Mathematics · Oct / Nov 2023 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. y 2 C O k x Figure 3 Figure 3 shows a sketch of the curve C with parametric equations π x 6 t 3sin 2 t y 2cos t 0 t 2 The curve meets the y-axis at 2 and the x-axis at k, where k is a constant.
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8(a)[1]
3π
8(b)[4]
d
y
−
2sin
t
=
d
x
6
−
6cos2
t
−
2sin
t
−
2sin
t
1 cosec
t
=
=
= −
(
)
6
6
−
6 1 2sin
−
2
t
12sin
2
t
8(c)[3]
P = 3
π
−
3,
2
2
2
3
π
y
−
2
= −
0
−
−
3
y
=
...
E.g.
6
2
or
2 3
π
2
3
c
c
...
= −
−
+
=
6
2
π
2
2
+
o.e
4
2
(d)(i)
d
x
(
)
2
2
Attempts
y
t =
4cos
t
6
−
6cos2
t
d
(
)(
)
(
)
2
2cos2
t
2
6
6cos2
t
=
+
−
=
6 2
−
2cos 2
t
=
6 2
−
1 cos4
+
t
π
π
2
2
d d
x
(
)
2
Volume =
π
y
t
=
6
π
1 cos4 d
−
t
t
d
t
0
0
(ii)
π
6 sin 4
π
t
2
2
3π
6
π
t
−
,
=
4
0
(a)
B1: For 3π. Condone (
)
3
π
0
and allow x = 3π. Remember to check the diagram as the 3π may appear there.
Do not allow decimals here.
(b)
x by using d
y
d
x
M1: Attempts to differentiate x(t) and y(t) and calculates d
y
÷
.
d
d
t
d
t
Condone poor differentiation but one of the terms must be of the correct form so it requires
x
→
α
+
β
cos2
t
or
y
→
A
sin
t
A1: d
y
−
2sin
t
d
y
−
2sin
t
or equivalent e.g.
=
=
(
)
8(d)[6]
x
6
−
6cos2
t
d
x
6
6 cos
2
sin
2
−
t
−
t
x = α
cosec
t
d
Condone sign errors only with any identities that are used.
A1: d
y
1 cosec
x = −
t
d
6
d
y
2sin
t
−
You can condone poor notation along the way e.g.
=
as long as the intention is clear.
(
)
d
x
6
−
6 cos
2
−
sin
2
−
2sin
t
−
2sin
2
t
−
2sin
2
t
An alternative for the final M1A1:
=
λ
cosec
t
λ
=
=
6
−
6cos2
t
6
−
6cos2
t
12sin
2
t
(c)
B1: At P, x = 3
π−
3,
y
=
2
or exact equivalents
2
May be implied by being seen embedded in their tangent equation.
M1: Full method of finding the y coordinate of N
Requires:
t = πinto their d
y
cosec
substitution of= α
t
to find gradient of tangent
4
d
x
use of their 3π
3,
2
−
(which may be inexact) in the formation of the equation of the tangent or the
2
substitution of these values into y = mx + c with the coordinates correctly placed.
the setting of x = 0 as well as finding the y value or rearranging to find “c”π
1
2
2
π
+
A1:
+
or exact simplified equivalent. E.g.
.
4
2
2 2
2
Does not need to be identified as a value for y so just look for the correct exact expression (may be c = …).
2 d
x
(d) (i) Note that the first 3 marks only involve the consideration of
y
t i.e. not an integral.
d
2
2
Condone poor notation e.g.
cost for
cos t as long as it is recovered and the intention is clear.
2 d
x
t with their d
x
M1: Attempts
y
t condoning slips
d
d
d
x
(
)
oe. Allow e.g. (
) (
)
2
2
A1:
y
4cos
2
t
6
6cos2
t
2cos
6
6cos2
with both sets of brackets present unless
t =
−
t
−
t
d
they are implied by subsequent work.
dM1: A full method to form an expression in terms of cos4 .t
Condone sign errors only with any identities that are used. Depends on the first method mark.
There are many ways of achieving the result so you will need to check their work carefully.
(
)
2
2
E.g.:
4cos
2
t
6
−
6cos2
t
=
4cos
2
t
12sin
t
=
12sin 2
t
=
6
−
6cos4
t
or
(
)
2
2
4
4cos
2
t
6
−
6cos2
t
=
4cos
2
t
12sin
t
=
48cos
t
−
48cos
t
2
2
2
2
2
4
2
cos4
t
=
cos 2
t
−
sin 2
t
=
2cos
t
−
1
−
4sin
t
cos
2
t
=
8cos
t
−
8cos
t
+
1
(
)
2
4
48cos
t
−
48cos
t
=
6 1 cos4
−
t
2
(
)
A1: Volume =
6
π
1 cos4
−
t
d
t
. All correct with correct limits and including the “dt”.
0
(d)(ii)
(
)
sin 4
t
M1: For
β
1 cos4
−
t
d
t
→
β
t
−
.
Allow with their β, a made up β or the letter β.
4
2
3π following full marks in (d)(i) and correct integration and correct limits.
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