Mathematics · May / June 2022 · Paper 41 · Question 4

YMA01

As printed in the paper

The question as text

4. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A curve has equation 3 – 9kx 2y + 8y 3 = 875 16x where k is a constant.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

4(a)[4]
y
=
(oe in correct form)
5(1
−
x
)
3(a)
d
A
t = −
0.5
d
d
A
A
=
π
x
2
=
2
π
x
d
x
d
x
d
A
d
A
" 0.5"
−
−
1
=
÷
=
=
d
t
d
t
d
x
"2
π
x
"
4 x
π
d
x
t = −
0.011368...
d
V
=
π
x
2
(3 )
x
=
3
π
x
3
d
V
=
9
π
x
2
d
x
d
V
d
V
d
x
1
(
)
=
×
=
9
π
x
2
×−
"
"
= −
2.25
x
d
t
d
x
d
t
4
π
x
d
V
(
)
= −
9
(Rate of decrease
=
) 9
mm s−
3
1
d
t
Notes
(a)
d
A
B1:
t = −
0.5
seen or implied from working
d
d
A
B1:
=
2
π
x
seen or implied from working. Must be in terms of x, but allow recovery if in terms of r and
d
x
later work uses r = 7 to achieve a solution.
Attempts to use an appropriate chain rule with their d
A
and d
A
e.g. d
x
d
A
d
A
M1:
=
÷
=
...
d
t
d
x
d
t
d
t
d
x
1
awrt
−
0.0114
or
cso (must have the negative sign)
A1:
−
28π
(b)
V
x
2(3 )
x
or
V
=
3
π
x
3
B1:
=
π
d
V
=
9
π
x
2
or ft from their equation for V in one variable
B1ft:
d
x
. Note the d
x
Their d
V
d
x
×
their
must be in terms of x or with x = 4 substituted first, M0 if they use
M1:
d
x
d
t
d
t
their answer to (a).
(Rate of decrease = ) 9 (
)
A1:
mm s−
3
1
(with or without the negative sign). May be scored following
d
A
t =
0.5
in part (a)
d
4(a)
16
x
3
−
9
kx y
2
+
8
y
3
=
875
d
y
( )
8
y
3
(
8
)
3
y
2
→
×
d
x
2 d
y
−
9
kx y
2
→
...
kxy
±
...
−
9
kx
d
x
d
y
d
y
d
y
d
y
48
x
2
−
18
kxy
−
9
kx
2
+
24
y
2
=
0
(24
y
2
−
9
kx
2
)
=
18
kxy
−
48
x
2
=
...
d
x
d
x
d
x
d
x
d
y
6
kxy
−
16
x
2
*
=
d
x
8
y
2
−
3
kx
2
(b)
5
5
2
6
k
y
−
16
d
y
5
2
2
=
0,
x
=
=
0
or
d
2
2
x
5
8
y
2
−
3
k
2
3
2
5
5
5
3
x
=
16
−
9
k
y
+
8
y
=
875
2
2
2
225
3
15ky −
100
=
0
or
2
50
−
4 ky
+
8
y =
87
5
5
3
5
2
20
20
3
64
E.g.
16
9
k
8
875
k
3
...
k
...
−
+
=
=
=
=
2
2
3
k
3
k
27
4
k =
3
Notes
(a)
2 d
y
B1:
For
y
3
→
3
y
. Allow if seen in aside working without the 8.
d
x
M1:
Correct attempt at implicit differentiation on the
−
9kx y
2
. Look for
−
9
kx y
2
→
...
kxy
±
...
−
9
kx
Collects both of their d
x terms together, collects non d
y
y
M1:
x terms the other side of the equation, factorises
d
d
and divides to achieve d
y
Must have two d
y
x =
...
terms, one from the attempt at differentiating
d
d
x
y , but condone if an extra d
y
and one from the attempt at differentiating
3
x =
...
term has been included.
d
d
y
6
kxy
−
16
x
2
A1*:
Achieves
with no errors
=
d
x
8
y
2
−
3
kx
2
(b)
M1: Uses the information to produce one equation in k and y, e.g. sets the d
y
equal to 0 and substitutes x = 5
d
x
5
or substitutes
x =
into the given equation. Allow one slip substituting.
2
A correct equation without fraction and with simplified coefficients, so
15ky −
100
=
0
oe or
A1:
225
2
50
−
4 ky
+
8
y =
3
87
5
oe
M1: For a complete method to find k so solves the equations simultaneously to achieve a value for k. May find y
20
first e.g substitutes their
into the original equation, solves to find y and substitutes this back into
k
=
3
y
20
20
to find k via
250
−
375 8
+
y
3
=
875
y
=
5
k
=
=
...
k
=
3
y
3 5
×
4
k =
A1
3
Alt:
5
x =
If they do not substitute
initially then score
2
M1: Uses numerator of d
y
equal to 0 to find y in terms of x and k and substitute into original equation (allowing
d
x
one slip)
A1: Correct equation:
3
8
x
2
8
x
2
8
x
2
6
kxy
−
16
x
2
=
0
y
=
16
x
3
−
9
kx
2
+
8
=
875
oe
3
k
x
3
kx
3
kx
5
M1: Substitutes
x =
and solves to find k
2
4
A1:
k =
3
4(b)[4]
V
=
π
x
2
(3 )
x
=
3
π
x
3
d
V
=
9
π
x
2
d
x
d
V
d
V
d
x
1
(
)
=
×
=
9
π
x
2
×−
"
"
= −
2.25
x
d
t
d
x
d
t
4
π
x
d
V
(
)
= −
9
(Rate of decrease
=
) 9
mm s−
3
1
d
t
Notes
(a)
d
A
B1:
t = −
0.5
seen or implied from working
d
d
A
B1:
=
2
π
x
seen or implied from working. Must be in terms of x, but allow recovery if in terms of r and
d
x
later work uses r = 7 to achieve a solution.
Attempts to use an appropriate chain rule with their d
A
and d
A
e.g. d
x
d
A
d
A
M1:
=
÷
=
...
d
t
d
x
d
t
d
t
d
x
1
awrt
−
0.0114
or
cso (must have the negative sign)
A1:
−
28π
(b)
V
x
2(3 )
x
or
V
=
3
π
x
3
B1:
=
π
d
V
=
9
π
x
2
or ft from their equation for V in one variable
B1ft:
d
x
. Note the d
x
Their d
V
d
x
×
their
must be in terms of x or with x = 4 substituted first, M0 if they use
M1:
d
x
d
t
d
t
their answer to (a).
(Rate of decrease = ) 9 (
)
A1:
mm s−
3
1
(with or without the negative sign). May be scored following
d
A
t =
0.5
in part (a)
d
4(a)
16
x
3
−
9
kx y
2
+
8
y
3
=
875
d
y
( )
8
y
3
(
8
)
3
y
2
→
×
d
x
2 d
y
−
9
kx y
2
→
...
kxy
±
...
−
9
kx
d
x
d
y
d
y
d
y
d
y
48
x
2
−
18
kxy
−
9
kx
2
+
24
y
2
=
0
(24
y
2
−
9
kx
2
)
=
18
kxy
−
48
x
2
=
...
d
x
d
x
d
x
d
x
d
y
6
kxy
−
16
x
2
*
=
d
x
8
y
2
−
3
kx
2
(b)
5
5
2
6
k
y
−
16
d
y
5
2
2
=
0,
x
=
=
0
or
d
2
2
x
5
8
y
2
−
3
k
2
3
2
5
5
5
3
x
=
16
−
9
k
y
+
8
y
=
875
2
2
2
225
3
15ky −
100
=
0
or
2
50
−
4 ky
+
8
y =
87
5
5
3
5
2
20
20
3
64
E.g.
16
9
k
8
875
k
3
...
k
...
−
+
=
=
=
=
2
2
3
k
3
k
27
4
k =
3
Notes
(a)
2 d
y
B1:
For
y
3
→
3
y
. Allow if seen in aside working without the 8.
d
x
M1:
Correct attempt at implicit differentiation on the
−
9kx y
2
. Look for
−
9
kx y
2
→
...
kxy
±
...
−
9
kx
Collects both of their d
x terms together, collects non d
y
y
M1:
x terms the other side of the equation, factorises
d
d
and divides to achieve d
y
Must have two d
y
x =
...
terms, one from the attempt at differentiating
d
d
x
y , but condone if an extra d
y
and one from the attempt at differentiating
3
x =
...
term has been included.
d
d
y
6
kxy
−
16
x
2
A1*:
Achieves
with no errors
=
d
x
8
y
2
−
3
kx
2
(b)
M1: Uses the information to produce one equation in k and y, e.g. sets the d
y
equal to 0 and substitutes x = 5
d
x
5
or substitutes
x =
into the given equation. Allow one slip substituting.
2
A correct equation without fraction and with simplified coefficients, so
15ky −
100
=
0
oe or
A1:
225
2
50
−
4 ky
+
8
y =
3
87
5
oe
M1: For a complete method to find k so solves the equations simultaneously to achieve a value for k. May find y
20
first e.g substitutes their
into the original equation, solves to find y and substitutes this back into
k
=
3
y
20
20
to find k via
250
−
375 8
+
y
3
=
875
y
=
5
k
=
=
...
k
=
3
y
3 5
×
4
k =
A1
3
Alt:
5
x =
If they do not substitute
initially then score
2
M1: Uses numerator of d
y
equal to 0 to find y in terms of x and k and substitute into original equation (allowing
d
x
one slip)
A1: Correct equation:
3
8
x
2
8
x
2
8
x
2
6
kxy
−
16
x
2
=
0
y
=
16
x
3
−
9
kx
2
+
8
=
875
oe
3
k
x
3
kx
3
kx
5
M1: Substitutes
x =
and solves to find k
2
4
A1:
k =
3

Practise this paper →

Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.