Question Point Value Your Score No Calc Total 50

2 NO CALCULATOR PORTION Mat 05-A (Salomone) Exam Sow all your work! Name: Score (50 possible): Problem -NC. (5 points) Use te limit definition of derivative to compute f (x) for te function f (x) x e x. e Hint: Simplify using properties of exponentials. You will also need to know tat lim. 0 Te definition of f (x), if it exists, is tat it takes te value given by te limit lim 0 f (x + ) f (x) (x + )e x+ xe x lim 0 xe x+ + e x+ xe x lim 0 xe x e + e x e xe x lim 0 lim xex 0 ( e ) + e x e lim xe x e + e x e 0 Now we just need to determine te beavior of te igligted terms above. We are given in te problem tat te first term approaces. Te ond term does as well, since e is a continuous function of and tis allows us to substitute 0. Tus f (x) lim 0 f (x + ) f (x) lim xe x e +e x e 0 }{{} }{{} xe x + e x.

3 Problem -NC. (5 points) In tis question, you will determine enoug properties of te function f (x) ex + e x e to be able to grap it. Note: remember, e is just a number. Its particular value, is not important to tis problem. (a) (5 points) Wat is te vertical intercept of tis function? Te vertical intercept is just te value of f (0): f (0) e(0) + e 0 e e. (b) (7 points) Determine on wat interval(s) tis function is increasing, and on wat interval(s) it is decreasing. f f f To do tis, we need to know were te derivative f is positive and were it s negative. We can find tis by first determining were f is zero: f (x) e e x 0 e e x x If x is very large and positive, e x 0 and f (x) e 0 is positive. If x is very large and negative, e x + and f (x) e is negative. decreasing:(, ) increasing:(, ) (c) (5 points) Is tis function always concave up? Always concave down? Neiter? Wy? f f f Tis depends on te sign of te ond derivative. But f (x) e x is always a positive quantity; tus f is always concave up. y f x (d) (8 points) On te axes provided, sketc te grap of f using your answers to parts (a) (c).

4 Mat 05 Exam Problem. (5 points) Tis problem concerns te function (a) (8 points) Determine te domain of tis function. g(t) (t )(t t + ) t t +. Tis function will be defined anywere, so long as its denominator is not zero: Te domain consists of all real numbers except and : (, ) (, ) (, ) t t + 0 (t )(t ) 0 t 0 and t 0 t and t or { t R : t and t }. (b) (0 points) Using algebra, compute lim t a g(t) for eac value of a not in te domain of g. Explain wat eac result means about te continuity of g. According to te result of part (a), we must compute lim g(t) and lim g(t). We begin by factoring te t t denominator: g(t) (t )(t t + ) t t + (t ) (t )(t ) t We can use tis simpler form to compute te limits we want; after all, limits only see te beavior of g(t) NEAR te t values in question. lim g(t) t t t + lim t t ±0 ± Te grap of g(t) as a vertical asymptote at t. t t + lim g(t) lim t t t Te grap of g(t) as a ole at t. y g(t) t (c) (7 points) At left is a partial grap of g(t). Fill in te gap, clearly indicating te nature of any discontinuities.

5 Mat 05 Exam Problem. (5 points) A weigt is attaced to a spring and suspended in a container of motor oil. If it is allowed to oscillate, its vertical position (measured in cm above equilibrium) as a function of time t in onds migt be given by te function p(t) e t cos t. (a) (0 points) Complete te data table below, and use your results to estimate te values of p (0.9), p (), and p (.). Include units in your answers. t () p (cm) Estimate te derivatives using average rates of cange: p (0.9) p(0.95) p(0.85) cm 0.7 cm.7 cm p () p(.05) p(0.95) cm 0.55 cm.55 cm p (.) p(.5) p(.05) cm 0. cm. cm (b) (0 points) Use your answers to part (a) to estimate p (), wit units. Wat does tis answer mean in practical terms? Again, we ll use an average rate of cange but not te rate of cange of p, rater te rate of cange of p. p () p (.) p (0.9) cm/ cm/ cm/ (c) (5 points) Is it reasonable, based on your answers, to expect tat p(t) satisfies te differential equation Wy or wy not? p + p p? We cannot predict tis for all values of t, since tis would require us to compute p and p but we don t know ow to do tat just yet. (Tis will require someting called te product rule. ) Instead, let s see weter our estimates from parts (a) and (b) indicate tat tis differential equation is satisfied at t. Te left-and side gives Meanwile, te rigt-and side is p ().85 + p () (.55) p () + p ().85 + (.55).05 p() (0.596).96 Tese numbers are reasonably close to one anoter; witin roundoff and estimation error, te differential equation appears to be satisfied at t. (Answers may vary.)

6 Mat 05 Exam y x Problem. (5 points) Sown at left is a grap of te derivative of a function. Use te grap to answer te following questions. (a) (6 points) On wat interval(s) is (x) decreasing? f f f will be decreasing werever is negative: (, 0) (0, ) (b) (6 points) List te x values of all local minimum and local maximum point(s) of, and justify your answers in one sentence. f f f will ave a local minimum werever switces from negative to positive. Tis appens at x. f f f will ave a local maximum werever switces from positive to negative. Tis appens at x, even toug does not exist at x! (c) (6 points) Te grap of is concave up on te interval (.5,.0). Wat does tis mean about te grap of (x) on tat interval? y x f f f Te answer: absolutely noting at least, noting immediately visible. Te concavity of will tell us weter is increasing or decreasing, and ence weter te grap of is getting more or less concave. Tis is nearly impossible to spot wit te naked eye. (d) (7 points) Using te axes provided at left, sketc a possible grap of te function (x). Te vertical scale is not important, only te sape and orizontal location.

7 Mat 05 Exam Problem. (5 points) Te latest press booklet for te 009 Lotus Exige S-0 sports car claims it can accelerate from 0 60 mp in.0 onds flat. A recent test-track run sowed tat under full trottle, te velocity of te car is modeled by te function v(t) 0 t 5t, were v is measured in mp and t in onds. (a) (5 points) According to tis model, after ow many onds will te car reac its maximum velocity, and wat is te maximum velocity? Note: do tis symbolically, sowing your work. You may include a grap or data table if you wis, but your answer must be exact. f f f We wis to find a local maximum of v(t) a point at wic v (t) 0 and v (t) is negative. v (t) 0 t / t 5 t t 6 v (t) 0 t / 0 t / v (6) 0 6 / 0 6 < 0 According to tis work, te car s velocity as a local maximum after t 6 onds. Te car s velocity at tis point will be v(6) 0 6 5(6) 80 mp. (b) (0 points) Determine te car s distance function d(t) an antiderivative of its velocity and use it to find te distance te car traveled during te first 0 onds of tis time trial. Note: write out te units of te antiderivative in your answer. Convert tem if you wis. An antiderivative of our function v(t) will be d(t) 0 t/ / 5t +C 80 t/ 5 t +C Does te +C matter? Not if we only care about ow far te car travels between t 0 and t 0, as measured by te difference d(0) d(0). (Te +C will cancel out.) We may as well drop it, or assume tat d(0) 0. We ten interpret d(t) to be te distance te car as traveled since t 0, and te total distance during te first 0 onds of te test will be merely d(0) 80 (0)/ 5 (0) 59.7 mp Te units of tis answer will be te product of te units of input and output of v namely, mp. To convert, we remember tat tere are 600 onds in one our: 59.7 mi r mi mi. 600