Math 131, Exam 2 Solutions, Fall 2010

Transcription

1 Math 131, Exam 2 Solutions, Fall 2010 Part I consists of 14 multiple choice questions (orth 5 points each) and 5 true/false questions (orth 1 point each), for a total of 75 points. Mark the correct anser on the anser card. For Parts I, only the anser on the card ill be graded. & 1. If D >. D > "!, hat is l? >" " "( "" " ) "& D) % ( ( "! % % A) B) C) % E) F)! G) H) I) J) "Î &.D " "Î "& %. D " Î '! " " "! "! % "! % > D> >, so > >, so > > Î '>Þ Therefore. D " " % >" % % % l ' Þ 2. If 1ÐBÑ sec B tan B, hat is 1 Ð Ñ? A)! B) C) " D) E) 1 % ( " ) ( ) F) " G) H) I) J)...B.B 1 ÐBÑ Ðse- BÑ Ðtan BÑ Ðtan BÑ Ðsec BÑ Ðsec BÑÐsec BÑ Ðtan BÑÐsec B tan BÑ % % % % sec Bsec Btan BÞ Therefore 1 Ð Ñ sec sec tan ) % " ' ' Ð Ñ Ð ÑÐ"Ñ Þ

2 3. To functions 0ÐBÑ and 1ÐBÑ are pictured belo. Let J ÐBÑ 0Ð1ÐBÑÑÞ What is J ÐÑ? % A) " B) C Ñ D) & E) " F)! G) H) I) J) " J ÐBÑ 0 Ð1ÐBÑÑ1 ÐBÑ, so J ÐÑ 0 Ð1ÐÑÑ 1 ÐÑ 0 Ð Ñ 1 ÐÑ " " Ð"Ñ Þ 4. For a certain function 0ÐBÑ you kno that 0 ÐÑ Þ Using this information, you sa, in a WebWorK problem, ho you can calculate an approximate value for 0ÐÞ**Ñ 0ÐÑÞ Suppose you also kno that 0ÐÑ (Þ Then calculate an approximate value for 0ÐÞ**ÑÞ A) 'Þ"& B) 'Þ*) C) 'Þ& D) 'Þ& E) 'Þ" F) 7.0 " G) (Þ!& H) (Þ I) (Þ' J) (Þ 0Ð 2Ñ 0ÐÑ 0Ð 2Ñ 0ÐÑ ÐÑ lim, so hen 2 is close to!. 0Ð!Þ!"Ñ 0ÐÑ!Þ!" Using 2!Þ!" gives Þ 0ÐÞ**Ñ 0ÐÑ!Þ!" Therefore 0ÐÞ**Ñ 0ÐÑ ÐÑÐ!Þ!"Ñ!Þ!ß so 0ÐÞ**Ñ 0ÐÑ!Þ! (!Þ! 'Þ*)Þ

3 5. At the point here B!, the tangent line to the graph of C0ÐBÑ / +ÐB "Ñ is horizontal. What is +? A) & B) % C) D) E) " F)! G) " H) I) J) % +B. +B +B. +ÐB Ñ Ð/ Ñ/ Ð+ÐBÑ Ñ +B +B.B.B +ÐB Ñ Ð+/ Ñ/ Ð+ÐBÑ Ñ 0 ÐBÑ Ð+ÐB "Ñ Ñ + ÐB "Ñ' +B +B +/ Ð+ÐB ÑÑ Ñ / Ð+ÐB ÑÑ Ñ + ÐB"Ñ % +ÐB"Ñ% Þ We are told that! / Ð+ Ñ 0 Ð!Ñ + +!, so + Þ 6. At time > (years) the size T of a certain population of animals is TÐ>Ñ "!!!! "*/. Ho fast is!þ!)> the size of the population changing at time >0? ÐAll ansers belo have units animals/year. Ñ A) & B) "( C) % D) ( E) *% F) *) G) "!' H) ""! I) ""% J) "") We ant to find.t hen >!Þ!Þ!)>!Þ!)>!Þ!)>.T Ð" */ Ñ!"!!!!Ð!Þ(/ Ñ (!! / Ñ Ð" */ Ñ Ð" */ Ñ.T (!! >! "!! l (Þ!Þ!)>!Þ!)>, so

4 7. A point is moving along a straight line. At time >, its position is = 0Ð>Ñand its &>6Þ Exactly to of the folloing statements are true. Which are the true statements? i) for times > &, the point's velocity is increasing ii) for times >, the point is moving in the positive direction along the line.= iii) hen is increasing, the point must be moving in the positive direction along the line. iv) ith the given information, it is possible to calculate the position of the particle hen > " v) the graph of the position function =0Ð>Ñhas exactly one inflection point. A) i) and ii) B) i) and iii) C) i) and iv) D) i) and v) E) ii) and iii) F) ii) and iv) G) ii) and v) H) iii) and iv) I) iii) and v) J) iv) and v).@.@ & True is increasing hen +! and >&! hen >..= False > &>'Ð>ÑÐ>Ñ! if >, so, for these > 's, = is decreasing (point is moving in the negative direction along the line)..=.@. = False iii) is increasing, then!, that is, the graph of =0Ð>Ñ is concave up. But a concave up graph =0Ð>Ñcan be either increasing or decreasing (that is, e can't decide if the point is moving left or right Ñ In other ords, the fact that.@ +! tells us the velocity is increasing, but that says nothing about the direction of the motion. False iv) All e kno about the position function =0Ð>Ñis > &>'Þ If e have any function =0Ð>Ñith this property, then the.= function =0Ð>ÑGÐhere Gis a constant) also satisfies > &>'. The position is not determined just by knoing the velocity..=.@ & & True v) 0 Ð>Ñ >&ß so 0 Ð>Ñ! if > ß and 0 Ð>Ñ! if > Þ Therefore 0 changes concavity at > & (and nohere else). This is the only inflection point. So i) and v) are the true statements.

5 B 8. Let C0ÐBÑB/ Þ The statement 0ÐBÑis increasing on the interval Ð ß,Ñ is true for some values of, and false for other values of,. What is the largest value for, that makes the statement true? " A),! B)," C), D),/ E), / " F) / G), H), " I), J), / 0ÐBÑ is increasing hen B B B B " " 0 ÐBÑBÐ / Ñ/ / Ð B"Ñ!. Since /! for all B, 0 ÐBÑ! hen " B"!, that is, hen BÞ BB "B 9. Let 2ÐBÑ Þ Find 2 Ð"Ñ A) B) C) " D)! E) " F) G) " H) I) J) " & Ð" B ÑÐ" BÑ ÐBB ÑÐBÑ ÐÑÐÑÐÑÐÑ " Ð"B Ñ % 2ÐBÑ ßso 2Ð"Ñ

6 10. Simplify tan Ð arcsin BÑ B 1 B " A) B) C) D) "*B " *B *B " *B " B " "*B " *B E) " *B F) " *B G) B H) B " I) B J) B This can be done using trig the trig identities "tan ) sec ) and sin ) cos ) "Þ But it's easier just to dra a picture illustrating an angle ) arcsin B Ð an angle ) hose sin is B) and compute its tangent: tanðñtanð arcsin BÑÑ B ) "*B,B / / 11. Suppose, is a constant. What is lim. ( ) BÄ! B Hint: this limit is a derivative.,, A) / B), C), D) / E) / F),/ G)! H) %, I),/, J) does not exist 0ÐBÑ0Ð!Ñ BÄ! B!,B / / If e choose 0ÐBÑ /, then 0 Ð!Ñ lim lim,b Þ BÄ!,B But e kno (from differentiation rules) that 0 ÐBÑ,/ ß so,b / / lim BÄ! B 0 Ð!Ñ,/ Þ B

7 1 12. A small cannonball is fired from the ground at an angle of % from the horizontal. It leaves the cannon ith a velocity of 200 ft/sec. If e assume that the only force acting on the cannonball is gravity, then it follos the path pictured belo. We can describe the motion of the projectile along this path using time > as a parameter:: B > C "'> "!! > for!ÿ>ÿ-, here - is the time hen the projectile hits the ground. At hat time, >, does the tangent line to this path have slope!? A) > sec B) >"!! & sec C) > sec D) > & ) sec &! "& && E) >& sec F) > sec G) > sec H) > sec %& % I) >) sec J) > sec Î.B.B.BÎ.B.B.B!ß Á!! "!! here > "!!!, and that happens hen > & ) sec. We ant to kno:! at hat time >? Since, e see that! is true henever provided that at the same time. Since is never, e only need to see

8 13. The point Ð!ß "Ñ is on the curve determined by the equation B sin C C cos B "Þ ÐPart of this curve is shon belo.) What is the slope of the tangent line to the curve at Ð!ß "Ñ? A) 0 B) 1 C) " D) E) " " " F) cos Ð"Ñ G) sinð"ñ H) tan Ð"Ñ I) sin Ð"Ñ J) cosð"ñ The picture is just to help make the problem visible hoever, the picture does indicate that you anser should be negative.. If e apply.b to both sides of the equation Bsin CCcos B ", e get.b.b Bcos C sin CCsin B cos C!Þ When B! and C"ßthis gives.b.b! sin "! "!, so sin " Ð!Þ)%"& Ñ

9 14. The figure shos the graph of Carcsin BÐalso ritten as Csin BÑß " What is the slope of the tangent line at the point TÐ ß 1 ' Ñ? Write your anser as a decimal rounded to 2 decimal places. " A) "Þ!% B) "Þ"" C) "Þ"& D) "Þ% E) "Þ* F) "Þ G) "Þ' H) "Þ%( I) "Þ& J) "Þ&(.B " " " % Þ The slope is "B.B l "Ð Ñ B " " % "Þ"&Þ

10 Questions 15) - 19) are trueîfalse questions 15. Suppose 0 has a derivative at the point +. Then lim0ðbñ 0Ð+ÑÞ BÄ+ A) True B) False True: lim0ðbñ 0Ð+Ñ means that 0 is continuous at +, and this BÄ+ is true if 0 has a derivative at The figure shos the graph of the derivative 0 ÐBÑ for some function 0ÐBÑÞ The function 0ÐBÑ is concave up over the interval Ð"ß,ÑÞ A) True B) False False: Over the interval Ð "ß,Ñ, 0 ÐBÑ is decreasing, so 0 ÐBÑ!Þ Therefore 0ÐBÑ is concave don over the interval Ð"ß,ÑÞ

11 Questions all refer to the folloing graph hich shos the derivative continuous function 2ÐBÑ 2ÐBÑ of some 17. 2ÐBÑ has a local minimum at B "Þ A) True B) False True: For B ", 2 ÐBÑ is negative, so 2ÐBÑ is decreasing. For B "ß 2 ÐBÑ is positive, so 2ÐBÑ is increasing. Therefore 2ÐBÑ has a local minimum at B "Þ 18. 2ÐBÑ has exactly one inflection point. A) True B) False False: An inflection point occurs here 2ÐBÑ changes from concave don to concave up (or vice-versa). This happens hen 2ÐBÑchanges from increasing to decreasing (or vice-versa). But 2 ÐBÑ is alays increasing. So 2ÐBÑ has no inflection points ÐBÑ has only one antiderivative, 2ÐBÑÞ A) True B) False False: If 2ÐBÑ is one antiderivative for 2 ÐBÑß then 2ÐBÑ G ß here G is any constant ß is another antiderivative. This is because Ð2ÐBÑ GÑ 2 ÐBÑÞ Since G could be chosen in infinitely many different ays, 2ÐBÑhas infinitely many antiderivatives.

12 Part II: (TOTAL 25 points) Generally, a correct anser ithout some ork indicating ho you got it is not enough. 20. This question simply tests your mechanical ability to use the differentiation formulas. After the differentiation is complete, you do not need to simply further. For example, ÐBÑÐB"Ñ leaving an anser in the form B &B ÐÑB ould be OK. ( Each part is orth 5 points.) ÐB"Ñ B B/ a) Find 0 ÐBÑ if 0ÐBÑ sin B. B B. Ðsin BÑ ÐB/ ÑÐB/ Ñ Ðsin BÑ B B B Ðsin BÑÐB/ / ÑÐB/ ÑÐcos BÑ Ðsin BÑ sin B.B.B 0ÐBÑ [sufficient anser] B B B B/ sin B / sin B B/ cos B sin B B B B B/ csc B/ csc BB/ cot Bcsc B Ðvarious other rearrangements).d b) Find.A if Dcos ÐAÑ.D.D.D.A.A.A sin ÐAÑ, so * cos Ð AÑ, so ( sin ÐAÑ c) Find.B if Ctan Ð B Ñ sec B. B.B.B sec B B. Ð Ñ Ð Ñ Ð Ñ Ð Ñ lnðñ.bb B B sec Ð Ñ Ð Ñ lnðñ ÐBÑ d) Find.B if B C cos ) sec Ð ) Ñ.B Î.) sec Ð Ñ Ð Ñ.BÎ. ) tan ) ) sin )

13 21. ( 5 points ) Use the definition of derivative (as a certain limit) to find 0Ð+Ñif 0ÐBÑ B " Þ 0ÐBÑ0Ð+Ñ BÄ+ B+ BÄ+ B+ BÄ+ B+ B" +" Method 1: 0 Ð+Ñ lim lim lim Ð+ "Ñ ÐB "Ñ ÐB "ÑÐ+ "Ñ 'Ð+ BÑ ' ' BÄ+ ÐB +ÑÐB "ÑÐ+ "Ñ BÄ+ ÐB "ÑÐ+ "Ñ Ð+ "Ñ lim lim Ð+ "Ñ ÐÐ+ 2Ñ "Ñ Ð+2Ñ" +" ÐÐ+2Ñ"ÑÐ+"Ñ 0Ð+2Ñ0Ð+Ñ Method 2: 0 Ð+Ñ lim lim lim '2 '2 ' 2 2ÐÐÐ+ 2Ñ "ÑÐ+ "Ñ ÐÐÐ+ 2Ñ "ÑÐ+ "Ñ ÐÐ+ 2Ñ "ÑÐ+ "Ñ lim lim lim ' Ð+ "Ñ