Mathematics · May / June 2021 · Paper 41 · Question 3
YMA01
As printed in the paper
The question as text
3. 20 cm h cm Figure 2 A bowl with circular cross section and height 20 cm is shown in Figure 2. The bowl is initially empty and water starts flowing into the bowl. When the depth of water is h cm, the volume of water in the bowl, V cm3, is modelled by the equation V = 1 2 (h + 4) 0 h 20 3 h Given that the water flows into the bowl at a constant rate of 160 cm3 s−1, find, according to the model,
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3(a)[2]
Attempts to find
=
20
seconds
160
d
V
8
2
3(b)[5]
Attempts
h =
h
+
h
d
3
d
V
d
V
d
h
8
d
h
2
Attempts to use
=
×
⇒
160
=
h
+
h
×
d
t
d
h
d
t
3
d
t
40
d
h
d
h
96
2
160
=
5
+
×
⇒
=
Substitutes
h =
5
⇒
(4.2) cm s 1
3
d
t
d
t
23
(a)
M1: Attempts to find the volume, using the given formula, at
h =
20
and dividing by 160. Implied by 20
Condone slips. For example they may have incorrectly multiplied out/calculated the expression for V
A1: 20 seconds. This requires the correct units as well
(b)
M1: For an attempt to differentiate V
d
V
Scored for an attempt to multiply out and then differentiate term by term to achieve
h =
d
1
d
V
or via the product rule to achieve
V
h
2
(
h
4
)
(
h
4
)
ph
qh
2
=
+
⇒
=
+
×
+
3
d
h
M1: Attempts to use d
V
d
V
d
h
(or equivalent) with d
V
and their d
V
160
=
×
t =
h .
d
t
d
h
d
t
d
d
A1: Correct expression involving d
h
8
d
h
2
t and h E.g.
160
=
h
+
h
×
d
3
d
t
and proceeding to a value for d
dM1: Dependent upon the previous M only. It is for substituting
h =
5
A1: Either 96
23 or awrt 4.2 cm s 1 There is no requirement for the units here
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