Question 3:
--- 3(a) ---
3(a) | d x
x 3 s i n 3 9 s i n 2 c o s    =  = o.e.
d  | B1
d y 2 s2 i n 2  −
=
d x 9 s i n c o s   | M1
dy −4sincos 4
= =− cosec
dx 9sin2cos 9 | A1
(3)
(b) | d y 8 1 3 1   
" " , x 3 . .. , y 1 . ..  =  = − = = = + =
6 d x 9 2 2 | M1
3 8  3 
y − = − x −
2 9 8 | dM1
16x+18y−33=0 | A1
(3)
(c) | x 23
e.g. x 3 s i n 3 s i n 3   =  =  y 1 c o s 2 1 1 2 s i n 2 2 2 x   = + = + − = −  
3 3 | M1
23 23
 x   x 
y = 2 − 2  2 = 2 − y  8 x 2 = 9 ( 2 − y ) 3 *
3 3 | A1*
q = 3 | B1
(3)
Alt(c) | e.g. 8 x 2 7 2 s i n 6 , 9 ( 2 y ) 3 9 ( 2 c o s 2 1 ) 3 9 ( 1 ( 1 2 s i n 2 ) ) 3    = − = − − = − − | M1
7 2 s i n 6 9 8 s i n 6 8 x 2 9 ( 2 y ) 3 *   =   = − | A1*
q = 3 | B1
Total 9
Alt (a) Implicit differentiation – note that some candidates may find the cartesian equation
and differentiate implicitly or explicitly.
B1: A correctly differentiated cartesian equation. (Do not be concerned with where the cartesian
equation has come from for this mark)
e.g. 1 6 x = 9  3  ( − 1 )  ( 2 − y ) 2 
d
d
y
x
M1: Substitutes in for x and y into their differentiated equation and rearranges to achieve an
expression for
d
d
y
x
in terms of . Condone poor differentiation.
A1: Achieves
d
d
y
x
4
9
c o s e c  = − with no errors but condone invisible brackets and poor notation
being recovered. Allow equivalent fractions to −
4
9
. They must have shown correctly how
they found the cartesian equation. i.e. they cannot just use the given result in (c).
(b)
M1: Attempts to find
d
d
y
x
, x and y when
6

 = . Must achieve values for all three which may be
implied by further work. Condone slips but if values are only stated then their
d
d
y
x
must be
correct for their k (look for 2 their k) and one of x or y must be correct.
Note that they may use their
d
d
y
x
in terms of x and y

= " −
2 7
1
( 2
6
−
x
y ) 2
"

by either finding the
differentiated equation in (a) or by converting in (b).
dM1: Complete method for the tangent using their values. It is dependent on the previous method
mark. Condone one sign error when substituting in the coordinates. If they use y=mx+c
they must proceed as far as c = ...
A1: 1 6 x + 1 8 y − 3 3 = 0
PMT
or any integer multiple of this equation. Must have = 0.
(c)
M1: A complete method to eliminate θ. Any identities used must be correct when substituted in.
Do not allow the parametric equations to be expressed as θ = and equated. May also be
awarded if seen in (a).
A1*: Correct proof with no errors including brackets although condone missing trailing brackets if
it does not affect the expression. Must be seen in (c).
B1: Correct value of q. Do not accept x=3 Can be scored for the domain −3„ x„ 3 which must
be in terms of x
------------------------------------------------------------------------------------------------------------------------
Alt(c)
M1: A complete method to express lhs and rhs both in terms of sin θ only or both in terms cos 2θ
only. Any identities used must be correct when substituted in. They cannot proceed from
9(2−y)3 =72sin6 without an intermediate stage but condone 8x2 =72sin6
A1*: Fully correct work showing lhs = rhs and gives a conclusion. There should be no errors
including brackets but condone missing trailing brackets if it does not affect the expression.
B1: As above in the notes for (c)