Mathematics · Oct / Nov 2020 · Paper 41 · Question 9
YMA01
As printed in the paper
The question as text
9. Bacteria are growing on the surface of a dish in a laboratory. The area of the dish, A cm2, covered by the bacteria, t days after the bacteria start to grow, is modelled by the differential equation 3 A A d 2 t > 0 = t t 2 d 5 Given that A = 2.25 when t = 3
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[7]
=
⇒
=
oe
d
t
5
t
2
3
5
t
2
A
2
1
1
−
−
2
A
2
= −
t
−
1
(
+
c
)
5
19
Substitutes
t
=
3,
A
=
2.25
⇒
c
=
...
−
15
2
1
19
30
2
t
Uses their
=
+
⇒
A
=
A
5
t
15
19
t
+
3
2
30
900
9(b)[2]
As
t
→∞
,
A
→
=
or awrt 2.49 cm2
19
361
(a)
∫
d
A
∫
d
t
∫
... A
d
∫
d
t
M1: Attempts to separate the variables either
...
oe such as
=
=
3
3
t
2
t
2
A
A
2
2
3
Condone without the integral signs but the
A , dA,
2
2t and dt must be in the correct positions.
Don't be too concerned on the position of the ''5''
1
−or
M1: Integrates one side correctly. Look for
pA
2
qt −
1
1
−and
dM1: Integrates both sides correctly. Look for
pA
qt −
1
2
A1: Correct intermediate stage (without +c ).
1
1
1
A
−
0.5
1
−
−
Either
−
2
A
= −
t
−
1
(
+
c
)
or
−
10
A
= −
t
−
1
(
+
c
)
or equivalent such as
−
= −
t
2
2
5
0.5
5
M1: Must have a ''+c '' now. It is for using
t
=
3,
A
=
2.25
⇒
c
=
...
.
There must be some evidence for this award which may be implied by their value for c if no working is seen.
One index must have started out correct
This may be awarded after an incorrect change of subject has occurred.
2
pt
from an equation of the form p
q
M1: It is for proceeding to
A
=
+ with correct operations.
r
=
qt
+
r
A
t
Candidates do not need to have found a value for r (their constant) to score this mark
The attempt must involve
using a common factor on the rhs of pq
=
+
r
A
t
''inverting'' both sides not each terman attempt to square both sides and not each term to reach A = ...30
t
2
A1:
A
=
cso. This mark can be awarded independent of the first M1
19
t
+
3
190
t
+
30
(b)
2
2
a
at
M1:
t
→∞
,
A
→
The form of part (a) must be
A
=
where a, b and c are all positive.
b
bt
+
c
c
The reason for this is that if we had ''
bt
−
c
''
, the area would be infinite at
t
=
and a limit would not exist
b
A1: 900
361 or awrt 2.49 cm2 following correct work.
900
Condone for both marks
A <
following correct work.
361
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