AP® CALCULUS BC
2011 SCORING GUIDELINES (Form B)
Question 2
The polar curve r is given by r (θ ) = 3θ + sin θ , where 0 ≤ θ ≤ 2π .
(a) Find the area in the second quadrant enclosed by the coordinate axes and the graph of r.
(b) For
π ≤ θ ≤ π , there is one point P on the polar curve r with x-coordinate
−3. Find the angle θ that
2
corresponds to point P. Find the y-coordinate of point P. Show the work that leads to your answers.
(c) A particle is traveling along the polar curve r so that its position at time t is ( x( t ) , y ( t ) ) and such that
dy
2π
dθ
, and interpret the meaning of your answer in the context of
= 2. Find
at the instant that θ =
3
dt
dt
the problem.
1 π
(a) Area = ∫ ( r (θ ) )2 dθ = 47.513
2 π 2
⎧ 1 : integrand
⎪
3 : ⎨ 1 : limits and constant
⎪⎩ 1 : answer
(b) −3 = r (θ ) cos θ = ( 3θ + sin θ ) cos θ
θ = 2.01692
y = r (θ ) sin (θ ) = 6.272
⎧ 1 : equation
⎪
3 : ⎨ 1 : value of θ
⎪⎩ 1 : y -coordinate
(c)
y = r (θ ) sin θ = ( 3θ + sin θ ) sin θ
dy
dy dθ ⎤
=⎡
⋅
= −2.819
dt θ = 2π 3 ⎣⎢ dθ dt ⎦⎥ θ = 2π 3
⎧ 1 : uses chain rule
⎪
3 : ⎨ 1 : answer
⎪⎩ 1 : interpretation
The y-coordinate of the particle is decreasing at a rate of 2.819.
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AP® CALCULUS BC
2013 SCORING GUIDELINES
Question 2
The graphs of the polar curves r = 3 and r= 4 − 2sin θ are shown in the figure
π and θ = 5π .
above. The curves intersect when θ =
6
6
(a) Let S be the shaded region that is inside the graph of r = 3 and also inside the
graph of r= 4 − 2sin θ . Find the area of S.
(b) A particle moves along the polar curve r= 4 − 2sin θ so that at time t
seconds, θ = t 2 . Find the time t in the interval 1 ≤ t ≤ 2 for which
the x-coordinate of the particle’s position is −1.
(c) For the particle described in part (b), find the position vector in terms of t. Find the velocity vector at
time t = 1.5.
(a) Area = 6π +
1 5π 6
( 4 − 2sin θ )2 dθ = 24.709 (or 24.708)
2 ∫π 6
 1 : x(θ ) or x( t )

3 :  1 : x(θ ) =
−1 or x( t ) =
−1
 1 : answer
(b) x = r cos θ ⇒ x(θ ) =
( 4 − 2sin θ ) cos θ
x( t=
)
( 4 − 2sin ( t ) ) cos ( t )
2
2
x( t ) = −1 when t = 1.428 (or 1.427)
(c) y = r sin θ ⇒ y (θ ) =
( 4 − 2sin θ ) sin θ
y ( t=
)
3:
( 4 − 2sin ( t 2 ) ) sin ( t 2 )
Position vector = x( t ) , y ( t )
(
( )) ( ) (
 1 : integrand

3 :  1 : limits and constant
 1 : answer
{
2 : position vector
1 : velocity vector
( )) ( )
=
4 − 2 sin t 2 cos t 2 , 4 − 2 sin t 2 sin t 2
v(1.5 ) = x′(1.5 ) , y′(1.5 )
=
−8.072, −1.673 ( or −8.072, −1.672 )
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AP® CALCULUS BC
2014 SCORING GUIDELINES
Question 2
The graphs of the polar curves r = 3 and r= 3 − 2sin ( 2θ ) are
shown in the figure above for 0 ≤ θ ≤ π .
(a) Let R be the shaded region that is inside the graph of r = 3
and inside the graph of r= 3 − 2sin ( 2θ ) . Find the area of R.
(b) For the curve r= 3 − 2sin ( 2θ ) , find the value of
θ =
π
6
dx
at
dθ
.
(c) The distance between the two curves changes for 0 < θ <
π
.
2
Find the rate at which the distance between the two curves is changing with respect to θ when θ =
(d) A particle is moving along the curve r= 3 − 2sin ( 2θ ) so that
of
dr
π
at θ = .
dt
6
9π 1 π 2
+ ∫
( 3 − 2sin ( 2θ ) )2 dθ
4
2 0
= 9.708 (or 9.707)
(a) Area =
(b) =
x ( 3 − 2 sin ( 2θ ) ) cos θ
dx
= −2.366
dθ θ = π 6
(c) The distance between the two curves is
D =3 − ( 3 − 2 sin ( 2θ ) ) = 2sin ( 2θ ) .
dD
dθ
(d)
3
⋅
dθ
dr
=
⋅3
dt
dθ
=
−6
( −2 )( 3) =
.
dθ
= 3 for all times t ≥ 0. Find the value
dt
 1 : integrand

3 :  1 : limits
 1 : answer
2:
{
1 : expression for x
1 : answer
2:
{
1 : expression for distance
1 : answer
2:
{
1 : chain rule with respect to t
1 : answer
= −2
θ =π 3
dr
dr
=
dt
dθ
dr
dt θ =π 6
π
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