Sol-161E3-F2009

Sol-161E3-F2009 - Q/W‘fl/vr .' MATH 161 — FALL 2009 ~...

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Unformatted text preview: Q/W‘fl/vr .' MATH 161 — FALL 2009 ~ THIRD EXAM — NOVEMBER 2009 ' TEST NUMBER 01 STUDENT NAME STUDENT ID LECTURE TIME RECITATION INSTRUCTOR ‘ RECITATION TIME INSTRUCTIONS 1. Fill in all the information requested above and the version number of the test on your scantron sheet. 2. This booklet contains 14 problems, each worth 7 points. There are two free points. The maximum score is 100 points. 3. For each problem mark your answer on the scantron sheet and also circle it is this booklet. 4. Work only on the pages of this booklet. 5. Books, notes, calculators are not to be used on this test. 6. At the end turn in your exam and scantron Sheet to your recitation instructor. (1) The position function of a particle after t seconds is given by s = 4:21?2 — t3. After how many seconds is the (acceleration equal to zero? 7-\ (a)1sec. (b) 5860. :- V/(‘f/S (c) 7sec. fl $15 a (d) 14 sec, (e) 28 sec. (2) A material has a half—life of 12 hours. If initially there are 4 grams of the material, how much is present after 8 hours? ’ (a) 22/3 /”/%/= Pa 5 d f (b) 23/4 pa :9/ =9 PZié/z {flat ('7 (c) 24/3 i (d) 23/2 (e) 8/3 3 (3) TWO people start from the same 'point. One walks east at 4 mi. / hr. and the other [814 er vflflmm pr—Mng/WN’ >6 . aw“? (b) mum/hr. ‘ ? I: p 4 x4 (9 V”, éfl/I/qflexrflm (0) 2m mi./hr. A (020) d: Y(0~/-xt) ’“+ fir!” (d) 6m mi. /hr. (ego) ” — BEE;- a’W/z s‘fx'fizf/ “(A {Rxafllfl‘ifi (e) 10m mi./hr. A #64 r- /0 mm 'X 5.10,. g 34/0, : a» V Std/(2a): .’ .~«zz”ézo.o ram-a a; . ' mm ) ‘ ‘53 A balloon is rising vertically from a point on the ground that is 60 feet from a ground~level observer. If the balloon is rising at a rate of 24 feet / see, how fast is the angle of elevation between the observer and the balloon increasing when this angle is g? ‘ a: l M A. . g W’a (a) 1/10 radians/secs 3 J '3 "O V. w “ad/éTZLQ/j’xl (” . SO ’0 6’ ‘22? QC) 416‘ (m LLWLM 49' 3 TV}, $446950? /M’ w 522.. 3,1 (e) 8/5 radians/sec. / £0 /0 (b) 1/15~ radians/sec, (c) 3/10 radians/sec. (d) 4«/§ / 15 radians / sec. (5) Use a linearization to approximate the value of {727101; .flj/ ' ‘. ‘/ . 51, J_ (a) 3+3}6 X _ ~£ :3 X "<§—)-:a/_3 “93% W301“: Wan #74») (xm), 3%11327) (0) “£5 :3 3Y3???" 5—" 737+ 31.2773 ‘ Am) (cw-2% ’ .. I .1 w 3 «4L “*7? ' ‘ ”“ n am If?) (e)3+_27%6 7 3 if / '7?) '7L "”'"“"'” 3760 (6) Let f 2: :03 —~ 39:2 + 3. Find all values of so where f has a local maximum. : exwx- —: 2 MM.) (0) a; = 2 3N - 4 >0 “<0 +’>O . «MWMFWWf—w «PM? (d) a: = 0 0 (e) 53 = 1’ 2 flimifimmfimpflj; 0 3’ i/«(imfl/(mmx a? 3/ 3 (9 5 (7) Find all open intervals in [0, 27r] Where the function f (t) = sin t+cos t is decreasing. / (it) : ([25 { 64% Z‘ 7: (J 2% Cm a» :mv-w Z" WW :9 2‘ 24277; W €517“ \__ I, / ' 1/70 %I/< 0 7/390 (6) (0,%)U Z",27r) 0 I 577 3/7“ <7 77 4: . W R? ~f~ W377“ 4’ ‘7; “c; Dec/L Wt (55/ (8) If f is continuous on [5, 7] and difierentiable on (5, 7) and its derivative satisfies 3 2 f’ > 2 for every :1: in the interval (5,7), we can conclude that f(7).— f (5) is in the following interval: WWW?) = 41%») (74* J :a nil/ft») ) Q 2‘ JAK/(V) 5 3 ::D 2/ <w) 3 Q rfl/flwfi/s’) («Ml ' (a) (4, 6) (b) (3, 7) ‘~---? (0) (4, 6] ¥<n<7 (d) [3, 7] (8) (0,11 (9) The graph of the first derivative of a function f is sketched below. We can con— clude that f is concave upward in the following intervals 9C) WMM m w») a (w) (10) If f is a function such that the graph of f’ is as sketched below, we can conclude that the following are local minimum values of f. (a) f(2) and its) (b) M) and f(5) (c) fa), f(3), f(5) and W) . (d) 116(3) and f(9) (e) M) and K3) _(Walnut,_.w.w._J;_W..-M_ .W__‘+fi__._+_._/.o 0 l '3 S 7 l (,7 0/ A - 4 a: ML ; Lamaze . 3: .7 q Z0 ciao/mm 3’ 91 7 , f {3) V 19/ (/1 41-2412 .Ae‘iawaz/z/M W», $52 $90 tea-v P >41»er pa m (11) The limit _ ,7.) ———-——~ ‘ J 1 fl (8") “1/5 - . ~ /7/u/U‘/ (b) 1/ 3 X—9 (,7 X a» s d (0)1 . , (d) 1/4 ~Qi/de X é; (8) —1/3 8 ( 12) The graph of f = 2333 + 3932 ~ 12:1: + 1 looks most like which of the following? a) w '7 W 0 > *—= ' 37W” V: -.,...o.._..(._ ' : w : £01” a J éé‘igxfl.) Kw) W 7 0 a d/u Ml” W/z. ., 5’3??? “7’41 \w—w... . now/w: [awa 3’20, W757 gfiiajr _. “‘sz [/‘L/V KIM/11} fl? 5? ML //‘/c<.fl // V. mam [b ) WNW-WWW w /'LI) 5 Y 7 9 (13) The point on the line ylr— a: ’7 which is closest to (1, 2) is: E (Y/ M / Wyn—«V anmlw. . _ ’ lav/)9) (Ly-w" " ' Y‘ y "N ‘ ’9‘ w J ' ' l low” I 1W3“ lit/42%) : / ><"”/> + Mn?) / H D ()0 “: (ya/)1; [)I’)“. D (X) : D. (xv) + c)‘.(i‘x+gi) it a (Q 57C" a?) )(gr‘ius if? L/ : (SH 7:? (4 DJ <35) (14) The maximum and minimum values of f = 2:2 + 4110 —~ 3 on the interval [—3, 3] are respectively 13/05) 1 336+ 9 :0 X‘fi'fi‘ fl T W<0mé3ry ~3'1 '3 f/>C)MZ~>/3) (a) 18 and «5 . (b) 18 and —6 (c) 18 and —7 (d) 24 and —6 L jam /: </ « flw g4?“ 4%. m a? >< (e) 24 and ——7 MN} l WNW") 74/9g r l: Z74 /V‘l z» _, u We”: F7 wt? ‘: W / ...
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Sol-161E3-F2009 - Q/W‘fl/vr .' MATH 161 — FALL 2009 ~...

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