MARK BOX problem poits HAND IN PART 0 10 1 10 2 5 NAME: Solutios 3 10 PIN: 17 4 16 65=13x5 % 100 INSTRUCTIONS This exam comes i two parts. (1) HAND IN PART. Had i oly this part. (2) STATEMENT OF MULTIPLE CHOICE PROBLEMS. Do ot had i this part. You ca take this part home to lear from ad to check your aswers oce the solutios are posted. O Problem 0, fill i the blaks. As you kow, if you do ot make at least half of the poits o Problem 0, the your score for the etire exam will be whatever you made o Problem 0. Durig this exam, do ot leave your seat uless you have permissio. If you have a questio, raise your had. Whe you fiish: tur your exam over, put your pecil dow ad raise your had. Durig the exam, the use of uauthorized materials is prohibited. Uauthorized materials iclude: books, persoal otes, electroic devices, ad ay device with which you ca coect to the iteret. Uauthorized materials (icludig cell phoes, as well as your watch) must be i a secured (e.g. zipped up, sapped closed) bag placed completely uder your desk or, if you did ot brig such a bag, give to Prof. Girardi to hold durig the exam (it will be retured whe you leave the exam). This meas o electroic devices (such as cell phoes) allowed i your pockets. Cheatig is grouds for a F i the course. At a studet s request, I will project my watch upo the projector scree. Upo request, you will be give as much (blak) scratch paper as you eed. The mark box above idicates the problems alog with their poits. Check that your copy of the exam has all of the problems. This exam covers (from Calculus by Thomas, 13 th ed., ET): 10.7 10.10, 11.1, 11.2. Hoor Code Statemet I uderstad that it is the resposibility of every member of the Carolia commuity to uphold ad maitai the Uiversity of South Carolia s Hoor Code. As a Caroliia, I certify that I have either give or received uauthorized aid o this exam. I uderstad that if it is determied that I used ay uauthorized assistace or otherwise violated the Uiversity s Hoor Code the I will receive a failig grade for this course ad be referred to the academic Dea ad the Office of Academic Itegrity for additioal discipliary actios. Furthermore, I have ot oly read but will also follow the istructios o the exam. Sigature : Prof. Girardi Page 1 of 11 Math 142
0. Fill-i the boxes. 0A. Power Series. Cosider a (formal) power series h(x) = a (x x 0 ), with radius of covergece R [0, ]. (Here x 0 R is fixed ad {a } is a fixed sequece of real umbers.) Without ay other further iformatio of {a }, aswer the followig questios. The choices for the ext 4 boxes are: AC, CC, DIVG, or aythig. AC stads for: is always absolutely coverget. CC stads for: is always coditioally coverget. DIVG stads for: is always diverget. aythig stads for: ca do aythig i.e., there are examples showig that it ca AC, CC, or DIVG. 0A.1. At the ceter x = x 0, the power series i (0A) AC. (0A) 0A.2. For x R such that x x 0 < R, the power series i (0A) AC. 0A.3. For x R such that x x 0 > R, the power series i (0A) DIVG. 0A.4. If 0 < R <, the for the edpoits x = x 0 ± R, the power series i (0A) aythig. 0A.5. Furthermore, if α ad β are i the iterval (x 0 R, x 0 + R), the (Hit: ote the x=β x=α h(x) dx = a + 1 (x x 0) +1 x=β x=α. x=β x=α already i there.) 0B. Taylor/Maclauri Polyomials ad Series. Let y = f(x) be a fuctio with derivatives of all orders i a iterval I cotaiig x 0. 0B.1. The th Taylor coefficiet of y = f(x) about x 0 is c = f () (x 0 )! 0B.2. The N th -order Taylor polyomial of y = f(x) about x 0, i ope form (so with... ad without a -sig), is P N (x) = f(x 0 ) + f (1) (x 0 )(x x 0 ) + f (2) (x 0 ) 2! (x x 0 ) 2 + f (3) (x 0 ) 3! (x x 0 ) 3 + + f (N) (x 0 ) (x x 0 ) N N! 0B.3. The Taylor series of y = f(x) about x 0, i closed form (so, with a -sig ad without... ), is P (x) = f () (x 0 )! (x x 0 ) Prof. Girardi Page 2 of 11 Math 142
0C. Parametric Curves. Cosider the curve C parameterized by x = x (t) 0C.1. 0C.2. for a t b. y = y (t) Express dy dx i terms of derivatives with respect to t. Aswer: dy dx = The arc legth of C, expressed as o itegral with respect to t, is dy dx Arc Legth = t=b t=a (dx ) 2 + ( ) 2 dy 1. Commoly Used Taylor Series Fill i Problem 1 s blak boxes with the choices a l, which are provided below. You may use a choice more tha oce or ot at all. A sample questio is already doe for you. a. b. c. x d. x! ( 1) =1 +1 x e. f. ( 1) x2 (2)! ( 1) x 2+1 (2 + 1)! ( 1) x2+1 2 + 1 g. x R j. ( 1, 1] h. ( 1, 1) k. [ 1, 1) i. [ 1, 1] l. oe of the others sample. A power series expasio for y = 1 1 x is a ad is valid precisely whe h. 1.1. A power series expasio for y = cos x is d ad is valid precisely whe g. 1.2. A power series expasio for y = si x is e ad is valid precisely whe g. 1.3. A power series expasio for y = e x is b ad is valid precisely whe g. 1.4. A power series expasio for y = l (1 + x) is c ad is valid precisely whe j. 1.5. A power series expasio for y = ta 1 x is f ad is valid precisely whe i. Prof. Girardi Page 3 of 11 Math 142
2. Idetify a parametrizatio of a circle with ceter at (0, 0) ad radius 1, which is traced i the various below described fashios, by fill i the blaks boxes with the below choices A I. You may use a choice more tha oce or ot at all. 2.1. traced oce, startig at (1, 0), i the couterclockwise directio. A parameterizatio is A 2.2. traced oce, startig at (1, 0), i the clockwise directio. A parameterizatio is B 2.3. traced oce, startig at ( 1, 0), i the couterclockwise directio. A parameterizatio is D 2.4. traced oce, startig at ( 1, 0), i the clockwise directio. A parameterizatio is C 2.5. traced oce, startig at (0, 1), i the couterclockwise directio. A parameterizatio is F x (t) = cos t y (t) = si t x (t) = cos t y (t) = si t x (t) = cos t y (t) = si t x (t) = cos t y (t) = si t x (t) = si t y (t) = cos t x (t) = si t y (t) = cos t x (t) = si t y (t) = cos t x (t) = si t y (t) = cos t (A) (B) (C) (D) (E) (F) (G) (H) Noe of the other choices. (I) 2sol. Note that for each of the give choices of a parameterizatio, [x (t)] 2 + [y (t)] 2 = 1 ad so the puffo is ideed ruig alog, i some fashio, the curve x 2 + y 2 = 1, which is a circle with ceter (0, 0) ad radius 1. Pluggig i t = 0 will give you the startig poit. The use the behaviour of the trig fuctios ivolved to get the directio. Prof. Girardi Page 4 of 11 Math 142
3. Taylor s Remaider Theorem. 3a. Let y = f(x) be a fuctio with derivatives of all orders i a iterval I cotaiig x 0. Let y = P N (x) be the N th -order Taylor polyomial of y = f(x) about x 0. Let y = R N (x) be the N th -order Taylor remaider of y = f(x) about x 0. Thus, f(x) = P N (x) + R N (x). Taylor s BIG Remaider Theorem tells us that, for each x I, R N (x) = f (N+1) (c) (N + 1)! (x x 0 ) (N+1) for some c betwee x ad x 0. 3b. I this problem, you must show your work ad clearly explai your thought process. Usig Taylor s (Big) Remaider Theorem (ad ot usig the facts o the Commoly Used Taylor Series hadout), show that 2 = e 2.!. Hit. Ideed, the facts listed o the Commoly Used Taylor Series are show by usig Taylor s Remaider Theorem so you ca thik of this problem as verifyig/showig oe of these facts listed o the Commoly Used Taylor Series hadout.. Hit. Cosider the fuctio f (x) = e x. Note that the f (2) = e 2. 3bsol. Prof. Girardi Page 5 of 11 Math 142
MULTIPLE CHOICE PROBLEMS Idicate (by circlig) directly i the table below your solutio to the multiple choice problems. You may choice up to 2 aswers for each multiple choice problem. The scorig is as follows. For a problem with precisely oe aswer marked ad the aswer is correct, 5 poits. For a problem with precisely two aswers marked, oe of which is correct, 2 poits. All other cases, 0 poits. Fill i the umber of solutios circled colum. Table for Your Muliple Choice Solutios Do Not Write Below problem 4 4a 4b 4c 4d 4e umber of solutios circled 1 2 B x 5 5a 5b 5c 5d 5e 6 6a 6b 6c 6d 6e 7 7a 7b 7c 7d 7e 8 8a 8b 8c 8d 8e 9 9a 9b 9c 9d 9e 10 10a 10b 10c 10d 10e 11 11a 11b 11c 11d 11e 12 12a 12b 12c 12d 12e 13 13a 13b 13c 13d 13e 14 14a 14b 14c 14d 14e 15 15a 15b 15c 15d 15e 16 16a 16b 16c 16d 16e 5 2 0 0 Total: Prof. Girardi Page 6 of 11 Math 142
STATEMENT OF MULTIPLE CHOICE PROBLEMS These sheets of paper are ot collected. 4. Let the fuctio y = f (x) have a power series power series represetatio c x, which is valid i some iterval ( R, R) where R > 0. 4sol. If a fuctio ca be represeted by a power series cetered at 0 o some iterval ( R, R), with R > 0, the that power series must be the Taylor series cetered at 0. So c 0 = f (0) (0) = f (0). 0! 5. Let the fuctio y = f (x) have a power series power series represetatio a x, which is valid i some iterval J cotaiig 0 ad the raduis of J striclty positive. Cosider the two statemets: (A) If y = f (x) is a eve fuctio (i.e., f ( x) = f (x)), the a 1 = a 3 = a 5 = = 0. (B) If y = f (x) is a odd fuctio (i.e., f ( x) = f (x)), the a 0 = a 2 = a 4 = = 0. 5sol. Both (A) ad (B) are true. *2. Suppose that f(x) = ax coverges for all x i a ope iterval ( R, R). = 0 a. Show that if f is eve, the a 1 = a 3 = a 5 = = 0, i.e., the Taylor series for f at x 0 cotais oly eve powers of x. = b. Show that if f is odd, the a 0 = a 2 = a 4 = = 0, i.e., the Taylor series for f at x 0 cotais oly odd powers of x. = It is kow that all power series that coverge to a fuctio f(x) o a iterval ( R,R) are the same. This is a key property of power series that will be eeded to complete this proof. a. If f(x) is eve, the f( x) = (1). Substitute x for x i the series ax. What are the coefficiets of the resultig power series for odd? = 0 The coefficiets for odd are (2). How does this show that the Taylor series for a eve fuctio f at x = 0 cotais oly eve powers of x? A. The coefficiets of the odd- terms i the series for f( x) must equal both a ad a. The oly solutio to a = a is a = 0. B. The coefficiets of the odd- terms i the series for f( x) must equal both a ad 2 a. The oly solutio to a = 2a is a = 0. C. The substitutio of x resulted i a coefficiet of 0 for all odd, so the statemet has bee prove. D. 1 The coefficiets of the odd- terms i the series for f( x) must equal both a ad. The oly 2 a solutio to a = 1 is 0. 2 a a = b. If f(x) is odd, the f( x) = (3). Substitute x for x i the series ax. What are the coefficiets of the resultig power series for eve? = 0 The coefficiets for eve are (4). How does this show that the Taylor series for a odd fuctio f at x = 0 cotais oly odd powers of x? A. The substitutio of x resulted i a coefficiet of 0 for all eve, so the statemet has bee prove. B. 1 The coefficiets of the eve- terms i the series for f( x) must equal both a ad. The oly 2 a solutio to a = 1 is 0. 2 a a = C. The coefficiets of the eve- terms i the series for f( x) must equal both a ad 2 a. The oly solutio to a = 2a is a = 0. D. The coefficiets of the eve- terms i the series for f( x) must equal both a ad a. The oly solutio to a = a is a = 0. (1) f(x) (2) a 1 (3) f(x) (4) a f(x) a 2 f(x) 0 2a a a 0 1 a 2 ID: 9.9.52. Problems 4 ad 5 were meat to help you with Problem 1. Prof. Girardi Page 7 of 11 Math 142
6. Fid the iterval of covergece of the power series (x 2) 10 =1 Recall that the iterval of covergece is the set of x s for which the power series coverges, either absolutely or coditioally. 6sol. The iterval of covergece is ( 8, 12). 7. What is the LARGEST iterval for which the power series (2x + 6) 7sol. 4 =1 is absolutely coverget? Prof. Girardi Page 8 of 11 Math 142
8. Suppose that the radius of covergece of a power series c x is 16. What is the radius of covergece of the power series c x 2? 8sol. c x 2 = c (x 2 ) coverges precisely whe x 2 < 16, or equivaletly, whe x 2 < 16, or equivaletly, whe x < 16. 9. Usig a kow (commoly used) Taylor series, fid a power series represetatio of the fuctio f(x) = 2 3 x 9sol. about the ceter x 0 = 0 ad state whe this represetatio is valid. Hit, by simple algebra, ( ) ( ) 2 2 1 f(x) = 3 x = 3 1 x. 3 10. Usig a kow (commoly used) Taylor series, fid a power series represetatio of the fuctio f(x) = 1 (1 x) 4 about the ceter x 0 = 0 which is valid for x < 1. Hit. Start with the Geometric Series (Prof. Girardi sometimes called him the work moose) ad differetiate (as may times as eeded). Be careful ad do t forget the chai rule: D x (1 x) 1 = ( 1)(1 x) 2 D x (1 x) = ( 1)(1 x) 2 ( 1) = (1 x) 2. 10sol. Prof. Girardi Page 9 of 11 Math 142
11. Fid the 3 rd order Taylor polyomial, about the ceter x 0 = 1, for the fuctio f(x) = x 5 x 2 + 5. 11sol. The computatios below show that the 3 rd order Taylor polyomial, about the ceter x 0 = 1, for the fuctio f(x) = x 5 x 2 + 5 is P 3 (x) = 5 + 3(x 1) + 9(x 1) 2 + 10(x 1) 3. f () (x) f () (x 0 ) we were give x 0 = 1 f () (x 0 )! 0 x 5 x 2 + 5 5 5 0! = 5 1 = 5 1 5x 4 2x 5 2 = 3 3 1! = 3 1 = 3 2 5 4x 3 2 20 2 = 18 18 2! = 18 2 = 9 3 5 4 3x 2 (5)(4)(3) (5)(4)(3) 3! = (5)(4)(3) (3)(2) = (5)(4) 2 = 10 12. Fid the 17 th order Taylor polyomial, about the ceter x 0 = 1, for the fuctio f(x) = 6x 7 + 5. Hit: We kow f (x) = P 17 (x) + R 17 (x). Thik about what Taylor s (big) Remaider Theorem says about y = R 17 (x) for the fuctio f(x) = 6x 7 + 5. 12sol. Note that the th -derivative of a polyomial of degree N is zero if > N. So if 8, the f () (x) = 0 for all x R. Let s follow the otatio from Problem 0B. If N 7, the f (N+1) (c) = 0 for ay c R ad so R N (x) = f (N+1) (c) (N + 1)! ad so P N (x) = f (x). So P 17 (x) = f (x). (x 1)N+1 = 0 (N + 1)! x 1 N+1 = 0, 13. Fid a parameterizatio for the lie segmet from ( 1, 2) to (10, 6) for 0 t 1. 13sol. as: x = 1 + 11t ad y = 2 8t 14. Fid dy for the parameterized curve give by dx x = 2t 2 + 1 14sol. y = 3t 3 + 2 for 1 < t <. x (t) = 1 + ( 10 ( 1 )) t = 1 + 11t y (t) = 2 + ( 6 2 ) t = 2 8t. dy dy dx = dx = 9t2 4t = 9t 4 Prof. Girardi Page 10 of 11 Math 142
15. Fid a equatio of the taget lie to the curve at the poit correspodig to t = 11π. x = t si t y = t cos t. 15sol. as: y = x 11π 11π (x (11π), y (11π)) = (0, 11π) dy dy dx = t=11π dx t=11π So equatio of taget lie to curve whe t = 11π is ( y 11π ) = 1 (x 0) 11π y + 11π = 1 11π x = cos t t si t si t + t cos t = 1 0 t=11π 0 11π = 1 11π. y = 1 11π x 11π. 16. Describe the motio of a puffo whose positio (x, y) is parameterized by x = 6 si t y = 3 cos t. 16sol. as: Moves oce clockwise alog the ellipse x2 36 + y2 = 1 startig ad edig at (0, 3). 9 [ ] 2 [ ] 2 Sice x(t) 6 + y(t) 3 = [si t] 2 + [cos t] 2 = 1, the puffo is movig alog the ellipse [ x 2 [ 6] + y ] 2 3 = 1, i.e., alog the ellipse x 2 + y2 = 1. 36 9 He starts at (x (0), y (0)) = (6 si 0, 3 cos 0) = (0, 3). He fiishes at (x (2π), y (2π)) = (6 si 2π, 3 cos 2π) = (0, 3). As he moves from t = 0 to t = 2π, he traces out the ellipse oe time. To figure out if he is movig CW or CCW, ote ( x ( ) ( π 2, y π )) ( 2 = 6 si π, 3 cos ) π 2 2 = (6, 0). So Mr. Puffo is movig clockwise. Prof. Girardi Page 11 of 11 Math 142