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Unformatted text preview: APE Calculus Ell:
HIDDE Scaring Guidelines AP” EALGULUS EIL':
20:15 EGDHIHG GUIDELINES ﬂmsﬁm Letfandghethnﬁnﬁmsgiwnhyﬂx}=%+sin[lrr}audg{1]=4'x.[ﬂ Rmmmmgimmmemmqnmmmmwmmmmgmpm nffandg, mullet Shattestﬂdedmgimintteﬁrstqmdtﬂmﬂmdhytha
graphsnffatﬂg: asimniuﬂmﬂgmabm (1}Findthnamufﬂ.
fn}FindﬂleaIEaufS. {c} Findthnnhmm ﬁfth: snﬁdgenmtﬁlwhm S ismwlvarlahnntitehmizmial
lint y = —l. ﬁx} = ﬁx} when i+ signj} = 4". film! 3 Mementwtm :r = DJTEllE and Emuhen I =1.
Let a = 0.1?32111 {a} _l':llgllx} — ﬁx}! dx = 0.054 urﬂﬂﬁj (In) Kim} — 34:11:} :h = 114111 1 "I
{c} JrLﬁﬂx} +1}1 — {g{x}+ 1r} d1 = 4.553 014.559 :iﬂIEgt'aﬂd : limits, constant, and answer BZJ'L: AP” CALCULUS so
zoos scoemo emoeunes ﬂmstionZ The enn‘e above is drawn in the {LI—plane and is described by the equation in
polar coordinates. r = E + sin{:'3_'.t?:} for t] E t? E Jr, wtﬂ'e r is measured in
meters and t? is measured inradiam. The desimtive of : Wi‘ﬂl respect to t? is r = E+si1[lﬂ'l _/ _ d; ..
£11:me E= 1+ Eeostltl}. fa?! I
(a) Find the area hounded by the mute and the I—aJris. ,. [13} First the angle 5‘ that conespomts to the point on the mine with Ieo ordinate —E_ E c " — isnegatit‘e. What does thisfaet say.r about r'?‘What doesthisfaet sayabonttteemve? tn) For 3 . . (it? [:1] Find the value oft? in the interval 0 E E3 1" % that conespends to the point onﬁle em‘eintheﬁrstquadrant
with greatest distance from the origin. hmﬁﬁr your answer. [3] Am = HS? d5 ‘ . .  ...1 .
= 3'3: w + smiLIH 1r d3 = 4.333 {b} —2 = runﬂag} = {H + sin{.?.f3}}cnsl:ﬂ}
I? = 2.?36 ".I . _
[c] Emceiiﬂfuriiﬁé‘f—x risdecreasingunthis l:mfumnimnahnutr :13 3 3 ’ _ _
ﬁltenral This IEEBIIE the. EMT»: is getﬁng closertu the. origin. 1 ' mfmﬂun about the. nave. 3 __{1:E=_£url_ﬂ47
_ 1 :amwer with justiﬁcation The greatest disﬂame DCCDIE whm I? = E. 3 AP” CALCULUS er:
2005 see RING GUIDELINES I''I. meta] wire of length 3 centimeters (c m} is heated at one end. The table above g'vﬁ selected 1iraluﬁ uf Ihe tempenture
fo:I:_ in degreﬁ Cﬂius (”CL of the wire .1: cm from the heated end. The funclien T 'E decreasing and. twice
differentiable. (a) Estimate T1?! Shaw the work that leads to your answer. Indicateunits ufmeasure. {b} 1'Write an integral expremien in terms of Fix} for the average temperature of the wire. Estimate the average temperature
of the wire using a Inpezeida] sum with the fun: suhinter'I‘als indicated In; the data inthe table. Indicate units of measure. I . E
(c) Find Ll T11} it; and indicate unit of measure. Emlain the meaning at J: Tiff} di in terms cf the temperature cf the 1IIIire.
{d} Are the data in the table consistent with the asserhnnihat This} 2: U for every x iuthe interval U n". x x'. 3 '3 Explain YEHII EDE‘IVEE. rm] — r1353: _ 35 _ s: (“3' s — a 5: [b]: %_[:T{x} (II II . .
Trapezoidal approximation for In T111} dr: ' ' ' H 1.1 1.1 ‘2'
=1m+931+93+ml4+ ”0+6“.1+5“:3.21 ‘1 '1 '1 Amuse tat11mm == $4 = T5.EET5'='C A I: ﬁx} d: = Tie} — rm} = 55 — 100 = —45=e TIE. tempetatme drops 45cc inn: ﬂle heated end. efthe wire to the
011:2: earl efthe wire. Average rate nfehange effemper‘atme on [L 5] is T”; : 5:3 — —5_'II'3_ _ _ J' l ItEUD slopes efseeani lines ﬁE—Tﬂ=_ﬂ_ I'll:answer'thiihesplanatien
5 — 5 No. thhe MVI', Tic'1] = —5_T5 for some {.1 in the interval [1, 5] and I’ll5'2} = —E for some :1 inthe interval [5, 6]. It follows that 1"" mt decrease somewhere in the interval [01: :3}. Therefore T" is net positive for ever].r .1' in [Ch 3]. Average rate of change oftemper'atme en [5, 6] is [hits of C'Cfem in [a], and. DC. in [b]: and {c} 1 : units. in (a), {b}: and {e} AP‘" CALCULUS so
zoos soo RING GUIDELINES Gilestiond Consider the diﬂhrentzial equation $ = 1r — y.
{a} On the axes provided, slzetch a slope ﬁeld for the given diﬂ'ermtzial equation at the.
twelve points indicated and sketch the solution curve that passes through the point.
[I], 1]. (Fete: Use the axes provided in the pink test booklet.) I [3' I {b} The solutionemve that passes throughthepoiut. [It], 1] has a local minI'mmnat .1' = lnlé]. What is the _v—co ordinate of this local minimum? {e} Let y = _I‘"{:_r:} he the particular solution to the given diﬂhmtial equation with It: initial condition
H'ﬂ] = 1. Use Euler’s method, starting at :I: = D with him steps of equal size, to approximate _,f'{:—D.4].
Show the work that leads to your answer. d2}? ‘11.! greater than ﬂ—ﬂﬁ] _ Explain your reasoning. Find in terms of I and y. Determine whether the approximation found in part [c] is less than or l : zero slopes
3 : l :nunzero slopes
[ l : cums Enough {ﬂ 1:! {b} $=Dmﬁn2x=f E:{l:ssts%=ﬂ Larisaer TIE. y—coot‘dimto is 2 hﬁl' (c) Iii—03] == H11} + .f’{:D:}{ﬂ1'} j . J’ 1 :Elﬂsr's metlmd with two steps
= l + [—lH—DE] = 1.2. .. I L l :Euler approximation to fl:—ﬂ_:'l}
11—0—4} = Iii—us} + .Idliﬂlili—ﬂl!
a 1.1 +{—1_6]{—{}_2} =13: _ d2?
_ ctr:
" . _ _ _ _ . . _ l : answerwithreason
dill Epomutromquadtantﬂhecame I 1.1] and y ,.. U.
1.52 .:_ ﬂ 4214:! since all solution slaves in quadrant II are
concave. up. 1 '1" . AP” CALCULUS er:
21:95 scemne GUIDELINES ﬂmetion 5
A car is traveling on a straight road. For I] E t E 24 seconds, the car’s velocity ﬁt}, in meters per second, is modeled by the piecewise—linear
function deﬁmd by the graph above. {4. 2411'] { In. El]? M
lung 24 24 (a) First ju vljrjy dr. Using cement units, eaplainthe Inag ef jﬂ my Jr. (b) For each of v14] and 1‘"I:ED:} _. ﬁnd the value or esp[min 1.1.th it does not
exist. Indicate units of measure. Tim. {wands} In ‘v'elln'ilgu
{metersper amend}
E 4 3 12 lt'l ED 24 (1:) Let a{:t:} he the car’s aeceleratien at time t, in meters per securulpet' second. For {I :. t < 24, 1write a
piecewisedeﬁned ﬁmtiun fur aﬁ}. {:1} Find the average rate ufehange of 1‘ over the inten'al E E t 5 EH. Does the. Mean 1'LI'alue Theat'an guarantee
a value of (3., for E d: c.‘ er. ED: such that v15} is equal to this average rate of change? Why 01' why nut? ‘ 21.. 1...,1. ._1.._1. 1..._‘.
(a; In 11_r_!dr=§l_4}l_2{}_l+£12.ll.2ﬂ}+§{3}l_2ﬂj =
'I'teearttm‘els 364] Maehﬁlese 21'1 amends. [h]: 2'14] ﬂees net exist because ' : v14} does not exist with cremation
I rue—$14: _ 5 ﬂ} _ "j ' ' : :v’lj2u}
1+1" ~.. 1‘4 2 _ :um'ts
2n — u a. =
" “m: 15—24 [c] 3 ifD<r:.4
ﬂ ifriczrrzlﬁ :ﬁumthevames 5= a, —% j _ : identiﬁes comm with {tenant intervals
—— if 16 a: r :. E4 ‘1‘ aﬁ} dueemtexistai 1 =4 and r'=llS_ HR} = [d] The average rate of change of 1' en [3, EH] is . _ ' : average rate efchauge of v on [3, ED] “[303 — 11:3} _ 5 1 2
20— a ‘ ‘E "If”: '
Ne, the MeanVahle Theoremdees not apply to 1” [3, ED] because I: is net diﬂ'erenﬁable at r = 16. : anm'er with lamination AP“ CALCULUS so
zoos seo RING GUIDELINES Question 6
Letfhe a funetionwiﬂs derivatives ofsll orders and for which HI] = 7. When :n is odd, 1t: nﬂsderivreaiis
— I
off at J: = 2 isD. When n isevenand 11' 111 the nthderivativeofa‘ at J: = 2 isgit‘erl'lznluT f4”]{_1}=hi—ﬂu.
{a} 1Write the sixthdegree Taylor polymmial for f about J: = E.
{h} Inﬁll. Taylor series for f about :I: = IL what is the coefﬁcient. of {I — if" ﬁJr Jr 31 l '3'
{c} Find the interval of emsret'gezte of the Taylor series for f about :I: = 2. Stew the work that leads to your answer. la} ﬁl_.1r}='_?+ 3%}; ' " : —! 1:]Joljrnonrialahom1':
 2:133:11!
{—1} each incorrect term
{—1} max fot'allexh'atemls,
+_ mismeofequalitjr L111 mtenI—2i3.
Thus theseries converges when—1:: I d: 5. When I = 5. the setiesis 1+2”? ”=1u" which diverges, became. E %, the hammnie saies: disages. Fl=1 _ 2.11
Wheux=—l, these:1e:1.s'7+v[3)_ﬂ = +%T‘ll_
::=12ﬂ. 3 Raﬁ“ which diverges became 3—1—1 the hannenie series divages. II—l
i:=1:l'I The interval ufemva‘geme is [—1, 5:}. : sets up ratio
: mines limit of ratio
: idalﬁﬁes interior of interval of eent‘ergmee : considets both endpoints
: whrsis"eenelmien for both endpoints ...
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 Spring '09
 Johnson
 Algebra, AP Calculus

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