AP Calculus BC - Parametric equations and vectors

Chapter 9- AP Exam Problems solutions

1.

2.

4.

3.

5.

D22 22

0 00

Area = 1 1 1− c( os 41θ θ =) θ−d ⎛

⎜1 sin 41sin (2θ)

2 2 2 4 4 8θ =d

ππ π π⎞θ⎟ =

⎠⎝∫ ∫

E6

r = 0 when cos3 0θ = ⇒ θ = ± . The region is for the interval from θ = – t o6 6

ππ πθ = .

6 ( 2

6

4cos32

π

π−

Area 1= θ d) θ∫

A Area =2 22 2 2

0 0 2 3cos

2 θ θd

π π1∫⋅ ((2 θcos ) co− s )θ θd = ∫

D Find the points of intersection of the two curves to determine the limits of integration.

4sin θ = 2 when sinθ = 0.5; this is at and 5π .6 6π

θ = ( 4sin( )θ − (2) )5

2 26Area = 12

dπ

π∫ θ

6.

Solution

(a)

(b) The intersection occurs when 3cosθ 1= + cosθ

cosθ = 12

3π

θ = ±

Area ((3cosθ) (1 cos− )+ θ )3 2 23

12

dπ

−π ∫= θ

3 23

3

3

3

3

33

121212

21 2sin sπ

+2 in

3 3 3 3

π

−π

π

−π

π

−π

π−π

= (8cos 1 2θ c− os− )θ dθ

= (4 4cos+ 2 1θ 2− cos− )θ dθ

= (3 4co+ s 2 2θ co− s )θ dθ

1 (3= θ+ 2sin 2θ 2− sin )θ

2sin 2−2 sinπ π )π ⎞⎛⎜= π

2+ (− −π− ⎟

⎝ ⎠= π

∫

∫

∫

1984 BC5

Consider the curves 3cos= θ and r r 1= c+ os . θ

(a) Sketch the curves on the same set of axes.

(b) Find the area of the region inside the curve r = 3cosθ and outside the curve r = +1 cosθ by setting up and evaluating a definite integral. Your work must include an antiderivative.

7.

1990 BC4 Solution

(a)

−5−5

5

yy

x

R

O−−55 55

5

x

(2 1(2 )

∫ ( )

]

22

0

0

00

00

1 2 sin2

2 2si θn( sin

4 sin 1 cos− 2

4cos2

4 1− +4 1 0

8

A θ θ)) d

d

θ θdθ θ −d

π

π

π π

ππ

θ θ

−θ θ − 1 sin 2 θ

π

− = −

= −

=

= −

= − ( ) ( ) − π[ −

= −

⌠⌡

∫∫

(b)

8. 1993 BC4 Solution

( )( )

22 2 2 3

2 2 2 2

6

22 2 6 12 6

At , =1 and x y 1= so4

+4 2 2 6 12 6

+x y x y − 2y

y dy x dy + xy − y dy+x ydx dx dx

dy dy dydx dx dx

=

x +

=

θ π=

= + −

(a)

(b) (2

0 00

2

0

2

0

sin 61∫ 4sin 31

2 6

3or 4∫π / 3

sin 32

∫ or 6 / 64sin 3

2

Aπ

π π

π

dθ θ = =π

dθ θ = =π

= θ θd = 1 cos 6− θ θ) θd = −∫ θ π =

−2 −1 1 2

−2

−1

1

2

y

xO

(c) = 2sin 3θ cos2sin 3

2sin 3

2sin 3

At , dx = 2 a− nd 4,= − so4

2 14 2

xydxddyd

dyd d

dy

θsinθ θ

sinθ θ + 6cos3 cθ osθθ

cosθ θ 6cos3+ sθ inθθ

θ θ

=

= −

=

θ π=

−= = dx −

or

12dx

8+8 =12dx

dy dy =

12.

9.

10.

11.

AP® CALCULUS BC 2003 SCORING GUIDELINES (Form B)

Question 2

The figure above shows the graphs of the circles x y2 2+ = 2 and

( )x y2 2+ =1 1 The . graphs intersect at the points (1,1) and (1, 1).

Let R be the shaded region in the first quadrant bounded by the two

circles and the x-axis.

(a) Set up an expression involving one or more integrals with respect to

x that represents the area of R.

(b) Set up an expression involving one or more integrals with respect to

y that represents the area of R.

(c) The polar equations of the circles are r = 2 and r = 2 cos , respectively. Set up an expression

involving one or more integrals with respect to the polar angle that represents the area of R.

(a) Area = 1 2

22

0 11 ( 1)x d +x 2 x dx

OR

)Area = (2

22

1

11 2

4x dx+

1 : integrand or geometric area3 :

1 : integrand for larger circle

for smaller circle

1 : limits on integral(s)

Note: < –1 > if no addition of terms

(b) Area (= 2 1y y1( ) dy)1

2 2

0

3 :

1 : limits

2 : integrand

1 > reversal

1 > algebra error

in solving for

1 > add rather

than subtract

2 > other errors

x

<

<

<

<

(c) Area = ( )2 24 2

0 4

112 (2 cos )

22dd +

OR

Area = ( )2 2 2

4

1 12 (2 cos )

28d+

for larger circle3 :

1 : integrand or geometric area

1 : integrand for smaller circle

1 : limits on integral(s)

Note: < –1 > if no addition of terms

10.

AP® CALCULUS BC 2005 SCORING GUIDELINES

Question 2

The curve above is drawn in the xy-plane and is described by the equation in polar coordinates r = + sin 2θ θ( ) for ≤ ≤θ π0 , where r is measured in meters and θ is measured in radians. The derivative of r with respect to θ is

given by dr = +1 2cos 2θ( ).dθ(a) Find the area bounded by the curve and the x-axis. (b) Find the angle θ that corresponds to the point on the curve with

x-coordinate 2− .

(c) For θ< < 23 3π π , d

drθ is negative. What does this fact say about r ? What does this fact say about the curve?

(d) Find the value of θ in the interval 0 θ≤ ≤ 2π that corresponds to the point on the curve in the first quadrant

with greatest distance from the origin. Justify your answer.

(a) Area 20

20

1212

r dπ

π

θ =

= + si( n 2θ θ( ) d) θ 4.382=

∫

∫3 :

1 : limits and constant 1 : integrand 1 : answer

⎧⎪⎨⎪⎩

(b) − = r2 cos( ) =θ θ + si (n 2( ))θ cos( ) θθ = 2.786

2 : { 1 : equation1 : answer

(c) Since ddrθ < 0 for π

3 θ< < 2 , r3 iπ s decreasing on this

interval. This means the curve is getting closer to the origin. 2 : { 1 : information about r

1 : information about the curve

(d) The only value in ⎢⎡0, 2

π⎣ ⎦

⎤⎥ where d

drθ = 0 is θ = 3

π .

θ r 00 0π3 1.913

π2 1.571

The greatest distance occurs when θ = 3π .

2 : 3 or 1.047

1 : answer with justification

1 : θ = π⎧⎪⎨⎪⎩

11.

AP® CALCULUS BC 2007 SCORING GUIDELINES

Question 3

The graphs of the polar curves r = 2 and r 3= + 2cos θ are shown in

the figure above. The curves intersect when 23θ = π and 3 .θ = 4π

(a) Let R be the region that is inside the graph of r = 2 and also inside the graph of r 3= + 2cos θ , as shaded in the figure above. Find the area of R.

(b) A particle moving with nonzero velocity along the polar curve given by r 3= + 2cos θ has position x( ) ,t y( )t( ) at time t, with θ = 0

when 0.t = This particle moves along the curve so that dr dr .dt = dθ

Find the value of dtdr

3 at θ = π and interpret your answer in terms of the motion of the particle.

(c) For the particle described in part (b), dy dy . Find the value of dtdy

3 at θ = π and interpret yourdt = dθanswer in terms of the motion of the particle.

(a) Area 34 22

2 312 2= +π ( ) ( +3 2cos3 2

10.370

π

π

=∫

4 :

d) θ θ ⎧ 1 : area of circular sector2 : integral for section of limaçon

1 : integrand 1 : limits and constant 1 : answer

⎪⎪⎨⎪⎪⎩

(b) 33

1.732dr drdt θ π dθ=

= =θ π= −

The particle is moving closer to the origin, since drdt < 0

and r 0> when θ = 3π .

2 : 3 1 :

1 : interpretation

drdt =θ π

⎧⎪⎨⎪⎩

(c) y r sin= = 3 2c+ osθ θ( ) sin θ

3 3dt θ π= dθ= =θ π

dy dy = 0.5

The particle is moving away from the x-axis, since

dtdy > 0 and y > 0 when θ = 3

π .

3 : 3

1 : expression for y in terms of

1 : dt θ π

1: interpretation

dy θ

=

⎧⎪⎪⎨⎪⎪⎩

12.AP® CALCULUS BC

2009 SCORING GUIDELINES (Form B)

Question 4

The graph of the polar curve r 1= − 2cos θ 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.

(b) Write expressions for ddxθ and d

dyθ in terms of θ .

(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 = 2π .

Show the computations that lead to your answer.

(a) r 0( ) 1;= − r θ( ) = 0 when θ = 3π .

Area of ( θ θ d)3 2

01 −1 2cos2=S

π∫

2 : { 1 : limits and constant 1 : integrand

(b) =x r cos θ and =y r sin θ

ddrθ = 2sin θ

d dxd =θ d

r r sin− θ 4si= n cosθ sθ in− θθ cos θ

d dyd =θ d

r r cos+ θ 2sin= 2θ sin θ θ 1+ 2co− s( )cosθ θ

4 :

1 : uses cos θ and

1 :

2 : answer

=x rdrd

y sr in= θ

θ

⎧⎪⎨⎪⎩

(c) When π ,2 θ = we have 0,x y= = 1.

2 2

dy ddydx θ =π dx d θ =π

θθ= = −2

The tangent line is given by 1= − 2y x.

3 :

1 : values for x and

1 : expression for

1 : tangent line equation

ydydx

⎧⎪⎪⎨⎪⎪⎩

13.

AP® CALCULUS BC 2011 SCORING GUIDELINES (Form B)

Question 2

The polar curve r is given by r(θ ) = +3 siθ θn , where ≤ θ ≤0 2π .

(a) Find the area in the second quadrant enclosed by the coordinate axes and the graph of r.

(b) For 2π θ≤ ≤ π , there is one point P on the polar curve r with x-coordinate −3. Find the angle θ that

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

ddtθ = 2. Find dt

dy at the instant that θ = 23π , and interpret the meaning of your answer in the context of

the problem.

(a) Area θr( )( )2

21 d =θ 47.2π

π∫= 513 3 : 1 : integrand1 : limits and constant

1 : answer

⎧⎪⎨⎪⎩

(b) − = r(θ )3 cos = (θ θ3 + si )n θ co sθθ = 2.01692y r θ( )sin θ( )= = 6.272

3 : 1 : equation 1 : value of θ 1 : y-coordinate

⎧⎪⎨⎪⎩

(c) y r(θ )sin= = (3θ θ sin+ )sθ in θ

=θ π 32 2 32.819dy dy ⋅ d

dtθ

θd dt θ π⎦ == −= ⎡ ⎤

⎥⎣⎢

The y-coordinate of the particle is decreasing at a rate of 2.819.

3 : 1 : uses chain rule

1 : answer1 : interpretation

⎧⎪⎨⎪⎩

14. 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

above. The curves intersect when θ = 6π and 6 .θ = 5π

(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 = t2. Find the time t in the interval t1 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) (5 6 2

61 4 − 2sin ) dθ θ = 2Area 6 2

π

π= π ∫+ 4.709 (or 24.708) 3 :

1 : integrand 1 : limits and constant 1 : answer

(b) ( )θ = () ( ( )) (t )

(t )

2 2

cos 4 2− sin )cθ θos

2sin c

1 w

os

hen 1.428 (or 1.427)

x

x(t t

t

x r

x

θ ⇒

=

=

= −

= 4 − 3 : θ )( x(t )(θ ) )

1 : or 1 : 1 or 1 : answer

xx x(t

= − = −

1

(c)

() ( 2 )) (t )2

sin ⇒θ 4 2− sin θθ )sin

2sin sin

y

y(tyr

t(θ ) (=

= 4 −

=

)

(( (2 )) ( ) (2 2 )) ( )2

Position vector

4 2sin t cos t , 4 2sin− sit tn

= x(t ) , y(t= −

( ) (1.5) , ′y (1.5) (

1.58.072, 1.673 or 8.072, 1.672

v = x′−= − − − )

3 : { 2 : position vector 1 : velocity vector

15. 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 ddxθ at

6θ = π .

(c) The distance between the two curves changes for 0 < θ < 2π .

Find the rate at which the distance between the two curves is changing with respect to θ when θ = 3π .

(d) A particle is moving along the curve r = 3 − 2sin (2θ ) so that ddtθ = 3 for all times t ≥ 0. Find the value

of drdt at 6 θ = π .

(a) ( θ ))2 2

09 1 3 − 2sin (24 29.708 (or 9.707

Area

)

ππ d θ

=

= ∫+

1 : answer

1 : integrand3 : 1 : limits

(b)

6

sin (2θ ))cos

2.3

2

66dxx

d π

θ

θ θ == −

= (3 − {2 : 1 : expression for 1 : answer

x

(c) The distance between the two curves is D = 3 − (3 − 2sin (2 )) = 2θ θsin (2 ).

3

dDdθ θ =π

= −2

{2 : 1 : expression for distance 1 : answer

(d)

( )− ( )6

2 3

3

6

dr = dr d drdt d dtθ θddrdt θ =π

θ =⋅ ⋅

= = −{2 : 1 : chain rule with respect to 1 : answer

t