Johanna Jogs Along a Straight Path

AP® CALCULUS AB/CALCULUS BC 2015 SCORING GUIDELINES Question 3 t (minutes) v( t ) (meters per minute) 0 12 20 24 40 0 200 240 –220 150 Johanna jogs along a straight path. For 0 ≤ t ≤ 40, Johanna’s velocity is given by a differentiable function v. Selected values of v( t ) , where t is measured in minutes and v( t ) is measured in meters per minute, are given in the table above. (a) Use the data in the table to estimate the value of v′(16 ) . (b) Using correct units, explain the meaning of the definite integral Approximate the value of table. 40 ∫0 40 ∫0 v( t ) dt in the context of the problem. v( t ) dt using a right Riemann sum with the four subintervals indicated in the (c) Bob is riding his bicycle along the same path. For 0 ≤ t ≤ 10, Bob’s velocity is modeled by B( t ) =t 3 − 6t 2 + 300, where t is measured in minutes and B( t ) is measured in meters per minute. Find Bob’s acceleration at time t = 5. (d) Based on the model B from part (c), find Bob’s average velocity during the interval 0 ≤ t ≤ 10. (a) v′(16 ) ≈ (b) 40 ∫0 240 − 200 = 5 meters/min 2 20 − 12 v( t ) dt is the total distance Johanna jogs, in meters, over the time interval 0 ≤ t ≤ 40 minutes. 40 ∫0 1 : approximation  1 : explanation  3 :  1 : right Riemann sum  1 : approximation v( t ) dt ≈ 12 · v(12 ) + 8 · v( 20 ) + 4 · v( 24 ) + 16 · v( 40 ) = 12 · 200 + 8 · 240 + 4 · 220 + 16 · 150 = 2400 + 1920 + 880 + 2400 = 7600 meters (c) Bob’s acceleration is B′(= t ) 3t 2 − 12t. B′( 5 ) = 3 ( 25 ) − 12 ( 5 ) = 15 meters/min 2 (d) Avg vel = 1 10 3 t − 6t 2 + 300 dt 10 ∫0 ( ) 10  1  t4 = − 2t 3 + 300t   10  4 0 =  1 : uses B′( t ) 2:  1 : answer  1 : integral  3 :  1 : antiderivative  1 : answer 1  10000  350 meters/min − 2000 + 3000 = 10  4  © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. AP® CALCULUS AB/CALCULUS BC 2015 SCORING GUIDELINES Question 1  t2  The rate at which rainwater flows into a drainpipe is modeled by the function R, where R( t ) = 20sin   cubic  35  feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by D( t ) = −0.04t 3 + 0.4t 2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8. There are 30 cubic feet of water in the pipe at time t = 0. (a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8 ? (b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your answer. (c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer. (d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the pipe will begin to overflow. 8 { 1 : integrand 1 : answer (a) ∫0 R( t ) dt = 76.570 2: (b) R ( 3) − D( 3) = −0.313632 < 0 Since R( 3) < D( 3) , the amount of water in the pipe is decreasing at time t = 3 hours.  1 : considers R( 3) and D( 3) 2:  1 : answer and reason (c) The amount of water in the pipe at time t, 0 ≤ t ≤ 8, is 0  1 : considers R( t ) − D( t ) =  3 :  1 : answer  1 : justification 30 + t ∫0 [ R( x ) − D( x )] dx. R( t ) − D( t ) = 0 ⇒ t = 0, 3.271658 t 0 3.271658 8 Amount of water in the pipe 30 27.964561 48.543686 The amount of water in the pipe is a minimum at time t = 3.272 (or 3.271) hours. (d) 30 + w 50 ∫0 [ R( t ) − D( t )] dt = 2: { 1 : integral 1 : equation © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. AP® CALCULUS AB 2013 SCORING GUIDELINES Question 6 dy = e y 3 x 2 − 6 x . Let y = f ( x ) be the particular solution to the Consider the differential equation dx differential equation that passes through (1, 0 ) . (a) Write an equation for the line tangent to the graph of f at the point (1, 0 ) . Use the tangent line to approximate f (1.2 ) . (b) Find y = f ( x ) , the particular solution to the differential equation that passes through (1, 0 ) . ( (a) ) dy = e0 3 ·12 − 6 ·1 = −3 dx ( x, y ) =(1, 0 ) ( ) −3 ( x − 1) . An equation for the tangent line is y =  1 : dy at the point ( x, y ) = (1, 0 )  dx 3:  1 : tangent line equation   1 : approximation f (1.2 ) ≈ −3 (1.2 − 1) = −0.6 dy (b) = 3 x 2 − 6 x dx ey dy = 3 x 2 − 6 x dx ey −e − y = x 3 − 3 x 2 + C ( ∫ ) ∫( ) −e −0 = 13 − 3 ·12 + C ⇒ C = 1 −e − y = x 3 − 3 x 2 + 1 e −y 3  1 : separation of variables  2 : antiderivatives  6 :  1 : constant of integration  1 : uses initial condition   1 : solves for y Note: max 3 6 [1-2-0-0-0] if no constant of integration 2 = − x + 3x − 1 ( ) 3 2 − ln ( − x + 3 x − 1) y= − y = ln − x3 + 3 x 2 − 1 Note: 0 6 if no separation of variables Note: This solution is valid on an interval containing x = 1 for which − x3 + 3 x 2 − 1 > 0. © 2013 The College Board. Visit the College Board on the Web: www.collegeboard.org. AP® CALCULUS AB 2013 SCORING GUIDELINES Question 4 The figure above shows the graph of f ′, the derivative of a twice-differentiable function f, on the closed interval 0 ≤ x ≤ 8. The graph of f ′ has horizontal tangent lines at x = 1, x = 3, and x = 5. The areas of the regions between the graph of f ′ and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f ( 8 ) = 4. (a) Find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer. (b) Determine the absolute minimum value of f on the closed interval 0 ≤ x ≤ 8. Justify your answer. (c) On what open intervals contained in 0 < x < 8 is the graph of f both concave down and increasing? Explain your reasoning. 5 3 (d) The function g is defined by g ( x ) = ( f ( x ) ) . If f ( 3) = − , find the slope of the line tangent to the 2 graph of g at x = 3. (a) x = 6 is the only critical point at which f ′ changes sign from negative to positive. Therefore, f has a local minimum at x = 6. 1 : answer with justification (b) From part (a), the absolute minimum occurs either at x = 6 or at an endpoint. = x 0= and x 6  1 : considers  3 :  1 : answer  1 : justification f= ( 0) 6 ∫8 f ′( x ) dx 8 ′( x ) dx f (8) − ∫ f = 6 f ( 6 ) = f (8) + = 0 ∫8 f ′( x ) dx 8 f ( 8 ) − ∫ f ′( x ) dx =− 4 12 = = −8 0 f (8) + 4 − 7 = −3 f (8) = 4 The absolute minimum value of f on the closed interval [ 0, 8] is −8. (c) The graph of f is concave down and increasing on 0 < x < 1 and 3 < x < 4, because f ′ is decreasing and positive on these intervals. (d) g ′( x ) = 3[ f ( x )]2 · f ′( x ) ( 52 ) ·4 = 75 g ′( 3= ) 3[ f ( 3)]2 · f ′( 3= ) 3 − 2 2: { 1 : answer 1 : explanation  2 : g ′( x ) 3:   1 : answer © 2013 The College Board. Visit the College Board on the Web: www.collegeboard.org.