Mathematics · Oct / Nov 2024 · Paper 31 · Question 8
YMA01
As printed in the paper
The question as text
8. H O t Figure 4 The heart rate of a horse is being monitored. The heart rate H, measured in beats per minute (bpm), is modelled by the equation –0.2t – 20 e –0.9t H = 32 + 40e where t minutes is the time after monitoring began. Figure 4 is a sketch of H against t. Use the equation of the model to answer parts (a) to (e).
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8(a)[1]
52 b.p.m.
8(b)[1]
32 b.p.m.
8(c)[5]
d
H
−
0.2
t
−
0.9
t
= −
8 e
+
18 e
d
t
0.2
0.9
0.7
−
t
−
t
t
−
8 e
+
18 e
=
0
4e
=
9
Sets
9
0.7
t
=
ln
=
t
...
4
T = 1.158 (minutes)
8(d)[2]
−
0.9
t
1
+
4 e
−
0.2
t
−
0.9
t
−
0.2
t
37
=
32
+
40 e
−
20 e
e
=
8
8
8
0.2
t
e
=
=
5ln
t
−
0.9
t
−
0.9
t
1 4 e
+
1 4 e
+
8(e)[3]
8
5ln
....
t
=
=
2
−
0.9 10
×
1
+
4e
( )
=
awrt 10.3947
t
2
(M )= 10.3955
(a)
B1: 52 b.p.m. Units are not required. Check for answer given in the body of the question
(b)
B1: 32 b.p.m. . Units are not required. Check for answer given in the body of the question
(c)
d
H
−
0.2
−
0.9
M1: Achieves
=
e
t
+
e
t
where P and Q are constants
P
Q
d
t
d
H
−
0.2
−
0.9
t
t
= −
8 e
+
18 e
A1:
but allow unsimplified versions
d
t
−
0.2
t
−
0.9
t
0.7
t
e
+
e
=
0
e
=
0.7
dM1: Sets
P
Q
m
n
or
e
−
t
where
0
m
=
n
m n
×>
±
M1: Solves an equation of the form
e
kt
where
0
using correct ln work.
m
=
n
m n
×>
It must proceed from a correct form of the derivative (so the first M1).
n
n
Look for
±
kt
±
kt
m
e
=
n
e
=
±
kt
=
ln
t
=
...
m
m
A1: T = awrt 1.158 minutes following the award of all previous marks. You can ignore units
…………………………………………………………………………………….
An alternative for the last three marks in part (c) are;
−
0.9
−
0.2
e
t
=
e
t
ln
−
0.9
=
ln
−
0.2
dM1:
P
Q
P
t
Q
t
M1: Collects terms in t and proceeds to t =… Under this method it is dependent upon both previous M’s
A1: T = awrt 1.158 minutes following the award of all previous marks. Units are not required
…………………………………………………………………………………………………………
There may be answers produced from a calculator. In cases such as this we are going to apply the rule
‘answers without full working may not gain credit.’
d
H
−
0.2
t
−
0.9
t
= −
8 e
+
18 e
=
0
t
=
1.158
scores 1,1,0,0,0
d
t
………………………………………………………………………………………………….
(d)
±
0.2
t
M1: Sets
H =
37
and proceeds to make
e
the subject
A1*: Proceeds to the given answer showing all necessary steps.
Note that this is a given answer so the steps must be logical without huge jumps
Acceptable proofs:
Example 1
−
0.9
t
−
0.9
t
1
+
4 e
1
+
4 e
−
0.2
−
0.9
−
0.2
37
=
32
+
40 e
t
−
20 e
t
e
t
=
−
0.2
=
ln
t
8
8
−
0.9
1
4 e
t
+
= −
5ln
t
8
8
t
=
5ln
−
0.9
t
1
+
4 e
Example 2
−
0.9
t
8
1 4 e
+
8
8
−
0.2
t
−
0.9
t
−
0.2
t
0.2
t
37
=
32
+
40 e
−
20 e
e
=
e
=
0.2
t
=
ln
=
t
5ln
0.9
0.9
8
1 4 e
+
−
t
1 4 e
+
−
t
1
+
4 e
Unacceptable proof:
−
0.9
−
0.9
1
4 e
t
1
4 e
t
+
+
−
0.2
t
−
0.9
t
−
0.2
t
37
=
32
+
40 e
−
20 e
e
=
−
0.2
t
=
ln
8
8
8
=
5ln
t
−
0.9
1
4 e
t
+
They need to deal with 0.2 and −sign in separate steps
…………………………………………………………………………………………………………
Alt (d) lns can be taken earlier. The M1 is scored for making 0.2 t the subject. See below
0.2
0.9
−
t
−
t
37
=
32
+
40 e
−
20 e
(
)
(
)
−
0.2
−
0.9
−
0.9
−
0.9
40e
t
5
20 e
t
ln 40
0.2
ln 5
20 e
t
0.2
ln 40
ln 5
20 e
t
=
+
−
t
=
+
t
=
−
+
We would need to see three further steps. E.g.
40
40
8
0.2
t
=
ln
=
t
5ln
=
t
5ln
−
0.9
t
−
0.9
t
−
0.9
t
5
+
20 e
5
+
20 e
1 4 e
+
……………………………………………………………………………………………………………………………
…………..
(e)
M1: Uses the iteration formula once. Allow for the embedded 10 leading to a value or sight of 2
A1: ( )
awrt 10.3947
t
=
2
(
)
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