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Calc 023 2011 Exam 2 Solutions

Calc 023 2011 Exam 2 Solutions - NAME Signature Section...

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Unformatted text preview: NAME Signature Section Instructor Mathematics 23 Midterm November 10, 2011 I. /14 points 2. /_14 points . . 3. /14 points L/ E Y ‘ , 4. /15 points ‘5. /14 points 6., /14 points _ 7. ,- /15"points TOTAL: ______/100 points Do not turn the page until instructed to do so. You, have one hour to do the exam. Do all of the work on the test paper (use the back and/or request more paper. if you need extra space). Show all work. One may receive no credit, even for a correct answer, if no work. is shown or if the work is incorrect. Do not simplify answers unless tOId to do so. Do not write'answers as decimal approximations. For example, if V2 is the answer, leave it that way. - On the other hand, you must evaluate elementary functions as standard values. > For example, simplify sin(7t/2) to 1. Do not use a calculator, computer, cell phone, notes, books, etc. Mathematics 23 Midterm Exam; November 10, 2011 Answer all questions and be sure to Show all work. N 0 notes, books, or calculators are alloWed. /, 1. Find the directional derivative of f (any) 2: 1136‘” at the point (2,0) in the direction of the angle 6 = 7r/3. ' v , Vi: : < {“34 7- axe-19> =3 VFW, 0) ‘9‘?)- \/\/4?_ ’y Remix "in ,,qulflm)r€ ’ 2. Use the chain rule to find (32/33 if z = 6“” Where a: = ‘s/t and y = t/s. ’33 2X 99 ‘aj EDS VXA'Z. + =€~ji+ ale/X 23C~£> a t V 5?, ,- “LR . 3 s11_ S 'i? ‘ ': t 5 i T34“??? «4: .‘ ' "i:.> 2 5L , 3. Evaluate the integral fol f3: 6(“2)dwdy. ‘ 3 2“ L X 3 f .6 0%“er if“; X X; :* IEXZ 3 l ”‘7 ' e a 5 Zfi~l> 4. A lamina occupies the region 17— — {O < :1: _<_ 9,0 < y < fl}, with density function p(ao ,y)— — 3/(332 + 1). a. Set up a double integral giving the Inoment of D with respect to the y axis. Your integral needs to have explicit limits of integration. lixfarm 7W r-‘r 3 w '- 9:33)) X XL laiéj‘dkpr, ”i! c2, b. Evaluate your integral in part (a). You may not receive partial credit for this part if your answer to part (a) is incorrect. " 5. Set up, but do not evaluate, an integral giving the volume of the solid above the cone 2 = V332 + y2 and below the sphere :02 + 3/2 + 22 = 1. Your integral needs to - have explicit limits of integration. (5,); 95 N; 5:; l /; L/ (Drfifiev‘*£ema £5 KER-{‘2’ é 52: 6 Consider the solid E enclosed by the paraboloid z— — $2 + 33/2 and the planes :3: p O, y— — 1, y— — 31,2: = 0. Set up, but do not evaluate, a triple integral giving the volurrie of the solid E. Your integral needs to have explicit limits of integration. 7. Use Lagrange multipliers to find the maximum value of f(a:, y, z) = 143: +183; + 122 subject to 3:2 + y2 + Z2 = 101. How do you know that your answer yields a maximum and not a minimum? PM}? ‘ aCY,y/2)= X +3 Jra WWl.‘ glA/xj I'Bllgié‘e \[cplrw—w E‘n'l‘zi‘ jCX/y/g) 2.0. % {jg—L z , “L :2 E?» #6:)» Ci) w .;v|i"\ r , J 3 A 4; j VE ~ MB ’ Um; QU\ WE. gel? _‘ Jr W i X :1 7 IN ‘ _ __ :JE NH; / 5 0’ ' é Mg _TL+‘: mohw'\mw“g Qatar's lkllmfrxé HQ, {GEM/533‘?» 3% “a“: gl‘va‘l L974—lyol'l-6IZ/Ifif; ngllgt“ l Q th WWW» ’K: \jalmea “\S‘ git/{“0119 L93 Cl’Lv/G'J‘tqg 6:51} , [-1.me 131: give; 3&le Smefllarisf Vail/legs) igé; ”ZEZJ if; ...
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