Mathematics · May / June 2023 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The temperature, θ °C, of a car engine, t minutes after the engine is turned off, is modelled by the differential equation d 2 k 15 t d where k is a constant. Given that the temperature of the car engine • is 85 °C at the instant the engine is turned off • is 40 °C exactly 10 minutes after the engine is turned off
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[7]
d
θ
d
θ
(
)
2
t = −
k
θ
−
15
=
−
k t
d
(
)
d
θ
−
15
2
d
θ
(
15
)
−
1
= −
θ
−
(
)
2
θ
−
15
1
−
= −
kt
+
c
θ
−
15
1
t
0,
85
c
=
θ
=
−
=
70
1
1
9
t
=
10,
θ
=
40
=
10
k
+
k
=
...
25
70
3500
1
9
t
1
=
+
θ
=
...
θ
−
15
3500
70
135
t
+
4250
θ
=
9
t
+
50
6(b)[2]
135
t
+
4250
20
=
=
t
...
9
t
+
50
t =
awrt 72
(a) Note that candidates cannot work backwards from the answer using differentiation
B1: Correct separation of variables.
The integral signs do not have to be present, but the dθ and dt do, and be in the correct positions
1
α
M1: Integrates
d
θ
to
(
)
2
θ
−
15
θ
−
15
A1: Correct equation including a k and another different constant.
M1: Uses
t
=
0,
θ
=
85
to find “c”. May be awarded after incorrect integration. May be also awarded if the
constant k had been assigned a value using given values of θ and t .
Their initial equation may have been (incorrectly) adapted but it must be solvable with
t
=
0,
θ
=
85
used leading to c =...
M1: Uses t = 10 and θ = 40 with their value of ''c'' to find k.
May be awarded after incorrect integration but there must have been two constants. Their initial
equation may have been adapted but it must be solvable with t = 10, θ = 40 AND their c =...
M1: Rearranges using correct algebra to obtain θ in terms of t.
For this to be awarded the integral must be in the correct form and both constants found using a
correct method . Condone slips in working but the overall process should be sound.
1
ct
+
d
Look for
=
α
t
+
β
θ
=
θ
−
15
et
+
f
297500
+
9450
t
A1: Correct expression. Allow integer multiples of the given answer E.g.
θ
=
3500
+
630
t
(b)
M1: Uses their answer to part (a) or equivalent (or possibly an earlier equation) to find t when θ = 20.
1
For this to be awarded, substitute
θ
=
20
into a ''correct'' form such as
=
α
t
+
β
=
θ
−
15
ct
d
+
Or substitute
θ
=
20
into a ''correct'' form such as
θ
=
=
t
...
where , , ,
c d e f ≠
0
et
+
f
They cannot just make values up for the c, d, e and f for example. It must follow their work in (a)
A1: Correct value which must have come from a correct equation. Exact value is 650
9 but allow awrt 72
Of course candidates can move the ''k'' or ''−k'' term over to the LHS or use limits .
6(a)
d
θ
d
θ
(
)
2
= −
k
θ
−
15
=
d
t
d
t
(
)
2
−
k
θ
−
15
d
θ
(
)
−
1
=
αθ
−
15
(
15
)
2
θ
−
1
= +
t
β
(
)
k
θ
−
15
1
t
=
0,
θ
=
85
=
β
70
k
1
9
50
t
=
10,
θ
=
40
=
β
+
10
k
=
...
β
=
...
25
k
3500
9
3500
50
= +
t
θ
=
...
(
)
9
θ
−
15
9
135
t
+
4250
θ
=
9
t
+
50
6(a)
d
θ
d
θ
(
)
2
t = −
k
θ
−
15
=
−
k t
d
d
(
15
)
2
θ
−
d
θ
(
)
−
1
= −
θ
−
15
(
)
θ
−
15
2
40
40
1
1
[
]
[]
Correct equation using limits
kt
10
kt
10
−
= −
Or
=
θ
−
15
0
θ
−
15
0
85
85
1
1
9
10
k
k
...
Correct attempt to find k
−
+
= −
=
25
70
3500
θ
1
[
]0
Correct attempt to find equation linking θ and t E.g.
kt
t
−
= −
θ
−
15
85
1
1
9
t
−
+
= −
θ
=
...
θ
−
15
70
3500
135
t
+
4250
θ
=
9
t
+
50
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