| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Cone: related rates of dimensions |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring chain rule application. Part (a) involves straightforward differentiation of V with respect to h, then substitution of h at t=3 (found from V=8t). Part (b) applies dh/dt = (dV/dt)/(dV/dh). While it requires multiple steps, each is routine calculus technique commonly practiced in A-level, making it slightly easier than average. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| 8(a) | dV 3πh2 πh2 |
| Answer | Marks | Guidance |
|---|---|---|
| labels | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| eg 288 or 96 × 3 | 3.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| h2 = 20.3 | 1.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| working in the manipulation | 2.1 | R1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 8(b) | States any correct chain rule |
| Answer | Marks | Guidance |
|---|---|---|
| 96𝑡𝑡 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| unit cm/s or cm s-1 | 3.2a | A1 |
| Subtotal | 3 | |
| Question 8 Total | 7 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | dV 3πh2 πh2
=
Obtains or OE
dh 12 4
Condone missing or incorrect
labels | 1.1b | B1 | dV 3πh2 πh2
= =
dh 12 4
When t = 3
πh3
V = =24
12
1
2883
⇒h=
π
2
dV π2883
=
dh 4 π
1 1
1
=π3× ×829443
4
=31296π
= 3 216×6π
=63 6π
Obtains v = 8 × 3 or 24
Can be embedded
eg 288 or 96 × 3 | 3.1b | B1
Equates their 24 to
3
𝜋𝜋ℎ
12
24×12
to obtain h= 3
π
or h2
2
288
3
=
Can be e(m𝜋𝜋bed)ded
Condone decimal values
h = 4.51
h2 = 20.3 | 1.1b | M1
Completes reasoned argument
to show given result AG
dV
Must include with at least
dh
one intermediate step without
288
Must not include incorrect
working in the manipulation | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | States any correct chain rule
dV dV dh
connecting , and
dt dh dt
8
PI by or correct answer
63 6π
or
states that
3
96𝑡𝑡 | 3.1b | M1 | dh dV dh
= ×
dt dt dV
dh 8
⇒ =
dt 63 6π
= 0.501 cm s-1
ℎ = � 𝜋𝜋
dV
Substitutes =8 and
dt
dV
=63 6π in their chain rule
dh
8
PI by or correct answer
63 6π
or
substitutes t = 3 in their
dh
= × ACF
dt 1 −2/3
96 𝑡𝑡
3
3 | 1.1a | M1
(𝜋𝜋)
dh
Obtains correct
dt
AWRT 0.501 cm/s
Must be at least 3sf with correct
unit cm/s or cm s-1 | 3.2a | A1
Subtotal | 3
Question 8 Total | 7
Q | Marking instructions | AO | Marks | Typical solutionWater is poured into an empty cone at a constant rate of 8 cm³/s
After $t$ seconds the depth of the water in the inverted cone is $h$ cm, as shown in the diagram below.
\includegraphics{figure_8}
When the depth of the water in the inverted cone is $h$ cm, the volume, $V$ cm³, is given by
$$V = \frac{\pi h^3}{12}$$
\begin{enumerate}[label=(\alph*)]
\item Show that when $t = 3$
$$\frac{dV}{dh} = 6 \sqrt[3]{6\pi}$$
[4 marks]
\item Hence, find the rate at which the depth is increasing when $t = 3$
Give your answer to three significant figures.
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2022 Q8 [10]}}