A. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.


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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 ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c How to find it: Try nd find limits by trditionl methods (plugging in). If you get 0 0 or!!, pply L Hopitl s rule, which sys tht lim x! c ( ) ( ) = lim x! c f x g x ( ) ( ) f " x g " x x! c. L Hopitl s rule cn be pplied whenever plugging in cretes n indeterminte form: 0 0,!!,0"!,! #!,!,0 0, nd! 0. A limit involving 0! " or " # " is found by creting quotient out of tht expression. A limit involving exponents (!,0 0, or! 0 ) involves tking nturl log of the expression to move the exponent down. e x + cos x " x ". Find lim x! 0 x 4 " x 3 A.! 30 B.! 4 C.! 6 D. 0 E. nonexistent. Find lim ( x ")ln( x ") x! + A. 0 B.! C. D.  E. nonexistent Demystifying the BC Clculus MC Exm
2 3. Find lim x! x e t " # dt x 3 " 4x A. 0 B. e4 C. e3 8 D.!e 6 E. nonexistent 4. A prticle moves in the xyplne so tht the position of the prticle t ny time t is given by x( t) = cost nd y( t) = sin 4t. Find lim dx. t! 0 dy A. 3 B. 6 C. 8 D. 4 E. 0 t. Find the! 5. A popultion of bcteri is growing nd t ny time t, the popultion is given by " t mximum limit of the popultion. A. 500 B e C. 500e D. + ln500 E. 500e # $ Demystifying the BC Clculus MC Exm
3 B. Integrtion by Prts Wht you re finding: When you ttempt to integrte n expression, 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 (exponentil nd trig) re typiclly the dv. 6.! x cos4x dx = A. x sin4x + cos4x + C B. x sin4x + 8 cos4x + C C. 8x sin4x + 3cos4x + C D. 8x sin4x! 3cos4x + C E. x sin4x! cos4x + C 7.! x e x dx = A. e x x! 8x +6 ( ) + C B. e x ( x! 8x + ) + C " x C. e x! x! % " x $ ' + C D. e x # &! x + %! x 3 $ $ ' + C E. e x # & + C # 4 & " 3 % Demystifying the BC Clculus MC Exm
4 8. Let R be the region bounded by the grph of y = x ln x, the xxis nd the line x = e, s shown by the figure to the right. Find the re of R. A. e + C. e B. e! D. e E. e + 9. The shded region between the grph of y = tn! x nd the xxis for 0 x s shown in the figure is the bse of solid whose crosssections perpendiculr to the xxis re squres. Find the volume of the solid. A.! + ln " B.! + e " ln 4 C.! " ln D.! 4 " ln E.! " ln 0. The function f is twicedifferentible nd its derivtives re continuous. The tble below gives the vlue of f, f! nd f!! for x = 0 nd x =. Find the vlue of " x f!! ( x ) dx. x ( ) f!( x) f!! ( x) f x "5 " 4 A. 0 B. 8 C. 6 D. 4 E Demystifying the BC Clculus MC Exm
5 B. Integrtion using Prtil Frctions Wht you re finding: When you ttempt to integrte frction, typiclly you let u be the expression in the denomintor nd hope tht du will be in the numertor. When this doesn t hppen, the technique of prtil dx frctions my work. One form of this type of problem is! where x + mx + n fctors into two x + mx + n nonrepeting binomils. dx How to find it: Use the Heviside method. Fctor your denomintor to get! ( x + ) ( x + b). You need to write ( x + ) ( x + b) s x + +. To find the numertor of the x + expression, cover up the x + in x + b expression, nd plug in x =!. To find the numertor of the x + b expression, cover up the ( x + ) ( x + b) x + b in expression, nd plug in x =!b. From there, ech expression cn be integrted. x + ( )( x + b).! 4x + x + 4x + 3 dx = A. ln x + 4 x C B. ln x + 4x C C. 5ln x + 3! ln x + + C D. ln x +! 5ln x C E. ln x C x +. Use the substitution u = cos x to find " sin x cos x cos x! dx. ( ) cos x! A. lncos x! + C B.!lncos x! + C C.!ln + C cos x! cos x D. ln + C E. ln cos x cos x cos x! + C Demystifying the BC Clculus MC Exm
6 3. x " 3 x! dx = A. x x + ln x + + ln x x!! + C B. + C C. x! ln x! + C D. x + ( ln x! + ln x +) + C E. ln x C x + 4. Region R is defined s the region between the grph of 9 y =, x = nd the xxis s shown in the x + x! figure to the right. Find the re of region R. A. ln B. + 3ln4 C. 3ln 4 D. 6ln E. infinite Demystifying the BC Clculus MC Exm
7 C. Improper Integrls Wht you re finding: An improper integrl is in the form be in the form continuous. b! " f ( x ) dx or " f ( x) dx or " f ( x) dx. It lso cn! f ( x ) dx where there is t lest one vlue c such tht c b for which f x How to find it: Improper integrls re just limit problems in disguise: b # f ( x ) dx = lim!" into two pieces: b $!" b with re nd volume problems.! #!! #! ( ) is not " f ( x ) dx = lim f ( x) b#! " dx or # f ( x ) dx. In the cse where there is discontinuity t x = c, the improper integrl is split! f ( x ) dx = lim! f ( x) dx + lim! f ( x) dx. Improper integrls usully go hndinhnd k" c # k" c + 5. Which of the following re convergent? k b k b I.! " dx II. x! dx III. x 0! " dx x 3 A. I only B. II only C. III only D. II nd III only E. I, II nd III " 6. # xe!4x dx = 0 A.! 6 B. 6 C. 6 D. 6 E. infinite Demystifying the BC Clculus MC Exm
8 7. The region bounded by the grph of y = 4, the line x = 4 nd the x  xis is rotted bout the xxis. x Find the volume of the solid. A. π B. π C. 4π D. 6π E. infinite 8.! " x ( x +) dx = A.! 4 B.! C. π D. π E. infinite 4 9. To the right is grph of f ( x) =. Find the vlue of 3 " f ( x ) x! dx. ( )! A. 0 B. 3 C.! 3 D. 6 E. Divergent Demystifying the BC Clculus MC Exm
9 D. Euler s Method Wht you re finding: Euler s Method provides numericl procedure to pproximte the solution of differentil eqution with given initil vlue. How to find it: ) Strt with given initil point (x, y) on the grph of the function nd given!x = dx. ) Clculte the slope using the DEQ t the point. 3) Clculte the vlue of dy using the fct tht dy! dy dx "x. 4) Find the new vlues of y nd x: y new = y old + dy nd x new = x old +!x 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. 0. Let y = f ( x) be the solution to the differentil eqution dy f 3 size of 0.5? dx = x + y with the initil condition tht ( ) =!. Wht is the pproximtion for f ( 4) if Euler s Method is used, strting t x = 3 with step A..5 B. 3.5 C. 4.5 D. 5.5 E..5. Let y = f ( x) be solution to the differentil eqution dy dx = y x with initil condition f 0 k constnt, k! 0. If Euler s method with 3 steps of equl size strting t x = 0 gives the pproximtion f 3 ( )! 0, find the vlue of k. ( ) = k, A.! B. C. D.  E.! Demystifying the BC Clculus MC Exm
10 . Consider the differentil eqution dy dx = y x the exct vlue of f 8 with initil condition f ( ) =!4. Find the difference between ( ) nd n Euler pproximtion of f ( 8) using step of 0.5. A. 0 B. C. D. 5 E (Clc) Consider the differentil eqution dy dx "!% "!% between the exct vlue of f $ ' nd n Euler pproximtion of f $ ' using two equl steps. # & # & = cos x with initil condition f ( 0) = 0. Find the difference A. 0 B C D E Demystifying the BC Clculus MC Exm
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. This constntly incresing curve hs distinctive Sshpe where the initil stge of growth is exponentil, then slows, nd eventully the growth essentilly stops. How to find it: Logistic growth is signled by the differentil eqution dp dt = kp ( P! t ). While this DEQ C cn be solved into P( t) =, 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 = 0. Also students dt need to know tht the curve hs horizontl symptote mening limp t t!" ( ) = C ( the crrying cpcity). 4. A popultion of students hving contrcted the flu in school yer is modeled by function P tht stisfies the logistic differentil eqution with dp dt = P " 600! P % $ '. If P( 0) =00, find lim # 800& P( t). t!" A. 400 B. 800 C.,600 D.,400 E. 4, A popultion is modeled by function G tht stisfies the logistic differentil eqution dg dt = G " e! G % $ '. If G 0 # 4e& A. 4 B. e C. e D. 4e E. 4e ( ) =, for wht vlue of G is the popultion growing the fstest? 6. Consider the differentil eqution dy dx = ky ( L! y ). Let y = f ( x) be the prticulr solution to the differentil eqution with f ( 0) =. If x! 0, find the rnge of f ( x). A. 0,L ( ) B. ( 0,) C. ( L,] D. [,L) E. [,kl) Demystifying the BC Clculus MC Exm
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 x = to x = 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 x b [ ] ( ) over n intervl [, b] is given by L = " + f! ( x ) dx. Most problems involving rc length need clcultors becuse of the difficulty of integrting the expression. 7. (Clc) An nt wlks round the first qudrnt region R bounded by the yxis, the line y = x nd the curve f ( x) = 6! 4x 3 s shown in the figure to the right. Find the distnce the nt wlked. A B. 0.3 C..485 D..88 E If the length of curve from x = to x = 8 is given by! + 8x 4 dx, nd the curve psses through the point (, 4), which of the following could be the eqution for the curve? 8 A. y =3! 9x B. y = 4! 3x 3 C. y = 7 + 3x 3 D. y =!! 3x 3 E. y = 9x! Demystifying the BC Clculus MC Exm
13 9. (Clc) The yellow bird in the populr gme Angry Birds flies long the pth y = 4 + 3x! x when x 0. When x = 4 (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 xxis, find the totl distnce the bird flies. A B..34 C..000 D E (Clc) The grphs of i) y = x, ii) y = x nd iii) y = x! ll pss through the points (0,0) nd (,). Find the difference in rc length from the lrgest rc length to the shortest rc length of these functions on the intervl [0,]. A B. 008 C D E Find the rc length of the grph of x = ( 3 y + ) 3 for 0! y! 3. A. B. 0 C. 6 D. 3 3 E Demystifying the BC Clculus MC Exm
14 G. Prmetric Equtions Wht you re finding: Prmetric equtions re continuous functions of t in the form x = f ( t) nd y = g( t). Tken together, the prmetric equtions crete grph where the points x 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 study of vectors prllels the study 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 x = slope of C t the point (x, y) is given by dy dy dx = dt,dx dx dt! 0. dt f ( t) nd y = g( t), then the d! dy $ The nd derivtive of the curve is given by d y dx = d # &! dy $ dt " dx % # & =. dx " dx % dx dt t = b! dx# The rc length is given by L = + dy! # % dt. The curve must be smooth nd my not intersect itself. " dt $ " dt $ t = 3. Wht is the re under the curve described by the prmetric equtions x = cost nd y = 3sin t for 0! t! "? A. 4 B. 8 C. 4 D. E A position of prticle moving in the xyplne is given by x = t 3! 6t + 9t + nd y = t 3! 9t!. For wht vlues of t is the prticle t rest? A. 0 only B. only C. 3 only D. 0 nd only E. 0, nd Demystifying the BC Clculus MC Exm
15 34. A curve C is defined by the prmetric equtions x = t! t! 4 nd y = t 3! 7t!. Which of the following is the eqution of the line tngent to the grph of C t the point (, 4)? A. y = 6! x B. x! 4y +4 = 0 C. 5x! 3y + = 0 D. y = 4x! 4 E. No tngent line t (, 4) 35. Describe the behvior of curve C defined by the prmetric equtions x = + t nd y = t 3 + t! t! t t =. A. Incresing, concve up B. Decresing, concve up C. Incresing, concve down D. Decresing, concve down E. Incresing, no concvity 36. Find the expression which represents the length L of the pth described by the prmetric equtions x = sin ( t) nd y = cos( 3t ) for 0! t! ". " A. L = # sint cost! 3sin3t dt B. L = " 4sin 4t + 9sin 9t dt 0! C. L = " 6sin 4t cos 4t + 9sin 9t dt D. L = " 4 sin t cos t + 9sin 3t dt 0! " E. L = 6sin t cos t + 9sin 3t dt 0! 0! Demystifying the BC Clculus MC Exm
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 x( t) nd y( t), its velocity vector x!( t) nd y!( t) or its ccelertion vector x!! ( 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: v t ( ) = x!( t) [ ] + [ y!( t) ]. The speed is sclr, not vector. 37. A prticle moves on plne curve such tht t ny time t > 0, its xcoordinte is t! t + t 3 while its ycoordinte is! t ( ). Find the mgnitude of the prticle s ccelertion t t =. A. 4 B. C. D. 3 E The position of n object moving in the xyplne with position function r( t) = + sint,t + cost, t 0. Wht is the mximum speed ttined by the object? A. B. C. D. 4 E. 39. A xyplne hs both its x nd ycoordintes mesured in inches. An nt is wlking long this plne with its position vector s t 3,3t!, t mesured in minutes. Wht is the verge speed of the nt mesured in inches per minute from t = 0 to t = 3 minutes? A. B. 4 3 C. 3! D. 3 E Demystifying the BC Clculus MC Exm
17 40. An object moving in the xyplne hs position function r( t) = the motion of the object. ( t +),t! 6ln t + ( ), t 0. Describe A. Left nd up B. Left nd down C. Right nd up D. Right nd down E. Depend on the vlue of t 4. An object moving long curve in the xyplne hs position x( t),y t dx dy = 8t + nd = sint for t! 0. dt dt ( ( )) t time t with At time t = 0, the object is t position (5, π). Where is the object t t =!? ( ) B. (! +! + 5,! +) C.! + 5,! + A.! +! + 5,! " D.!, % $ ' E.! + 5,! # & ( ) ( ) ( ) t time t with 4. (Clc) An object moving long curve in the xyplne hs position x( t),y ( t) dx dt = t dy + 3t + nd dt = et! for t " 0. At time t = 0, the object is t position (6, 7). Find the position of the object t t =. A. (4.667, 3.053) B. (3.683,.78) C. (.37, 9.78) D. (4.47, 6.053) E. (.573, ) Demystifying the BC Clculus MC Exm
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