June 2025 Paper 3 Q11
11 A block of ice is melting.
At time \(t\) minutes, the block is in the shape of a cuboid with dimensions of \(4x\), \(2x\) and \(x\), as shown in the diagram.
All measurements are in centimetres.
(a) The volume, \(V\ \text{cm}^3\), of the block of ice decreases at a rate which is proportional to its surface area.
When \(x = 4\) the volume of the block of ice is decreasing at a rate of \(7\ \text{cm}^3\) per minute.
Show that
\[\frac{\mathrm{d}V}{\mathrm{d}t} = -0.4375x^2\] [4 marks](b) Find \(\dfrac{\mathrm{d}V}{\mathrm{d}x}\) in terms of \(x\) [2 marks]
(c)
(i) Using the results from parts (a) and (b), find \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) [2 marks]
(ii) Interpret, in context, your answer to part (c)(i). [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms a correct expression for the surface area of the block of ice Can be unsimplified terms or states that the surface area is proportional to \(x^2\) | B1 | 3.1b |
| States \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = m(SA)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -m(SA)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = m(\text{their } 28x^2)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -m(\text{their } 28x^2)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = mx^2\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -mx^2\) Accept any letter for \(m\) | M1 | 3.3 |
| Substitutes \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -7\) and SA = their ‘28’ \(\times\, 4^2\) into their differential equation \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = m(SA)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -m(SA)\) or substitutes \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -7\) and \(x = 4\) into their differential equation of the form \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = m(28x^2)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -m(28x^2)\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = mx^2\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -mx^2\) Accept any letter for \(m\) | M1 | 3.3 |
| Completes reasoned argument to show the given result with no incorrect expression for surface area and no incorrect substitution seen. AG To be awarded R1, marks M1M1 must be scored as the minimum Ignore units | R1 | 2.1 |
| (4) |
Typical solution
\[S.A. = 2\left(8x^2 + 2x^2 + 4x^2\right) = 28x^2\]\[\frac{\mathrm{d}V}{\mathrm{d}t} = -28kx^2\]\[x = 4,\ \frac{\mathrm{d}V}{\mathrm{d}t} = -7\]\[\Rightarrow -7 = -28k \times 16\]\[\Rightarrow k = \frac{7}{28 \times 16} = \frac{1}{64}\]\[\Rightarrow \frac{\mathrm{d}V}{\mathrm{d}t} = -\frac{28}{64}x^2 = -0.4375x^2\]| Scheme | Marks | AO |
|---|---|---|
| Forms a correct expression for the volume of the block of ice or states \(8x^3\) | M1 | 3.1b |
| Obtains \(24x^2\) | A1 | 1.1b |
| (2) |
Typical solution
\[V = 4x \times 2x \times x\]\[= 8x^3\]\[\frac{\mathrm{d}V}{\mathrm{d}x} = 24x^2\]| Scheme | Marks | AO |
|---|---|---|
| (i) Uses the chain rule to connect \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -0.4375x^2\), their \(\dfrac{\mathrm{d}V}{\mathrm{d}x} = 24x^2\) from part (b) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) | M1 | 3.1a |
| Obtains \(-\dfrac{7}{384}\) AWFW \(-[0.018, 0.0183]\) | A1 | 1.1b |
| (2) | ||
| (ii) Identifies the height/width/length/edge/side/dimension/measurement is decreasing at a constant rate | E1 | 3.5a |
| States it occurs at the rate of \(\dfrac{7}{384}\) or [0.018, 0.0183] or [0.036, 0.0366] or [0.072, 0.0732] cm per minute FT their constant \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) from part c(i) | E1F | 3.5a |
| (2) | ||
| (10 marks) |
Typical solution
(c)(i)
\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}V} \times \frac{\mathrm{d}V}{\mathrm{d}t}\]\[= \frac{1}{24x^2} \times -0.4375x^2\]\[= -\frac{7}{384}\](c)(ii)
The height is decreasing at a constant rate of 0.018 cm per minute.