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Q. Fundamental Theorem of Calculus (FTC) / Accumulation Function b What you are finding: Students should certainly know that the FTC says that " f ( x ) dx = F (b) # F ( a) a b where F ( x ) is an antiderivative of f ( x ) . This leads to the fact that F ( a) + " f ( x ) dx . Students should be a prepared to use substitution methods to integrate and be able to change ! the limits of integration using this b ! substitution. Students should ! also know that " a a f ( x ) dx = # " f ( x ) dx . !b x The accumulation function looks like this: F ( x ) = " f (t ) dt . It represents the accumulated area under the 0 curve f starting at zero (or some value) ! and going out to the value of x. The variable t is a dummy variable. It is important to believe that this is a function of x. ! " 92. #e cosx sin x dx = 0 A. 1 e B. e " ! ! 1 e C. e "1 D. 1" 1 e E. e B. u = cos x,du = "sin x dx x = 0,u = 1, x = #,u = "1 ! ! # "1 1 cosx u "1 1 ! $ e sin x dx = " $ e du = "(e " e ) = e " e 0 1 ! 3 93. # 20 " 4 x dx is equivalent to A. "1 ! u du # 4 "1 5 1 15 B. "1 3 # u du 4 1 C. 1 16 " u du 4 8 16 16 D. 4 " u du 8 E. " u du 8 ! 3 # C. 20 " 4 x dx ! u = 20 " 4 x,du = "4 dx ! ! 3 # 20 " 4 x dx = 1 x 94. " 1 "1 8 1 16 u du = # # u du 4 16 4 8 ln t dt = t x 2 "1 ! A. 2 x2 B. 2 ! ! ! ! x = 1,u = 16, x = 3,u = 8 1 C. (ln x ) 2 2 D. (ln x ) 2 "1 2 E. ln x 1 C. u = ln t,du = dt t = 1,u = 0,t = x,u = ln x! t ! ! ln x 2 x ln x ln x ) ln t u2 ( " t dt = " u du = 2 = 2 1 0 0 ! © 2011 www.mastermathmentor.com ! - 41 - Demystifying the MC AB Calculus Exam 95. The graph of the piecewise linear function f is shown in the x figure to right. If g( x ) = " f (t ) dt , which of the following is the 1 greatest? A. g("4 ) ! D. g( 4 ) B. g("3) E. g(5) C. g(1) -3 ! ! ! of the triangle # f (t) dt : the area !)= B. g("3 ! 1 4 g( 4 ) = # f (t ) dt : the area of the triangle = 1 96. (Calc) Let F ( x ) be an antiderivative of ! 1 ("4)("3) = 6 2 = A. -15.879 B. -11.629 ! 4 " E. 1 ( 3)(2) = 3 all others will be less. 2 x 3 + x + 1 . If F (1) = "2.125, then F ( 4 ) = C. 7.274 D. 15.879 ! ! x + x + 1 dx = F ( 4 ) # F (1) $ F ( 4 ) = F (1) + 3 1 E. 11.629 4 " x 3 + x + 1 dx 1 4 F ( 4 ) = #2.125 + " x 3 + x + 1 dx = #2.125 + 13.754 = 11.629 1 2 97. If f is a continuous function and F "( x ) = f ( x ) for all real numbers x, then ! 1 1 A. 3F ("2) " 3F (2) B. F ("2) " 3F (2) 3 3 1 1 ! D. 3F ("5) " 3F ( 7) E. F ( 7) " F ("5) ! 3 3 ! ! dx E. u = 1" 3x,du = "3 ! " x = "2,u = 7, x = 2,u = "5 C. 1 1 F (2) " 3F ("2) 3 3 ! 7 1 2 1 "5 1 1 1 "3 f (1" 3x!) dx = " # f ( u) du = F ( u) "5 = F ( 7) " F ("5) # 3 "2 37 3 3 3 4 6 98. If f is continuous for all real numbers x and " f ( x ) dx = 10, then 1 ! A. 37 B. 39 C. 35 D. 57 6 A. # f (1" 3x ) dx = "2 6 " [ f ( x # 2) + 2x ] dx = 3 E. 25 6 # [ f ( x " 2) + 2] dx = # !f ( x " 2) dx + # 2 dx 3 3 3 u = x " 2,du = dx x = 3,u = 1, x = 6,u = 4 6 # 4 f ( x " 2) dx = 3 © 2011 www.mastermathmentor.com ! # 1 6 f ( x ) dx = 10 # 2x dx = x 3 - 42 - 2 6 3 = 36 " 9 = 27 $ 10 + 27 = 37 Demystifying the MC AB Calculus Exam R. Derivative of Accumulation Function (2nd FTC) d x What you are finding: You are looking at problems in the form of " f (t) dt . This is asking for the rate dx a of change with respect to x of the accumulation function starting at some constant (which is irrelevant) and ending at that variable x. It is important to understand that this expression is a function of x, not the variable t. In fact, the variable t in this expression could be any variable (except x). ! d x nd How to find it: You are using the 2 Fundamental Theorem of Calculus that says: " f (t) dt = f ( x ). dx a Occasionally you may have to use the chain rule that says d dx g ( x) " f (t) dt = f (g( x )) # g$( x ) . a ! "2 99. Let f be the function given by f ( x ) = # t sin2t dt for $ " % x % ". In what interval(s) is f ( x ) x decreasing? ! $ # #' $ "# ' $ # ' B. &" , ) C. &"#, ) * & ,#) ! % 2 2( % 2 ( %2 ( ! E. ("#,# ) # 2 ( d% +# # B. f "( x ) = '' $ t sin2t dt ** = +t sin2t = 0 , t = 0, , dx & ! 2 ! 2 ) x ! f "( x ) : + + + + + + 0 + + + + + + + 0 + + + + + + + +0 + + + + + + A. (0,") D. ("#,0) ! ! -# 2 +# 0 # 2 # % # ( % #( % # #( f "( x ) < 0 : ' + ,0* - ' 0, * and f decreasing on ' + , * & 2 ) & 2) & 2 2) 100. The graph of the function f shown to the right has a horizontal tangent at x = 4. Let g be the continuous function defined by !t g( x ) = " f ( t ) dt . For what value(s) of x does the graph of g have a 0 point of inflection? ! A. 3 and 5 B. 0, 3, and 5 D. 2 and 4 E. 4 only C. 2 only t D. g( x ) = " f (t ) dt # g$( x ) = f ( x ) # g$$( x ) = f $( x ) 0 % & % g$$( x ) : + + + + + ' ' ' ' ' ' ' + + + + + g changes concavity at x = 2, x = 4 !!!!! © 2011 www.mastermathmentor.com ! 2 4 - 43 - Demystifying the MC AB Calculus Exam d 101. dx ! x3 # tan(t 4 1 "1) dt = A. sec 2 ( x12 "1) B. tan( x 4 "1) D. 3x 2 tan( x12 "1) E. 12x11 tan( x12 "1) d D. dx ! ! x3 # tan(t 1 C. tan( x12 "1) 4 ! ! 4 "1) dt = tan ( x 3 ) "1 ( 3x 2 ) = 3x 2 tan( x12 "1) ! [ ] x 102. (Calc) Let f be the function given by f ( x ) = " cos( t 2 + t ) dt for # 2 $ x $ 2. Approximately, for what 0 ! percentage of values of x for " 2 # x # 2 is f ( x ) decreasing? A. 30% B. 26% ! d B. f "( x ) = dx C. 44% ! D. 50% E. 59% ! # cos(t 2 + t) dt = cos( x 2 + x ) x 0 By the graphs below, f "( x ) < 0 for $ 2 < x < 1.849 and 0.849 < x < 1.728 So f "( x ) < 0 for 1.849 + 2 + 1.728 $ 0.849 = 1.030 values out of 4 = 26% ! x 103. Let f ( x ) = " g(t ) dt where g has the graph shown to the right. Which a of the following could be the graph of f ? ! ! A. B. C. D. E. A. f ( x ) will have a horizontal tangent at x = b and x = c. The accumulated area will be negative until slightly to the right of the origin when it becomes positive. © 2011 www.mastermathmentor.com ! - 44 - Demystifying the MC AB Calculus Exam S. Interpretation of a Derivative as a Rate of Change What you are finding: As mentioned in section I (Related Rates), a quantity that is given as a rate of change needs to be interpreted as a derivative of some function. Typical problems ask for the value of the function at a given time. These problems can be handled several ways: a) solving a Differential Equation with initial condition (although DEQ’s may not have even been formally mentioned yet) b) Integral of the rate of change to give accumulated change. This uses the fact that: b b # R"(t) dt = R(b) $ R(a) or R(b) = R(a) + # R"(t) dt a a 104. The table below gives values of a function f and its derivative at selected values of x. If f " is 2 ! continuous on the interval [-4, 4], what is the value of $ f "(2x ) dx ? #2 x f ( x) f "( x ) A. -2 -4 8 -6 B. -8 ! A. ! u = 2x,du = 2dx 2 -2 -1 -6 -4 -4 ! -2 C. 5 2 -4 2 4 4 4 ! D. 20 E. 1 x = "2,u = "4, x = 2,u = 4 4 1 1 $ f #(2x ) dx = 2 $ f #(u) du = 2 f (u) "2 1 -2 0 "4 4 "4 = 1 1 f ( 4 ) " f ("4 )] = ( 4 " 8) = "2 [ 2 2 105. The graph of y = f "( x ) , the derivative of f , is shown in the ! figure to the right. If f (0) = "1, then f (1) = A. 1 ! B. 2 ! D. 4 C. 3 E. 5 1 B. Method 1: # 1 f "( x ) dx = f (1) $ f (0) % f (1) = f (0) + 0 # f "( x ) dx 0 1 1 # f "( x ) dx = area of trapezoid = 2 (5 + 1) = 3 % f (1) = $1+ 3 = 2 0 Method 2 : f "( x ) = $4 x + 5 % f ( x ) = $2x 2 + 5x + C f (0) = 0 + C = $1 % f ( x ) = $2x 2 + 5x $1 % f ( x ) = $2 + 5 $1 = 2 ! © 2011 www.mastermathmentor.com - 45 - Demystifying the MC AB Calculus Exam 106. (Calc) The rate of change of people waiting in line to buy tickets to a concert is given by w ( t ) = 100( t 3 " 4t 2 " t + 7) for 0 # t # 4. 800 people are already waiting in line when the box office opens at t = 0. Which of the following expressions give the change in people waiting in line when the line is getting shorter? ! 3.773 3.773 # w"(t) dt A. B. 1.480 ! C. 800 " 1.480 2.786 D. 3.773 " w(t) dt $ w#(t) dt 1.480 2.786 # w"(t) dt E. ! 0 ! " w(t) dt ! 0 ! t= b B. Since w ( t ) represents the rate of people waiting, " w(t) dt represents the change of t= a the number of people in line between 2 times. The line is getting shorter when its rate of change is negative. The 800 people already in line doesn't enter into it. 107. (Calc) A cup of coffee is heated to boiling ( 212°F ) and taken out of a microwave and placed in a 72°F room at time t = 0 minutes. The coffee cools at the rate of 16e"0.112t degrees Fahrenheit per ! minute. To the nearest degree, what is the temperature of the coffee at time t = 5 minutes? A. 105°F ! B. 133°F ! C. 166°F 5 ! ! D. T (5) " T (0) = ! ! D. 151°F E. 203°F 5 # "16e"0.112t dt $ T (5) = T (0) " # 16e"0.112t dt % 151 ! 0 0 ! ! 108. (Calc) The Cheesesteak Factory at a Philadelphia stadium has 25 steak sandwiches ready to sell when it opens for business. It cooks cheesesteaks at the rate of 4 steaks per minute and sells cheesesteaks at # 2"t & the rate of 1+ 6sin% ( steaks per minute. For 0 ≤ t ≤ 20, at what time is the number of steaks ready $ 41 ' to sell at a minimum (nearest tenth of a minute)? A. 0 ! B. 3.4 C. 10.3 D. 17.1 E. 20 t $ $ 2#x ' ' $ 2#t '0 D. C( t) = 25+ * 4 " &1+ 6sin& ) ) dt + C ,( t) = 4 " /1+ 6sin& )2 = 0 % 41 ( ( % 41 (1 . % 0 $ 2#t ' 4 = 1+ 6sin& ) + Critical values : t = 3.417,t = 17.083 % 41 ( t C( t) © 2011 www.mastermathmentor.com ! 0 3.417 17.083 20 25 30 3.2 6.8 - 46 - Demystifying the MC AB Calculus Exam T. Average Value of a Function What you are finding: You are given a continuous function f ( x ) , either an algebraic formula or a graph, as well as an interval [ a,b] . You wish to find the average value of the function on that interval. b How to find it: f avg = ! " f ( x ) dx a b#a . The units will be!whatever the function f is measured in. Again, be careful. The average rate of change of a function F on [ a,b] is not the same as the average value of the function F on [ a,b] . ! b Average value of the function: ! " f ( x ) dx ! a b#a b Average rate of change of the function (average value of the rate of change): ! # F "( x ) dx a b$a F (b) $ F ( a) b$a = #" 3" & 109. The average value of sin 2 x cos x on the interval % , ( is $2 2 ' ! "2 2 A. B. C. 0 D. -1 E. 1 3# 3" ! ! 3! 2 2 " sin x cos x dx # 3# ! ! A. ! f avg = 2 u = sin x,du = cos x dx x = ,u = 1 x = ,u = $1 3# # 2 2 $ 2 2 -1 "u f avg = 2 du 1 # $1 u3 $2 = = 3 1 3# 2 110. The velocity of a particle moving along the x-axis is given by the function v ( t ) = te t . What is the ! average velocity of the particle from time t = 1 to t = 3 ? A. 19e 9 " 3e 2 B. 4e 9 C. 3 " te ! ! D. v avg = v avg = © 2011 www.mastermathmentor.com ! 1 1 2 3e 9 " e 2 D. t2 dt ! 3 #1 3 u = t ,du = 2t ! dt 2 2 " 2te t dt 1 2 = 1 2 9 "e u du 1 2 - 47 - e9 " e ! 4 E. e 9 " e ! t = 1,u = 1 t = 3,u = 9 1 u9 e e9 # e 1 =2 = 2 4 Demystifying the MC AB Calculus Exam sin(2x + 3) on the interval [-2, 3] ? x 2 + 2x + 3 111. (Calc) What is the average value of y = A. -0.057 B. 0.051 C. 0.061 ! 3 B. f avg = D. 0.256 E. 0.303 sin(2x + 3) dx 2 + 2x + 3 0.256 = = 0.051 3+2 5 "x -2 112. Newton the cat begins to walk along a ledge at time t = 0. His velocity at time t, 0 ≤ t ≤ 8, is given by the function whose graph is given in the ! What is Newton’s average speed from t = 0 to t = 8? figure to the right. A. 0 B. 2 9 4 E. 5 D. C. 3 8 2 4 6 8 " v (t) dt " v (t ) dt + " v (t ) dt + " v (t ) dt # " v (t ) dt ! = 0 8#0 6 + 10 + 5 + 3 24 !!!!!Average Speed = = =3 8 8 C. Average Speed = 0 2 4 6 8#0 $ "( x # 9) ' 113. (Calc) Barstow, California has the daily summer temperature modeled by T ( t ) = 85 + 27sin& ) % 12 ( ! where t is measured in hours and t = 0 corresponds to 12:00 midnight. What is the average temperature in Barstow during the period from 1:30 PM to 6 PM? A. 60° B. 89° ! D. 110° C. 104° 18 ! ! ! ! D. Tavg = E. 115° $ " ( x # 9) ' ) dt ( + 110° 4.5 * 85+ 27sin&% ! 12 13.5 114. Let f ( x ) = 3x 2 " 6x . For how many positive values of b does the average value of f ( x ) on the ! interval [0, b] equal the average rate of change of f ( x ) on the interval [0, b] ? ! A. 4 B. 3 C. 2 # (3x 2 " 6x ) dx 0 E. none b"0 f (b) " f (0) b 3 " 3b 2 3b 2 " 6b 6 ± 12 = $ = $ b 2 " 6b + 6 = 0 $ b = b b b 2 © 2011 www.mastermathmentor.com ! ! ! b C. D. 1 - 48 - Demystifying the MC AB Calculus Exam U. Straight-Line Motion - Integrals What you are finding: In section J, we looked at straight-line motion by the derivative process. You were typically given a position function x ( t ) and took its derivative to find the velocity function v ( t ) and took the velocity’s derivative to get the acceleration function a( t ) . Questions like finding maximum velocity or minimum acceleration could then be answered. ! ! Using integrals, we can work backwards. Typical questions involve knowing the velocity function, the ! position of the particle at the start of the problem, and integrating to find the position function. How to find it: x ( t ) = " v (t ) dt + C and v (t ) = " a(t ) dt + C . Using these equations is essentially solving DEQ’s with an initial condition, usually the position at time t = 0 or velocity at t = 0. Two other concepts that come into play are displacement and distance over some time interval [ t1,t 2 ] . ! Displacement: the difference in position over [ t1,t 2 ] : x ( t 2 ) " x ( t1 ) = t2 # v (t ) dt . Displacement can be t1 positive, negative or zero. ! t2 Distance: how far the particle traveled over [ t1,t 2 ] : ! " v (t ) dt . Distance is always positive. t1 115. A particle moves along the x-axis so that at any time t ≥ 0, its velocity is given by v ( t ) = 5 " 2t " 3t 2 . If the particle is at position x = "3 at ! time t = 2, where was the particle at t = 1? A. -5 B. 2 C. 3 ! B. x ( t ) = D. 4 # (5 " 2t " 3t ) dt = 5t " t 2 2 ! E. 11 " t3 + C x (2) = 10 " 4 " 8 + C = "3 $ C = "1 x ( t ) = 5t " t 2 " t 3 "1 $ x (1) = 5 "1"1"1 = 2. 116. A spider is walking up a vertical wall. At t = 0, the spider is 18 inches off the floor. The velocity v of the ! in inches per minute at time t spider, measured minutes, 0 ≤ t ≤ 10 is given by the function whose graph is shown to the right. How many inches off the floor is the spider at t = 10? A. 13.5 D. 31.5 B. 19.5 E. 37.5 C. 24 10 D. y ( t ) = 18 + " v (t ) dt = 18 + 9 + 4.5 # 3 + 3 = 31.5 0 ! © 2011 www.mastermathmentor.com - 49 - Demystifying the MC AB Calculus Exam 117. The table below gives selected values of v ( t ) , of a particle moving along the x-axis. At time t = 0, the particle is at the origin. Which of the following could be the graph of the position, x ( t ) , of the particle for 0 ≤ t ≤ 4 ? ! 0 3 t v(t) 1 0 2 -1 3 1 4 3 ! ! A. B. D. E. C. D. Velocity is zero at t = 1 and at 2 < t < 3 so the graph of x ( t ) must have horizontal tangents at t = 1 and at 2 < t < 3. 118. A particle moves along the y-axis with velocity v ( t ) for 0 ≤ t ≤ 5. ! The graph of v ( t ) is shown to the right with the positive areas indicated by P, Q, and R. What is the difference between the distance that the particle travels and its displacement for ! 0≤t≤5? ! A. P - R D. 2R B. 2(P + Q + R) E. 2(Q + R) C. 2Q 5 E. Distance = " v (t ) dt = P + Q + Q + R = P + 2Q + R 0 5 Displacement = " v (t) dt = P # Q + Q # R = P # R 0 P + 2Q + R # (P # R ) = 2Q + 2R © 2011 www.mastermathmentor.com ! - 50 - Demystifying the MC AB Calculus Exam 119. (Calc) A particle moves along the x-axis so that its velocity v ( t ) = 12te"2t " t + 1. At t = 0, the particle is at position x = 0.5. What is the total distance that the particle traveled from t = 0 to t = 3 ? A. 1.448 B. 1.948 C. 2.911 ! D. 4.181 E. 4.681 ! 1.69 3 3 # v (t ) dt " # v (t) dt or D. Distance = 0 1.69 # v (t ) dt = 4.181. 0 The starting position has no bearing on the answer. 120. (Calc) A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a( t ) = cos(1" 2 t ) . If the velocity of the particle is -2 at time t = 0, then the speed of the particle at ! t = 2 is A. 0.613 ! B. 0.669 C. 1.331 D. 1.387 E. 1.796 2 D. v ( t ) = v (0) + $ a(t) dt = "2 + 0.613 = "1.387 # Speed = 1.387 0 ! 121. A particle moves along the x-axis so that its velocity v ( t ) = 2e1"t " 2t , as shown in the figure to the right. Find the total distance that the particle traveled from t = 0 to t = 2. 2 "2 e 2 C. 2e " + 4 e 2 E. 2e + + 2 e 2 B. 2e " " 4 e A. 2e + ! ! D. 2e " 2 ! ! ! A. Distance = ! 1 2 # (2e1"t " 2t) dt " # (2e 0 1"t 1 1 " 2t ) dt 2 Distance = ["2e1"t " t 2 ] 0 " ["2e1"t " t 2 ]1 = "2 "1" ("2e) + 2e"1 + 4 " (2 + 1) Distance = "3 + 2e + © 2011 www.mastermathmentor.com ! 2 2 + 4 " 2 "1 = 2e + " 2 e e - 51 - Demystifying the MC AB Calculus Exam V. Area/Volume Problems What you are finding: Typically, these are problems with which students feel more comfortable because they are told exactly what to do or to find. Area and volume are lumped together because, almost always, they are both tested within the confines of a single A.P. free response question. Usually, but not always, they are on the calculator section of the free-response section. Area problems usually involve finding the area of a region under a curve or the area between two curves between two values of x. Volume problems usually involve finding the volume of a solid when rotating a curve about a line. How to find it: Area: Given two curves f ( x ) and g( x ) with f ( x ) " g( x ) on an interval [a, b], the area x= b between f and g on [a, b] is given by A = # [ f ( x ) " g( x )] dx . While integration is usually done with respect x= a ! ! to the x-axis, these problems sometimes show up in terms of y: A = y= d # [m( y ) " n( y )] dy . y= c Volume: Disks and Washers: ! The method I recommend is to establish the outside Radius R, the distance from the line of rotation to the outside curve, and, if it exists, the inside radius r, the distance from the line of rotation to the inside curve. The formula when rotating these curves about a line on an interval is given by: ! y= d x= b 2 2 2 2 V = " $ [ R( x )] # [ r( x )] dx or V = " $ [ R( y )] # [ r( y )] dy ( ) x= a ( ) y= c A favorite type of problem is creating a solid with the region R being the base of the solid. Cross sections perpendicular to an axis are typically squares, equilateral triangles, right triangles, or semi-circles. Rather than give formulas for this, it is suggested that you draw the figure, establish its area in terms of x or y, and !expression on the given interval. ! integrate that 122. What is the area of the region enclosed by the graphs of y = x " 4 x 2 and y = "7x ? A. 4 3 16 68 80 C. 8 D. E. 3 3 3 ! be able to sketch the line and parabola. B. Even without a calculator, students should B. x " 4 x 2 = "7x # 4 x 2 " 8x = 0 # 4 x ( x " 2) = 0 # x = 0, x = 2 !2 ! ! 2 % 2 4 x 3 (2 32 16 2 2 A = $ ( x " 4 x + 7x ) dx = $ (8x " 4 x ) dx = '4 x " = * = 16 " 3 )0 3 3 & 0 0 ! 123. (Calc) What is the area of the region in the first quadrant enclosed by the graph of y = 2cos x, y = x, and the x-axis? ! A. 0.816 B. 1.184 C. 1.529 1.03 A. A = #2 ! 1.03 " x dx + " 2cos x dx 0 D. 1.794 1.03 or A = + " -,cos 0 E. 1.999 % y( . ' * $ y0 dx = 0.816 & 2) / $1 ! © 2011 www.mastermathmentor.com - 52 - Demystifying the MC AB Calculus Exam 124. The three regions A, B, and C in the figure to the right are bounded by the graph of the function f and the x-axis. If the areas of A, B, and C are 5, 2, and 1 1 # [ f ( x ) + 2] dx ? respectively, what is the value of "4 A. 28 C. 6 E. -12 ! 1 C. B. 10 D. -4 1 1 # [ f ( x ) + 2] dx = # f ( x ) dx + # 2 dx "4 "4 "4 1 " 5 + 2 "1+ [2x ]"4 = "4 + [2 " ("8)] = "4 + 10 = 6 ! x"3 2 is shown in the figure to the right. Which of the following calculations would accurately compute the area of region R? ! 9 ! # x " 3& dx I. ) % x " ( $ 2 ' 0 1 9 $ x # 3' II. " 2 x dx + " & x # ) dx % 2 ( 0 1 125. Region R, enclosed by the graphs of y = ± x and y = 3 ! III. # (2x + 3 " x ) dx 2 "1 ! A. I only ! B. II only C. III only D. II and III E. I, II, and III D. I does not take into consideration y = " x II is accurate. y = x is always the top curve but y = " x is the bottom curve x"3 on [0,1] and y = is the bottom curve on [1,9]. 2 III is accurate. If we integrate with respect to y, we would use right minus left. 3 That gives # (2y + 3 " y ) dy. y is a dummy variable, You can use any variable. 2 "1 (when you calculate it with your calculator, you use x ). ! © 2011 www.mastermathmentor.com - 53 - Demystifying the MC AB Calculus Exam 126. (Calc) On the graph to the right, the line y = kx "10 , k a constant, is tangent to the graph of y = x 2 " 4 . Region R is enclosed by the graphs of y = kx "10 , y = x 2 " 4 , and the y-axis. Find the area of region R. A. 4.287 ! ! D. 9.798 ! B. 4.899 ! C. 6.124 E. 12.247 B. If the curves are tangent they intersect so x 2 " 4 = kx "10 and are tangent so 2x = k. x 2 " 4 = 2x 2 "10 # x = 6 which is the intersection point. Thus k = 2 6. 6 The line is therefore y = 2x 6 "10. A = $ [x 2 )] ( " 4 " 2x 6 "10 dx = 4.889. 0 127. If the region bounded by the y-axis, the line y = 4, and the curve y = x is revolved about the x-axis, ! the volume of the solid generated would be A. 88" 3 B. 56" ! D. 128" 3 E. 640" 3 C. Students should be able to sketch the graph without a calculator. 16 16 ! 2( % x = 4 " x = 16!so V = # + '4 2 $ ! x * = # + (16 $ x ) & ) 0 0 ! ! C. 128" ( ) 16 % , x2( 256 / V = #'16x $ * = #. 256 $ 1 = 128# 2 )0 2 0 & ! " 128. (Calc) If the region bounded by the x-axis, the line x = , and the curve y = sin x is revolved about 2 " the line x = , the volume of the solid generated would be 2 ! A. 1.142 B. 1.468 C.!1.793 D. 3.142 E. 3.586 ! " " # x = # sin#1 y 2 2 2 1 $" #1 ' V = " * & # sin y ) dy = 1.142" = 3.586 %2 ( 0 E. R = " 2 $" ' by shells (not in curriculum) :V = 2" * & # x ) sin x dx = 3.586 %2 ( 0 © 2011 www.mastermathmentor.com ! - 54 - Demystifying the MC AB Calculus Exam 129. A piece of candy is determined by the curves y = ± x 3 , for 1 ≤ x ≤ 2 as shown in the figure to the right. For this piece of candy, the cross sections perpendicular to the x-axis are equilateral triangles. What is the volume of the piece of candy in cubic units? ! 127 7 127 3 D. 21 15 3 2 127 3 E. 7 A. B. ! ! ! ! C. 254 3 7 ! 1 E. Area = (2) y y 3 = y 2 3 = x 6 3 2 2 2 #x7& 3 7 3 127 3 6 V = 3 " x dx = 3% ( = 2 )1) = (127) = ( 7 7 7 $ 7 '1 1 ( ) 130. A lake shaped in a right triangle is located next to a park as shown in the figure to the right. The depth of the lake at any point along a strip ! edge is f ( x ) feet. Which of the following x miles from the park’s expressions gives the total volume of the lake? 3 A. # 4 & ) %$ 4 " 3 x (' f ( x ) dx! B. 0 4 D. " f ( x ) dx E. 0 ! ! 4 3 # 3 & ) %$ 3 " 4 x (' f ( x ) dx 0 3 2 C. 2 " f ( x ) dx 0 4 " f ( x ) dx 0 ! 4 A. The hypotenuse is given by y = 4 " x. The dimensions of our 3 ! 4 rectangular solid will be height : 4 " x, width : dx and depth : f ( x ) 3 3 # 4 & So the volume is given by V = ) % 4 " x ( f ( x ) dx $ 3 ' 0 ! 131. (Calc) Let R be the region between the graphs of y = 2sin x and y = cos x as shown in the figure to the right. The region ! R is the base of a solid with cross sections perpendicular to the x-axis as rectangles that are twice as high as wide. Find the volume of the solid. ! A. 4.472 B. 7.854 D. 15.708 E. 31.416 C. 8.944 D. 2sin x = cos x " x = 0.464,3.605 3.605 2 s = 2sin x # cos x. Area = 2s( s) = 2s . V = 2 $ (2sin x # cos x )2dx = 15.708 0.464 © 2011 www.mastermathmentor.com ! - 55 - Demystifying the MC AB Calculus Exam W. Differential Equations dy = ( Algebraic Expression) . The dx goal of solving a DEQ is to work backwards from the derivative to the function; that is to write an equation in the form of y = f ( x ) + C (since the technique involves integration, the general solution will have a constant of integration). If the value of the function at some value of x is known (an initial condition), the ! value of C can be found. This is called a specific solution. What you are finding: A differential equation (DEQ) is in the form of How!to find it: In Calculus AB, the only types of DEQ’s studied are called separable. Separable DEQ’s are those than that can be in the form of f ( y ) dy = g( x ) dx . Once in that form, both sides can be integrated with the constant of integration on only one side, usually the right. Word problems involving change with respect to time are usually models of DEQ’s. A favorite type is a ! rate of change of y is proportional to some expression. The equation that problem using the words: The dy describes this statement is: = k " (expression) . The rate of change is usually with respect to time. dt Usually there is a problem that requires you to create a slope field. Simply calculate the slopes using the given derivative formula and plot them on the given graph. Usually the slopes will be integer values. 132. The slope!field for the equation in the figure to the right could be A. dy = x + y2 dx B. dy = x " y2 dx D. dy =x+y dx ! E. dy = x2 " y dx ! ! C. dy = xy dx E. Since the slope is 0 at (1,1), choices A and C are eliminated ! Since the slope ! is 0 at ("1,1), choice B is eliminated The slope field cannot be D because along the line y = "x, all slopes should be zero. 133. If ! A. ! dy = 3x 2 y 2 and if y = 1 when x = 2, then when x = -1, y = dx 1 10 B. A. ! ! 1 8 dy = 3x 2 dx " 2 y ! 1 19 D. "1 8 E. -1 1 = # 3x 2 dx " $ = x 3 + C y ! 1 $1 1 $1 = 8 + C " C = $9 $ = x 3 $ 9 " y = 3 = y x $ 9 9 $ x3 1 1 x = $1 " y = 3 = 9 $ ($1) 10 © 2011 www.mastermathmentor.com ! C. dy #y 2 - 56 - Demystifying the MC AB Calculus Exam 134. The rate of change of the volume V of the oil in a tank with respect to time t is inversely proportional to the square root of V. If the tank is empty at time t = 0 and V = 9 when t = 2, find the equation which describes the relationship between t and V. A. V = 9 2 t 4 B. V = D. ! ! t2 36 C. V 3 2 = 4 t 9 D. V 3 2 = 27 t 2 E. V = 3t dV k 2 = " V 1 2 dV = kt " V 3 2 = kt + C ! dt 3 V ! ! 2 t = 0,V = 0 : V 3 2 = kt 3 2 32 2 27 (9) = 2k " k = 9 V 3 2 = 9t " V 3 2 = t 3 3 2 135. (Calc) The Athabasca Glacier in Canada is one of the few glaciers to ! drive. Like many glaciers in the world, it is receding (in which you can this case its toe is moving away from the road). In the year 2000, its toe was 500 feet from the road and in the year 2010, its toe was 600 feet from the road. The rate the distance that the toe increases from the road is proportional to that distance. Assuming the glacier recedes at the same rate, what will be the average distance in feet per year (nearest integer) the glacier recedes from 2000 to the year 2050? A. 10 B. 13 C. 15 D. 16 E. 18 ds = ks " s = Ce kt . dt #6& 600 = 500e10k " k = ln% ( 10 $5' C. s is the distance from the road : 500 = Ce 0 " s = 500e kt The year 2050 : s = 500e 50k ) 1244. Average rate of change : s(50) * s(0) 1244 * 500 = ) 15 50 50 ! © 2011 www.mastermathmentor.com - 57 - Demystifying the MC AB Calculus Exam