Math 150B S22 Midterm 3 Solutions

California State University Northridge MATH 150B: Calculus II, Spring 2022 Midterm Exam 3 Solutions April 14, 2022. Duration: 65 Minutes. Instructor: Dr. Jing Li Student Name: Student number: Instructions: • Take your time to read the entire paper before you begin to write, and read each question carefully. Remember that certain questions are worth more points than others. Make a note of the questions that you feel confident you can do, and then do those first: you do not have to proceed through the paper in the order given. • This exam contains 6 problems on 7 pages (including this title page) for a total of 110 points. • You have 65 minutes to complete this exam. • This is a closed book exam, and no notes of any kind are allowed. No calculators are allowed. The use of cell phones, pagers or any text storage or communication device is not permitted. • The correct answer requires justification written legibly and logically: you must convince me that you know why your solution is correct. Answer these questions in the space provided. Show all work that is used to derive your answers. Use the backs of pages if necessary. • Where it is possible to check your work, do so. • Good Luck! Page 2 3 4 5 6 7 Total Points 20 20 20 20 15 15 110 Your Grades Math 150B Midterm 3 Student Name: Student ID: 2 Question 1. [40 points] Determine if the series absolutely converges, conditionally converge, or diverge. 1 (°1)k X a. (10 points) k=1 k 4/3 Solution Wefirst note that EI III LEE Theseries of Therefore É the absolute values is É Solution Wefirst note that by 43 1 absolutely converges 1 (°1)k X °1 k=1 tan k b. (10 points) So p series with p 4 converges We conclude that distantk a I 0 and fist to the Divergence test we conclude that É diverges Math 150B Midterm 3 Student Name: Question 1. cont’d 1 (°1)k X k=3 ln k c. (10 points) E II Solution Wefirst note that Wealso note that C K Ink Jk or Theseries E3 1 since However Ik 0Canty Hence I diverges E Ik diverges Student ID: Ez Ik for k 23 So by the comparison Test We conclude that E II K 3 12 is an alternatingseries dis Ik the alternating series test we by E Ik diverges X cos k d. (10 points) conclude 1 k=1 Solution We 3 Wenote that 0 conclude that that EÉg Entente converges conditionally converges k3 We first note that É of É 19 Wealso note that Icook El thus 19 sit The p series É with p 3 converges we conclude that É 19 Therefore É of converges absolutely converges so by the comparison test Math 150B Midterm 3 Student Name: Question 2. [20 points] Consider the following series 1 X (°1)k k=1 Student ID: 4 (x ° 3)k k6k a. (8 points) Find the radius of convergence. Solution Bythe Ratio Test relist testifies sotheseries convergesfor 1 31 6 51 1 1 ie 1 1 14 3429 We conclude that the radius of convergence R 6 b. (8 points) Test the endpoints for convergence/divergence. Solution To check at the endpoints ID At X 9 Ewitty É Et which is the z IÉÉÉ ÉI alternatingharmonicseries which is the harmonic series it is convergent it is divergent c. (4 points) What is the interval of convergence? Draw it. Solution the interval of convergence is 1 3 91 Thegraph is we i as follows Math 150B Midterm 3 Student Name: Student ID: Question 3. [10 points] Find the Maclaurin series expansion (4 nonzero terms). 5 f (x) = x 2 sin(2x) the Maclaurinseriesexpansionfor six is given asfollows x E É i t 4154 t É E1 FI for wks Solution six therefore the Maclaurinseriesexpansion forsin ex isgiven as follows sintax ex Eff 45,1 n É t É É t andthen the Maclaurin series expansion for fix fix for was x'sinex is X'sintax x ex II 4,1 t YE É É ÉÉ É t war for seethenextpageforanothermethod Question 4. [10 points] Find the function represented by the following series. 1 (x ° 3)k X k=1 Solution Note that 52k É EE É 1 1 EMET This is a geometric series with ratio so its sum is É Hence EI 43 É k 3 and the first term a Ef x'sin ex fix Question 3 fix x'sinex fly fix a sinex fro Xanax YEAT sinixt SX Wtf Yo o 8 wax 12 4 72 an ax t 8 sinex 4 8 4 next 8 as ex 8Xsinex 24xsinx e4 sin ex fibx x2 as ex 48 16 2 iz uz 8 4 Gesinax as ex t byxanax 2 48 16 six sax STnext 16xsinex 24sinex lbx'sinex bx wax v4 rusax Ho O 64 coax X asinex zz xsinex 96wax t 32x'coax 64 wax 160xsinex flex 32 2 160 sin ext X zcosex 480 64 4sinux t 64x wax 64x crux 384xanax v8 x si next 480 644262 128xsin ext 896xsinex 960 128x coax 1344 128 4 Max 128x sinex 1170 160 wax 160si next 320xcoax fix o x2sinex fYo 12 8 7 wax sinex 48x wax f'd Flo Xox x2sinex a sinex z ex as ex resinext x ras x 3 Tnx 25inext 4xosax t 4 xcosex 4 F f 0 to zsinxtgxasux f'd a 160 2 32 asinex 160 64 3in ex 320sinex 1 7 384 0 as ex 384cosex FRO X asinex 768Xsinex 1344 zgy 765 FX 896 sinex fix 2 3584 sinex 256 22688 sinex 2 3584 51227168 512xsinax 4608X six 512 2 81 2 2048 Xanax 512x Snax t 256 1344 1284 t rsinex 128 2x wax 1792Xanax 256x coax 8965inex 256 x2 as x as ex 9216 wax 264 0 2048 as ex 2048 coax f x asin ex 4096xsinex 9216 therefore fro flo x fig x't fÉ o to x EX v43 43 5 5 t 45 7 x 3 Ex f x4fÉ x5 f x6 411 7 1148 f4 x1 4 6 5 141 47 8 x 1341 7 8 x8 934 9 t Math 150B Midterm 3 Student Name: Question 5. [15 points] Student ID: 6 a. (10 points) Find the quadratic approximation of ln(1 + 4x) at a = 0. Solution To find the quadratic approximation calculatethefollowing fix In144x fix 4 fix 41445 4 Therefore Pax ln 144 o fro 4144x flo 1614452 the quadraticapproximation fool flo x F1X 0 h1 of h 144x at 0 4144.051 f'd 16144.05 16 44x Inc of 4x 4 at a x2 4 0 is given 8 2 £ § b. (5 points) Use the remainder formula to estimate the error on ° 18 , 18 . Solution given by Rex f C16 Therefore IRan by the remainderfor the quadratic approximation of hit 4x at a 3 x where c is between a f'd we a0 r 0 and x inclusive 14433.4 53 2 43144 5 74 for X E E t JJ HIP FIFER E IEEE HE É It 0 is Math 150B Midterm 3 Student Name: Question 6. [15 points] Student ID: a. (5 points) Find the 3rd order Taylor approximation of Solution p 7 x at a = 4. Tofindthe3rdorderTaylorapproximation we calculate the ofTx at a 4 A4 54 2 fix Tx t 4 t 4 fix L x 4 4 I fix E LHX 2 4x f4 ta 4É 4 52 52 51 f4 3 4 fil 4 t x 8 Ep 6 8 Therefore the3rdorderTaylorapproximation Tx at a 4 is givenby of 4 4 4 t fifth 4 3 a Paix fix flu ox 4 Ev 45 321 47 following 4 413 ex 4 4 p tax by fix b. (5 points) Use this to approximate 6. n t Solution For x 6 we have 8416 4 t 2 416 4 1316 t at t t 16413 84 264 1 4 84 15,7 c. (5 points) Use the remainder formula to estimate the error of the approximation. Theremainderfor the3rdorderTaylorapproximation Tx at a 4 of approximate 56 is givenby Solution R318 4 to 16 4 where c is between a 4 and x 6 inclusive fE and 154 3 1E X 14 4 E 16 414 Ygx 1 76.8.8 E 447 641 e 4 24 É 24