Mathematics · Oct / Nov 2022 · Paper 41 · Question 10
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As printed in the paper
The question as text
10. A spherical ball of ice of radius 12 cm is placed in a bucket of water. In a model of the situation, • the ball remains spherical as it melts • t minutes after the ball of ice is placed in the bucket, its radius is r cm • the rate of decrease of the radius of the ball of ice is inversely proportional to the square of the radius • the radius of the ball of ice is 6 cm after 15 minutes Using the model and the information given,
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10(a)[5]
d
r
1
d
r
k
∝±
= ±
States or uses
or
t
t
d
2
d
2
r
r
r
k
d
= ±
r
d
r
=
±
k t
d
2
d
t
r
2
1
3
3 r
= ±
kt
+
c
(
)
t
r
c
Substitutes
=
0,
=
12
=
576
(
)
(
)
t
=
r
=
c
=
k
= ±
AND
15,
6 and their
576
33.6
1
3
3
3 r
= −
t
+
or
r
= −
100.8
t
+
1728
o.e.
33.6
576
10(b)[2]
Sets
r
=
−
t
+
=
0
100.8
1728
0
120
7 minutes or awrt 17.1 minutes
10(c)[1]
12
O
(a)
r
r
k
r
d
1
d
d
1
∝±
= ±
= ±
Accept k λ or other constant. Condone
B1: States or uses
,
or
t
t
t
d
d
d
r
2
r
2
kr
2
r
d
1
= ±
t
d
r
2
(
)
3
M1: Integrates to achieve an expression of the correct form
...
r
=
...
t
+
c
o.e
There is no requirement for + c
1
3
A1: Correct integration with two different unknown constants. E.g. look for
3 r
= ±
kt
+ o.e
c
r
d
1
1
r
3
t
c
So if for example
= −
= −+ the score would be B1, M1, A0 then dM0, A0
t
d
r
2
3
M1: Uses both boundary conditions to form an equation involving r and t. For this to be awarded
there must be two different unknown constants that are initially correctly placed.
nr
= ±
k t
+
c
It is dependent upon having achieved ...
'' ''
'' ''
which may have been achieved from
an
d
r
t = ±
kr
2
incorrect assumption. E.g.
d
1
3
3 r
3
= −
t
+
r
= −
t
+
A1: Correct equation. E.g.
33.6
576
or
100.8
1728
o.e. such as
5
120
3
t
= −
r
+
504
7
(b)
M1: Sets
r
=
−
t
+
=
=
t
. Follow through on their equation. Condone if this
0
100.8
1728
0
...
produces a negative value for t. Alt sets V = 0 to find t = ...
A1: Awrt 17.1 minutes. Must include the units. Also allow 17 minutes 8 seconds or 17 minutes 9
seconds
(c)
B1: For the correct shaped curve only in quadrant 1 starting at (0,12), ignoring value of t.
d
r
k
= −
As
the gradient should appear to get increasingly steeper.
d
t
r
2
If there are two curves given on the axes, both need to be correct for this mark to be awarded
Do not be concerned if it is not infinite at the t axis.
The curve on the left would be at the limit of what we would
allow
as the gradient on the left hand side does appear to increase.
We condone its appearance as it approaches the t axis.
It does not get less steep
................................................................................................................................................................
..
d
r
k
4
d
V
3
2
= ±
Note that it is possible to use
V
=
π
r
=
π
r
with
to set up and solve
4
r
t
3
d
d
2
r
π
d
V
or d
V
672
t = ±π
k
t = ±
β
In this case the equation becomes
V
=
2304
π
−
t
4
d
d
5
(a)
B1: States or uses either d
V
or its exact value which is d
V
2016
π
β
t = ±
t = −
o.e such as
d
d
15
d
V
672
π
t = −
.
d
5
V
M1: For d
+ and attempts to find the values of β and c using the given
t = ±
β
V =
±
βt
c
d
conditions.
3
It is dependent upon knowing that
V
=α
r
and using this to find V at r = 6 and 12 with t = 15
and t = 0
672
π
V
=
π
−
t
A1:
2304
5
3
V
=α
r
3r with t.
M1: Substitutes
to form an equation linking
It is dependent on the previous M1 in this method
π
4
672
3
A1: Achieves
π
r
=
2304
π
−
t
3
5
(b)
(
)
3
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