Ouestion 1 f -^2\ The rate at which raw sewage enters â t eatment tank is given by E(r) = Tl5cosl $ gallons

Transcription

1 Ouestion 1 f -^2\ The rate at which raw sewage enters â t eatment tank is given by E(r) = Tl5cosl $ gallons \> ) per hour for 0 < t < 4 hours. Treated sewage is removed from the tank at the constant rate of 645 gallons per hour. The treâtment tank is empty at time I = 0. (a) How many gallons of sewage enter the treatment tank during the time interval 0 < I 4? Round your answer to the nearest gallon. (b) For 0 < / < 4, at what time I is the amount of sewage in the treatment rank greatest? To the nearest gallon, what is the maximum amount of sewage in the tank? Justiô/ your answers. 1$ For O < t < 4, rhe cost of treating the raw sewâge that enters the tank at rime I is ( r) 'dollars per gallon. To the nearest dolla, what is the total cost of feating all the sewâge that enters the tank during the time interval O < t < 4? øl ffzçt at = 3981 sallons integral ans\.ver (b) Let,S(t) be the amoùnt of sewage in the treatment tank at time f. Then S'(t) = B1 1-6a5 and S'þ) = I y "n E(t) = 6a5. On the interval O<t<4, E(t) = 645.when = atj,d t = r (hours) 0 0 amount of sewage in featment tank 4l 1 : sets E(l) = 645 1:identifies t = as a candidate I : amount of sewage aí. t = I : conclusion 2309 lo"otu,,) dr - 6a5(2.309) = t63i.t.b 3.559!"tn n1, a, - 64s(3.ss9) = (4) = The amount of sewage in the treatment tank is greatest at t = hoùrs. At that time, the amount of sewage in the tank, rounded to the ne est gallon, is 1637 gallons. (c) Total cosr = lr-to.tt - o.ozt) E(t) dr The total cost of treating the sewage that enters the tank during the time interval 0 < t < 4, to the nearest dollar, is $47 4.

2 Ouestion 2 A particle moving along a cuwe in the x)-plane has position ( ( ), l( )) at time / > 0, where & t6 ^\r/ - dv " dt-\t -) At time / = 3, the particle is at rhe point (5, 4). (a) Find the speed of fhe particle at time / = 3. (b) Write an equation for the line tangent to the path of the paficle at time / = 3.,. (c) Is there a time / at which the particle is farthest to the dght? Ifyes, explain why and give the value of / and the x-coordinate of the position of the particle at that time. If no, explain why not. (d) Desc ibe the behavior of the path of the particle as f increases without boùnd. I : answer (b) At =',*=#=+=-í = ". [1:slope ' l I: equation of tangent line An equation for the line tangent to the path at the poinr (5,4) is y -4 = -4t'- t) e (c) Since ff, o fo, 0 < t <2 una ff <0 lor >2, the particle is farthest to the right at t = 2. The -coordinate of the position is 4:.xrlt = 5 + f' E - t\"' dr = ).let o1 5.js2 Jt \ ) (d) since t\# = (-3)'/' un,,y# = o, the particle will continue moving farther to the left while approaching a horizontal asymptote.,. { t: behavior of -r ll:behaviorofy

3 Ouestion 3 I (minutes) t6 /1(r) ('c) go The temperature, in degrees Celsius ("C), ofan oven being heated is modeled by an increasing differentiable function 11 of time l, where f is measured in minutes. The table above gives the temperatùre as recorded every 4 minutes over a l6-rninute period. (a) Use the datâ in the table to estimale tfe instantaneous rate at which the temperature of the oven is changing at time t = 10. Show the crfìputations that lead to your answer. Indicate units of measure. (b) Write an integral expression in terms of' ll for the average temperature of the oven between time f = 0 and time / = i6. Estimate the average temperature of the oven using a left Riemann sum with four subintewals of equal length. Show the computations that lead to your answer. (c) Is your approximation in part (b) an uderestimate or an overestimate of the average temperature? Give a reason lor your answer. (d) Are the daø in the table consistent with or do they contradict the claim that the temperature of the oven is inüeasing at an increasing rate? Give a reason for your answer. (a) H'(ro) = Htll) _lrtst =Y = f,"c I ^;, (b) Average temperature is ltu n1, a, = 4.(6s ) Average temperature I Jn H(r) dt = 71.5"C 2: difference quotient answer with units r.ló *1, a(r)at left Riemann sum answer (c) The left Riemann sum approximation is an underestimate of the integral because the graph of Il is increasing. Dividing by 16 will not change the inequality, so 7l.5oC is an underestimate of the average temperature. I : answer with reason (d) If a continuous flnction is inc easing at an increasing rate, then the slopes of the secant lines of the graph of the function are increasing. The slopes of the secant lines for the four intervals in therableare i i i ana loq. respectively. Since the slopes are increasing, the data âre consistent with the claim. OR By the Mean Value Theorem, the slopes are also the values of H'(c ) fo some times I 1c2 1ca < c4, respectively. Since these derivâtive values are positive and increasing, the data are consistent with the claim. 3: I : considers slopes of four secant lines I : explanation I : conclusion consistent with explanation -39-

4 Gatculus BC Ouestion 4 Letlbe the function given by /(x) = (lnx)(sinx). The figure above shows the graph of/ for 0 <x.3ztt. The function g is defrned by gþ)=l: f(t)dtfor0<x3ztt. (a) Find s(1) and s'(i). (b) On what intervals, if any, is g increasing? Justify your answer. (c) For 0 < x 1 2x, find the value of at which g has an absolute minimum. Justify your ânswer. (d) For 0 < x <2x, is the e a value of ;r at which the graph of g is tangent to the;r-axis? Explain why or why not. Graph of /.t 1at s(r) = l,í(,),t, - 0 and s'(l) - "r(l) - 0 I : s(l) I : s'(l) g is increasing on the interval (b). Since e'g) = f(r), 13x3nbecause/(x) > 0 for 1<x<2. 2: (c) For 0 < x < 2tc, C'þ)= f(r) = 0 when ;r = l, z g' = / changes from negâtive to positive only at = 1. The absolute minimum must occur at r = I or at the right endpoint. Since g(1) = 0 and.)ì.n.1n s(2o\ = J, fq) ú = ), f(r) ù + J, flr) dt < 0 by comparison of the two areas, the absolute minimum occurs at r = 2n. (d) Yes, the graph of I is tangent to the x-axis at x : 1 since s(1) = 0 and C'(1) = 0. 2: I : identifies I and 2z as candidates indicates that the graph of g decreases, increases, then decreases 1 :justifies g(2ft) < g(1) 1 : answer 1 : answer of "yes" with = I I :explanation

5 Ouestion 5 Let / be the function satisfying f'(x) = 4x - zxf(x) for all real numbers.x, with /(0) = 5 an )yyf a) = z. (a. Find the value ot liç+* -Z* 1r)) rtr. Strow rhe work rhar leads to your rnr*".. JO (b) Use Euler's method to approximate /(-1), stafting ât ; = 0, with two stepsoofequal size. (c) Find the particul solution y = (x) to the differential eqùation! = +, 4ry wirh the initiaì dx: condition /(0) = I-Øx-zxf(x))dx = ff r'rð o, = ;y!!r'ata" = afrc,li] nmf(a)-/(o)) =2-5 = -3 I : use of FTC I : answer from limiting process (b) /( f( = /(o) /'(o) (-+) = '.0 (- j) = ' =,( +).'(-j) (-+)='., (-j) : Euler's method with two steps : Euler's approximation to.f (-l) {Ò U}rrav = xax -!nl+-zyl=," *e lnla-zyl= -x2 + B 2y-4=Ce-'2 C =6 Y =2 for all eal numbers 1 : separates variables 1 : antiderivatives 1 : constant of integration 1 : uses initial condition 1 : solves for y max2f 5 [l ] if no constant of integration

6 Calculus BG Ouestion 6 The function I is continuous for all real numbers.r and is defined by,1* =sr*(411 forx*0. x (a) Use L'Hospital's Rule to find the value of g(0). Show the work that leads ro your answer. (b) Let / be the function given by /(-r) = cos(2:r). Write the first four nonzero terms and the general (c) (d) term of the Taylor series for / about.r = 0. Use your answer from part (b) to write the first three nonzero terms and the general term of the o Taylor series for I about r = 0. a Determine whether g has a relative minimum, a relative maximum, or neither at = 0. Justify your answer. (a) Since g is continuous, s(0)=liqe{x) n+u =l'"lgú 1.r Itm* -2sin(Zx\ r-io x,. -4cos(2x) (b) cos(2rc). (2r)a (2r) ï4!x-6lx+. 16 q 64, ^,zti ç\'l#+.+(-t)"f;.x'z" ^2t1 +.,,1 t : first t\'r'o terms : next two lerms : general term (c) qlx) = tx- - 6't a ^2t (-r)'frr.x')"-'z + : first three terms : general term (d) s'( ) e"þ)= t -fi,-;t I.so8'(0) = t ii _ -!*,, I... so s,,(0) > 0. Therefore, g has a relative minimum at r = 0 by the Second Derivative Test. ; answer :justification