Math 132, Spring 2003 Exam 2: Solutions

Transcription

1 Math 132, Spring 2003 Exam 2: Solutions No calculators with a CAS are allowed. e sure your calculator is set for radians, not degrees if you do any calculus computations with trig functions. Part I, Multiple Choice, 5 points/problem: blacken your answers on the answer card. _ 1. Find the value of ' & /.! A) ) / C) / E) diverges to _ D)!/ &! & F) / G) / H) 2 / I) 2e J) diverges to _ Let? & ß.?!., so.?.þ Then ' &! /. '?? & /.? / G / GÞ!!! Then ' _ > >Ä_ > & &! & ' & & > /. lim /. lim / l lim Ð/ / Ñ / Þ >Ä_ >Ä_!! 2. For a certain radioactive isotope, the amount remaining after 3 hours is 78 of the original amount. What is its half-life? ( Round your answer to 2 decimal places. ) A) 6.21 hrs ) 8.37 hrs C) 8.85 hrs D) 9.12 hrs E) 9.24 hrs F) 9.56 hrs G) 9.75 hrs H) 9.93 hrs I) hrs J) hrs.c.>! 5> Let C amount of isotope present at time >. Then 5Cand CC/ ÞSince!! $5 $5 lnð!þ()ñ $ 0.78C C / ß we get that 0.78 /, so 5Þ 52 ln If 2half-life, we have C! C/! ÞThen lnð Ñ ln 52, so 2 5 $ ln )Þ$(hrs. lnð!þ()ñ

2 3. Suppose we perform the partial fraction decomposition What is the value of F? $ E F G Ð ÑÐ Ñ A) F! ) F C) F D) F E) F F) F G) F H) F I) F J) F $ $ $ $ E F G EÐ ÑÐ Ñ FÐ Ñ GÐ Ñ Ð ÑÐ Ñ Ð ÑÐ Ñ Equating numerators gives Þ $ EÐ ÑÐ Ñ FÐ Ñ GÐ Ñ Setting gives $ $F, so F Þ 4. During Lab 2, LoggerPro software collected the following data for > (time, in sec) (velocity, in m/sec). Use Simpson's Rule to estimate the change in position (displacement) of the moving person for! Ÿ > Ÿ!ÞÞ ( Round your answer to 2 decimal places.) A)!Þ! m )!Þ! m C)!Þ!' m D)!Þ! m E) 0.13 m F)!Þ& m G)!Þ& m H)!Þ m I)!Þ$ m J)!Þ22 m The displacement is given by The interval Ò!ß!ÞÓ is divided into 4 subintervals, with? > 0.1. Simpson's approximation is '!!Þ!Þ W Ð!Þ& Ð!Þ!Ñ Ð!Þ!Ñ Ð!Þ!Ñ Ð!Þ!ÑÑ!Þ! m

3 ' /! / 5. Find the value of. (if the integral converges). / / A) ) C) D) / E) diverges to _ / / F) G) / H) I) / J) diverges to _ / / The integral is improper (type II) because / / has a vertical asymptote at!þ If we let?/ ß.?/.ß so ' /.'.? lnl?l G ln l/ l GÞ Then ' / /! / >Ä! > / > /?. lim '. lim ln l/ l l lim lnð/ Ñ lnðe Ñ _Þ >Ä! >Ä! > 6. Suppose a radioactive isotope is emitting radiation at a rate of!þ!/ > millirems/min, beginning at time >!. What is the total amount of radiation it will emit over all time that is, if the process continues forever? A)!Þ! millirems )!Þ!!& millirems C)!Þ! millirems D)!Þ!& millirems E)!Þ! millirems F)!Þ!& millirems G) millirems H) 10 millirems I) 40 millirems J) an infinite amount _ +! +Ä_! > + lim +! +Ä_ Total amount emitted ' >!Þ!/.> ' > lim!þ!/.>!þ! lim / l!!þ! Ð / / Ñ!Þ!Ð! Ñ!Þ! ÐmilliremsÑ +Ä_

4 ' 7. You use Simpson's rule with 8! to estimate the value of.. Using the error ' control formula, we get ERROR l. W! l Ÿ??? ( Find the best possible answer with this information; round your answer to 6 decimal places.) A) ) C) D) E) F) G) H) I) J) ' QÐ Ñ )!Ð!Ñ l. W l Ÿ, where Q is chosen so that l0 ÐÑl Ÿ Q on Òß Ó.! & w ÐÑ & We have 0ÐÑ ß0 ÐÑ ß ÞÞÞß0 ÐÑ &. Since & is positive ÐÑ and decreasing on Òß Ó, 0 ÐÑ has its max value at the point ß so ÐÑ $ 0 ÐÑÓ Ÿ. This is the best choice of Q we can make. $ ' $ & Ð Ñ! )!Ð!Ñ )!Ð!Ñ Therefore l. W l Ÿ!Þ!!!!$ ÐÑ 8. According to one simple physiological model, an adult male athlete needs 20 calories per day per pound of body weight to maintain his weight. If he consumes more or fewer calories than those required to maintain his weight, his weight changes at a rate proportional to the difference between the number of calories consumed and the number needed to maintain his current weight. If an athlete weighs 160 lbs. and consumes 3000 calories/day, which differential equation describes [Ð>Ñ the athlete's weight at time >? ( In each equation, k is a constant. ).[.[.[.>.>.> A) 5Ð 3000 [!Ñ ) 5Ð[!Ñ C) 29)! 5[.[.[.[ $!!!.>.>.> [ D) 5Ð150 8 [ Ñ E) 5Ð3000![ Ñ F) 5Ð![ Ñ.[ [.[ [.[.>!.> '!.> G) 5Ð3000 Ñ H) Ð$!!! Ñ I) 5Ð160![ Ñ.[ J).> 5Ð [Ñ.[.> 5Ðnumber of calories consumed number needed to maintain current weight [Ñ 5Ð$!!!![ Ñ

5 9. Find the area in the first quadrant bounded by the curves Cln, C and 2. $ & Ð Ñ A) 4 ln ) ln C) 1 D) ln E) FÑ ln G) ln $ H) 2 ln I) 2 ln J) $ ln The figure shows the positions of the curves and the region (shaded): The area of the region is E' Ð Ñ ln.. Using integration by parts, we get ' ln. ' ln. ln '.ln GÞ Then ' Ð Ñ ln. Ð Ð ln Ñ Ñl Ð ln Ñ l $ Ð ln Ñ Ð!Ñ ln Þ

6 '! 10. Find W ( the Simpson approximation with 4 subdivisions) for.þ!( ( &$ )$ *$ A) ' ) ' C) ' D) ' E) ' F) && G) )) H) *( I) ( J) (! $ $ $ $ $! Method I (longer): We have?, so W $ ÐÐ!!Ñ Ð Ñ Ð Ñ Ð$ $Ñ Ð ÑÑ )) Ð! ) )!Ñ Ð *Þ$$Ñ $ $ Method II (shorter): Simpson's Rule approximates the integrand with pieces of parabolas. Since the integrand is itself a parabola, Simpson's approximation W (with any choice of even 8) will actually give the exact value of the integral: W '! $ ' ' ) ) )). Ð Ñl Þ $! $ ' $ Suppose the (m/sec) of a point moving along a line satisfies.= $=.> > Ð>!Ñ =ÐÑ 1 What is the position = of the particle when > *? A) m )! m C) & m D)! m E) 35 m F) 27 m G) 22 m H) 17 m I) 12 m J) 8 m Separating variables in the d.e. and integrating both sides, we get ' $ $Î H $Î ln l=l ln l>l H ln l>l H, so that l=l / l>l. Then H $Î $Î H = / l>l Gl>l (where G / ÑÞ '.= $.> = >, so Since =ÐÑ, we have G, that is, G. Therefore =l>l $Î, so $Î =Ð*Ñ* ( (m)

7 12. The temperature X of a body (C ) at time > (min) is changing at a rate of.x X.> > >. The slope field ( direction field ) for the differential equation is shown below ( some solution curves were added here on the solutions sheet. ). Over the time interval & > &, which of the following are true ( you may need to use either the slope field or the differential equation itself to answer). i) If the temperature is 3 when >, then the body is cooler when >. ii) No matter what the temperature is when >, the body is cooler after time >. iii) The temperature of the body at time > depends only on >, not on the initial temperature X! of the body. iv) If the initial temperature of the body is 1 ß then the body never changes temperature. A) only i) ) only ii) C) only iii) D) only iv) E) only i), ii) F) only i),iii) G) only i), iv) H) only ii), iii) I) only ii), iv) J) only iii), iv)

8 i) is true: look at the curve Solution 1 in the figure ii) is false: look at the curve Solution 2 in the figure iii) is false: In the figure, Solution 1 and Solution 2 (just for example) show different temperatures for the same time ( >, say). They differ at > because they had different initial temperatures. X! iv) is true: you can see (approximately) the figure that the solution curve through Ð!ß Ñ is horizontal. More rigorously: the constant curve X has X! XÐ!Ñ and it obviously satisfies the d.e., (plug it in!) 13. Using Euler's method, with stepsize 2Ð.Ñ!Þ&, estimate CÐ0.5 Ñif.C. C C CÐ!Ñ (Round your answer to 2 decimal places. The slope field pictured below is not needed for solving the problem but if you understand it, it will help you narrow down your choices.)

9 A) ) C) D) E) F) G) H) I) J) The slope field (with solution curve through Ð!ß Ñ now included on the solution version) indicates that CÐ!Þ&Ñ should be just a little larger than Þ ( This rules out answers E-J ). Format 1 for solution: for Euler's method we have 2.!Þ& and (in text's.c notation) the d.e.. JÐßCÑC CÞ We begin at Ð! ßC! ÑÐ!ßÑ. The next two steps in Euler's method would give us Ð, CÑÐ ßCÑ and ÐßCÑÐ ßCÑ, and we need to know C Þ Euler's method gives C C! 2J Ð! ß C! Ñ J Ð!ß Ñ Ð!Ñ, and then C C 2J Ð ß C Ñ ÐJ Ð ß ÑÑ Ð Ñ Þ!'& ' Format 2 for solution: C..CÐC CÑ.. C.C!! (!Þ& Þ!'& ' '

10 ' / w 14. Suppose that 0 ÐÑ. and that 0 Ð/Ñ!. What is the value / of ' ww 0 ÐÑln.? ( Hint: integration by parts. ) w A) ) C)! D) E) 1 $ & ( F) G) 2 H) I) 3 J) Let?ln ß.@0 ÐÑ. 0 ÐÑ. ww Using the formula for integration by parts gives ' / / ww w / ln ln ' w 0 ÐÑ. Ð ÑÐ0 ÐÑÑl 0 ÐÑ.. Substituting gives ' / ww w w 0 ÐÑln. (ln /Ñ 0 Ð/Ñ Ð ln Ñ 0 ÐÑ w!! 0 ÐÑ Þ Part II, True or False, 1 point each. lacken your answers on the answer card. 15. In Lab 2 you used a motion detector to gather position and velocity data. When you backed away from the sensor, LoggerPro recorded negative velocities. A) True ) False False: LoggerPro measured distance from the sensor. When you backed away, this distance was increasing, so your velocity was positive.

11 _ '. É È 16. We can conclude that diverges by comparing it to the _ integral ' È.Þ A) True ) False True: since in these integrals, we know that È Ÿ, so Þ We know '. ' Î. diverges (why?). É È È È È È _ Therefore because ' _ '. É È È. diverges by the Comparison Test., we conclude that the first integral.a A 17. Let A0Ð?Ñbe a solution of the differential equation.??. Then 0ÐÑ 0Ð$ÑÞ A) True ) False.A A.A False: From the d.e..??, we see.? is always positive. Therefore any solution A 0Ð?Ñ is an increasing function so 0ÐÑ 0Ð$ÑÞ _ 18. ' $.> converges if and only if 5!Þ Ð>Ñ 5 A) True ) False True: ' Ð>Ñ > > ' $ ' $ $ '.>.>.>Þ ut we know that >.> converges if an only if : 5, that is, if and only if 5!Þ : 19. Let C be the amount of a radioactive isotope present at time >. If the isotope has halflife of 0.39 years, then ( rounded to 2 decimal places).c.> 1.78 CÞ A) True ) False.C.>! 5> True: We know that 5C, which has solution CC/. If the half-life is!þ$*, then!! 5Ð!Þ$*Ñ ln C C /, which gives 5!Þ$* Þ() (rounded to two places), so 5C Þ()CÞ.C.>

12 Part III À These are two free response problems 20, 21) worth a total of 25 points. Write your answers on the test pages. Show your work neatly and cross out irrelevant scratchwork, false starts, etc. Please put your name on each of the following pages, since they may be separated during grading. Also, please add your Discussion Section Letter (available on your exam front cover sheet) on each page so that we can return papers through discussion sections. Name Discussion Section Letter 20. ÐNote: there may be several ways to do some of these integrals--some harder, some easier.) ' * $ a) Find sin cos. Let? sin ß so.? cos.. Then ' * $ sin cos. ' * sin cos cos.! ' * sin sin cos ' * ' *?? Ð Ñ.? Ð? Ñ.???.? G! sin! sin GÞ b) Evaluate ' È!. È! Method I (easy): the integral represents the area under C and above Ò!ß Ó. This region is a quarter circle with radius 2, so it has area 1ÐÑ 1Þ Method II (trig substitution): let sin ),. cos ).). Then ' È. ' È sin 2 cos. ' Ècos. ' ) ) ) ) cos ) ) cos ).) ' cos ). ' ) cos ).) Ð) sin ) Ñ G ) sin ) G. Since ) arcsin Ð Ñ, ) sin ) arcsinð Ñ sinð arcsin Ð ÑÑ. Therefore ' È!. arcsinð Ñ sinð arcsinð ÑÑl! 1 1 sin Ð Ñ!!. 1 (Note that sinð arcsin Ð ÑÑ can be simplified to È, but that's not needed to complete the problem.)

13 c) Find ' Ð ÑÐ Ñ. Method I (easiest): The function 0ÐÑ Ð ÑÐ Ñ is an odd function Ði.e., its graph is symmetric with respect to the origin), so '.!Þ Ð ÑÐ Ñ Method II (not much harder): ' ' Ð ÑÐ Ñ. Ð Ñ. Let? ß.?., so that..?. $ Ð Ñ $? Then (adjusting the limits) ' '..?!.. EÐ Ñ FÐ Ñ Ð ÑÐ Ñ Ð Ñ Ð Ñ Ð ÑÐ Ñ E F Method III: (use partial fractions) Equating numerators gives EÐ Ñ FÐ ÑÀ Let to get F so F Let to get E so E Ð ÑÐ Ñ Then ' ' Î Î.. Ðlnl l lnl lñl ( ln $ ln ln ln $Ñ!Þ

14 21. A tank initially contains 200 L of pure water. rine containing 5g of salt/l enters the tank at a rate of 2L/min. There is instant mixing and solution leaves the tank at 2L/min. Let C denote the amount of salt in the tank at time >..C a) Write a differential equation that describes this situation:.>....c C.>!!! g/min C!! g/min.c C so.>!!! (g/min) Ðrate salt entersñ Ðrate salt exitsñð& g/l) (2 L/min) Ð g/l) (2 L/min) b) Solve the differential equation and find the amount of salt in the tank after 100 minutes..c C.>!!!!! Ð!!! CÑ.C!!! C!!.> so '.C '.>!!! C!!!!!! so ln l!!! Cl > H, or ln l!!! Cl > H. > H H > Then l!!! Cl /!! / /!!, H > > H so!!! C / /!! G/!! (where G / constant). Therefore C!!! G/!! >. > Since CÐ!Ñ!, we get G!!!, so C!!!!!!/!! Þ!! Therefore CÐ!!Ñ!!!!!!/!!!!!!!!/ '$Þ g. c) What is the steady state amount that is, the limit of the amount of salt C as >Ä_? ( If you can't do parts a) and b), you should nevertheless be able to give a reasonable explanation that arrives at the correct answer.) lim C lim!!!!!!/!!!!! (g). >Ä_ >Ä_ > This is completely plausible. As time passes, the brine in the tank looks more and more like the brine entering the tank, so, in the long run, the amount of salt in the tank should approach Ð!! L Ñ Ð& g/l Ñ!!! g of salt.