A. Limits  L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. 1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.


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1 A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by trditionl methods (plugging in). If you get or, pply L Hopitl s rule, which sys tht lim c = lim c f g f g. L Hopitl s rule cn be pplied whenever plugging in cretes n indeterminte form:,,,,,, nd. A limit involving or is found by creting quotient out of tht epression. A limit involving eponents (,, or ) involves tking nturl log of the epression to move the eponent down. e + cos. Find lim A. B. C. 6 D. E. noneistent e + cos + C. lim = = e sin lim = = e cos lim 6 = = e sin lim 6 = 6 = 6. Find lim ( )ln + A. B. C. D.  E. noneistent A. lim ( )ln + lim + = ln ( ) = lim + ( ) = = lim ( +) ( ) + = Demystifying the BC Clculus MC Em
2 . Find lim e t A. B. e C. e 8 D. e 6 E. noneistent C. lim lim e t = e t e = e =. A prticle moves in the yplne so tht the position of the prticle t ny time t is given by t = cost nd y( t) = sin t. Find lim t. A.  B. 6 C. 8 D.  E. A. 8sin cos 8sin cos = lim = sin t sin 8( sin + cos ) lim = t cos 5. A popultion of bcteri is growing nd t ny time t, the popultion is given by 5 + t mimum limit of the popultion. A. 5 B. 5 + e C. 5e D. + ln5 E. 5e E. y = 5 + t ln y = ln5 + t ln + t t lim ln5 + t ln + ln + t t t = ln5 + lim = ln5 + lim t t + t t ln y = ln5 + y = e ln 5 + = e ln 5 e = 5e t t = ln5 + t. Find the Demystifying the BC Clculus MC Em
3 B. Integrtion by Prts Wht you re finding: When you ttempt to integrte n epression, you try ll the rules you hve been given to tht point  typiclly power, substitution, nd the like. But if these don t work, integrtion by prts my do the trick. Integrtion by prts is usully used when you re need to find the integrl of product. How to find it: Integrtion by prts sttes tht u dv = uv v du + C. To perform integrtion by prts, set up: u = v =. You need to fill in the u nd the dv from the originl problem. Determine du = dv = du nd v, then substitute into the formul. You re replcing one integrtion problem with nother tht might more esily be done with simple methods. The trick is to determine the u nd the dv. Functions tht cn be powered down re typiclly the u nd functions tht hve repetitive derivtives (eponentil nd trig) re typiclly the dv. 6. cos = A. sin + cos + C B. sin + 8 cos + C C. 8 sin + cos + C D. 8 sin cos + C E. sin cos + C B. u = v = sin du = dv = cos cos = sin cos = sin + 8 cos + C = sin sin 7. e = A. e ( 8 +6) + C B. e ( 8 + ) + C C. e + C D. e + + C E. e + C D. u = v = e du = dv = e e = ( e ) e u = v = e du = dv = e e = ( e ) ( e ) = ( e ) e + e = e + + C e Demystifying the BC Clculus MC Em
4 8. Let R be the region bounded by the grph of y = ln, the is nd the line = e, s shown by the figure to the right. Find the re of R. A. e + C. e e A. A = ln u = ln v = B. e D. e E. e + du = dv = A = ln e 9. The shded region between the grph of y = tn nd the is for s shown in the figure is the bse of solid whose crosssections perpendiculr to the is re squres. Find the volume of the solid. e = e e ln = e + A. + ln B. + e ln C. ln D. ln E. ln E. V = tn = tn u = tn v = du = + dv = tn = tn + = tn ln( + ) = tn ln = ln. The function f is twicedifferentible nd its derivtives re continuous. The tble below gives the vlue of f, f nd f for = nd =. Find the vlue of f ( ). f ( ) f ( ) f 6 5 A.  B. 8 C. 6 D. E. 6 C. f ( ) = f [ ] f ( ) = [ ] = + 8 = 6 5 [ ] f f f f Demystifying the BC Clculus MC Em
5 B. Integrtion using Prtil Frctions Wht you re finding: When you ttempt to integrte frction, typiclly you let u be the epression in the denomintor nd hope tht du will be in the numertor. When this doesn t hppen, the technique of prtil frctions my work. One form of this type of problem is where + m + n fctors into two + m + n nonrepeting binomils. How to find it: Use the Heviside method. Fctor your denomintor to get. You need to ( + ) ( + b) write ( + ) ( + b) s + +. To find the numertor of the + epression, cover up the + in + b epression, nd plug in =. To find the numertor of the + b epression, cover up the ( + ) ( + b) + b in epression, nd plug in = b. From there, ech epression cn be integrted. + ( + b) = A. ln C B. ln C C. 5ln + ln + + C D. ln + 5ln + + C E. ln C + C. u  substitution doesn't work here. u = + +,du = ( + ) = + + ( +) = + + ( +) = 5ln + ln + + C. Use the substitution u = cos to find sin cos ( cos ). cos A. lncos + C B. lncos + C C. ln cos cos D. ln cos + C E. ln cos cos + C D. u = cos,du = sin du uu u du = ln u ln u = ln u u u cos = ln + C cos + C Demystifying the BC Clculus MC Em
6 . = A. + ln + + ln + C B. + C C. ln + C D. + ( ln + ln +) + C E. ln C + ) D. = + + = = + ln + ln + = + ln + ln + + C ( could lso be done with u  sub). Region R is defined s the region between the grph 9 of y =, = nd the is s shown in the + figure to the right. Find the re of region R. A. ln B. + ln C. ln D. 6ln E. infinite 9 D. A = + = ( + ) = 9 ( + ) ( + ) = ln ln + = ln + = ln ln = ln ( ln ) = ln = ln = 6ln Demystifying the BC Clculus MC Em
7 C. Improper Integrls Wht you re finding: An improper integrl is in the form be in the form continuous. b f ( ) or f ( ) or f ( ). It lso cn f ( ) where there is t lest one vlue c such tht c b for which f How to find it: Improper integrls re just limit problems in disguise: b f ( ) = lim into two pieces: b b is not f ( ) = lim f ( ) or b b f ( ). In the cse where there is discontinuity t = c, the improper integrl is split f ( ) = lim f ( ) + lim k k c with re nd volume problems. 5. Which of the following re convergent? b k c + k f ( ). Improper integrls usully go hndinhnd I. II. III. A. I only B. II only C. III only D. II nd III only E. I, II nd III 6. e = A. 6 B. 6 D. = which is divergent = = = = C. 6 D. 6 E. infinite B. u = v = e du = dv = e e = e 6e = 6 = 6 e + e = Demystifying the BC Clculus MC Em
8 7. The region bounded by the grph of y =, the line = nd the  is is rotted bout the is. Find the volume of the solid. A. B. C. D. 6 E. infinite C. V = = 6 = 6 V = + = 8. ( +) = A. B. C. D. E. infinite B. u = du = + = =, u =, =,u = + = = tn tn = = ( +) = du u + = tn u 9. To the right is grph of f ( ) =. Find the vlue of f ( ). A. B. C. D. 6 E. Divergent = E. = ( ) + + Both clcultions re divergent Demystifying the BC Clculus MC Em
9 D. Euler s Method Wht you re finding: Euler s Method provides numericl procedure to pproimte the solution of differentil eqution with given initil vlue. How to find it: ) Strt with given initil point (, y) on the grph of the function nd given =. ) Clculte the slope using the DEQ t the point. ) Clculte the vlue of using the fct tht. ) Find the new vlues of y nd : y new = y old + nd new = old + 5) Repet the process t step ). There re clcultor progrms vilble to perform Euler s Method. Typiclly, Euler Method problems occur in the nonclcultor section where only one or two steps of the method need to be performed.. Let y = f ( ) be the solution to the differentil eqution f size of.5? = + y with the initil condition tht =. Wht is the pproimtion for f ( ) if Euler s Method is used, strting t = with step A..5 B..5 C..5 D. 5.5 E..5 B. y Let y = f ( ) be solution to the differentil eqution = y with initil condition f k constnt, k. If Euler s method with steps of equl size strting t = gives the pproimtion f ( ), find the vlue of k. = k, A. B. C. D.  E. A. y k k k k k k k k + k k + k = k( + k) = k =,k = Demystifying the BC Clculus MC Em
10 . Consider the differentil eqution = y between the ect vlue of f 8 with initil condition f =. Find the difference nd n Euler pproimtion of f ( 8) using step of.5. A. B. C. D. 5 E. 5 A. y y = ln y = ln + C y = e ln +C y = C = C C = y = y = 5 = 5 Difference = 5 5 = Students should see tht ech y will be twice the vlue of nd not perform Euler over nd over gin.. (Clc) Consider the differentil eqution = cos with initil condition f =. Find the difference between the ect vlue of f nd n Euler pproimtion of f using two equl steps. A. B.. C.. D..555 E..77 C. y = = = cos y = sin + C = sin + C C = y = sin y = sin = Difference = + 8 = Demystifying the BC Clculus MC Em
11 E. Logistic Curves Wht you re finding: Logistic curves occur when quntity is growing t rte proportionl to itself nd the room vilble for growth. This room vilble is clled the crrying cpcity. The curve hs distinctive Sshpe where the initil stge of growth is eponentil, then slows, nd eventully the growth essentilly stops. How to find it: Logistic growth is signled by the differentil eqution dp = kp( P t). While this DEQ C cn be solved into Pt =, students re not responsible for tht eqution. They need to know how to Ckt + de determine the time when the logistic growth is the fstest. This is ccomplished by d P =. Also students need to know tht the curve hs horizontl symptote mening limpt = C ( the crrying cpcity). t. A popultion of students hving contrcted the flu in school yer is modeled by function P tht stisfies the logistic differentil eqution with dp = P 6 P. If P =, find limpt. 8 t A. B. 8 C.,6 D., E.,8 C. Since this is logistic, there is horizontl symptote to P( t) nd thus dp dp = P 6 P = P P = = 5. A popultion is modeled by function G tht stisfies the logistic differentil eqution dg = G e G. If G e A. B. e C. e D. e E. e =, for wht vlue of G is the popultion growing the fstest? C. dg = G e G = G e e d G 6. Consider the differentil eqution differentil eqution with f A.,L G e = e G e = eg = e P = e hlf the crrying cpcity = ky( L y). Let y = f ( ) be the prticulr solution to the =. If, find the rnge of f ( ). B. (,) C. ( L,] D. [,L) E. [,kl) D. You must recognize this DEQ s logistic one. So = ky L y = y = nd y = L. So the grph of f line y = L. But since, the rnge is [,L). Not recognizing the logistic eqution would involve you solving the DEQ which is time consuming nd unnecessry. hs horizontl symptotes long the  is nd the Demystifying the BC Clculus MC Em
12 F. Arc Length Wht you re finding: Given function on n intervl [, b], the rc length is defined s the totl length of the function from = to = b. For this section, we will only concentrte on curves tht re defined in function form. Functions defined prmetriclly, in polr or in vectorvlued forms hve their own formuls. How to find it: The rc length of continuous function f b [ ] over n intervl [, b] is given by L = + f ( ). Most problems involving rc length need clcultors becuse of the difficulty of integrting the epression. 7. (Clc) An nt wlks round the first qudrnt region R bounded by the yis, the line y = nd the curve f ( ) = 6 s shown in the figure to the right. Find the distnce the nt wlked. A..9 B.. C..85 D..88 E..85 E. First, it is necessry to find the intersection point of the two curves. 6 = t =. The distnce between (,) nd (,6) is 6 nd from (,) to (,) is 5 [ ] D = f ( ) + 5 [ ] D = = D = = If the length of curve from = to = 8 is given by + 8, nd the curve psses through the point (, ), which of the following could be the eqution for the curve? 8 A. y = 9 B. y = C. y = 7 + D. y = E. y = 9 5 [ ] = 8 f ( ) =±9 nd f ( ) =± + C C. f f =± + C = + C = nd C = 7 or + C = nd C =. So y = 7 + or y = Demystifying the BC Clculus MC Em
13 9. (Clc) The yellow bird in the populr gme Angry Birds flies long the pth y = + when. When = (the point on the figure to the right), the plyer touches the screen nd the bird leves the pth nd trvels long the line tngent to the pth t tht point. If the bird crshes into the is, find the totl distnce the bird flies. A. 6.8 B.. C.. D. 5.6 E. 8. =. y( ) = + 8 = 8 E. y = y Tngent line : y 8 = ( ) y = Length : L = + ( ) + + = 8.. (Clc) The grphs of i) y =, ii) y = nd iii) y = ll pss through the points (,) nd (,). Find the difference in rc length from the lrgest rc length to the shortest rc length of these functions on the intervl [,]. A..58 B. 8 C..8 D..6 E..9 A. i) d = + =.79 ii) d = + =.79 iii) + ln =..79. =.58.. Find the rc length of the grph of = ( y + ) for y. A. B. C. 6 D. E. A. = ( y + ) ( y) = yy ( + ) L = + y y + = y + y + = y + = y + L = y + y = 9 + = Demystifying the BC Clculus MC Em
14 G. Prmetric Equtions Wht you re finding: Prmetric equtions re continuous functions of t in the form = f ( t) nd y = gt. Tken together, the prmetric equtions crete grph where the points nd y re independent of ech other nd both dependent on the prmeter t (which is usully time). Prmetric curves when grphed do not hve to be functions. Typiclly, it is necessry to tke derivtives of prmetrics. Since the stu of vectors prllels the stu of prmetrics, in this section we will only nlyze the very few problems tht re not ssocited with motion in the plne. How to find it: If smooth curve C is given by the prmetric equtions = slope of C t the point (, y) is given by =,. The nd derivtive of the curve is given by d y = The rc length is given by L = t = b t = + d = d. f ( t) nd y = gt, then the. The curve must be smooth nd my not intersect itself.. Wht is the re under the curve described by the prmetric equtions = cost nd y = sin t for t? A. B. 8 C. D. E. B. cost = cos t = nd sin t = y Since sin t + cos t = + y = y = When t =, = nd when t =, = A = = Demystifying the BC Clculus MC Em = 6 6 ( + ) = 8. A position of prticle moving in the yplne is given by = t 6t + 9t + nd y = t 9t. For wht vlues of t is the prticle t rest? A. only B. only C. only D. nd only E., nd = t C. = t t + 9 = t t + = 6t 8t = 6t t = nd = t t =. = t =,t = ( t ) = t =, t =
15 . A curve C is defined by the prmetric equtions = t t nd y = t 7t. Which of the following is the eqution of the line tngent to the grph of C t the point (, )? A. y = 6 B. y + = C. 5 y + = D. y = E. No tngent line t (, ) D. = t t = t t 6 = ( t ) ( t + ) = t =,t = y = t 7t = t 7t 6 = = Tngent line : y = = t 7 t 7 7 = [ t = ] 6 = y = 5. Describe the behvior of curve C defined by the prmetric equtions = + t nd y = t + t t t t =. Only t = stisfies this eqution. A. Incresing, concve up B. Decresing, concve up C. Incresing, concve down D. Decresing, concve down E. Incresing, no concvity B. = d y = d = t + t t ( ) = t = t t + t t =t t + t < so C is decresing [ t =] 5 +t t d y > so C is concve up [ t =] 6. Find the epression which represents the length L of the pth described by the prmetric equtions = sin ( t) nd y = cos( t ) for t. A. L = sint cost sint B. L = sin t + 9sin 9t C. L = 6sin t cos t + 9sin 9t D. L = sin t cos t + 9sin t E. L = 6sin t cos t + 9sin t E. L = = sint cost = sint Demystifying the BC Clculus MC Em 6sin t cos t + 9sin t
16 H. VectorVlued Functions Wht you re finding: While concepts like unit vectors, dot products, nd ngles between vectors re importnt for multivrible clculus, vectors in BC clculus re little more thn prmetric equtions in disguise. How to find it: Typiclly, you will be given sitution where n object is moving in the plne. You could be given either its position vector t nd y( t), its velocity vector ( t) nd y ( t) or its ccelertion vector ( t) nd y ( t) nd use the bsic derivtive or integrl reltionships tht hve been tught in AB clculus to find the other vectors. The one formul tht students should know is tht the speed of the object is defined s the bsolute vlue of the velocity: vt = ( t) [ ] + [ y ( t) ]. The speed is sclr, not vector. 7. A prticle moves on plne curve such tht t ny time t >, its coordinte is t t + t while its ycoordinte is t. Find the mgnitude of the prticle s ccelertion t t =. A. B. C. D. E. E. vt = t + t,( t )( t) = t + t,t 8t t = 6t,t 8 =, = + = 8. The position of n object moving in the yplne with position function r( t) = + sint,t + cost, t. Wht is the mimum speed ttined by the object? A. B. C. D. E. C. vt = ( t) [ ] + [ y ( t) ] = cos t + sint vt = cos t + sint + sin t = sint Clculus techniques could be used to mimize this epression but since sint hs minimum/mimum vlues of nd, the speed will rnge from to =. 9. A yplne hs both its nd ycoordintes mesured in inches. An nt is wlking long this plne with its position vector s t,t, t mesured in minutes. Wht is the verge speed of the nt mesured in inches per minute from t = to t = minutes? A. B. C. D. E. 9 B. L = + = 9t + 9 = t + = t + [ ] = 8 = Avg velocity = L = in min Demystifying the BC Clculus MC Em
17 . An object moving in the yplne hs position function r( t) = Describe the motion of the object. ( t +),t 6ln t +, t. A. Left nd up B. Left nd down C. Right nd up D. Right nd down E. Depend on the vlue of t E. = which is lwys negtive when t so object moving left t + = ( t + t ) t + ( t ) 6 = t t + = t + t 6 = t + t + t + chnges sign t t = so y chnges direction t t =.. An object moving long curve in the yplne hs position t, yt = 8t + nd = sint for t. ( ) t time t with At time t =, the object is t position (5, ). Where is the object t t =? B. ( + + 5, +) C. + 5, + A , D., E. ( + 5,) B. = t + t + C = C = 5 = t + t + 5 y = cost + C y = + C = C = + y = cost + + = = + 5 y = (, y) t = = + + 5, = + t time t with. (Clc) An object moving long curve in the yplne hs position t, yt = t + t + nd = et for t. At time t =, the object is t position (6, 7). Find the position of the object t t =. A. (.667,.5) B. (.68,.78) C. (.7, 9.78) D. (.7, 6.5) E. (.57, .97) E. = + y = y + ( t ) = =.57 y ( t ) = 7 + e t =.97 (,y ) =.57, Demystifying the BC Clculus MC Em
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