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20d.2.The region R models a lake. The depth of the lake at any point x, y hPART Bate, an expression involving an integral that can be used to find theNumber of problems—4volume of water in the lake.NO CALCULATOR IS ALLOWED FOR THESE PROBLEMS.is described by the function h x h 10 - x 2 . Write, but do not evalu-dyyConsider the logistic differential equation dt 4 12 - y h . Lety f t h be the particular solution to the differential equationwith f 0 h 6 .a.21AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAMA slope field for this differential equation is given below. Sketchpossible solution curves through the points 0, 6 h and - 3, - 6 h .Time—60 minutesDuring the timed portion for Part B, you may continue to work on theproblems in Part A without the use of any calculator.TimetP t h3.910:30 12a.m. a.m. .m.4.567.5910.512804256031193 1329 1583 1291A free health care clinic was open to the public for 12 hours, beginning at 9 a.m. ( t 0 ) and ending at 9 p.m. The number of peoplesigning in to be seen for care at time 0 # t # 12 is modeled by atwice-differentiable function P. Values of P t h at various times t areshown in the table above.a.Use the data in the table to estimate the rate at which the number ofpeople seeking care was changing at 5 p.m. ( t 8 ). Show the computation that leads to your answer and indicate units of measure.b.b.c.d.to estimate the total number of people seeking care during theUse Euler’s method, starting at t 0 with two steps of equal size, toapproximate f 1 h .Write the third-degree Taylor polynomial for f about t 0 , anduse it to approximate f 1 h .Use a midpoint Riemann sum with four intervals of equal size12-hour period.c.If planners estimated that the rate, in people per hour, atwhich people could receive care was modeled by the functionr t h 400r sin cFind the particular solution y f t h with f 0 h 6 .r t 3hm for 0 # t # 12 , how many people could15receive care in the period from 11 a.m. ( t 2 ) to 9 p.m. ( t 12 )?d.At what time during the period 0 # t # 12 was the rate at which people were seen for care greatest? Estimate the number of people perhour that could be seen for care at the maximum rate. Was it higheror lower than the number of people signing in for care at that hour?

22AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAMAn object moving along a curve in the xy-plane is at positiondydxrtrt x t h, y t hh at time t, where dt 6 cos a 4 k and dt 12 sin a 4 k forSOLUTIONS: AP Calculus BC Practice Testt 0 . At time t 1 , object is at position 1, - 1 h .Multiple Choicea.Write an equation for the line tangent to the curve at time t 1 .Section 1 Part Ab.Find the acceleration vector and the speed of the object at time t 1 .c.Write an integral expression that can be used to find the total dis-4.23The graph of y f x h has two relative minima at x b and x fand one relative maximum at x d . The graph of y f l x h should have1. C.tance traveled by the object over the time interval 0 # t # 2 .d.Is there a time t 2 1 at which the object is on the x-axis? Explainyour reasoning.5.a.The twice-differentiable function f is defined for all real numbersand satisfies the conditions f 1 h 1 , f l 1 h - 1 , and f m 1 h 0 .The function g is given by g x h f x h ln x h . Find gl 1 h , and gm 1 h .Show the work that leads to your answers.b.The function h is given by h x h f x h sin rx h . Find hl x h and writean equation for the line tangent to the graph of h at x 1 .6.a.The derivative of a function f is given by f l x h 2x - 1 h e 2x - 1 andf 1h 3 .The function has a single critical point. Find this point. Is this a relative maximum, a relative minimum, or neither?b.On what intervals, if any, is the function decreasing and concaveup? Explain your reasoning.c.Find the value of f 5h .zeros at the corresponding x-values, which eliminates B, D, and E.y f l x h should change from negative to positive at each of the relativeminima, and from positive to negative at the relative maximum, whicheliminates A.2. E.If f has a relative minimum at c, then f l c h 0 and f m c h 2 0 .None of the statements are true.dy3. E. dx x 2 y would have all non-negative slopes above the x-axisdyand that is not the case. dx xy 2 would have all non-negative slopes todythe right of the y-axis, also not the case. dx xy xy 2 x y y 2 h shouldhave negative slopes throughout the second quadrant, since x is negativedyand y y 2 is positive. dx x 2 y xy y x 2 x h should be negative in thedyfourth quadrant. Neither of these is true. dx x 2 y xy 2 is the only remaining possibility.4. D.Use partial fraction decomposition. ## x - 9dxh x 2hAdxx-9 ##dx x 2 - 7x - 18Bdxx 2 . Since Ax 2A Bx - 9B 1 ,11A - B and 2A - 9B 1 . Solve to find A 11 and B - 11 , and then11dx1integrate. 11# x dx- 9 - 11 # x 2 11 6ln x - 9 h - ln x 2 h@ .

245. A.Use integration by parts with u x 2 , du 2xdx , dv sin xdx , andv - cos x .#x2sin xdx - x 2 cos x # 2x cos xdx . This requires that you useparts again, this time with u 2x , du 2dx , dv cos xdx , and v sin x .Then#x2sin xdx - x 2 cos x 2x cos xdx - x 2 cos x 2x sin x - 2 # sin xdx - x 2 cos x 2x sin x 2 cos x C .6. B.If the position vector is t - t , 2 - 3t h , the velocity vector is2232t - 3t , - 6 2 - 3t h , and the acceleration vector is 2 - 6t, 18 . At t 2 ,2the acceleration vector is - 10, 18 .k31dxp converges, which implies thatp is finite, then /xxn 1131p 2 1 . - p 1 - 1 so / -p diverges. p 2 2 1 and p 2 1 so the two ren 1 x11. C. If limk"3If f is the function defined by f x h 2x 6 - 9x 4 , then the firstderivative f l x h 12x 5 - 36x 3 0 when 12x 3 x 2 - 3 h 0 . Solving givesyou x 0 and x ! 3 . Check the sign of the first derivative for changeof sign.f l xhx 1- 3 - 3 - 3 1 x 1 0– 0001x1 33x2 30–0 9. E. If f is twice differentiable and if g x h f f x hh , thengl x h f l f x hh f l x h and gm x h f l f x hh f m x h f l x h f m f x hh f l x h f l f x hh f m x h 6 f l x h@2 f m f x hh .10. C. If3/a xnn 0nis a Taylor series that converges to f x h for all realx, then f l x h a 1 2a 2 x 3a 3 x 2 4a 4 x 3 5a 5 x 4 f / na n x n - 13and f m x h 2a 2 6a 3 x 12a 4 x 2 20a 5 x 3 f f m 2h / n n - 1ha 23nn 2n-2.n 1/ n n - 1ha x3nn 2n-2.#maining series converge.12. A. Solve x t 3 to get t x - 3 and substitute into y t - 2 h2 toget y x - 5 h2 x 2 - 10x 25 . The parabola has a y-intercept of 25.313. E.#kk4x 3 e 2x dx limk"30#xe32x 4dx . If u 2x 4 and du 8x 3 dx , limk"3#xe32x 4dx001 2x k111@6 2k99 e 2k - C lim lime C - limk"3 8k"3 888 k " 3 e - 1 which goes to positive in04dy7. B. If dx cos x sin 2 x , make a substitution with u sin x andsin 3 xdu cos x , and integrate to get y 3 C . If y 0 when x 0 ,r111rsin 3 xC 0 and y 3 . Substitute x 2 to get y 3 sin 3 2 3 1 h3 3 .8. D.25AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAM44finity. The improper integral diverges.2000PkP kaa14. D. dPdt P 500 - 4 500P 1 - 2000 integrates to P 1 Ce -500t .(If you don’t recall this, separate the variables and integrate using apartial fraction decomposition.) Use the initial condition to find that2000100 1 C , so 100 100C 2000 , 100C 1900 and C 19 . The equation for the population at time t is P t h 2000-500t and limP t h 2000 .t"31 19e15. C. The function is decreasing when the first derivative is negative. If f x h 3x 3 - 4x , then f l x h 9x 2 - 4 . First find the zeros of thederivative. If 9x 2 - 4 0 , x ! 23 . Check the sign of the derivative todetermine where it is negative. f l - 1 h 9 - 4 2 0 , f l 0 h - 4 1 0and f l 1 h 9 - 4 2 0 , so the function is decreasing on the interval22-3 1x1 3.dxd a1kdx2 dx16. D. Given that dtx k dt , differentiate to get - x dt k dt and1dxsubstitute x 3 , to get - 91 dxdt k dt . Solve to find k - 9 .

2617. B. Check each series. For3 1/ nn 3,n 1n 12 ka1 lim limn"3 n 3n"3n 3 12 n 3n 2 5n 6 2 1 , so the series diverges. Forand nn 4 n 1n 2 5n 43/ sin nnrh , sin nrh 0 for all n, so the series converges. The final series,n 1333/ n 2 1 2 / n 1 1 2 / n1 is a multiple of a harmonic series. The harn 1n 1n 232monic series diverges, therefore, / n 1 diverges.n 12-k18. A. y xx 1 will have a relative minimum at x 3 if the derivative x - 1 h 2x h - x 2 - k h 1 h x 2 - 2x kyl is equal to zero at x 3 , and x - 1 h2 x - 1 h2changes from negative to positive. Substitute x 3 into the derivativeand solve 3 - 2 3 h k 0 to get k - 3 . Check for the proper change27AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAM22. E. If f is continuous on the closed interval 6a, b@ , then the MeanValue Theorem guarantees that there exists c such that a 1 c 1 b andf bh - f ahb - a . While the other values might be possible, noneis certain.f l ch 33nn23. C. If f x h / a 1 - 1x k , then f 2h / a 21 k 1 .n 1n 1xxe -1e24. C. lim lim 1.x " 0 sin xx " 0 cos xdy25. C. If y arcsin e x h , then dx 21 2xe x h 21 - e x h2222xe x.1 - e 2x226. E. Use integration by parts, with u x 2 , du 2xdx , dv sec x tan xdx ,and v sec x . Then#x2sec x tan xdx x 2 sec x - 2 # x sec xdx .2of sign, being careful to avoid the discontinuity at x 1 . If k - 3 ,yl x 2 - 2x - 3so yl 2 h 4 - 42 - 3 - 3 and yl 4 h 16 - 82 - 3 59.31 x - 1 h219. C. If h x h f x h g x h 1 g x hh f x h g x h f x h6g x h@ , then2hl x h f x h gl x h g x h f l x h f x h 2g x h gl x h f l x h6g x h@2 and whenx 1 , hl 1 h f l 1 h6g 1 h 6g 1 h@ @ gl 1 h6 f 1 h 2f 1 h g 1 h@ . Substitutingr27. C. The first few choices can be estimated geometrically.r2sin xdx0r#and#sin xdx will be equal to each other, and to the given integral,2r#butrcos xdx will be the negative of the given integral. Integrate23rto check the remaining options.2r##03x1xcos 2 dx 2 2 sin 2r321sin 2xdx - 2 cos 2x3r20 1 andg 1 h 0 , hl 1 h f l 1 h60 0@ gl 1 h6 f 1 h 2f 1 h 0@ f 1 h gl 1 h .12dy20. D. If ln y cos x 0 , then y e - cosx and dx e - cosx sin x .dy28. A. If dx cos x , then the average rate of change of y with respect to x onr111rthe closed interval 60, r@ is r 0 # cos xdx r 6- sin x@0 r 0 0 h 0 .23n21. C. f x h e 3x Start with e x 1 x x2 x6 f nx! f 3x 2 h2 3x 2 h3 3x 2 hne 3x 1 3x 2 2 6 f n! f9x 4 27x 63 n x 2n9 1 3x 2 2 6 f n! f The coefficient of x 4 is 2 .22-r 3-r3 1.0

28Section I, Part B29. E. Find f l x h e 1 - 2x x 2 - 2x h , and consider when the derivative2is equal to zero. e1 - 2x x 2 2 - 2x h 0 only at x 1 , since e1 - 2x x 22 0 forall x. Check to see if this critical number is in fact a minimum. Whenx 1 1 , the derivative is positive, and negative when x 2 1 , so the criticalpoint is a relative maximum, not a minimum. Therefore the function30. D. The length of the path described by the parametric equations2#02 # t 1 h2 3t - 1 h2 dt2t 2 2t 1 9t 2 - 6t 1 dt 0#10t 2 - 4t 2 dt . 6.070 .031. A. The length of the curve determined by the equations x t - 1 and4y t from t 0 to t 4 is#0 t - 1h t dt 224#t - t 1 dt . 7.555 .2032. C. The volume of the solid generated by rotating about the x-axisthe region in the first quadrant enclosed by the curve y 1 - sin x andrthe x- and y-axes is r#02 1 - sin x h2 dx . 0.356r . 1.119 .33. A. Each point on the curve 2x 2 - y 1 is of the form x, 2x 2 - 1h . Consider the distance from x, 2x 2 - 1h to - 2, 0 h . d x 2 h2 2x 2 - 1 h2 x 2 4x 4 4x 4 - 4x 2 1 4x 4 - 3x 2 4x 5 .minimum distance, take the derivative. d l Tofindthe316x - 6x 4 2 4x 4 - 3x 2 4x 538x - 3x 2. Set the derivative equal to zero and solve by cal4x 4 - 3x 2 4x 5culator. x .- 0.824 . Check the change of sign of the derivative to verifythis point is a minimum. d l - 1 h d l 0h 34. D. Find the points of intersection by solving 2 2 sin i 2 and1i 0 or i r . The area of the region is A 2128x 3 - 3x 23 - 2 1 0 and4x 4 - 3x 2 4x 58x 3 - 3x 22 2 0 . The x-coordinate of the nearest4x 4 - 3x 2 4x 55point is - 0.824 .# 6 2 2 sin ih - 2 @di r220# 68 sin i 4 sin i@di # 64 sin i 2 sin i@di # 64 sin i 1 - cos 2i@dirrr220001r - 4 cos i i - 2 sin 2i 0 4 r h - - 4 h 8 r . 11.142 .x#ehas no relative minimum.x t 1 and y 3t - 1 , when 0 # t # 2 , is given by29AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAM-t 2dte -x12lim35. A. lim x"2x"2x-21 e4236. C. The approximate value of e 2 using the fourth-degree Taylor23x4polynomial about x 0 for e x is e x . 1 x x2 x6 24evaluated at4 8 16x 2 , or e 2 . 1 2 2 6 24 . 7 .111 1 - x21dx # c 1 dx x sin -1 x22 m1x1x-1-1rra1 k - a- 1 - k 2 r22#37. D.rk38. E.##sin xdx 3001-1 sin xdx means that - cos x 0 3 6- cos x k @ . Evalukrate and solve - cos k 1 3 61 cos k@ to find that cos k - 21 so k 5r6 .39. D. If the position vector is sin t, cos 2t h , then the velocity vectoris cos t, - 2 sin 2t h , and the acceleration vector is - sin t, - 4 cos 2t h . Attime t 1 , its acceleration vector is - sin 1, - 4 cos 2 h - 0.8414, 1.665 h .540. C.#2f x h dx 3#2 x - 2h2 dx 5#3dxx-2 3#2 x 2 - 4x 4 h dx 5#3dxx-253x38 3 - 2x 2 4x 2 ln x - 2 h 3 9 - 18 12 h - a 3 - 8 8 k ln 3 - ln 1 813 - 3 ln 3 3 ln 3 . 1.432 .

3041. C. To find the slope of the line normal to the curve x 2 y - xy 2 2x ,dydyfirst differentiate implicitly. x 2 dx 2xy - 2xy dx - y 2 2 , sodydydy2 - 2xy y 2x 2 dx - 2xy dx 2 - 2xy y 2 and dx . At 3, 1 h ,x 2 - 2xy2 - 2 3 h 1 h 1dy2-6 1-3dx 9 - 2 3 h 1 h 9 - 6 3 - 1 so slope of the tangent is- 1 and the slope of the normal is the negative reciprocal, or 1 .dydy2t42. B. Find dxdt 2t - 3 and dt 2t . Then dx 2t - 3 . When x 5 ,5 t 2 - 3t 1 . Solve t 2 - 3t - 4 0 to find t 4 or t - 1 , but rejectdy2t8t - 1 since t 0 . dx 2t - 3 1.6543. C. The basic relationship is the Pythagorean Theorem with c 20dbfeet so a 2 b 2 400 . Differentiate to get 2a dadt 2b dt 0 , and subdbdbdbdbakstitute, using dadt 2 dt . a 2 dt b dt 0 means that dt 2a b h 0 ,and since we know dbdt is not zero, because the ladder is moving, it mustbe true that 2a b 0 or b - 2a . Substitute into a 2 b 2 400 to geta 2 4a 2 400 and solve to find a 4 5 . 8.944 .44. B. If C h 2rr h 12 then h 12 - 2rr and V 31 rr 2 h be2 3comes V 31 rr 2 12 - 2rr h or V 4rr 2 - 23 r r . Differentiate and set thederivative equal to zero. V l 8rr - 2r2 r 2 0 factors to 2rr 4 - rr h 04. 1.273 .so r 0 or r r x - 2 hn 1x-22n 1 45. A. Find the radius of convergence. n 222 x - 2 hnx-2x-2and 2 1 1 when - 1 1 2 1 1 or 0 1 x 1 4 . Then check the end333 - 2hn - 1hn 2 n - 1 hn / 2 .points of the interval. When x 0 , / n 1 /n 12n 0 2n 0n 0n 1 - 1h2Since 1 , the series does not converge. When x 4 ,2 - 1 hnn333n/ 42-n 21 h / 22n 1 / 21 , which also diverges. The interval of conn 0n 0n 0vergence is the open interval 0, 4 h .31AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAMSection II, Part A1.# a cos a r2x k - x1a. A 2-12rxr sin a 2 kx3- c 3 - xm-11- 1 hk dx # cos a r2x kdx - # x11-1-12- 1 h dx 222 24 4 r 1 h - r - 1 h - a - 3 - 3 k r 3 c 2.607 .-114 4 3 c 2.607 and you can deb. From a., you know that the total area is r11444duce that # a cos a r2x kk dx rand # x 2 - 1 h dx - 43 . Since r 1 - 3 ,-1-1you can expect that k 1 0 . Therefore, the area of the region below y k is# k - x 1hdx c kx - x31 xm114 a k - 3 1 k - a - k 3 - 1 k 2k 3 .-114Then 2k 43 2 2.607 h so 2k 3 1.3035 and k - 0.015 .31-11#c. The volume of the solid is V -11# a cos a r2x k - xr8-122rx2r f cos a 2 k - x 1 pdx 222r 1 k dx 8 4.131 h 1.622 .# R h xhdx 1d. The volume of water in the lake is1#-1rx k2a cos a 2hk h2 - x - 1 10 - x dx .-12.a. The requested curves pass through the given points, follow the direction of the field, and approach zero as x " - 3 .

32b. Using Euler’s method, with the given initial point 0, 6 h and interdy16vals of width 21 , find dt 4 12 - 6 h 9 and y 1 6 9 2 10.5 .dyRepeat for the second interval. dt 104.5 12 - 10.5 h 154.75 3.9375and y 2 10.5 3.9375 21 12.46875 . f 1 h . 12.469 .d2 yydydyy1 dy 4 c - 1 dt m 12 - y h 4 dt . Substic. Given dt 4 12 - y h , finddx 23dyy9 2 yy 8 . Find the third detute dt 4 12 - y h and simplify to get 9y - 4dy93y 8 y 2 k dt . Evaluate the derivatives at x 0 and y 6 .rivative a 9 - 2dyd2 y66262, 126 9 h2 4dt16 12 - 6 h 16 12 - 6 h 0 , and the thirddx81kderivative is equal to a 9 - 27 272 9 - 2 . The third-degree Taylor poly810 2 - 2 33nomial for f about t 0 is f x h . 6 9x 2! x 3! x 6 9x - 274 x3or f x h . 6 9x - 274 x , so f 1 h . 8.25 .dyyyd. The logistic differential equation dt 4 12 - y h 3y a 1 - 12 k integrates, with the help of partial fractions, to y 12 -3t . Recall that1 edydydy1y 3dt , which integratesy 3dt becomes # y - 12 #1 - 12y a 1 - 12 ke 3t CeCto ln y - ln 12 - y h 3t C or y 12. Since f 0 h 6 , 6 123t CCe 1e 1and C 1 .Section II, Part B3.a. The rate at which the number of people seeking care was changing- 1291387at 5 p.m. was decreasing at a rate of 8049 - 7.5 - 1.5 - 258 peopleper hour.33AP CALCULUS BC PRACTICE EXAMAP CALCULUS BC PRACTICE EXAMb. Use a midpoint Riemann sum with four 3-hour intervals. Thetotal number of people seeking care during the 12-hour periodwas 3 728 h 3 1329 h 3 1291 h 3 256 h 2184 3987 3873 768 10813 people.c. The number of people who could receive care was12#2400r sin cr t 3hm dt 400r1512#2sin cr t 3hr t 3hm dt - 400r 15 cos cmr151512 2r1- 6000 cos r 6000 cos 3 6000 6000 2 9000 people.r t 3hmisatamaximumwhen15r t 3hr t 3hr80r2cm 0r l t h 400r 15 cos c 15 m 0 .Solve3 cos15r t 3h rr t 3h 2 , 2t 6 15 and t 4.5 .or cos c 15 m 0 . If15rr 4.5 h 400r sin a 2 k 400r . 1256.637 which is less than the 1329d.r t h 400r sin cpeople signing in for care at t 4.5 .4.dyrtrta k anda ka. Find dxdt 6 cos 4dt 12 sin 4 , and divide to findrt12 sin a 4 kdydyrtra kdx 6 cos a rt k 2 tan 4 . When t 1 , dt 2 tan 4 2 . The equa4tion for the line tangent to the curve at time t 1 is y 1 2 x - 1h ory 2x - 3 .b. If the velocity vec

AP Calculus BC Practice Exam CALCULUS BC SECTION I, Part A . After examining the choices, select the choice that best answers the question. No credit will be given for anything written in the test book. In this test: Unless otherwise . 14 AP CALCULUS BC PRACTICE EXAM AP CALCULUS BC PRACTICE EXAM 1