Section 2 Practice Tests

Transcription

1 Section Practice Tests This section gives 18 sample problems that mix and match concepts similar to problems on the free-response section of the AP exam. While any of these examples would be a good review for students, they have been designed in three groups (A, B, and C) of six problems. In each set of six problems, the first two allow the use of a graphing calculator, and the other four do not. If a teacher gives one of these sets of six as a practice test, he can be sure that most of the concepts usually covered in an AP free-response section will be covered. However, in my attempt to be comprehensive, some examples may have more than the usual number of subparts. I have also indicated (in parentheses) how many points out of the typical nine that I would allocate for each subpart. Topic Practice Test A Practice Test B Practice Test C A1 A A3 A A5 A6 B1 B B3 B B5 B6 C1 C C3 C C5 C6 A. Horizontal Asymptotes d B. Approximate Rate of Change a b b,c a,b C. Intermediate Value Theorem d D. Equation of a Tangent Line a c c d E. Local Linear Approximation c c F. Continuity and Differentiabilty a G. Mean Value Theorem d d H. Horizontal and Vertical Tangents b,e I. Related Rates a,b a J. Straight-Line Motion - Derivatives a a-c a,d K. Function Analysis b b,e b d d e c L. Second Derivative Test e b M. Absolute Extrema c c d c N. Computation of Riemann Sums b c d O. Accumulation function / Fundamental Theorem a c,d a b,c d,e a-e a-c Integral of a rate of P. change to give accumulated change c a-e a-c d Derivative of Q. Accumulation Function (nd FTC) b a d R. Average Value of a Function d e c d b S. Straight-Line Motion - Integrals e d,e b,c T. Area/Volume b-d a-d a-c U. Derivative of the inverse of a function c V. Differential Equations b a,c Limits a d Chain Rule a e a d d a b Implicit Differentiation a c Integration with Substitution d,e b graphing calculator required graphing calculator graphing calculator permitted/not necessary not permitted Demystifying the AB Exam illegal to post on Internet

2 Demystifying the AB Exam illegal to post on Internet

3 Example A1: (Cal The amount of gas in gallons in a gas station storage tank is modeled by a continuous function on the time interval t, where t is measured in minutes. In this model, rates are given as follows: (i) The rate at which the gas tank is being filled by a truck 3 f ( t) = sin t gallons per minute for t <. (ii) For the first minutes, no cars pump gas. After that gas is pumped into cars at the following rate: 175 for t < 1 g( t) = 15 for 1 t At t =, the amount of gas in the storage tank is 1 gallons. a) How many gallons of gas enter the storage tank in the time interval t <? Round your answer to the nearest gallon. (1) For t, find the time interval(s) during which the amount of gas in the storage tank is decreasing. Give a reason for each interval. () For t, at what time t is the amount of gas in the storage tank a maximum. To the nearest gallon, compute the amount of gas in the tank at this time. Justify answers. () For t, what is the average rate of change of gas in the tank? Specify units. () 3 a) sint dt = 3,81 gallons The amount of gas in the tank is decreasing on the intervals 8.99 < t <1 and 15.5 < t because f t ( ) < g( t) on those intervals. Since f ( t) g( t) changes from positive to negative at t = 8.99 and 15.5, the candidates for the absolute maximum are at t = 8.99, and t = ( ) gallons of gas t min f t f t ( ) dt 175(.99) ( ) dt 175( 8) 15( 3.5) The maximum amount of gas in the tank is gallons occuring at 15.5 minutes. [ f ( t) g( t) ] dt = ( ) 8( 15) = = 3.6 gal min Demystifying the AB Exam illegal to post on Internet

4 Example A: (Cal Let f be the function f ( x) = 3x + x x 3 whose roots are x =.33, x =, and Let line l be the tangent line to the graph of f at x =. Line l intersects the graph of f at x = -1. Let regions R, S and T be defined as shown in the figure to the right. a) Find the equation of line l. Show the reasoning that leads to your answer. () Find the volume of the solid generated when region T is revolved about the x-axis. () The region S is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of the solid. () Find the area of region R. (3) a) ( ) = 6x +1 3x f ( ) = =1 ( ) =1 + 8 = 6 ( ) y = x + f x f y 6 =1 x 3.33 ( ) V = π 3x + x x 3 dx = 55.13π V = s dx = x + 3x + x x 3 dx = x 3 3x + dx = [ ( )] ( ) 1 A = ( x + ) dx + 3x + x x 3 = = ( ) dx ( ) dx OR A = 1 ( ) ( ) x 3 3x + 3x + x x 3 `.33 ( ) dx = Demystifying the AB Exam illegal to post on Internet

5 Example A3: Linda has a new pony and she goes out from home on a 1-minute ride along a straight road. During the time interval t 1, her velocity v( t) in feet/min is modeled by the piecewiselinear function whose graph is shown to the right. a) Find Linda s acceleration at time t =.5 minutes. Indicate units of measure. (1) At what time(s) does Linda turn around and head towards home? Give a reason for your answer. (1) What is the maximum distance Linda gets from her home? Justify your answer. (3) After riding 1 minutes, how far from home is Linda? Justify your answer. () e) What is Linda s average speed for her 1-minute ride? () a) a(.5) = 5 ft min At t = 5 and t = 9 minutes. At those times, her velocity goes from positive to negative. Because of part, the maximum distance that Linda is away from home must occur at t = 5 or t = 9. She turns towards home at t = 5 and travels 6 feet then turns around and travels feet. So she is furthest away from home at t = 5. 5 v( t) dt = = 7 feet. 1 v( t) dt = = She is feet from home. e) 1 v( t) dt 1 = = ft min Demystifying the AB Exam illegal to post on Internet

6 Example A: The function f is defined and differentiable on the interval [-6, 8] and satisfies f ( ) = 3. The graph of y = f ( x), the derivative of f, consists of a semicircle and six line segments as shown in the figure below. a) Find f ( 6) and f ( 6). () Find the x coordinate of each point of inflection of the graph of y = f ( x) on -6 < x < 8. Explain your reasoning. (1) Find the equation of the tangent line to f at x = -6. Use this line to approximate f ( 5.9). Is this value an over-approximation or under-approximation? Explain. () Find the absolute maximum value of f ( x) on -6 x 8. Justify your answer. () e) The function g is defined as g( x) = x f x of g on -6 < x < 8. Explain your reasoning. () ( ). Find the values of x for each critical point a) ( ) = 3 + f ( x ) f 6 ( ) = 3 + f ( x ) f 6 6 dx = π 6 ( ) = π 7 dx = = 3 e) The graph of y = f ( x) has inflection points at x = and x = 5 because f changes from positive to negative at x = and f changes from negative to positive at x = 5. f 6 ( ) =, f ( 6) = 3 y 3 = ( x + 6) y = x +15 ( ) ( 5.9) +15 = = 3. f 5.9 f is positive so f is increasing, f is decreasing, so f is concave down. So 3. is an over - approximation for f ( 5.9). Candidates for points of rel. max. are at x = 5 and x = because f switches from positive to negative at those values. ( ) = 3 + f ( x ) f 5 f 8 5 ( ) = 3 + f ( x ) g x 8 dx = f ( ) = 3 + f ( x ) dx = π 3 f 6 dx = π 5 The maximum value of f is. ( ) = x f ( x) = x = f ( x) ( ) : x = and x. x = f x ( ) = Demystifying the AB Exam illegal to post on Internet

7 Example A5: Kurt and Sam are rolling a ball between them along the x-axis. The ball s velocity is modeled by a differentiable function v, where the position x is measured in yards and time t is measured in seconds. Selected values of v( t) are given in the table below. The ball is at position x = - when t = seconds. t (seconds) v t ( ) (yards/secon a) Estimate the acceleration at t = 1 seconds. Show your computations and indicate units of measure. (1) 3 Approximate v( t) dt using a trapezoidal sum with five subintervals and using correct units, explain its meaning in the context of the problem. (3) Approximate the ball s position along the x-axis after 3 seconds. (1) Explain why the ball must have accelerated to the left at yards/sec at least one time between t =3 and t = 8 seconds. () e) Suppose that the acceleration of the ball is positive for < t < 3 seconds. Explain why the position of the ball at t = 3 must be greater than x = 8 yards. () a) a 1 ( ) = v ( 1) = 1 yds sec 3 v( t) dt ( ) + 5 ( 6 ) + 8 ( ) ( ) + 6 ( 1 + 8) v( t) dt = 5 The ball's change in position from t = to t = 3 is approximately 5 yards. + 5 = 3. v is differentiable on (,3) By the Mean Value Theorem, v ( t) = a( t) = v( 8) v( 3) 8 3 = 6 5 =. e) Since a( t) > on [,3],v ( t) > on [,3] So the minimum position of the ball is x( 3) = x( ) + v( t) x( 3) > + 3( ) = 1 > 8 3 dt Demystifying the AB Exam illegal to post on Internet

8 Example A6: Let x be the height of water, measured in feet in the cubic-shaped rain tank as shown in the figure to the right. Let t be measured in hours. The volume of V of rain in the tank is increasing at the rate of 6 x cubic feet per hour. a) When there is 6 ft 3 of water in the tank, how fast is the volume of rain water changing? Specify units. () Show that dx dt = x 3 () Given that x = when t =, solve the differential equation dx dt = for x in terms of t. () 3 x At what time is the rain tank empty? (1) a) x 3 = 6 ft x = dv dt = 6 = 1 ft 3 hr V = x 3 dv dt = 3x dx dt = 6 dx dt = 6 x = 3x x 3 x x 3 dx = dt 5 x 5 = t + C t =, x = : 5 ( 3) = C = x 5 = t x 5 = 5 t + 6 = 3 5t 5 x = ( 3 5t) 5 x = ( 3 5t) 5 = 5t = 3 t = 3 5 hours Demystifying the AB Exam illegal to post on Internet

9 Example B1: (Cal During rush hour time, there is a backup of cars at a bridge s tollbooth. The numbers of vehicles in line at time t is modeled by a twice-differentiable function L for t 9. Values of L( t) are shown in the table below. t (minutes) L t ( ) (vehicles) The waiting time in minutes to get through the tollbooth at the bridge is given by W ( x) = x, where x is the number of vehicles that are lined up. a) Find the rate of change of wait time when 1 vehicles line up. Use units of measure. (1) Approximate the rate of change of the number of vehicles waiting in line at t = 1 minutes.(1) Use a trapezoidal sum to estimate the average number of vehicles waiting in line at the toll booth during the first hour of rush hour period. () For t 9, what is the fewest number of times at which L ( t) must equal zero. Give a reason for your answer. () e) Suppose that at t = 3 minutes, vehicles join the line at the rate of vehicles/minute. Find the rate of change of the waiting time in line at t = 3. Indicate units of measure and explain the meaning of your answer. (3) a) W ( x) = 1 x W x ( ) = 1 1 = 1 min vehicle or 3 sec vehicle 6 1 L ( 1) 1 = 5 vehicles min or = vehicles min or = 3 vehicles min 6 L( t) dt 6 1 ( ) + ( 6 +1 ) ( ) + 15 ( ) = 76.5 vehicles L is differentiable on [,9] so the Mean Value Theorem applies. L ( t) > for some t on (,3) and ( 5,6). L t Since L t ( ) < for some t on ( 3,5) and 6,8 ( ) is continuous on [,9], the Intermediate Value Theorem ( ) = for at least 3 values of t on [,9]. implies that L t ( W [ L( t) ]) = W L t [ ( )] L ( t) 1 e) W [ L( 3) ] L ( 3) = W ( 1) = 1 = 1 min vehicle =1 min vehicle min min This means that at t = 3 minutes, the waiting time is increasing by 1 minute every minute. ( ) Demystifying the AB Exam illegal to post on Internet

10 Example B: (Cal Jack is climbing down a vertical beanstalk that goes down to the ground and the Giant is chasing him. At time t =, Jack is 1,6 feet high and the Giant is, feet high. They both only have enough strength to climb for 6 minutes. Jack moves down the beanstalk at the rate of J( t) = 5 5t feet per minute. The Giant moves down the beanstalk at the rate of G( t) =15 + 5sin( t /) feet per minute. a) How far down does Jack climb in 15 minutes? (1) What is the distance between Jack and the Giant at time t = 15 minutes and is it increasing or decreasing at this time? Give a reason for your answer. (3) At what time is the distance between Jack and the Giant a maximum? Justify answer. () Write an equation using an integral to find the time k when Jack reaches the bottom. Do not solve. (1) e) Does the Giant reach the bottom of the beanstalk? Explain. () 15 a) J( t) dt = 5 5t dt = feet 15 Jack's height :16 Giant's height : 15 ( ) dt = J t 15 G t ( ) dt = = ft 9.9 = ft Distance between them = feet and increasing as J 15 Distance a maximum when J( t) G( t) = or J( t) = G( t) This occurs only at t = 9.66 min. Since J t ( ) > G( 15) ( ) G( t) > if t < 9.66 and J( t) G( t) < if t > 9.66 there is an absolute maximum at t = 9.66 min. k 16 J t 6 e) G t ( ) dt = ( ) dt = k or J( t) dt = = ft No. After 6 minutes, he is still ft. high Demystifying the AB Exam illegal to post on Internet

11 Example B3: Let R be the shaded region bounded by the graphs of y = x +,y = x, and the y-axis as shown in the figure to the right. a) Find the area of R. () Find the volume of the solid when R is revolved around the x-axis. () Write, but do not evaluate, an expression that represents the volume of the solid when R is revolved around the y-axis. (3) The region R is the base of a solid. For this solid, the cross sections perpendicular to the x-axis are semi-circles. Write, but do not evaluate an expression that represents the volume of this solid. () a) x + = x x = A = ( x + x) dx = 3 x 3 + x x = = 16 3 ( ) x V = π x + dx = π x + x + x dx = π x x 3 + x x 3 3 ( ) = π = π Bottom section uses disks and top section uses washers. If y = x x = y If y = x + x = ( y ) V = π y dy + [ y ( y ) ]dy ( ) :V = π x or by shells not in AB curriculum Radius = x + x V = π x + x dx ( ) ( ) x + x dx Demystifying the AB Exam illegal to post on Internet

12 Example B: A particle moves along the x-axis at any time t given by v( t) = e 1 t 1. At t =, the e particle is at x = 1. a) Find the acceleration of the particle at t =. (1) Is the speed of the particle increasing at t =? Give a reason for your answer. (1) Find all values of t where the particle changes direction. Justify your answer. () Find the position of the particle at t =. () e) Find the total distance traveled by the particle over the time interval t. (3) a) a( t) = e 1 t a( ) = e 3 = e 3 v( ) = e 3 1 e = 1 e 1 3 e < Since v( ) < and a( ) <, the speed is increasing. v( t) = e 1 t 1 e = e1 t = e 1 e) 1 t = 1 t =1 ( ) > when t <1 and v( t) < when t >1 so the particle v t changes direction at t =1 only. dt x( ) = x ( ) + e 1 t 1 e ( ) = x e1 t 1 e t = e 3 e e = e +1 1 e 3 e Dist = e 1 t 1 dt = e 1 t 1 e dt e 1 t 1 e dt e 1 e1 t 1 e t e1 t 1 e t 1 1 = 1 e1 t 1 e t 1 e 1 e + e + 1 e 3 + e 1 e 1 e = 1 e 3 1 e + e e1 t + 1 e t Demystifying the AB Exam illegal to post on Internet

13 Example B5: Let f be a function defined on the closed interval [,6]. The graph of f, consisting of five line segments is shown to the right. Let ( ) = f ( t) g x x dt. 1 a) Find g( ), g ( ), and g ( ) (3) Find the average rate of change of g on the interval x 6. (1) For how many values of c, where < c < 6, is g ( equal to the average rate found in part? (1) Find the x-coordinate of each point of inflection of g on - < x < 6. (1) e) Define h( x) = x g( x). Determine if there is a relative minimum, relative maximum, or neither for h at x =. Justify your answer. (3) a) ( ) = f ( t) g 1 g ( x) = d dx g x g 6 x dt = 3 +1= f ( t ) dt = f x 1 ( ) = f ( x) g ( ) = f ( ) = 1 ( ) g( ) = ( ) = = 3 ( ) g ( ) = f ( ) = 1. g ( = f ( = 3. There are values of c on [,6] where f ( = 3. e) ( ) = f ( x) : g x g > on (,), h x g < on (, 1), g > on 1, g < on,5 ( ) so inflection pt. at x = 1. ( ) so inflection pt. at x =. ( ) = x g ( x) + g( x) ( ) = g ( ) + g( ) = ( 1) + ( ) = so h has a critical point at x =. ( ) = x g ( x) + g ( x) + g ( x) h h x ( ) = g ( ) + g ( ) + g ( ) = = 3 ( ) = and h ( ) >, h has a relative minimum at x = h Since h Demystifying the AB Exam illegal to post on Internet

14 Example B6: Consider the closed curve in the xy-plane given by a) Show that dy dx = 3 x y 1 y 3 6y + 3x 18x =1 () Find any x-coordinates where the curve has a horizontal tangent. (1) Write an equation for the line tangent to the curve at the point (, 3) and use this line to approximate the value of y in the equation above when x = 3. () Determine the concavity of the curve at the point (, 3). () e) Is it possible for this curve to have a vertical tangent at points where it crosses the y-axis? Explain your reasoning. () a) dy 6y dx 6 dy + 6x 18 = dx dy ( dx 6y 6) =18 6x dy 18 6x = dx 6y 6 = 3 x y 1 dy dx = 3 x y 1 = x = 3 dy = 3 dx (,3) 9 1 = 1 8 y 3 = 1 8 ( x ) y = ( x ) y = 5 3 d y dx = ( y 1) ( 1) ( 3 x)y dy dx = ( y 1) d y dx = y 1 ( ) y( 3 x) d ( y 1) 3 y ( y 1) ( y) 3 x dx (,3) ( y 1) y 1 ( ) 3 x = < so the curve is concave down. Vertical tangents when y 1 = or y = ±1 The curve crosses the y - axis when x =. e) At y =1: y 3 6y + 3x 18x = x 18x =1 or 3x 18x =16 At y = 1: y 3 6y + 3x 18x = x 18x =1 or 3x 18x = 8 Neither of these equations have x = as solutions Demystifying the AB Exam illegal to post on Internet

15 Example C1: (Cal A company is offering a coupon for money off at a popular chain restaurant if people respond to an on-line survey in the month of June. The rate at which people answer surveys is 6 modeled by the function A defined by A( t) = t 8t + 7 The rate at which coupons are mailed out to people who answered the survey is modeled by 8 the function M defined by M ( t) = t 5t + 7 ( ) and M ( t) are measured in people per day and time t is measured in days. These Both A t functions are valid for the month of June ( t 3), the days at which the survey is valid. At time t =, one hundred surveys have already been taken but no coupons have been sent out. a) How many people answered surveys by June 15 (t = 15)? Round to the nearest integer. (1) The company will send a $1 coupon for all people who return surveys by June 15 and $5 for all surveys returned after June 15. What is the dollar value in coupons that the company will send for the month of June? Round your answer to the nearest dollar. () [ ] Let F ( t) = A( ) M ( ) + A( x) M ( x ) t dx. Find the value of F 15 ( ) and explain its meaning in the context of the problem. Round your answer to the nearest integer. () Find the value of F ( 15) and explain its meaning in the context of the problem. () e) At what time t for t 3 does the model predict that the number of people who have answered the survey but their coupon hasn t been mailed out is a maximum? () 15 a) 1 + A t ( ) dt A t = = 11 surveys. ( ) dt A( t) dt = 1 11 F ( t) = A( ) M ( ) + A( x) M ( x ) [ ] ( ) + 5( 53) = $13,675. dx = = 371 There are 371 people who have answered the surveys but their coupons haven t been sent out as of June 15. F ( t) = d dt t [ ] dx = A t A( ) M ( ) + A( x) M ( x ) ( ) M ( t) F ( 15) = A( 15) M ( 15) = = People who completed surveys but coupons haven't been sent is decreasing at the rate of people a day on June 15. e) F ( t) = A( t) M ( t) = A( t) = M ( t) t = 9.75 ( June 9) Demystifying the AB Exam illegal to post on Internet

16 Example C: (Cal The figure to the right shows the graph of f ( x) = ( 1 ln x) in the 1 st quadrant. Line l is given by y = 8 ex and is tangent to the 1 graph of f at the point e,. Region R is bounded by the graph of f, the line l, and the dotted vertical line in the figure. Region S is bounded by the graph of f, the dotted vertical line, and the x-axis. a) Find the area of region R. () Find the volume of the solid formed by rotating region S about the x-axis. () The region S is the base of a solid. For this solid, each cross section perpendicular to the x- axis is a rectangle whose height is the square of its base. Find the volume of the solid. () Show that the equation of line l is y = 8 ex. (3) a) 8 ex = x = e A = [ ( 1 ln x ) ( 8 ex) ] dx =.37 e 1 e f ( x) = 1 ln x = x = e e [ ] dx V = π ( 1 ln x ) =.71π =.31 base = V = e ( 1 ln x),height = ( 1 ln x) e [ ( 1 ln x ) 6 ] dx =.815 e f ( x) = ( 1 ln x) 1 x 1 f e = 1 ln 1 e e ( ) = ( 1 +1) ( e) = e Line l : y = e x 1 y = ex + y = 8 ex e Demystifying the AB Exam illegal to post on Internet

17 Example C3: (Cal A snowstorm hits a city. The amount of snow on the ground in inches is modeled by a differentiable function S for t < 8 where t is measured in hours. S( t) is constantly increasing and its graph is concave down. Values of S( t) at various times t are shown in the table below. t (hours) S( t) (inches of snow) a) Between which two consecutive hour entries on the chart above is the average snowfall rate the greatest? (1) Approximate the rate in which snow is falling at t = 8 hours. (1) Using the answer from part, approximate the amount of snow on the ground at t = 7.5 hrs. Is this an under-approximation or over-approximation? Justify your answer. (3) Using a midpoint sum with three subintervals given by the table, approximate the value of 8 S ( t ) dt. Explain the meaning of 1 8 S( t) d in terms of snowfall. () 8 ( ) S( 3) a. Hours 3 and. S 3 = = 3 inch hr b. ( ) S( 8) S( 7) S = =1 inch hr 1 c. ( )( t 8) ( ) S = t + 5 ( ) 1.5 inches S S 8 = S 8 S 13 =1 t 8 S 7.5 Since S is increasing and concave down, 1.5 inches is an over - approximation. d. 8 S ( t ) dt S( t) dt = 5 8 = 7 ( ) + 3( 8) + ( 1) = = 5 The average amount of snow on the ground over the 8 - hour period is 6 3 inches Demystifying the AB Exam illegal to post on Internet

18 Example C: Let f be the functions defined by f x a) Show that f is differentiable. (3) Find the average value of f on [, ] (3) ( ) = x ( π +1 ) π +1 for x < sinπx + x +1 for x Find the derivative of the inverse to y = sinπx + x +1 at x = 5. (3) a) The left piece is a line and the right piece a sine curve. Both are differentiable everywhere. ( ) = π + π +1= 3 lim lim f x x So f is continuous at x =. f ( x) = lim x π +1 for x < πcosπx +1 for x f x f x ( ) = π +1 lim x + x + ( ) = sinπ + +1= 3 f x ( ) = πcosπ +1= π +1 So f is differentiable at x = and thus is differentiable everywhere. f avg = [ x( π +1) π +1] dx + [ sinπx + x +1] dx πx + x πx + x + cosπx + x π + x f avg = π + π π π + + f avg = 1 π f avg = = 6 π Inverse : x = sinπy + y +1 If x = 5, 5 = sinπy + y +1 sinπy + y = y = 1 = ( πcosπy +1) dy dx dy dx = 1 πcosπy +1 dy 1 = dx y= πcosπ +1 = 1 π Demystifying the AB Exam illegal to post on Internet

19 Example C5: A tightrope walker climbs a pole and walks along a tightrope that extends in a straight line in both directions from the pole. For t 16, the tightrope walker s velocity is modeled by the piecewise-function defined by the graph below. The graph is formed by two quartercircles and straight lines. a) At what time in the interval t 16 is the tightrope walker stopped? (1) At what time in the interval t 16 is the tightrope walker farthest from the pole? How far is he from the pole at that time? Justify your answers. (3) Find the total distance the tightrope walker travels in the time interval t 16. () Write expressions of the tightrope walker s velocity v(t), acceleration a(t), and distance from the pole x(t) that are valid for the time interval 1 t 1. (3) a) t =,t =,t =11. At those times, the velocity is zero. At t =, he is v t 11 At t = 11, he is v t 16 At t = 16, he is v t ( ) dt ( ) dt ( ) dt = ( π) = 8 π from the pole = 8 π = 8.5 π from the pole = 8.5 π = 5 π from the pole He is farthest from the pole at t = 11 and is π from the pole 16 v t ( ) dt = 8 π = 38 π v( t) = 3( t 11) = 3t 33 a( t) = 3 ( ) = 3 π + 3s 33 ds x t t Demystifying the AB Exam illegal to post on Internet

20 Example C6: Consider the differential equation dy = x( y 1). dx a) On the axis provided, sketch a slope field for the given differential equation at the 1 points indicated. () Show that any point with initial condition that is along the y-axis where y > 1 creates a relative maximum for its particular solution. (3) ( ) to the given ( ) =. (3) Find the particular solution y = f x differential equation with initial condition f For the solution in part, find lim f x x ( ). (1) a) At any point where x =, dy = so there are critical pts. when x =. dx d y dy = x dx dx + ( y 1 )( ) = x( x) ( y 1) ( y 1) d y dx = ( y 1 )( x ) If x = and y >1, d y dx < By the nd derivative test, we have relative maxima when x = and y >1. dy = x dx y 1 ln y 1 = x + C y 1 = Ce x ( ) = : 1 = C C = 3 f y =1 + 3e x [ ] = lim lim 1 + x 3e x x e x = Demystifying the AB Exam illegal to post on Internet