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Unformatted text preview: ﬂ/W‘ﬂ/vt .’ 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 13 seconds is given by s = 42252  153. After
how many seconds is the acceleration, equal to zero? (a)lsec. My: (b)5S€C. (C) 7sec. :5 (—49 (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) 22/8 //%/= @556 (hm/4 W439 :9 lay/£766
(’7 (c) 24/3 (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
walks north at 2 mi. How fast is the distance between them changing after 10 minutes? 1? 1314 X: E. E #M’L PfM7RW/9/M/4 ﬂm V/A éﬂ/mﬂ'ﬁ/rr‘m } (c) [email protected] mi./hr. A (0/1)
4: Y 0“(—xl) *4. /cx—//‘“ (d) 6m mi. /hr. 6990) w ~ 55:5; ; l
d’/{/; éfxluﬁ'm [ﬂxgﬁ/V‘Q‘JLQEH: (a) m/2 mi./hr.
)4 (b) \/2—0 mi. / hr. « ? T: M
4 14 19 (e) 10¢2‘0 mi./hr. A #6141” /0 WWW .’ X 3;) (DJ 34/50” : 3,, $5 t
5‘ 1,» may: i «afﬁrm «; 9.420 a: ,
Ellison ) 77m ‘ (4) 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 3—"? ﬁx m 4 I . m L (a) 1/10 radians/sec. D J '0 _ {,0 “ta/érzihglﬁgrj;
@ m C ‘1‘} 72?" 20 42/5 60 l (b) 1/15 radians/sec. ‘
wﬁm 6’ 3 77/3) S’é/céi7o? (c) 3/10 radians/sec.
(d) 4«/§ / 15 radians/sec. (e) 8/5 radians/sec. / ('3 IV / (9 “1/2:
~  1/? / 7.1 “L, l
(a) 3+ 516 X ‘1; t) X "”§)_“"“3/:?
<b> Km 9 (a > 4 #7sz mm, @1537,
(0) 4"+93% :7 37f??? :3 1/37 4 3 27% *‘ Am)
1 r g + 3’ “‘7 AM
[ﬁn‘4)» (8) 3 + 2—756 I
3” 3.) 7A “WM”
3 76? K)
(6) Let f(x) = x3 — 3:32 + 3. Find all values of cc Where f has a local maximum.
(a) a: = 0, x = 2 y _/
:TE/I/kj: ﬁxkﬂéx @9321 :5ng X75/ X53 ,..,.»zi';r7:i¢J/¢/rmx a) y 3 c? 5 (7) Find all open intervals in [0, 27r] Where the function f (t) = sin tI—cos t is decreasing. ,_/ M
quj‘, LL V 71
sew
57 EC, .1— 9/7“
1r 7/
i)de nL (8) If f is continuous on [5, 7] and differentiable on (5, 7) and its derivative satisﬁes 3 Z f’(w) > 2 for every 1c in the interval (5, 7), we can conclude that f(7)_— f(5)
is in the following interval: (80(476) :aW/N) 6’<(‘<7
(b) (3,7) 1 ,
Q 2‘ (“/60 5 3
K—a (c) (4,61 :3 4 <wg) é é
(d)[3,7l ‘ ‘: 15/77) ’“ﬂ/S’)
(e) (071] (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 (a) (0,3) and (5,7) l
(b) (2,4) and (6,7)
(0) (1,3) and (5,7)
(d) (2, 4) and (6,7) (e) (0,2) and (4,6) 3‘0 %”>g) 571 /g{>) (I [4453) (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 f(8) (b) f (1) and f(5) (0) f0), f(3), f(5) and f(7) .
\9 (d) f(3) and f(9) (e) M) and f(3) , 7 , , \ , , L/ , 3/ v / a) (3/751
$32 Agiégmign.1.:.4L,...€2;,f 0 f ‘3 6 7 L7. {5/
4
A ' .x‘“ ).__L, l, *f ,5: , W, ,m J, / J ‘7 4‘
Z0 dim/144m zit/‘1‘" 3 9X I “ﬁr/[i] (l V12 14/; 2/», (11) The limit (a) —1/5
(b) 1/3
@1
(d) 1/4 (6) —1/3  _ / > , I
,I \ L
hm *— ls equad to ‘ ,. "'47; , » L "’f
xxx/w I; , mx w/ , Cam, 2%; g) 8% an “(gig/7h ~— ALF, A/
Xad 6x ( 12) The graph of f = 23:3 + 3502 — 12:13 + 1 looks most like Which of the following? 0‘) w "5 ﬁn m : [9x 4— Q ﬂea ravgimi;—~ 7“?!) L/ f3.) _/2/’Lr*)zi’f ﬂap/awawg’l of" /77 57 ‘1va /; kip [f ¢ . “7, ...mm.m.~am«wnM‘ magma //1, a y 4 7 9 /
(13) The point on the line a: /7 Which is Closest to (17 2) is: <a>(1,8> j 1/ WI
W77 (1» (~25) "
I (CM—1,6) '1‘
(d) (0’ 7) W I m) i/w)‘
(e) (3’10) /’ 2* (MW/W“) / .1 _ \ ‘
D (3"?) " 9* “WM :>~(’x+e‘f)
: a (:3 W9») it? a?) he; :f) 1:]:(L‘3)7L7:{ (433;?) l
I
E (14) The maximum and minimum values of f = 3:2 + 4:1; — 3 on the interval [—3, 3] I
are respectively A I /(/>{_ ) f 4/ :‘i 6) ‘Qx, I 73  ’ ‘9 a («3, “9/ (b) 18 and —6 T‘““‘“JT“““+ f C é ? t “'3 “A 3 12y) [LO] ! (c) 18 and —7 ’ (d) 24 and —6 (a) 18 and ~5 (e) 24 and ~7 ...
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 Fall '08
 Gabriel
 Derivative, Scantron Sheet, recitation instructor

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