FP2 POLAR COORDINATES: PAST QUESTIONS
1.
The curve C has polar equation
–

 
4
r = a cos2θ, 4
2
(a)
2
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
and
(a)
r = 2,

0   ,
2
r = 1.5 + sin 3 

0   .
2
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 form
aπ + 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
107
.
The area enclosed by the curve is 2
Find the value of a.
(Total 8 marks)
4.
A
C
R
P
O
N
In itia l lin e
The curve C shown in the diagram above has polar equation
City of London Academy
2

r  4(1  cos ), 0   .
2

 .
2
At the point P on C, the tangent to C is parallel to the line
(a)
 
 2, 
Show that P has polar coordinates  3  .
(5)

2 at the point A. The tangent to C at P meets the initial line at
The curve C meets the line
the point N. The finite region R, shown shaded in the diagram above, is bounded by the initial


2 , the arc AP of C and the line PN.
line, the line
 
(b)
Calculate the exact area of R.
(8)
(Total 13 marks)
5.
C
C
2
O
1
= 0
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 circle
C2.
(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)
City of London Academy
3
(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
C
P
R
O
In itia l lin e
The diagram above shows a sketch of the curve C with polar equation
2
r = 4sinθcos θ,

0≤θ< 2 .
The tangent to C at the point P is perpendicular to the initial line.
City of London Academy
4
(a)
3  
 , 
Show that P has polar coordinates  2 6  .
(6)


 2, 
4 .
The point Q on C has polar coordinates 
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

2
4
  sin

6

2cos2 
1 1

 cos 4  d
2 2

(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.
m
C
O
In itia l lin e
r 4 a cos 2 , 0  

4 , and a line m
The figure above shows a curve C with polar equation

 
8 . The shaded region, shown in the figure above, is bounded by C and
with polar equation
1 2
a (  2).
m. Use calculus to show that the area of the shaded region is 2
(Total 7 marks)
9.
City of London Academy
5
= 2
P
R
l
C
O
= 0

r 2  a 2 cos 2 , 0   .
4 The line l is parallel to the initial
A curve C has polar equation
line, and l is the tangent to C at the point P, as shown in the figure above.
(a)
(i)
2
2
Show that, for any point on C, r sin θ can be expressed in terms of sin θ and a
only.
(1)
(ii)
 a 

, 
6
2

Hence, using differentiation, show that the polar coordinates of P are
.
(6)
The shaded region R, shown in the figure above, is bounded by C, the line l and the half-line

  .
2
with equation
(b)
a2
3 3 4.
Show that the area of R is 16


(8)
(Total 15 marks)
10.
City of London Academy
6
l
P
C
R
O
= 0
Q
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 = 63a.
(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)
City of London Academy
7
11.
The curve C has polar equation
and the line D has polar equation
(a)


–2 < 2,
r = 6 cos ,


  
3
,
r = 3 sec 

5
–6 < 6 .
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.
C
1
B
C
2
O
In itia l lin e
A
The diagram above is a sketch of the two curves C1 and C2 with polar equations
C1 : r = 3a(1 – cos  ),
–   < 
C2 : r = a(1 + cos  ),
–   < .
and
The curves meet at the pole O, and at the points A and B.
City of London Academy
8
(a)
Find, in terms of a, the polar coordinates of the points A and B.
(4)
(b)
3 3
Show that the length of the line AB is 2 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)
City of London Academy
9
14.
D
B
O
A
In itia l lin e
C
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
a2
3 (49  483).
(8)
(Total 14 marks)
MARK SCHEMES
1.
(a)
B1B1
City of London Academy
10
2
B1 shape
B1 Labels
(b)
Tangent parallel to initial line when y = r sin θ is stationary
d
2
2
Consider therefore d (a cos2θ sin θ)
2
= –2 sin 2θ sin 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
 1
  1
– 


a, 
a,
6  2
6 
2

Coordinates of the points
(c)
Area =
1
2

4
–
4

1
r 2 d  a 2
2

4
–
4
 cos 2 d
A1A1
6
M1A1

=
1 2  sin 2  4
a2
a2
a 

[
1
–
(–
1
)]

2  2  –
4
2
M1A1
4
4
[12]
2.
(a)
1.5  sin 3  2
and
 
 sin3  0.5
 3 

5
or
18
18
 18

  (1.5  sin 3 ) 2 d  1
2
 18
 9   2
Area =
,-
  5 
 or

6 
6 ,
M1 A1,
A1
3
5
1
2
(b)
City of London Academy
M1 M1
11
 18

1
1

 1
(2.25

3sin
3


(1

cos
6

))d

2
2 
2
 18
 9   2
5
=
M1
5
=
1
 18
1 
1
sin 6 )) 
2  (2.25  cos3  2 ( 
6


18
1
  22
– 9
M1 A1
13 3 5

24
36

M1 A1
7
[10]
2x
3.
2x
1
A  (a  3 cos  ) 2 d
2 0

2

Applies
2
1
r 2 ( d )
2 0

with
correct limits.
Ignore dθ.
B1
2
(a + 3cosθ) = a + 6acosθ + 9 cos θ
=
 1  cos 2 
a 2  6a cos   9

2


1 cos 2
cos 2  
2
Correct underlined expression.
M1
A1
Integrated expression with at
least 3 out of 4 terms of the form
± Aθ ± Bsinθ ± Cθ ± Dsin2θ
M1 *
2x
1 
9 9

A   a 2  6a cos    cos 2 d
2 0
2 2


2
9
9
 1 

   a 2  6a sin     sin 2 
2
4
 2 
0
1
Ignore the 2 . Ignore limits.
2
a θ + 6asinθ + correct ft
integration.
Ignore the
1
 2a 2 0  9  0 – (0)
2

a 2 
Hence,

9
2
a 2 
City of London Academy
1
2
. Ignore limits.

a 2 
9 107


2
2
A1 ft
Integrated expression equal to
9
2
107
2
.
A1
dM1 *
12
9 107
a2  
2 2
a 2 49
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)
2
r cos  = 4(cos  – cos ) or rcos  = 4 cos  – 2 cos 2 – 2
B1
d( r cos  )
d( r cos  )
4( sin   2 cos  sin  ) or
4( sin   sin 2 )
d
d
M1A1
1
4(–sin  + 2 cos  sin ) = 0  cos  = 2 which is

satisfied by  = 3 and r = 2(*)
dM1A1
5
Alternative for first 3 marks:
dr
4 sin 
d
B1
dx
dr
 r sin   cos 
 4 sin   8 sin  cos 
d
d
M1A1
Substituting r = 2 and 0 = f into original equation scores 0 marks.
(b)
1 2
r d (8) (1  2cos  cos 2 )d
2


M1
sin 2  

(8)   2 sin  
 
4
2

M1A1
City of London Academy
13

  3
sin 2  2
 3
  

8  2 sin  

8

2
  
 4
4  
2
 2

3
3
3  
2  16  7 3
8  
M1
1
1
3
1 3 
2
Triangle: 2 (rcos )(r sin ) = 2
M1A1
Total area:
2  16  7 3  
3
15 3
(2  16) 
2
2
(A1)A1
8
2
M1 needs attempt to expand (1 – cos ) giving three
terms (allow slips)
2
Second M1 needs integration of cos  using cos 2 ± 1
Third M1 needs correct limits– may evaluate two areas and subtract
M1 needs attempt at area of triangle and A1 for cao
Next 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
1
Solve to obtain cos  = 2
M1


5
 = ± 3 and points are (4a, 3 ) and (4a, 3 )
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)
1 2
1
r d to give a 2 (3  2 cos  ) 2 d
2
Use area = 2


M1
Obtain (9 + 12cos + 2cos2 + 2)d
City of London Academy
14
A1
Integrate to give 11 + 12sin + sin 2
M1 A1


5
and
3 or theirs
Use limits 3 and , then double or 3
M1
16a 2
3
Find a third area of circle =
B1
38a 2 13 3a

3
2
Obtain required area =
2
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
2
Lose final A only if a missing in last line
(c)
O
2a
4a
5a
correct shape
B1
5a and 4a marked
B1
2a marked and passes through O
B1
3
First B for approximately symmetrical shape about initial line, only 1
loop which is convex strictly within shaded region
[15]
City of London Academy
15
6.
(a)
5
6
y
5
4
3
2
1
–4
5 – 3
–3 –2
–1
–1
x
1
2
3
4
5
–2
6
7
5 + 3
–3
–4
–5
5
Shape (close curve, approx. symmetrical about the initial line,
in all ‘quadrants’ and ‘centred’ to the right of the pole/origin).
Shape (at least one correct ‘intercept’ r value... shown on sketch
or perhaps seen in a table).
–6
B1
B1
2
(Also allow awrt 3.27 or awrt 6.73).
(b)
y = r sin  = 5 sin  + 3 sin  cos  M1
dy
d = 5 cos  – 3 sin2  + 3 cos2  (= 5 cos  + 3 cos 2)
2
2
5 cos  – 3(1 – cos ) + 3 cos  = 0
2
23 cos  + 5 cos  – 3 = 0
(23 cos  – 1)(cos  + 3) = 0
cos  = ... (0.288...)

1 
 Also allow arccos

2
3


 = 1.28 and 5.01 (awrt) (Allow 1.28 awrt)
 1  11
 
3 
2
3

 2 (Allow awrt 5.50)
r=5+
6
A1
M1
M1
A1
A1
nd
2 M: Forming a quadratic in cos .
rd
3 M: Solving a 3 term quadratic to find a value of cos  (even
if called ).
Speacial case: Working with r cos  instead of r sin :
st
2
1 M1 for r cos  = 5 cos  + 3 cos 
st
1 A1 for derivative – 5 sin  – 23 sin  cos , then no further marks.
(c)
2
2
r = 25 + 103 cos  + 3 cos 
25 10
3 cos   3 cos 2 d 
B1
53
 sin 2 
 10 3 sin   3

2
 4 
(ft for integration of (a + b cos ) and c cos 2 respectively)
M1 A1ft A1ft
2
1
3 sin 2 3 
25  10 3 sin  
  .....

2
4
2 0
M1
1
53
 (50  3 ) 
2
2 or equiv. in terms of .
A1
City of London Academy
16
6
st
1 M: Attempt to integrate at least one term.
1
2 M: Requires use of the 2 , correct limits (which could be 0
to 2, or – to , or ‘double’ 0 to ), and subtraction
(which could be implied).
nd
[14]
7.
(a)
3
x = r cos  = 4 sin  cos 
dx
4 cos4   12 cos 2  sin 2 
d
M1
any correct expression
dx
 dx

0
0  4 cos 2  (cos 2   3 sin 2  ) 0

 d

Solving d
1
3
1

sin   or cos 
or tan 

2
2
6
3
AG
r=
4 sin

 3
cos 2 
6
6 2
AG
6
M1A1
M1
A1 cso
A1cso
So many ways x may be expressing e.g.
2
2 sin 2 cos , sin 2(1 + cos 2), sin 2 + (1/2) sin 4
dx
leading to many results for d
Some relevant equations in solving
2
2
2
[(1 – 4 sin ) = 0, (4 cos  – 3) = 0, (1 – 3 tan ) = 0, cos 3 = 0]

dx

0
6 satisfies d
Showing that
, allow M1 A1
dx
providing d correct
Starting with x = r sin  can gain M0M1M1
A

4

4


1
1
r 2 d  .16 sin 2  cos 4 d
2
2 
(b)
6
2
6
4
2
2
2
2
2
8 sin  cos  = 2 cos (4 sin  cos ) = 2 cos  sin 2
2
= (cos2 + 1)sin 2
1  cos 4
cos 2 sin 2 2 
2
= Answer
AG
City of London Academy
M1
M1
A1 cso
17
3
First M1 for use of double angle formula for sin 2A
2
Second M1 for use of cos 2A = 2 cos A – 1
Answer given: must be intermediate step, as shown, and
no incorrect work
 
 
(c)
Area =
 sin 4   4 
1
3
 6 sin 2  2  8    

 
6
ignore limits
2 

sin


  sin    1 3  
1
3 
 sin 3  
sin
 

2 8
8  6
3 12
8 
6



 (sub. limits)
1  
   
6 8
 3 
3  1


 16  12  16   6 ,  24


both cao
5
M1A1
M1
A1, A1

b sin 4
2
For first M, of the form
(Allow if two of correct form)
On ePen the order of the As in answer is as written
a sin 3 2 
[14]
8.
1
2
Use of
r
2
d
B1


Limits are 8 and 4
B1
2
2
2
16a cos 2 = 8a (l + cos 4)
M1
(1  cos 4 )d  
sin 4
4
M1 A1

sin 4   4

A 4a  
4   8

2
   

a 2  4     (0  1)
 4 8

City of London Academy
18
M1
  1
a 2   1  a 2 (  2)
2  2
cso
A1
[7]
9.
(a)
2
(i)
2
2
2
2
2
2
r sin θ = a cos 2θsin  = a (1 – 2 sin ) sin 
2
2
B1
1
4
(= a (sin θ – 2 sin θ))
d
d (a2(sin2  – 2sin4 )) = a2(2sin  cos  – 8 sin2  cos ), = 0
M1, A1, M1
(ii)
2
2
2 = 8 sin 
(Proceed to a sin  = b)
1
sin θ = 2
(b)
a2
2
a


6
θ=
, r= 2

a2
cos2 d  4
2

 ...  4  a4
6
M1
sin 2
  3
1  2 
A1, A1 cso
1 r 2 d
M: Attempt 2
, to get k sin 2θ M1 A1

M: Using correct limits
M1 A1
1  a . 1   a   3   3a 2


16
 = 2  2 2  2 2 
M: Full method for rectangle or triangle
 3a 2  a 2
4
R = 16
6
  3 a 2
1  2   16
(33 – 4)
M: Subtracting, either way round
M1 A1
dM1 A1 cso
8
[15]
2
(a)
(ii)
(b)
Final M mark is dependent on the first and third M’s.
2
First A1: Correct derivative of a correct expression for r sin θ or r sin θ.
Attempts at the triangle area by integration: a full method is required for M1.
2
Missing a factors: (or a ) Maximum one mark penalty in the question.
City of London Academy
19
10.
(a)
P
R
N
O
= 0
Q
3a
4a(1 + cos) = cos 
or
M1
or
A1
or
M1
2
4cos  + 4cos – 3 = 0
(2cos – 1)(2cos + 3) = 0
1
cos = 2 ,


  
3

or
A1
3a 

1  
r 
r = 4a 
2
2
r – 4ar – 12a = 0
(r – 6a)(r + 2a) = 0
r = 6a
Note ON = 3a

PQ = 2 × ON tan 6 = 63a (*)
cso M1 A1
6
2
2
2
or PQ = 2× [(6a) – (3a) ] = 2(27a ) = 63a (*)
cso
or any complete equivalent
(b)
1
2× 2


3
0
...
r 2 d ... 16 a 2

...
2
(1 + cos) d
M1
r
2
d
1
...
1

1  2 cos  2  2 cos 2 d
=… 
...
M1
2
cos   cos2
City of London Academy
20
1
3

 2   2 sin   4 sin 2 

= …
A1

3
  3

8 
22
2
2
= 16a
(= 2a [4 + 93]  56.3a )
M1 A1

use of their 3 for M1
1
6 3
2
Area of POQ = 2
a × 3a or 9a 3
B1
2
R = a (8 + 93)
cao A1
7
[13]
11.
(a)
For C: Using polar/ Cartesian relationships to form Cartesian equation
M1
2
2
so x + y = 6x
A1
2
2
[Equation in any form: e.g. (x – 3) + y = 9 from sketch.
6x
x2  y2 
2
x  y2
or
]


r cos    3
3

For D:
and attempt to expand
M1
3y
x

2
2 = 3 (any form)
M1A1
5
(b)
City of London Academy
21
“Circle”, symmetric in initial line passing through pole
B1
Straight line
B1
Both 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 line
M1
Using correctly in Cartesian equation for straight line
M1
Correct Cartesian equation
City of London Academy
22
A1
(c)
Cartesian: Eliminate x or y to form quadratic in one variable
M1
2
2
[2x – 15x + 18 = 0, 4y – 6 3 y = 0]
Solve to find values of x or y
M1
Substitute to find values of other variable

3
3 3
 x  or 6; y 0 or

2
2 

B1A1

Points must be (6, 0) and (3, 3 )
B1A1
12.
(a)
3a(1 – cos) = a(1 + cos)
M1
1


2a = 4a cos  cos = 2  = 3 or – 3
M1
3a
r= 2
A1A1
4
[Co-ordinates of points are
(b)
( 32a , 3 ) and ( 32a ,  3 ) ]
3a 3
AB = 2rsin = 2
M1A1
2

3

1
 2
2
Area = 3 r d
1
[a 2 (1  cos ) 2  9a 2 (1  cos  ) 2 ]d
2
=

City of London Academy
23
M1 M1
2
a
= 2
[1  2 cos  cos
2
  9(1  2 cos  cos 2  )]d
A1
2
a
[ 8  20 cos   8 cos 2  )]d
2
=
= 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)
2
= a [–4 + 10 3 –
3 ] or = a2[–4 + 9 3 ] or 3.022a2
A1
7
(d)
3
2
3a
= 4.5  a =
3
B1
2
Area = 3[9 3 – 4], = 9.07 cm
M1, A1
3
[16]
3
13.
(a)
Shape + horiz. axis
B1
3
B1
City of London Academy
24
2
(b)
Area =
1
2
2
 r d
1
2
= 2  9 cos 2 d
M1
9 cos 4  1
d
2
= 2
M1
use of

1
2
r
2
2
use of cos4 = 2cos 2 – 1

9  sin 4   4
 
2  8
2
6

=
M1, A1
9 
3  

 

2  8 16 12 
=
M1
9
3
 

2  24 16 
=
or 0.103
A1
6
(c)
subst.

4
and

6
r sin  = 3 sin  cos 2
dy 
d = 3 cos  cos 2 – 6 sin  sin 2
(diff. r sin )
M1, A1
dy
dy
2
2
d = 0  6 cos  – 3 cos  – 12 sin  cos  = 0
use of d = 0
M1
2
2
6 cos  – 3 cos  – 12(1 – cos  )cos  = 0
use double angle formula
M1
3
18 cos  – 15 cos  = 0
solving
M1
2
cos  = 0 or cos  =
A1
\ r = 3(2 ×
=2
 r sin  = 2
M1
2 6
 d= 3
A1
8
5
6
5
6
2
or tan  =
1
5
2
or sin  =
1
6
)–1
1
6
use of d = 2r sin 
[16]
City of London Academy
25
14.
(a)
(5a, 0), (3a, 0)
B1, B1
2
allow on diagram
B1 only if no zeros or diagram with initial line
(b)
3 + 2 cos  = 5 – 2 cos 
M1
cos  =
1
2
M1
 5
,
= 3 3
A1

both, accept – 3

5
3
(4a, ) , (4a, 3 )
A1
4
(c)
Allow 2 M1s below for method shown in either calculation
2
2
(5 – 2cos) d
(3 + 2cos) d
2
2
= (25 – 25cos + 4cos )d
= (9 + 12cos + 4cos )d
M1
= (27 – 20cos + 2cos2)d
= (11 + 12cos + 2cos2)d
M1
= 27 – 20sin + sin2,
= 11 + 12sin + sin2
A1, A1
A=2×
1
2
2
2
a (5 – 2 cos) d + 2 ×
1
2
2
2
a (3 + 2cos) d adds
M1

3
2 0

=a
(5  2 cos ) 2 d  a 2

(3  2 cos )

3
2
d
correct limits
A1


3
3
2
= a [27 × 3 – 103 + 2 ] + a [11( – 3 ) – 63 – 2 ]
2
dM1
City of London Academy
26
a2
= 3 (49 – 483] (*)
A1 cso
8
[14]
15.
(a)
1 2
a 1  cos 2   2 cos d
2
M1 A1
correct with limits
1 2
cos 2  1
a  1
 2 cos d
2
= 2 
M1 A1
1
2
=2× 2a

sin 2 


  4  2  2 sin  

0
A1
 3  3a 2

2
=a  2  = 2
A1
6
(b)
2
x = a cos  + a cos 
r cos 
M1
dx
d = a sin   2a cos  sin 
A1
dx
1
d = 0  cos  =  2
finding 
M1
2
4
 = 3 or  = 3
City of London Academy
27
a
a
r = 2 or r = 2
finding r
M1
2
a
A: r = 2 ,  = 3
a
 2
B: r = 2 ,  = 3
both A and B
5
(c)
1
x=4
a
 WX = 2a +
1
4
A1
a
1
=24
a
M1 A1
2
(d)
27 3a 2
8
WXYZ =
B1 ft
1
(e)
3 100
27 3
2
2
Area = 8  100 
= 113.3 cm
M1 A1
2
[16]
16.
(a)
A
O
closed loop
City of London Academy
l
B1
28
symmetry in l
(b)
Area of loop =
B1
2
1
2



= 4






2
r2


2
0

2
d =
1
2
2
 4 cos 3 



2
d
M1

 2
3
 (1  sin 2  ) cos
cos 

d = 4  0
d
M1 A1


sin 3   2
sin




3 0
= 4
= 4(1 –
(c)
1
3
) =
M1 A1
8
3
M1 A1
7

y = r sin  = 2 sin  cos  and the value of  in (0, 2 ) for
dy
which d = 0 is required
3
2
dy
5
1
d = 2 cos 2   3sin 2  cos 2 
5
5
M1 A1
1
2
2
2
= 2 cos   3 cos   3 cos 
M1
1
2
2
= cos  (5 cos   3)

cos  ¹0 in (0, 2 ), so cos  =
1
2
A1
3
5
At A,  = 0.68, r = 1.36 (2 decimal places)
M1
A1, A1
7
[16]
1. No Report available for this question.
City of London Academy
29
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


5

from 0 to 2 minus the area from 0 to 18 minus the area from 18 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 was
well 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.
1 2
r d
Most candidates applied the formula 2
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
2
by candidates were algebraic; 6a sin θ was sometimes missed out when (a + 3cos θ) was
2
expanded. Sometimes a was “taken out” from some candidates’ integral. Most candidates knew
91  cos 2 
2
2
they needed to substitute
for 9cos θ in order to integrate the expression but

1
sometimes there were errors in dealing with the 9 and the 2 . There were a significant minority
2
of candidates who did not know the correct strategy to apply in order to integrate cos 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.
dx
This was dximplied as the tangent was perpendicular to the initial line and so d = 0. A
number of candidates chose to consider r sin θ instead and little credit was given for this.
1
(1  2 cos   cos 2  )d
In part (b) most realised that they needed to find 2
. There were a
number of sign slips but the methods were understood, and most tried to use double angle


and
3 and to subtract. This
formulae. To find the shaded area they needed to use the limits 2
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

City of London Academy
30
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 to
score 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 would
be exact. Even candidates who had a correct solution for one value of θ (1.28) were often unable
to 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
1
cos 2   (cos 2  1)
3
cos

2
squaring (5 +
), sign or numerical errors in applying
and
1
integration slips. Sometimes the limits for integration were wrong or the 2 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.
1 2
r d
Candidates were on the whole correct in the use of 2
and many of them were successful

2
in negotiating the expansion of (3 + 2cos θ) 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
2
needed was (16πa )/3. It was pleasing to note how many candidates made it correctly to the
76 38
to
3 .
final answer although some found it unnecessary to simplify to 6
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
3
2
2
rearrange r sin θ cos θ, using either sin θ + cos θ = 1, or a double angle formula or a
City of London Academy
31
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
1  cos 4
4
2
it was not uncommon to see incorrect double angle formulae; cos θ =
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
1
3
as 3 sin 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

4
9.

8
and
.
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
2
though they knew that cos 2 = 1 –2 sin .
Most knew in part (a)(ii) that they had to deal with rsin , but made it more difficult for
2
2
themselves by trying to differentiate a cos 2 sin  rather than a (1 – 2sin ) 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.
sin 2   2 sin 4  sin  
2 sin 2 
.
Completely correct solutions to part (b) were rare. While better candidates did realise that they
1 a
  a


cos  
sin 
2
6  2
6
had to use the area of the triangle  2
, there was much confusion over
City of London Academy
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1 2
a cos 2 d 
which limits to use for the integral 2
, and which area this represented. A


common mistake was to use 2 instead of 4 as the upper limit, even though the curve was

0  
4.
defined only for

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
 
6a tan  
 3  , but many gave a convincing proof. A sizeable minority of candidates made
PQ =
the unwarranted assumption in part (a) that the tangent at P was parallel to the initial line, whilst
this 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.
2
2
2
In part (b) most tried integrating 16a (1 + cos θ) and knew how to deal with the cos θ by using
the 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 r  6 cos  was
recognised, or plotted, as the correct circle in part (b) but its cartesian equation had often not

r  3 sec(   )
3
been found in part (a). Recognition of
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

cos(   ) 
3
generous first mark, but solving the resulting equation cos 
½ was generally not
well done; probably a sign that confidence had taken a knock in the earlier parts. Most

cos(   )
2
3
candidates expanded
to give cos   3 sin  cos 1 ; those who progressed
2
further to 3 sin  cos   sin  or 2 sin( 2   ) 1, usually completed the solution,
although cancelling sin  in the former case was quite common. A neat solution was to use the


cos
 cos(2  ) 1.
3
3
factor formulae to give
City of London Academy
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

 ) 
cos(   ) 
3
3
It was very disappointing to see “ cos 
½  cos  = ½ or
½”,
even though, more disappointingly in this case, they gave the correct answers !
cos(
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
cos 4  1
d
( r sin  ) 0
2
errors such as
. In part (c) many candidates who solved d
ignored the
instruction 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


 4a ,  
3  instead of
out of range, giving 
5

 4a ,
3



 . Part (c) is a very demanding question.
Many could gain some marks by showing that they were able to integrate
5
2 cos  
 3  2 cos  
2
and
2
but 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
City of London Academy
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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
3a
2 or
3a
4 and many candidates used WX = a + the projection rather than using 2a
given as
+ projection. In (d) a minority of students used 3a rather than 3a .
16. No Report available for this question.
15.
W
X
A
in itia l lin e
B
O
Z
Y
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)
3 3a
Given that the length of WZ is 2 ,
(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)
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16.
The curve C is given by
r=
3
2 cos 2


 
, 2
2,
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)
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