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Unformatted text preview: Prof. Girardi Math 142 Fall 2011 11.22.11 Exam 3 MARK BOX INSTRUCTIONS: (1) To receive credit you must:
(a) work in a logical fashion, show all your work, indicate your reasoning;
no credit will be given for an answer that just appears;
such explanations help with partial credit
(b) if a line/box is provided, then:
— show you work BELOW the line/box
—— put your answer on/ in the line/box
(c) if no such line/ box is provided, then box your answer
(2) The MARK BOX indicates the problems along with their points.
Check that your copy of the exam has all of the problems. (3) You may not use a calculator, books, personal notes. (4) During this exam, do not leave your seat unless you have permission. If you have a question,
raise your hand. When you ﬁnish: turn your exam over, put your pencil down, and raise
your hand. (5) This exam covers (from Calculus by Stewart, 6th ed., ET): 11.8, 6.1—6.3, 10.310.4 and the take home was over 11.911.11 . Honor Code Statement
I understand that it is the responsibility of every member of the Carolina community to uphold and maintain
the University of South Carolina’s Honor Code. As a Carolinian, I certify that I have neither given nor received unauthorized aid on this exam. I understand that if it is determined that I used any unauthorized assistance or otherwise violated the
University’s Honor Code then I will receive a failing grade for this course and be referred to the academic Dean and the Ofﬁce of Academic Integrity for additional disciplinary actions.
Furthermore, I have not only read but will also follow the above Instructions. /‘—'z,/’—_\ Signature : 10v) 31W? E’Xm it am; 1. Consider the formal power series i (5m + 10)n
 n=1 n
Hint: (5x + 10)" = [5 (a: +2)]” = 5" (a: +2)” = 5" (a: — (—2) )" '
The center is $0 = ‘1 and the radius of convergence is R = ‘5 .
As we did in class, make a number line indicating where the power series is: absolutely convergent,
conditionally convergent, and divergent. Indicate your reasoning and speciﬁcally specify what test(s) you are using. Don’t forget to check the endpoints, if there are any. &.
(Ogmvl obs. (oWV' (AM/Uses
‘11 v
nll 2. 10W 537mg Ecmm J j; 7, 2. Consider the formal power series ”2:22 (1n n)" Hint 1: (lairt); = [Ff—"F so would you rather use the root test or the ratio test? Hint 2: ln(a’) = rln(a.) but (ln(a))' aé rln(a) The center is $0 = 0 and the radius of convergence is R = M . As we did in class, make a. number line indicating where the power series is: absolutely convergent,
conditionally convergent, and divergent. Indicate your reasoning and speciﬁcally specify what
test(s) you are using. Don’t forget to check the endpoints, if there are any. 01) “V‘ 5
, g ”madly Qi 937;, 3g: TH'ISPROBIEMHASPARI‘S:7a, 7b,7e, 7d. TheragionRisthesameforandpam.
WRbetheregionintheﬁrstquadrmtenclosedbyy=2xandy=x+4andz=0. 3 la. Empress them ofR as integraKs) with respect to z. 3 12:. Using the'sheﬂ method, empress ,as inbegraKs) the volume of the solid generatgd by revolving R about the
y; axis. x="{
'3 1M [um—(2%)] ax
=0 Volume =
'X W) (W0 (knWB 2': (am) [ «mama, M ,(m) ( 9c) [ UH) “(7.701 M 5 3 ﬂd. «QMi a. q a "M Mr (mm? 55.4%) Mﬁm Mu H to alga r}. WWW =3 Pamﬁﬁ W at _,4_‘$
CBS the disk/W mm: “P1993 33 integraKa) the volume of the solid generated by revolving R about the line y = 1, " Fal’QOOj thaj E’quﬂ’ laéﬁ/gm'mﬂ H [If Gimme: the Curve in polar coordinate
r = 5  5 sin 0 . 4 la. Theperiodofr=5w5$in0is 2” ‘n'
L! 1a. Lhcmﬂofgm5—5iﬁn0 z 2. Ll /Ic. Make. a chart, as we did in class, to help you graph 1' 2 5  651x19. 1d. Graph 7' == 5  531219.
Clearly label the points, in polar mordimt (139), where the graph crosses the :zaxis or vy—axis. (513 T “3ka bike 3N We, ,1 Fbcprws the area enclosed by r = 5 — Emma as an integral with Impact to 9
(0k with rospmt to 9 means 3 d0 in there).
(You do not have to evaluate this integral.) A953? [#0)]? :19 K, < xww’é’Y‘H
There: mi» mama: magmas (4&5 —\'o  (3’3
m 2
A = I, i? ' 5 5‘ 9 C}
7’ ~30 [5 S r 1 9
E32 2..
95‘ “2'22 j 55—»Ew9] 0/9
—Tyz m z
wa/Z _ 2.» k i. a 52’ K [13'55‘h61 39
i " _ , 5.; i
57»? Qai Lbwswl an
" a:
333: 2
H w
M7 3’ l S [E'Sstné—quja A 1 131.8 [5’55171éj
1; i
pay W . 1 f! 4 n} s,
c)" 1 ‘ .L a” 1‘ LE” 55!“.9] Ob H\ 300%) ﬁiilfi
“:2 2. {It 26079 1
.. , Z __ « 4 >_ . '2. __ ﬁ 1. _a$m6+i
‘ '5 ‘ .6 5 J5 5535:): 9 4255m 9 ’ZBBmB
m: is w 1 2.3 d‘la/ (11>) CNN <4 so \MWM 4? Mb“
~—+———~——l————l————r —I’+ 0 WQ 5. Fillin the 6 blanks. Consider the power series
00 2 n=1 0 is absolutely convergent
o is conditionally convergent
o is divergent \_ r ' 4’" g
\QOIO SF m3) EXOLYVL 12,] 0 7 . )2; oi
WM. 9P”) m “3 21(4)an : 2(4)" an M)“ _n q‘ 'V,$o
x:i?%¢wZUlan'X amnion : Mnovd on > 0 MA ”C if a: = 17 then Z(—1)" ans,"1 conditionally converges
if 17 < :1: then Z(—1)" aux" diverges . Then this power series %/ ”’"MWW’
center at .730 = 0 and Fri __ .9de d— QNCYI so
Hyman» .. z:( (—0 a (X 0> h‘l where all of the an’s are positive. Let’s say that you know that
if 0 < a: < 17 then Z(—1)" ans" converges radius of convergence R = ’ :lL Also, what can you say about the following interval? Fill in the blanks below with: o inconclusive (not enough information given to decide in general). if — 17 < a: < 0 then Z(—1)" aux" abS. WV ’
if z < —17 then Z(—1)" aura" (Al \1 or ¥ '
if a: = 0 then 23—1)" aux” dbl) 3 W02 ' /"_' if x = ‘17 thén EH)" aux” M n n
1 Z [to ['17)] an = mm M n n a )7“ : Z Pym?) a omU’WOU/MJ n . 21H)“ Pr 2 2: (re an diic’t‘fg' ...
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 Fall '11
 KUSTIN
 Calculus

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