Exam 3 with Answer Key - Analytic Geometry and Calculus II |, Exams of Analytical Geometry and Calculus

Download Exam 3 with Answer Key - Analytic Geometry and Calculus II | and more Exams Analytical Geometry and Calculus in PDF only on Docsity! MATH 2108-003 EXAM # 3 NAME: _ (Signature) TUBSDAY = cae KEY NAME: November 27, 2007 (Print) °3:30 pan. - 4:45 poms STUDENT I. D ee ee INSTRUCTIONS: There are 5 questions on this exam worth a total of 100 points. In addition, there is an optional EXTRA-CREDIT question at the end of the exam worth 20 points, but any score over 100 will be truncated to 100 . Clarity of exposition (including proper spelling and punctuation is an integral part of a correct solution to any problem. In particular, DO NOT PUT EQUAL SIGNS BETWEEN THINGS THAT ARE NOT EQUAL. But do put them where they belong. It is necessary to show all your work. GOOD LUCK: KEEP COOL! PLEASE SIGN THE FOLLOWING STATEMENT: On my honor, I declare that the work that follows is entirely my own. With regard to all the questions on this exam, 1 have neither given nor received help from anyone [including yself, say, via any type of cheat sheet or device (e.g., cell phone}]. Nor have I used a programmable calculator. NAMES: — (Signature + Name of Calculator) 1. oa, 30 4 > 2 3. a. b 0 a. d. a cf 20 fa 5.8. aw |S GRADE: MATH 2108-003 11/27/2007 Page 1 NAME: Wey lL. a, Find the (arc) length of the curve defined by the equation , 238 y gh for » satisfying 35<2<80, S/o a 2 ye 3 x dv Yeo. ! ay . 2, 3 UX = ye Tx a x “3 3 ¥ yo a ley jel/4g\* 2. 1 » (] = 10 points] (ay) ~ ——T7 ds= \!* GQ =i a x= 8D ya VO X= fo “7 We Ls fas [ dew | Gent) Ay Ke3S \o3s X= BS a ~ _ us go tla Fl Jax Xr t Co a = ue } 4 dle dun, a duds Uzasel= 3 ay a ee 3)g_j 42 Sl _ ae | _ 3 | eae x (fen P- (ey 3 \ Bo Lo [7 oe (4 | WAN | x42 | 3 { it | —~, 4G ~ 23 u-e| MATH 2108-003 11/27/2007 Page 4 vame: KEY an a =F) 3. Determine the following limits: 2 73 ® [ BX = L 3x a bn tania) R l= +.) yo ! -(ary! by -L mae as Be Yeo Ur ® SCoe BIN EX) Note alee HA _ 3 ae (OO) | _ lle) 2 ele) "2. ee ao re Teetoe }°3 = KL aciseyal Mtb ae el £ ys Soo Ye cas vot Sin Xl ~xM L ye™ + & Leer + Ue _ ® yo “ye (es %) yp Cesxcl + €O8% . lim [= —— ro -X * = 8cosx + 4e7* ©) iy. en esi 1 [5 points] & . gp eS tel ~ ye St 1G = 8 =e eee aR e. li 0 15 points] id. ny = |x | (1 sins) = 3 x Te, _ Ss. Iwlt rsmx Le m= be xX MATH 2108-003 11/27/2007 Page 5 NAME: __ KEY. a 3. Determine the following limits: & @®) (ti\ire*4) d. lim miete ) = EL. M+ ee* noe 3 n —— (5 points] B Frc. Coutrn ( ROK 2x) reeset fuueti {Dias S71 og [uy = & bee & oF MATH 2108-003 11/27/2007 Page 6 NAME: __ kev 4. A certain type of bacteria increases continuously at a constant rate proportional to the number present in the existing population, Hence, P(t), the size of the population of bacteria in the colony at time t. is given as a function of time, t, measured in hours, by the formula P(t) = Ce for some (positive) constant k. If there are 125 bacteria present in the colony at noon (time t = 0) and 6 hours later there are 190 bacteria present, predict how many bacteria will be in the colony at noon the next day( ie., mere pity Ge jas = Poy ~ Ce 2 C= Cle [20 points] WV ! P&)=C-€ Lace kG LR 40 = P(t) ~ Jas-e = [as-@ bet bh 190 _ §-36 =38 e = it “eas 25 ~~ wz Ne 36 NW 35 | | ng p.06t 78S Oss h Gb [x [38] Wb bet ~ 126-2@ gt l Ba (3894 gd es al. us é@ = 129-3 P/aa)= |29 eb ree slmue 43s (3t lng eth (33 4u3 ne bb 1. 244 EL). 2Y WIN