AP® Calculus BC 2009 Scoring Guidelines
Form B
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AP® CALCULUS BC
2009 SCORING GUIDELINES (Form B)
Question 1
A baker is creating a birthday cake. The base of the cake is the region R in the first quadrant under the graph of y=f(x) for
πx. Both x and y are
0≤x≤30, where f(x)=20sin
30
measured in centimeters. The region R is shown in the figure
2ππxcos. above. The derivative of f is f′(x)=330()()(a) The region R is cut out of a 30-centimeter-by-20-centimeter rectangular sheet of cardboard, and the
remaining cardboard is discarded. Find the area of the discarded cardboard.
(b) The cake is a solid with base R. Cross sections of the cake perpendicular to the x-axis are
semicircles. If the baker uses 0.05 gram of unsweetened chocolate for each cubic centimeter of cake, how many grams of unsweetened chocolate will be in the cake? (c) Find the perimeter of the base of the cake.
(a) Area =30⋅20−∫030f(x)dx=218.028 cm2 3 : { 2 : integral 1 : answer⌠π⎛f(x)⎞=⎮(b) Volume dx=2356.194 cm3 ⎜⎟⌡02⎝2⎠Therefore, the baker needs 2356.194×0.05=117.809 or 117.810 grams of chocolate. 302 3 : { 2 : integral 1 : answer(c) Perimeter =30+ ∫0301+(f′(x))2dx=81.803 or 81.804 cm 3 : { 2 : integral 1 : answer© 2009 The College Board. All rights reserved.
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AP® CALCULUS BC
2009 SCORING GUIDELINES (Form B)
Question 2
A storm washed away sand from a beach, causing the edge of the water to get closer to a nearby road. The rate at which the distance between the road and the edge of the water was changing during the storm is modeled by f(t)=t+cost−3 meters per hour, t hours after the storm began. The edge of the water was 35 meters from the road when the storm began, and the storm lasted 5 hours. The derivative of f(t)
1−sint. is f′(t)=2t(a) What was the distance between the road and the edge of the water at the end of the storm? (b) Using correct units, interpret the value f′(4)=1.007 in terms of the distance between the road and
the edge of the water. (c) At what time during the 5 hours of the storm was the distance between the road and the edge of the
water decreasing most rapidly? Justify your answer. (d) After the storm, a machine pumped sand back onto the beach so that the distance between the road
and the edge of the water was growing at a rate of g(p) meters per day, where p is the number of days since pumping began. Write an equation involving an integral expression whose solution would give the number of days that sand must be pumped to restore the original distance between the road and the edge of the water.
(a) 35+∫05f(t)dt=26.494 or 26.495 meters 2 : {1 : integral 1 : answer(b) Four hours after the storm began, the rate of change of ⎧1 : interpretation of f′(4) 2 : ⎨the distance between the road and the edge of the water ⎩ 1 : unitsis increasing at a rate of 1.007 metershours2. (c) f′(t)=0 when t=0.66187 and t=2.84038 The minimum of f for 0≤t≤5 may occur at 0, 0.66187, 2.84038, or 5. f(0)=−2 f(0.66187)=−1.39760 f(2.84038)=−2.26963 f(5)=−0.48027 The distance between the road and the edge of the water was decreasing most rapidly at time t=2.840 hours after the storm began. (d) − ⎧1 : considers f′(t)=0⎪3 : ⎨ 1 : answer ⎪⎩ 1 : justification ∫05f(t)dt=∫0xg(p)dp 2 : {1 : integral of g 1 : answer© 2009 The College Board. All rights reserved.
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AP® CALCULUS BC
2009 SCORING GUIDELINES (Form B)
Question 3
A continuous function f is defined on the closed interval −4≤x≤6. The graph of f consists of a line segment and a curve that is tangent to the x-axis at x=3, as shown in the figure above. On the interval 0 (a) Is f differentiable at x=0? Use the definition of the derivative with one-sided limits to justify your answer. (b) For how many values of a, 4−≤a<6, is the average rate of change of f on the interval [a,6] equal to 0 ? Give a reason for your answer. (c) Is there a value of a, 4−≤a<6, for which the Mean Value Theorem, applied to the interval [a,6], 1 guarantees a value c, a ∫0 x f(t)dt for −4≤x≤6. On what intervals contained in [−4,6] is the graph of g concave up? Explain your reasoning. f(h)−f(0)2= (a) lim−h3h→0f(h)−f(0)lim<0 +hh→0Since the one-sided limits do not agree, f is not differentiable at x=0. 2 : {{{1 : sets up difference quotient at x=0 1 : answer with justificationf(6)−f(a)=0 when f(a)=f(6). There are (b) 6−atwo values of a for which this is true. (c) Yes, a=3. The function f is differentiable on the interval 3 Visit the College Board on the Web: www.collegeboard.com. AP® CALCULUS BC 2009 SCORING GUIDELINES (Form B) Question 4 The graph of the polar curve r=1−2cosθ for 0≤θ≤π is shown above. Let S be the shaded region in the third quadrant bounded by the curve and the x-axis. (a) Write an integral expression for the area of S. dydx (b) Write expressions for and in terms of θ. dθdθ(c) Write an equation in terms of x and y for the line tangent to the graph of the polar curve at the point where θ=Show the computations that lead to your answer. π. 2 π(a) r(0)=−1; r(θ)=0 when θ=. 3Area of S= (b) x=rcosθ and y=rsinθ 2 : 12∫0π3(1−2cosθ)2dθ {1 : limits and constant 1 : integrand dr=2sinθ dθdxdr=cosθ−rsinθ=4sinθcosθ−sinθ dθdθdydr=sinθ+rcosθ=2sin2θ+(1−2cosθ)cosθ dθdθ⎧1 : uses x=rcosθ and y=rsinθ⎪ 4 : ⎨ 1 : drθd⎪⎩ 2 : answer θ=π, we have x=0,y=1. (c) When dydxθ=π 22dydθ=dxdθθ=π=−2 2The tangent line is given by y=1−2x. ⎧ 1 : values for x and y⎪⎪dy3 : ⎨ 1 : expression for dx⎪⎪1 : tangent line equation⎩ © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® CALCULUS BC 2009 SCORING GUIDELINES (Form B) Question 5 Let f be a twice-differentiable function defined on the interval −1.2 at 1.x= (b) For −1.2 ′(x))2+f′′(x)⎤. Is g′′(−1) positive, negative, or f(c) The second derivative of g is g′′(x)=ef(x)⎡(⎣⎦zero? Justify your answer. (d) Find the average rate of change of g′, the derivative of g, over the interval [1,3]. (a) g(1)=ef(1)=e 2 g′(x)=ef(x)f′(x), g′(1)=ef(1)f′(1)=−4e2 The tangent line is given by y=e2−4e2(x−1). (b) g′(x)=ef(x)⎧ 1 : g′(x)⎪ 3 : ⎨ 1 : g(1) and g′(1)⎪⎩1 : tangent line equation f′(x) ef(x)>0 for all x So, g′ changes from positive to negative only when f′ changes from positive to negative. This occurs at x=−1 only. Thus, g has a local maximum at x=−1. 2 : { 1 : answer 1 : justification(c) g′′(−1)=e⎡(f′(−1))+f′′(−1)⎤ ⎣⎦ef(−1)>0 and f′(−1)=0 Since f′ is decreasing on a neighborhood of −1, f′′(−1)<0. Therefore, g′′(−1)<0. g′(3)−g′(1)e=3−1f(3)f(−1)2 2 : { 1 : answer 1 : justification(d) f′(3)−e2f(1)f′(1) =2e2 2 : {1 : difference quotient 1 : answer© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® CALCULUS BC 2009 SCORING GUIDELINES (Form B) Question 6 The function f is defined by the power series f(x)=1+(x+1)+(x+1)+\"+(x+1)+\"= for all real numbers x for which the series converges. (a) Find the interval of convergence of the power series for f. Justify your answer. (b) The power series above is the Taylor series for f about x=−1. Find the sum of the series for f. (c) Let g be the function defined by g(x)= why g− cannot be determined. (12)x2 n (x+1)n ∑n=0 ∞ ∫−1 f(t)dt. Find the value of g− ()1 , if it exists, or explain 2 (d) Let h be the function defined by h(x)=fx2−1. Find the first three nonzero terms and the general term of the Taylor series for h about x=0, and find the value of h(a) The power series is geometric with ratio (x+1). The series converges if and only if x+1<1. Therefore, the interval of convergence is −2 © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. 因篇幅问题不能全部显示，请点此查看更多更全内容