FP2 POLAR COORDINATES: PAST QUESTIONS - … · FP2 POLAR COORDINATES: PAST QUESTIONS 1. The curve C...

FP2 POLAR COORDINATES: PAST QUESTIONS

1. The curve C has polar equation

r2 = a2 cos2θ, 44

–

(a) Sketch the curve C.

(2)

(b) Find the polar coordinates of the points where tangents to C are parallel to the initial line.

(6)

(c) Find the area of the region bounded by C.

(4)(Total 12 marks)

2.

The diagram above shows the curves given by the polar equations

r = 2,,

20

and r = 1.5 + sin 3 .

20

(a) Find the coordinates of the points where the curves intersect.

(3)

The region S, between the curves, for which r >2 and for which r < (1.5 + sin 3θ), is shown shaded in the diagram above.

City of London Academy 1

(b) Find, by integration, the area of the shaded region S, giving your answer in the formaπ + b√3, where a and b are simplified fractions.

(7)(Total 10 marks)

3.

The diagram above shows a sketch of the curve with polar equation

r = a + 3 cos θ, a > 0, 0 ≤ θ < 2π

The area enclosed by the curve is .

2

107

Find the value of a.

(Total 8 marks)

4.

A

NO

C

P

R

I n i t i a l l i n e

The curve C shown in the diagram above has polar equation


.2

0),cos1(4

r

At the point P on C, the tangent to C is parallel to the line .

2

(a) Show that P has polar coordinates

3,2

.

(5)

The curve C meets the line 2

at the point A. The tangent to C at P meets the initial line at the point N. The finite region R, shown shaded in the diagram above, is bounded by the initial

line, the line 2

, the arc AP of C and the line PN.

(b) Calculate the exact area of R.

(8)(Total 13 marks)

5.

= 0O

C 2 C 1

The diagram above shows the curve C1 which has polar equation r = a(3 + 2 cos θ), 0 ≤ θ < 2π and the circle C2 with equation r = 4a, 0 ≤ θ < 2π, where a is a positive constant.

(a) Find, in terms of a, the polar coordinates of the points where the curve C1 meets the circleC2.

(4)

The regions enclosed by the curves C1 and C2 overlap and this common region R is shaded in the figure.

(b) Find, in terms of a, an exact expression for the area of the shaded region R.

(8)


(c) In a single diagram, copy the two curves in the diagram above and also sketch the curve C3 with polar equation r = 2acosθ, 0 ≤ θ < 2π Show clearly the coordinates of the points of intersection of C1, C2 and C3 with the initial line, θ = 0.

(3)(Total 15 marks)

6. (a) Sketch the curve C with polar equation

r = 5 + √3 cos θ, 0 ≤ θ ≤ 2π.

(2)

(b) Find the polar coordinates of the points where the tangents to C are parallel to the initial line θ = 0. Give your answers to 3 significant figures where appropriate.

(6)

(c) Using integration, find the area enclosed by the curve C, giving your answer in terms of π.

(6)(Total 14 marks)

7.

Q

RC

O

P


The diagram above shows a sketch of the curve C with polar equation

r = 4sinθcos2θ, 0 ≤ θ < 2

.

The tangent to C at the point P is perpendicular to the initial line.


(a) Show that P has polar coordinates

6,

2

3

.

(6)

The point Q on C has polar coordinates

4,2

.

The shaded region R is bounded by OP, OQ and C, as shown in the diagram above.

(b) Show that the area of R is given by

4

6

2 d4cos2

1

2

1cos22sin

(3)

(c) Hence, or otherwise, find the area of R, giving your answer in the form a + bπ, where a and b are rational numbers.

(5)(Total 14 marks)

8.

O

C

m


The figure above shows a curve C with polar equation 40,2cos4

ar

, and a line m

with polar equation 8

. The shaded region, shown in the figure above, is bounded by C and

m. Use calculus to show that the area of the shaded region is ).2(

2

1 2 a

(Total 7 marks)

9.


O

P

R C

l

= 0

= 2

A curve C has polar equation .

40,2cos22

ar The line l is parallel to the initial

line, and l is the tangent to C at the point P, as shown in the figure above.

(a) (i) Show that, for any point on C, r2 sin2 θ can be expressed in terms of sin θ and a only.

(1)

(ii) Hence, using differentiation, show that the polar coordinates of P are

6,

2

a

.

(6)

The shaded region R, shown in the figure above, is bounded by C, the line l and the half-line

with equation .

2

(b) Show that the area of R is .433

16

2

a

(8)(Total 15 marks)

10.


P

Q

O

C

= 0

l

R

The curve C which passes through O has polar equation

r = 4a(1 + cos ), – < .

The line l has polar equation

r = 3a sec , – 2

< < 2

.

The line l cuts C at the points P and Q, as shown in the diagram.

(a) Prove that PQ = 63a.

(6)

The region R, shown shaded in the diagram, is bounded by l and C.

(b) Use calculus to find the exact area of R.

(7)(Total 13 marks)


11. The curve C has polar equation r = 6 cos , – 2

< 2

,

and the line D has polar equation r = 3 sec

3 , – 6

< 6

5

.

(a) Find a cartesian equation of C and a cartesian equation of D.

(5)

(b) Sketch on the same diagram the graphs of C and D, indicating where each cuts the initial line.

(3)

The graphs of C and D intersect at the points P and Q.

(c) Find the polar coordinates of P and Q.

(5)(Total 13 marks)

12.

B

A

OI n i t i a l l i n e

C

C 1

2

The diagram above is a sketch of the two curves C1 and C2 with polar equations

C1 : r = 3a(1 – cos ), – <

and

C2 : r = a(1 + cos ), – < .

The curves meet at the pole O, and at the points A and B.


(a) Find, in terms of a, the polar coordinates of the points A and B.

(4)

(b) Show that the length of the line AB is 2

33

a.

(2)

The region inside C2 and outside C1 is shown shaded in the diagram above.

(c) Find, in terms of a, the area of this region.

(7)

A badge is designed which has the shape of the shaded region.

Given that the length of the line AB is 4.5 cm,

(d) calculate the area of this badge, giving your answer to three significant figures.

(3)(Total 16 marks)

13. (a) Sketch the curve with polar equation

r = 3 cos 2, – 4

< 4

(2)

(b) Find the area of the smaller finite region enclosed between the curve and the half-line

= 6

(6)

(c) Find the exact distance between the two tangents which are parallel to the initial line.

(8)(Total 16 marks)


14.


D

O

C

B A

A logo is designed which consists of two overlapping closed curves.

The polar equations of these curves are

r = a(3 + 2cos ) and

r = a(5 – 2 cos ), 0 < 2.

The diagram above is a sketch (not to scale) of these two curves.

(a) Write down the polar coordinates of the points A and B where the curves meet the initial line.

(2)

(b) Find the polar coordinates of the points C and D where the two curves meet.

(4)

(c) Show that the area of the overlapping region, which is shaded in the figure, is

3

2a

(49 483).

(8)(Total 14 marks)

MARK SCHEMES

1. (a)

B1B1


2

B1 shapeB1 Labels

(b) Tangent parallel to initial line when y = r sin θ is stationary

Consider therefore d

d

(a2 cos2θ sin2 θ)

= –2 sin 2θ sin 2 θ + cos 2θ(2 sin θ cos θ) = 0 M1A1

2 sin θ[cos 2θ cos θ - sin 2θ sin θ] = 0

sinθ ≠0 cos3θ = 0 θ = 6

or 6

–

M1A1

Coordinates of the points

6

–,

2

1

6,

2

1 aa

A1A16

(c) Area = 4

4

–24

4

–2 d2cos

2

1d

2

1

ar

M1A1

= 2

)]1(––1[42

2sin

2

1 224

4

–

2 aaa

M1A14

[12]

2. (a) 1.5 sin3 2 sin3 0.5 5

3 or 6 6

, M1 A1,

and

5 or

18 18

A1

3

(b) Area =

518

18

212 (1.5 sin 3 ) d

, -

212

9

M1 M1


=

518

18

1 12 2(2.25 3sin 3 (1 cos6 ))d

-

212

9

M1

=

5

181 12 2

18

1(2.25 cos3 ( sin 6 ))

6

–

212

9

M1 A1

13 3 5

24 36

M1 A1

7

[10]

3. x

aA2

0

2 d)cos3(2

1

Applies

)d(2

12

0

2x

r

with B1correct limits.

Ignore dθ.

(a + 3cosθ)2 = a2 + 6acosθ + 9 cos2θ

=

2

2cos19cos62

aa2

2cos1cos2

M1

Correct underlined expression. A1

d2cos2

9

2

9cos6

2

12

0

2

x

aaA

Integrated expression with atleast 3 out of 4 terms of the form

± Aθ ± Bsinθ ± Cθ ± Dsin2θ M1 *

2

0

2 2sin4

9

2

9sin6

2

1

aa

Ignore the 21

. Ignore limits.

a2θ + 6asinθ + correct ftintegration. A1 ft

Ignore the 21

. Ignore limits.

)0(–09022

1 2 a

2

92 a

2

92 a

A1

Hence,

2

107

2

92 aIntegrated expression equal to .2

107 dM1 *


2

107

2

92 a

492 a

As a > 0, a = 7 a = 7 A1 cso

Some candidates may achieve a = 7 from incorrect

working. Such candidates will not get full marks

[8]

4. (a) r cos = 4(cos – cos2 ) or rcos = 4 cos – 2 cos 2 – 2

B1

)2sinsin(4d

)cosd(or )sincos2sin(4

d

)cosd(

rr

M1A1

4(–sin + 2 cos sin ) = 0 cos = 2

1

which is

satisfied by = 3

and r = 2(*)

dM1A1

5

Alternative for first 3 marks:

sin4d

d

r

B1

cossin8sin4d

dcossin

d

d

rr

x

M1A1Substituting r = 2 and 0 = f into original equation scores 0 marks.

(b) d)cos2cos(1(8)d

2

1 22r

M1

24

2sinsin2)8(

M1A1


371628

33

22

4

38

4

2sinsin2

2

38

2

3

M1

Triangle: 2

1

(rcos )(r sin ) = 2

331

2

1

M1A1

Total area:

2

315)162(

2

337162

(A1)A1

8

M1 needs attempt to expand (1 – cos )2 giving threeterms (allow slips)

Second M1 needs integration of cos2 using cos 2 ± 1Third M1 needs correct limits– may evaluate two areas and subtractM1 needs attempt at area of triangle and A1 for caoNext A1 is for value of area within curve, then final A1 is cao,must be exact but allow 4 terms and isw for incorrect collection of terms

Special case for use of r sin gives B0M1A0M0A0

[13]

5. (a) a(3 + 2cos) = 4a

M1

Solve to obtain cos = 2

1

M1

= ± 3

and points are (4a, 3

) and (4a, 3

5

)

A1, A1

4

First A for r = 4a second for both values in radians.Accept 1.0471… and 5.2359….2 dp or better for final A

(b) Use area = dadr 222 )cos23(

2

1 give to

2

1

M1Obtain (9 + 12cos + 2cos2 + 2)d


A1Integrate to give 11 + 12sin + sin 2

M1 A1

Use limits 3

and , then double or 3

5 and

3

or theirs

M1

Find a third area of circle = 3

16 2a

B1

Obtain required area = 2

313

3

3822 aa

A1 , A1

8

First M for substitution, expansion and attempt to use double angles.

Second M for integrating expression of the form a + bcos + ccos2

Lose final A only if a2 missing in last line

(c)

4 a 5 a 2 a O

correct shape

B15a and 4a marked

B12a marked and passes through O

B1

3

First B for approximately symmetrical shape about initial line, only 1loop which is convex strictly within shaded region

[15]


6. (a)

5 – 3

5

5 + 3

5

y

x

– 4 – 3 – 2 – 1 1 2 3 4 5 6 7

6

5

4

3

2

1

– 1

– 2

– 3

– 4

– 5

– 6

Shape (close curve, approx. symmetrical about the initial line,in all ‘quadrants’ and ‘centred’ to the right of the pole/origin). B1Shape (at least one correct ‘intercept’ r value... shown on sketchor perhaps seen in a table). B1

2(Also allow awrt 3.27 or awrt 6.73).

(b) y = r sin = 5 sin + 3 sin cos M1

d

dy

= 5 cos – 3 sin2 + 3 cos2 (= 5 cos + 3 cos 2) A1

5 cos – 3(1 – cos2 ) + 3 cos2 = 0 M1

23 cos2 + 5 cos – 3 = 0(23 cos – 1)(cos + 3) = 0 cos = ... (0.288...) M1

= 1.28 and 5.01 (awrt) (Allow 1.28 awrt)

32

1arccos allow Also

A1

r = 5 + 2

11

32

13

(Allow awrt 5.50) A16

2nd M: Forming a quadratic in cos .

3rd M: Solving a 3 term quadratic to find a value of cos (evenif called ).

Speacial case: Working with r cos instead of r sin :

1st M1 for r cos = 5 cos + 3 cos2 1st A1 for derivative – 5 sin – 23 sin cos , then no further marks.

(c) r2 = 25 + 103 cos + 3 cos2 B1

4

2sin3sin310

2

53dcos3cos31025 2

M1 A1ft A1ft(ft for integration of (a + b cos ) and c cos 2 respectively)

.....2

3

4

2sin3sin31025

2

12

0

M1

2

53)350(

2

1

or equiv. in terms of . A1


6

1st M: Attempt to integrate at least one term.

2nd M: Requires use of the 2

1

, correct limits (which could be 0to 2, or – to , or ‘double’ 0 to ), and subtraction(which could be implied).

[14]

7. (a) x = r cos = 4 sin cos3 M1

224 sincos12cos4d

d

x

any correct expression M1A1

Solving

0)sin3(coscos40

d

d0

d

d 222

xx

M1

63

1or tan

2

3cosor

2

1sin

AG A1 cso

r = 2

3

6cos

6sin4 2

AG A1cso6

So many ways x may be expressing e.g.

2 sin 2 cos2 , sin 2(1 + cos 2), sin 2 + (1/2) sin 4

leading to many results for d

dx

Some relevant equations in solving

[(1 – 4 sin2 ) = 0, (4 cos2 – 3) = 0, (1 – 3 tan2 ) = 0, cos 3 = 0]

Showing that 6

satisfies 0

d

d

x

, allow M1 A1

providing d

dx

correctStarting with x = r sin can gain M0M1M1

(b)

dcossin16.2

1d

2

1 44

6

24

6

2 rA

8 sin2 cos4 = 2 cos2 (4 sin2 cos2) = 2 cos2 sin2 2 M1

= (cos2 + 1)sin2 2 M1

2

4cos12sin2cos 2

= Answer AG A1 cso


3

First M1 for use of double angle formula for sin 2A

Second M1 for use of cos 2A = 2 cos2 A – 1Answer given: must be intermediate step, as shown, andno incorrect work

(c) Area =

4

6

3

8

4sin

22sin

6

1

ignore limits M1A1

83

2sin

123sin

6

1

8

sin

82sin

6

1 33

(sub. limits) M1

24,

6

1

16

3

1216

3

86

1

both cao A1, A15

For first M, of the form

4sin

22sin 3 ba

(Allow if two of correct form)On ePen the order of the As in answer is as written

[14]

8. Use of dr 221

B1

Limits are 8

and 4

B1

16a2 cos2 2 = 8a2 (l + cos 4)

M1

4

4sin)4cos1(

d

M1 A1

4

84

4sin4 2

aA

)10(

8442

a


M1

)2(2

11

222

aa

csoA1

[7]

9. (a) (i) r2 sin2θ = a2 cos 2θsin2 = a2(1 – 2 sin2) sin2

B1 1

(= a2(sin2θ – 2 sin4θ))

(ii) dd

(a2(sin2 – 2sin4 )) = a2(2sin cos – 8 sin2 cos ), = 0M1, A1, M1

2 = 8 sin2 (Proceed to a sin2 = b) M1

sin θ = 21

θ = 6

, r = 2

a

A1, A1 cso 6

(b) 4d2cos

2

22 aa sin 2 M: Attempt

d21 2r

, to get k sin 2θ M1 A1

231

4...

24

6

a

M: Using correct limits M1 A1

= 163

23

221.

221 2aaa

M: Full method for rectangle or triangle M1 A1

R = 16231

4163 222 aaa

(33 – 4)M: Subtracting, either way round dM1 A1 cso 8

[15]

(a) (ii) First A1: Correct derivative of a correct expression for r2sin2θ or r sin θ.

(b) Final M mark is dependent on the first and third M’s.

Attempts at the triangle area by integration: a full method is required for M1.

Missing a factors: (or a2) Maximum one mark penalty in the question.


10. (a)

P

Q

O= 0

RN

4a(1 + cos) = cos

3a

or r = 4a

r

a31

M1

4cos2 + 4cos – 3 = 0 or r2 – 4ar – 12a2 = 0A1

(2cos – 1)(2cos + 3) = 0 or (r – 6a)(r + 2a) = 0M1

cos = 2

1

,

3

or r = 6aA1

Note ON = 3a

PQ = 2 × ON tan 6

= 63a (*)

cso M1 A1

6

or PQ = 2× [(6a)2 – (3a)2] = 2(27a2) = 63a (*)

csoor any complete equivalent

(b) 2 ×

...

...

23

0

2 16...d2

1ar

(1 + cos)2d

M1

dr 2

= … d2cos

2

1

2

1cos21

...

...

M1

cos2 cos2


= …

2sin

4

1sin2

2

3

A1

= 16a2

8

33

2

(= 2a2[4 + 93] 56.3a2)

M1 A1

use of their 3

for M1

Area of POQ = 36

2

1

a × 3a or 9a2 3

B1

R = a2(8 + 93)

cao A1

7

[13]

11. (a) For C: Using polar/ Cartesian relationships to form Cartesian equation

M1

so x2 + y2 = 6x

A1

[Equation in any form: e.g. (x – 3)2 + y2 = 9 from sketch.

or 22

22 6

yx

xyx

]

For D: 3

3cos

r

and attempt to expand

M1

2

3

2

yx

= 3 (any form)

M1A1

5

(b)


“Circle”, symmetric in initial line passing through pole

B1Straight line

B1Both passing through (6, 0)

B1

3

(c) Polars: Meet where 6cos cos( 3

– ) = 3

M1

3 sin cos = sin2

M1

sin = 0 or tan = 3 [ = 0 or 3

]

M1

Points are (6, 0) and (3, 3

)

B1, A1

5

[13]

Alternatives (only more common):

(a) Equation of D:Finding two points on lineM1Using correctly in Cartesian equation for straight lineM1Correct Cartesian equation


A1

(c) Cartesian: Eliminate x or y to form quadratic in one variable

M1

[2x2 – 15x + 18 = 0, 4y2 – 36 y = 0]Solve to find values of x or y

M1

Substitute to find values of other variable

2

33or 0 6;or

2

3yx

B1A1

Points must be (6, 0) and (3, 3

)

B1A1

12. (a) 3a(1 – cos) = a(1 + cos)

M1

2a = 4a cos cos = 2

1

= 3

or – 3

M1

r = 2

3a

A1A1

4

[Co-ordinates of points are 32 3( , )a

and 32 3( , )a ]

(b) AB = 2rsin = 2

33a

M1A12

Area =

3

3

2

1

r2d

= d])cos1(9)cos1([

2

1 2222 aa


M1 M1

= d)]coscos21(9coscos21[

222

2a

A1

= d)]cos8cos208[

22

2a

= k[–8 + 20sin …B1….. – 2sin2 – 4]B1

Uses limits 3

and – 3

correctly or uses twice smaller area

and uses limits 3

and 0 correctly.(Need not see 0 substituted)

= a2[–4 + 10 3 – 3 ] or = a2[–4 + 9 3 ] or 3.022a2

A1

7

(d) 3a 2

3

= 4.5 a = 3

B1

Area = 3[9 3 – 4], = 9.07 cm2

M1, A1

3

[16]

13. (a)

3

Shape + horiz. axis

B1 3

B1


2

(b) Area = 21

r2 d

= 21

9 cos22 d use of 21

r2

M1

=

d

2

14cos

2

9

use of cos4 = 2cos22 – 1M1

=

4

628

4sin

2

9

M1, A1

=

1216

3

82

9

subst. 4

and 6

M1

=

16

3

242

9

or 0.103A16

(c) r sin = 3 sin cos 2

d

d y

= 3 cos cos 2 – 6 sin sin 2 (diff. r sin )M1, A1

d

dy

= 0 6 cos2 – 3 cos – 12 sin2

cos = 0 use of d

dy

= 0M1

6 cos2 – 3 cos – 12(1 – cos2

)cos = 0 use double angle formulaM1

18 cos3 – 15 cos = 0 solving

M1

cos = 0 or cos 2 = 65

or tan2 = 5

1

or sin2 = 6

1

A1

\ r = 3(2 × 65

) – 1= 2

r sin = 2 61

use of d = 2r sin M1

d = 3

62

A18

[16]


14. (a) (5a, 0), (3a, 0)

B1, B1

2

allow on diagramB1 only if no zeros or diagram with initial line

(b) 3 + 2 cos = 5 – 2 cos

M1

cos = 21

M1

= 3

5,

3

A1

both, accept – 3

(4a, 3

) , (4a, 3

5

)

A1

4

(c) Allow 2 M1s below for method shown in either calculation

(5 – 2cos)2 d (3 + 2cos)2 d

= (25 – 25cos + 4cos2 )d = (9 + 12cos + 4cos2)dM1

= (27 – 20cos + 2cos2)d = (11 + 12cos + 2cos2)dM1

= 27 – 20sin + sin2, = 11 + 12sin + sin2A1, A1

A = 2 × 21

a2(5 – 2 cos)2 d + 2 × 21

a2(3 + 2cos)2 d adds

M1

= a2 3

03

222 d)cos23(d)cos25(

a correct limits

A1

= a2[27 × 3

– 103 + 2

3

] + a2[11( – 3

) – 63 – 2

3

]

dM1


= 3

2a

(49 – 483] (*)

A1 cso

8

[14]

15. (a) dcos2cos1

2

1 22 a

M1 A1

correct with limits

=

dcos2

2

12cos1

2

1 2a

M1 A1

= 2 × 2

1

a2

0

sin224

2sin

A1

= a2

2

3

= 2

3 2a

A1

6

(b) x = a cos + a cos2

r cos M1

d

dx

= a sin 2a cos sin

A1

d

dx

= 0 cos = 21

finding M1

= 3

2

or = 3

4


r = 2

a

or r = 2

a

finding r M1

A: r = 2

a

, = 3

2

B: r = 2

a

, = 3

2

both A and B A15

(c) x = a41

WX = 2a + a41

= 2 a41

M1 A1

2

(d) WXYZ = 8

327 2a

B1 ft

1

(e) Area = 8

327

100 2

1003

= 113.3 cm2

M1 A1

2

[16]

16. (a)

A

O l

closed loop B1


symmetry in l B12

(b) Area of loop = 21

2

2

2

r

d = 21

2

2

3cos4

d M1

= 4

2

0

3cos

d = 4

2

0

2 cos)sin1(

d M1 A1

= 4

2

0

3

3

sinsin

M1 A1

= 4(1 – 31

) = 38

M1 A17

(c) y = r sin = 2 sin 23

cos and the value of in (0, 2

) for

which d

dy

= 0 is required

d

dy

= 21

25

cossin3cos2 2 M1 A1

= 21

25

25

cos3cos3cos2 M1

= )3cos5(cos 221

A1

21

cos ¹ 0 in (0, 2

), so cos = 53

M1

At A, = 0.68, r = 1.36 (2 decimal places) A1, A17

[16]

1. No Report available for this question.


2. The finding of the two values of in part (a) was usually correctly done but some candidates

then wasted valuable time in taking their two values and working in a circle to find the r values - some of which were not equal to 2

In part (b) the vast majority of the candidates knew how to find the area enclosed between the two radius vectors and the curve simply as one integral. Others chose to split it into one area

from 0 to 2

minus the area from 0 to 18

minus the area from 18

5

to 2

- a long way round. Most of the candidates decided to use the integration method to find the area of the sector as well. Errors in integration were usually in the use of the double angle formula although most used it correctly. A minority of candidates forgot to square the function before integrating, but where this squaring had been done the subsequent integration of the trigonometric functions waswell done. Where candidates fell down was in the careful application of the limits to their functions – writing things more neatly would have helped in a number of cases. Some forgot about the area of the sector of the circle completely. It was, however, pleasing to see many completely correct solutions.

3. This question was well answered by candidates and statistics showed that around 50% of the candidature gained all 8 marks available for this question.

Most candidates applied the formula d

2

1 2r

but a number of them struggled to write down the correct limits to find the relevant area for their expression. The majority of the errors made

by candidates were algebraic; 6a sin θ was sometimes missed out when (a + 3cos θ)2 was

expanded. Sometimes a2 was “taken out” from some candidates’ integral. Most candidates knew

they needed to substitute

2

2cos19

for 9cos2 θ in order to integrate the expression but

sometimes there were errors in dealing with the 9 and the 21

. There were a significant minority

of candidates who did not know the correct strategy to apply in order to integrate cos2 θ. These candidates usually lost 7 of the 8 marks available for this question.

4. Part (a) required candidates to consider r cos θ and to differentiate to find a maximum value.

This was dximplied as the tangent was perpendicular to the initial line and so d

dx

= 0. A number of candidates chose to consider r sin θ instead and little credit was given for this.

In part (b) most realised that they needed to find d)coscos21(

2

1 2

. There were a number of sign slips but the methods were understood, and most tried to use double angle

formulae. To find the shaded area they needed to use the limits 3 and

2

and to subtract. This gave the finite area enclosed by the arc AP and the straight lines OP and OA. They then needed to find the area of triangle OPN. The sum of these two areas gave them the total area required. A


number of candidates found the area of a rectangle or used a triangle where the sides had incorrect lengths, particularly where the base was 2 units instead of 1 unit. Many answers for the triangle area included a pi term instead of a root 3.Full marks in question 8 was indicative of a good grade A candidate.

5. There were many good solutions to this question, although few scored full marks.It was part (b) that caused the most difficulty and that was sometimes omitted.Some candidates spent too long plotting points for their sketch in part (a), but most were able to produce a closed curve in approximately the correct position. Indication of scale was required toscore the second mark here.In part (b), a few candidates started to differentiate r cos θ instead of r sin θ, but most were able to produce a correct derivative and to proceed to find a quadratic in cos θ. Despite many correct quadratic equations, mistakes frequently occurred from that point onwards. Some of these mistakes were careless, others stemmed from an apparent expectation that the values of θ wouldbe exact. Even candidates who had a correct solution for one value of θ (1.28) were often unableto produce the second value. The r coordinate of the required points was often missing or wrong.In general, candidates were much happier with part (c), where there were many excellent solutions. There was, however, plenty of scope for mistakes, and these included slips in

squaring (5 + cos3 ), sign or numerical errors in applying )12(cos

2

1cos2

and

integration slips. Sometimes the limits for integration were wrong or the 2

1

was omitted from the area formula, losing the last two marks

6. Most candidates started this polar coordinates question successfully only losing an accuracy mark for the values of θ, writing them as π/3 and –π/3 rather than π/3 and 5π/3.

Candidates were on the whole correct in the use of dr 2

2

1

and many of them were successful

in negotiating the expansion of (3 + 2cos θ)2 including the use of double angles. The mistakes usually consisted of identifying appropriate limits and valuable marks were lost here. Some examples of this included finding the whole area of C1 using limits of 0 and 2π and then subtracting that part of C1 between –π/3 and π/3. This inevitably resulted in errors as some candidates forgot about the circle part of the problem and proceeded to subtract the area rather than add it. Some candidates struggled with integration to establish that the 1/3 of the circle

needed was (16πa2)/3. It was pleasing to note how many candidates made it correctly to the

final answer although some found it unnecessary to simplify to 3

38 to

6

76

.

7. In part (a) there was some confusion as to whether it was r cos θ or r sin θ, or indeed r, that needed to differentiated with respect to θ. However, many correct solutions were seen, sometimes clearly aided by the given answers for r and θ.Again this was a challenge to mark as there were so many different approaches with a large number of different correct equations that led to the given results. Some candidates chose to

rearrange r sin θ cos3 θ, using either sin2 θ + cos2 θ = 1, or a double angle formula or a


combination of both, before they differentiated, and some waited until they had differentiated before making a similar move. Solutions were, therefore, often not as concise as they might have been.

Part (b) was not done well by a large number of candidates, who had little structure to their work. One mark was often awarded, perhaps generously at times, for correct use of the cosine double angle formula, but a complete method was usually only seen from the better candidates. Many candidates spent much time on this part, often producing very unwieldy expressions, and

it was not uncommon to see incorrect double angle formulae; cos4 θ = 2

4cos1

was also seen too often.

Many candidates went on to gain marks for integration in part (c) although the downfall here came often in not realising that the first term could be integrated directly, or in giving the result

as 3

1

sin3 3θ; that error still allowed candidates to gain 3 marks, however, and that was a common score.

8. The method needed was well understood and most could use an appropriate formula and identify the correct double angle formula to carry out the indefinite integration. The limits,

however, proved testing and only about 60% of the candidates used the correct limits of 8

and

4

.

9. There were some very poor and rushed solutions to this question. In part (a)(i), surprisingly many candidates were unable to link the required expression with the given polar equation, even

though they knew that cos 2 = 1 –2 sin2.

Most knew in part (a)(ii) that they had to deal with rsin , but made it more difficult for

themselves by trying to differentiate sin2cosa rather than a2(1 – 2sin2 ) sin2 (as suggested by part (a)(i)). Consequentially many mistakes, often due to the fractional power, were seen in differentiation, and few candidates proceeded legitimately to the given polar coordinates of P. Particularly disappointing at this level was the (not uncommon) mistake of

square rooting separate terms of a sum, e.g. 242 sin2sinsin2sin .

Completely correct solutions to part (b) were rare. While better candidates did realise that they

had to use the area of the triangle

6sin

26cos

22

1 aa

, there was much confusion over


which limits to use for the integral d2cos

2

1 2a, and which area this represented. A

common mistake was to use 2

instead of 4

as the upper limit, even though the curve was

defined only for 40

.

10. This question was often answered very well. In part (a) the majority of the candidates were able to find the values of θ at P and Q and usually they were then able to prove the required result. There was some dubious trigonometry seen here, such as OP = 6a followed by

PQ = 6 tan

3a

, but many gave a convincing proof. A sizeable minority of candidates made

the unwarranted assumption in part (a) that the tangent at P was parallel to the initial line, whilstthis yielded a correct value for θ it did not, of course, receive any credit although such candidates were allowed to use their value in the following part.

In part (b) most tried integrating 16a2(1 + cos θ)2 and knew how to deal with the cos2 θ by usingthe double angle formula. Most chose suitable limits but common errors were to use

, or 2 6

instead of 3

. Some forgot to subtract the area of the triangle and a few only

calculated the area above the initial line but, apart from careless slips, many candidates gave very good solutions.

11. There is no doubt that this was the most challenging and least productive question for most candidates. In trying to convert from polar equations to cartesian equations many candidates

often took a page of working for very little, if any, reward. The equation cos6r was recognised, or plotted, as the correct circle in part (b) but its cartesian equation had often not

been found in part (a). Recognition of )

3sec(3

r

as a straight line was relatively rare, and its cartesian equation only found by the better candidates.

Some candidates who had been successful in part (a) used the cartesian equations of the graphs to find their points of intersection and convert them to polars coordinates, but for most candidates part (c) involved solving a trigonometric equation. Most candidates gained a very

generous first mark, but solving the resulting equation cos )

3cos(

½ was generally notwell done; probably a sign that confidence had taken a knock in the earlier parts. Most

candidates expanded )

3cos(

to give 1cossin3cos2 ; those who progressed

further to 2sincossin3 or ,1)2sin(2 usually completed the solution, although cancelling sin in the former case was quite common. A neat solution was to use the

factor formulae to give

3cos

.1)

32cos(


It was very disappointing to see “ cos )

3cos(

½ cos = ½ or )

3cos(

½”, even though, more disappointingly in this case, they gave the correct answers !

12. Good candidates were able to score full marks in their solutions to this question. For the majority of candidates cos = ½ usually appeared in (a); a few ignored the range of and gave A as (3/2a, 5/3) or (-3/2a, -1/3). Having an answer to aim for in (b) certainly helped many candidates to work out what they should do, but many omitted this part or attempted to find an

arc length. In part (c), most used r2 for one or both curves, even if the ½ was missing or limits

were wrong. There were good answers for the integration of trig functions cos and cos2. There was often a muddle over whether to use one or both curves in finding the required area – the significance of the values of θ from (a) was not realised by many. A mixture of limits 0 to for C2 and 0 to /3 for C1 was not uncommon, although some retrieved the situation by then

subtracting the area /3 to for C2. Perhaps because time was running out there was evidence

of carelessness over signs as candidates attempted to tidy up their solutions. A very few

candidates used r instead of r2, or attempted to integrate ‘a’, thus producing answers in a or a3

– clearly not dimensionally correct. In part (d) the value of a was usually found correctly, although some candidates mistakenly used a = 4.5. Some candidates left their final answer in the exact form, rather than as a decimal to 3 significant figures as requested in the question.

13. Part (a) was generally correct although a minority of students did not restrict their values of to those stated in the question.

In parts (b) and (c) many candidates used the double angle formulae although some had slight

errors such as 2

14cos

. In part (c) many candidates who solved 0)sin(

r

d

d

ignored theinstruction to give the exact distance or used d = 2r not 2rsin.

14. Parts (a) and (b) were generally well done although many gave an answer for point C which was

out of range, giving 4 ,

3a

instead of

54 ,

3a

. Part (c) is a very demanding question.

Many could gain some marks by showing that they were able to integrate

23 2cos

and

2

5 2cosbut finding the right limits and putting together the correct areas proved difficult.

Correct solutions were, however, seen and some of these were very stylish and commendably brief.

15. Most candidates made a good attempt at the area, including the correct use of the double angle


formula in the integral. Most candidates attempted to differentiate r cos but several candidates

lost marks by giving a value for outside the range. Many candidates had difficulty in finding the length WX. Incorrect trigonometry led to the projection of OA onto the initial line being

given as 2

3a

or 4

3a

and many candidates used WX = a + the projection rather than using 2a

+ projection. In (d) a minority of students used a3 rather than a3 .

16. No Report available for this question.

15.

O

W

A

B

Z Y

X

i n i t i a l l i n e

The diagram above shows a sketch of the cardioid C with equation r = a(1 + cos ), < . Also shown are the tangents to C that are parallel and perpendicular to the initial line. These tangents form a rectangle WXYZ.

(a) Find the area of the finite region, shaded in the diagram above, bounded by the curve C.

(6)

(b) Find the polar coordinates of the points A and B where WZ touches the curve C.

(5)

(c) Hence find the length of WX.

(2)

Given that the length of WZ is 2

33 a

,

(d) find the area of the rectangle WXYZ.

(1)

A heart-shape is modelled by the cardioid C, where a = 10 cm. The heart shape is cut from the rectangular card WXYZ, shown in the diagram above.

(e) Find a numerical value for the area of card wasted in making this heart shape.

(2)(Total 16 marks)


16. The curve C is given by

r = 23

cos2 , 22

,

where (r, ) are polar coordinates.

(a) Sketch the curve C.

(2)

(b) Find the area of the region enclosed by C.

(7)

At the point A on C, = , where 0 < < 2

, and the tangent at A to C is parallel to the initial line.

(c) Find, to 2 decimal places, the value of and the corresponding value of r.

(7)(Total 16 marks)


FP2 POLAR COORDINATES: PAST QUESTIONS - … · FP2 POLAR COORDINATES: PAST QUESTIONS 1. The curve C...

Documents

Transcript of FP2 POLAR COORDINATES: PAST QUESTIONS - … · FP2 POLAR COORDINATES: PAST QUESTIONS 1. The curve C...