Doc1 - CE 1. solution on gr anhing calculator to 1 14E...

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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 .‘FII-EIJ + 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 anMfiiM-tla rm?“ ‘1'“: Hana-1' darn-r1:- Fur-m Hm: n'lnirr'iv. h-u lrfl _r-'I'I h-u Ird. _r-'I'I+n Ird. Inll'l HIE SEHfl-CJI'ClE 15 T Z I": and HIE BIEEUJI HIE rectangle 15 U. 23111123 Elle-E are under the x-axis, 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 -::- vttjzn-ICI. 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—2-3 is 2 iii] The T.ralue efthe function f at 3-: : El must equal the limit eff as x—za-EI. 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 [53-3le %[% 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 + LL-E—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:§ “32-11: :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.

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Doc1 - CE 1. solution on gr anhing calculator to 1 14E...

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