AP Calculus BC. Sample Student Responses and Scoring Commentary. Inside: Free Response Question 5. Scoring Guideline.

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1 018 AP Calculus BC Sample Student Responses and Scoring Commentary Inside: Free Response Question 5 RR Scoring Guideline RR Student Samples RR Scoring Commentary 018 The College Board. College Board, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks of the College Board. AP Central is the official online home for the AP Program: apcentral.collegeboard.org

2 AP CALCULUS BC 018 SCORING GUIDELINES Question 5 5 ( ) : { (a) 1 π Area = 4 ( + cos θ) π 1 : constant and limits : integrand (b) dr dr = sin θ = θ= π r( π ) = + cos( π ) = dy dr y = rsin θ = sin θ + rcos θ dx dr x = rcos θ = cos θ rsin θ π π sin ( ) + cos dy dy ( ) = = = dx θ= π dx θ= π π π cos( ) sin ( ) dy dr 1 : = sin θ + r cos θ dx dr or = cos θ r sin θ : dr dy sin θ + r cos θ 1 : = dx dr cos θ r sin θ 1 : answer The slope of the line tangent to the graph of r = + cos θ at π θ = is. OR dy y = rsin θ = ( + cos θ) sin θ = cos θ +cos θ sin θ dx x = rcos θ = ( + cos θ) cos θ = sin θ 4sin θ cos θ π π π cos( ) +cos ( ) sin dy dy ( ) = = = dx θ= π dx θ= π π π π sin ( ) 4sin ( ) cos ( ) The slope of the line tangent to the graph of r = + cos θ at π θ = is. (c) dr dr dr 1 = = sin θ = sin θ θ = π 1 = = = radians per second π sin ( ) dr dr 1 : = : dr 1 1 : = sin θ 1 : answer with units 018 The College Board.

5 5 5 5 5 SJ 5 5 5 5 y 6A 5. The graphs of the polar curves r = 4 and r = + cos 0 are shown in the figure above. The curves intersect n 5n at 0 = - and 0 = -.

3 SJ y 6A 5. The graphs of the polar curves r = 4 and r = + cos 0 are shown in the figure above. The curves intersect n 5n at 0 = - and 0 = -. (a) Let R be the shaded region that is inside the graph of r = 4 and also outside the graph of r = + cos B, as shown in the figure above. Write an expression involving an integral for the area of R. R IT/ tm'-1/ - ( t aole)) Jd (9- U- n, 5tr/, q ½ 5u - () a.<,o-sg-jjd-b l1 i Unuthorlzed copying or reuse of,,y part of this pogo Is illegal. - J 8- Continue question 5 on page The College Board.

$5 5 5 5 5 5 5 1! (b) Find the slope of the line tangent to the graph of r = + cos 8 at 0 =. r =..\-uo s e ':,( = (,+d.w., \9) c,a B 'fj -:::: ( \- a<..p e-> \ \"\ e SA d - J 1!$ The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of units per second.

The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of units per second.

4 ! (b) Find the slope of the line tangent to the graph of r = + cos 8 at 0 =. r =..\-uo s e ':,( = (,+d.w., \9) c,a B 'fj -:::: ( \- a<..p e-> \ \"\ e SA d - J 1! (c) A particle moves along the portion of the curve r = + cos 8 for O < 0 <. The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of units per second. Find the rate at which the angle 0 changes with respect to time at the instant when the position 1! of the particle corresponds to 0 =. Indicate units of measure. dy" - cff - - a1,, '"'l c V\ e d e ctt == -ol,siv18 Gff- chcl I'" ,_ = S'lV') & ct 8 ol(; Q {o; \ per c::: r o.a i a. V\, $ db -Q - <;ec. Cl oe _19 -B Unauthorized copying or reuse of any part of this page Is Illegal GO ON TO THE NEXT PAGE. 018 The College Board.

5 y 5. The graphs of the polar cuves r = 4 and r = + cos 0 are shown in the figure above. The curves intersect. n Sn at e = - and e =. (a,) Let R be th shaded rgion that :is inside the graph of r = 4 and also outside tb.e graph of r = + cos B, as shown in the figure above. Write an expression involving an integral for the area of R. Unauthorized copying or reuse of any part of this page Is Illegal Continue question 5 on page The College Board.

$y'( 5\r.,G -c &e-t(1 -x:::.::, r ca { _-:::" cosfr\»-1-c a) cr)$g+ co?$ co1e nit) 11: ( c) A particle m9ves along the portion of the curve r = +

The particle moves in such a way that the distance between the particle and

Find the rate at which the angle 0 changes with respect to time at the

6 (b) Find the slope of the line tgent to the gr 't h of r = + cos 0 at 0 =!:. y'( 5\r.,G -c &e-t(1 -x:::.::, r ca { _-:::" cosfr\»-1-c a) cr)$g+ co?b-s\, z. e c)\1 _ ,.,,._...,..,_. clx -:;\t'l+j - 4.co1e nit) 11: ( c) A particle m9ves along the portion of the curve r = + cos 0 for O < 0 <. The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of units per second. Find the rate at which the angle 0 changes with respect to time at the instant when the position <if the particle c01xesponds to 0 =!!:.. Indicate units of measure. ' Jr Jn _ A. 1 l:= ( -e >Jt Unauthorized copying or reuse of any part of this page Is Illegal GO ON TO THE NEXT PAGE. 018 The College Board.

5 5 5 5 5 l s 5 [I 5(., y 5. The graphs of the polar curves r = 4 and r = + cos 0 are shown in the figure above.

n that is inside the graph of r = 4 and also outside the graph of r = + cos 0, as shown in the figure above.

7 l s 5 [I 5(., y 5. The graphs of the polar curves r = 4 and r = + cos 0 are shown in the figure above. The curves intersect n: 5n: at 0 = - and 0 = -. (a) Let R be the shaded regi?n that is inside the graph of r = 4 and also outside the graph of r = + cos 0, as shown in the figure above. Write an expression involving an integral for the area of R. Unauthorized copying or reuse ot any part of this page Is Illegal Continue question 5 on page The College Board.

8 5 [5.5 [5 5 C_s_ EI 5(.,.. - j (b) Find the slope of the line tangent to the graph of r = + cos e at 0 = t ( c) A particle moves along the portion of the curve r = + cos 0 for O < 0 <. The pa1ticle moves in such a way that the distance between the particle and the origin increases at a constant rate of units per second. Find the rate at which the angle 0 changes with respect to time at the instant when the position 71: of the particle corresponds to 0 = -. Indicate units of measore.. '- d(( I Unauthorized copying or reuse of! any part of this page Is Illegal. I -19- GO ON TO THE NEXT PAGE. 018 The College Board.

9 AP CALCULUS BC 018 SCORING COMMENTARY Question 5 Overview In this problem a polar graph is provided for polar curves r 4 and r cos. It was given that the curves 5 intersect at and. In part (a) students were asked for an integral expression that gives the area of the region R that is inside the graph of r 4 and outside the graph of r cos. A correct response should resource the formula for the area of a simple polar region as half of a definite integral of the square of the radius function. The area of R is given by 4 cos 4 cos. d d d In part (b) students were asked for the slope of the line tangent to the graph of r cos at. A correct response should deal with the conversion between polar and rectangular coordinate systems given by y rsin and x rcos, differentiate these with respect to using the product rule, and find the slope of the line tangent to dy dy d the graph as the value of at. In part (c) the motion of a particle along the portion of the curve dx dx d r cos for 0 is such that the distance between the particle and the origin increases at a constant rate of units per second. Students were asked for the rate at which the angle changes with respect to time at the instant when the position of the particle corresponds to and to indicate units of measure. A correct response dr d should use the chain rule to relate the rates of r and with respect to time t: sin. Recognizing that dr d from the problem statement, it follows that radians per second. For part (a) see LO.4D/EK.4D1 (BC). For part (b) see LO.1C/EK.1C7 (BC), LO.A/EK.A4 (BC), LO.B/EK.B1. For part (c) see LO.1C/EK.1C7 (BC), LO.A/EK.A4 (BC), LO.C/EK.C. This problem incorporates the following Mathematical Practices for AP Calculus (MPACs): reasoning with definitions and theorems, connecting concepts, implementing algebraic/computational processes, connecting multiple representations, building notational fluency, and communicating. Sample: 5A Score: 9 The response earned all 9 points: points in part (a), points in part (b), and points in part (c). In part (a) the response earned the first point with the constant of 1 and the limits of integration of 5 and in line 1. The response would have earned the integrand points with 4 cos in line 1 with no simplification. In this case, the correct simplification in line earned the points. In part (b) the response earned the first point for one of the following: dx dy is computed in line 5 on the left, and is computed in line 4 on the d d left. The response earned the second point with the assembly of dy in line 1 on the right. The response would dx have earned the third point with the expression on the right with no simplification. In this The College Board.

10 AP CALCULUS BC 018 SCORING COMMENTARY Question 5 (continued) case, the correct simplification to earned the point. In part (c) the response earned the first point for application dr d of the chain rule with sin in line 1. Although the response substitutes for dr and for d before isolating, the equation d in the last line on the left earned the second point. The response sin earned the third point with radians per second in the last line on the right. Sample: 5B Score: 6 The response earned 6 points: 1 point in part (a), points in part (b), and points in part (c). In part (a) the response earned the first point with the constant of 1 and the limits of integration of 5 and in line 1. The response did not earn either of the integrand points because the integrand is incorrect. In part (b) the response earned the first point for one of the following: dy dx is computed in the numerator of line, and is computed d d in the denominator of line. The response earned the second point with the assembly of dy in line. The dx response would have earned the third point with the expression 0 0 in line with no simplification. In this 0 case, the correct simplification to earned the point. In part (c) the response earned the first point for application dr d of the chain rule with sin in line. The response earned the second point because although sin is incorrectly evaluated as 1 in line, d d is isolated correctly with a consistent result in the equation 1 in the last line. The response did not earn the third point because the evaluation of sin as makes the response not eligible for the third point. The units are correct. Sample: 5C Score: The response earned points: 1 point in part (a), no points in part (b), and points in part (c). In part (a) the response earned the first point with the constant of 1 and the limits of integration of 5 and. The response did not earn either of the integrand points because the integrand is incorrect. In part (b) the response did not earn any points. The response did not earn the first point because neither dx dy nor is presented. Therefore, the d d response is not eligible for the second and third points. In part (c) the response earned the first point for dr d application of the chain rule with sin in line 1 on the right. Although the response substitutes for 018 The College Board.

11 AP CALCULUS BC 018 SCORING COMMENTARY Question 5 (continued) dr d d and for before isolating, the equation in line 4 on the right earned the second point. The response did not earn the third point because the answer is incorrect. The units are correct. 018 The College Board.

AP Calculus AB. Sample Student Responses and Scoring Commentary. Inside: Free Response Question 1. Scoring Guideline.

AP Calculus AB. Sample Student Responses and Scoring Commentary. Inside: Free Response Question 1. Scoring Guideline. 217 AP Calculus AB Sample Student Responses and Scoring Commentary Inside: RR Free Response Question 1 RR Scoring Guideline RR Student Samples RR Scoring Commentary 217 The College Board. College Board,