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Unformatted text preview: CE 1. solution on gr anhing calculator to
1 14E —e“F'lx:C whichis .339 Area: JEE—ewax) dx:
. El 01:» a) Intersection point of x5: and e43“: : 1.424 cuhic units 1 .
h) Um'ng washers: :rc J'(I:l—ewax)2—I:l—1ﬁllz)dx :
239 c) ICross section area: (ﬁ—eﬁxﬂﬁj («H—e4“)
1 1.554 cuhic units Volume of solid : A 5 (ﬁ—e—3KF dx :
9 a) f is increasing when 1" ' {x}: 0 or b) The second derivative is positive when the first derivative is increasing and
neg ative when the first derivative is decreasing. The first derivative goes from decreasing to increasing or the second derivative changes sign at and the
first derivative goes from increasing to decream'ng or the second derivative changes sign at changes c) From the graph, 1" '{xj : —2 when x : CI. The derivative is the slope function, so the
slope ofthe curve at {0.3) is — 2. The equation ofthe tangent line then would he wares—mm d) If '{xj dx : f {x} + C The integral ofacurve is the areaunderthe curve, III .‘FIIEIJ + A .‘F'txl dx = .‘FIIC'J El
A f ' {x} d:: hycalmlating areaunder curve gives +% — 2  g : El :: ff—Elj : 93 or 4.500 4
5111111313. .‘FIIC'H [J .‘F'txil dx = .‘FHJ Lalo: Thus, ﬂ—Elj — The integral computed by finding the area ofthe ﬁgure subtracting the area ofthe anMﬁiMtla rm?“ ‘1'“: Hana1' darnr1: Furm Hm: n'lnirr'iv. hu lrﬂ _r'I'I hu Ird. _r'I'I+n Ird. Inll'l HIE SEHﬂCJI'ClE 15 T Z I": and HIE BIEEUJI HIE rectangle 15 U. 23111123 ElleE are under the xaxis, both are negative so it is [— 8 — {—2 :Itﬁl] : 2st — 8. The value ofthe
4 integralis 2:rc —8 so fI{O:I+ ﬂ]. ﬁx) dx : H4) :3 +2:rt—B : 2:rE—5 or 1.283 5.
2.
a) Acceleration is the derivative ofvitj. Udng graphing calculator to calculate the
derivative of v{t:I att : 2 gives Since v{2jis neg ative, att : 2, ~ .13.
speed : {t + lJm'nl F . The derivative of speed at t = 2 is neg ative, therefore
hent : 2, the speed would he decreasing . h) The particle changes direction when the velocitvchanges sign. For the velocity to
change sign, it must equal CI. Graphing vI{t:I and making a chart ofwhere it is El,
ahove the axis or pom‘tive, and below the axis or neg ative. ﬂax£2.50? := v{t:IEO 2.50ch3 :: vttjznICI. Thus,the particle changes direction one time at t = 2. 50'? .
2.55? 3 c] — Jvﬁt] dt + $m; dt =
2 2.55?
d) The particle begins at l and goes negatively Jvﬁtj dt where it comes to rest. Its 255'?
pom‘tion then would he 1 + [J‘va dt : — 2.255. It then goes in the positive 3
direction Sign) dt from — 2.255. or ends at position
2
3 — 2.255 + Elwﬁtj dt : — 119?. ThusI ance distance fromthe originis the
2 Ipositionl , the furthest it is from the origin is 2.255 a) R '{cj can he approximated hythe following difference quotient w many—M40) , . 55—40
ecu—4:3 "’ R {‘15} "2' 50—40 ~ R '{45} ~ 1.50:) gallonsl'minutes2 h) The rate ofchange of fuel consumption of Mt) would he R '{tﬁL Since the rate of
change of fuel consunmtion, R '{t} , is at its greatest at t : 45 minutes, then this
must he alocal maximum for R. I“). Since R {t} and R. '{tj are differentiable, and since R ' {t} has alocal maximum at 45 minutes, then R ' '{453 must equal c] {20"EIO:I+{33*10: + {40*10) +{55*e:n;. + {ewes} = m Since Risincreasing on
the entire intervalI the left sums are each lower than the actual area under the curve. Thus the actual integral is greater than the left Riematm sum Note: Here the area of each of the Riematm rectangles must he ccmputed
individually as the intervals are not constant in length. h
d) JRﬂ) dt would he the total gallons used from time CI minutes to time h: minutes. h:
% J Rl{t:I dt is the average value ofthe function Rm fromCI to h and would he the average g allons used per minute from time '3 to time h. a) For afunction to he continuous, i) The function must he deﬁned atEl. AtElthe functionfisp'El +1 : 2 ii) the limit must exist from both the + side and the  side. e] Ar". lirn x_p_3_ 5—325—322 er the limit eff as 3—23 is 2 iii] The T.ralue efthe function f at 3: : El must equal the limit eff as x—zaEI. It does. he function f is continuous at x : El 3 5
:Pweragevalueeff *q'x+1 dx + J {5—3) :13 :: H IE Esﬂiﬁﬂl + IS [533le %[% The function g must he mntinueus at x : El sinee g is differentiable. 3:: kﬁJEI + :
melj +2 32 2k: Elm +2 2 ~ “n 25~ , g“ 1 + LLE—a—Lw—a} ]= g [g The derivatives efthe eernpenents of g must be equal at 3: : El er k
2 3+1 : In :: k = 4m Substituting into ﬁrst equatienI 9 s
Bm:Elrn+E:m:§ “3211: :E ...
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This note was uploaded on 11/17/2010 for the course CALC 151 taught by Professor Steve during the Spring '10 term at Universidad Nacional de Asuncion.
 Spring '10
 STEVE

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