A2 June 2025 Paper 1 Q8
8. A rambler starts a walk at the bottom of a hill at 10 am.
The rambler walks all the way to the top of the hill and then turns around and walks back down to the bottom of the hill.
The differential equation
\[\sin t\,\frac{\mathrm{d}x}{\mathrm{d}t} - x\cos t = A\sin 2t\sin t \qquad t \geqslant 0\]where \(A\) is a constant, is used to model the vertical displacement, \(x\) metres, of the rambler from the bottom of the hill, \(t\) hours after the start of the walk.
Given that after 1 hour
- the rambler has a vertical displacement of 213 m
- \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 273.2\)
Given that the vertical displacement of the top of the hill is 300.68 m
| Scheme | Marks | AO |
|---|---|---|
| \((\sin 1)(273.2) - 213\cos 1 = A(\sin 2)(\sin 1) \Rightarrow A = 150\) | B1 | 3.3 |
| (1) |
Notes
B1: \(A =\) awrt 150
Note \(A = 150.0436\ldots\)
| Scheme | Marks | AO |
|---|---|---|
| \(\sin t\,\dfrac{\mathrm{d}x}{\mathrm{d}t} - x\cos t = 150\sin 2t\sin t \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} - x\dfrac{\cos t}{\sin t} = 150\sin 2t\) \(IF = \mathrm{e}^{-\int \frac{\cos t}{\sin t}\,\mathrm{d}t} = \mathrm{e}^{-\ln\sin t} = \operatorname{cosec} t\) \(\Rightarrow \operatorname{cosec} t\,\dfrac{\mathrm{d}x}{\mathrm{d}t} - x\operatorname{cosec} t\cot t = 150\dfrac{\sin 2t}{\sin t}\) oe or \(x\operatorname{cosec} t = \displaystyle\int 150\dfrac{\sin 2t}{\sin t}\,\mathrm{d}t\) oe | M1 | 3.1b |
| Uses \(\sin 2t = 2\sin t\cos t\) to arrange into an integrable form and then integrates \(x\operatorname{cosec} t = 300\displaystyle\int \cos t\,\mathrm{d}t \Rightarrow x\operatorname{cosec} t = \ldots\) | dM1 | 3.1a |
| \(x\operatorname{cosec} t = 300\sin t\ \{+c\}\) | A1 | 1.1b |
| Uses \(t = 1,\ x = 213\) to find the value of \(c\) \(\dfrac{213}{\sin 1} = \text{``}300\text{''}\sin 1 + c\) \(\Rightarrow c = \ldots\) | M1 | 3.4 |
| \(x = \text{``}300\text{''}\sin^2 t + \text{``}0.687\text{''}\sin t\) or \(\mathrm{f}(t) = \text{``}300\text{''}\sin^2 t + \text{``}0.687\text{''}\sin t\) | A1 | 1.1b |
| (5) |
Notes
You may mark (b) (c) and (d) together so do not be concerned about the labelling.
M1: Finds an integrating factor of the form \(\mathrm{e}^{\pm\int \frac{\cos t}{\sin t}\,\mathrm{d}t} = \mathrm{e}^{\pm\ln\sin t}\) or \(\operatorname{cosec} t\) (or \(\sin t\)) and proceeds to multiply through by their IF oe
Look for
\(I.F. = \mathrm{e}^{\pm\int \frac{\cos t}{\sin t}\,\mathrm{d}t} \Rightarrow x \times \text{`their } I.F.\text{'} = \displaystyle\int A\sin 2t \times \text{`their } I.F.\text{'}\,\mathrm{d}t\)
Where they must be using a numerical \(A\)
dM1: Dependent on the previous method mark. Uses the identity \(\sin 2t = 2\sin t\cos t\) to arrange RHS into an integrable form with a correct integrating factor of \(\operatorname{cosec} t\) and then attempts to integrate using their numerical \(A\). Condone a sign slip on their RHS.
\(x\operatorname{cosec} t = 2A\displaystyle\int \cos t\,\mathrm{d}t \Rightarrow x\operatorname{cosec} t = \pm\alpha\sin t\ \ (+c)\)
A1: Correct general solution, condone missing \(+ c\)
M1: Using the model, \(t = 1,\ x = 213\) to find the value of their constant \(c\). Substitution if not seen is implied by sight of 213 for \(x\) and at least one correct value for \(\sin t\). They must use radians and not degrees; you may need to check.
A1: Correct particular solution of \(x = 300\sin^2 t + 0.687\sin t\)
\(c =\) awrt 0.69 (when using \(A = 150\))
Alternatively:
When using answers for \(A\) which have not been rounded to 3sf, the equation will instead be \(x = 300.0872..\sin^2 t + 0.613\sin t\) in which case allow \(c =\) awrt 0.61
(corrected from the printed mark scheme: the integrating factor exponent is printed with d\(x\) in place of d\(t\) in the scheme and in the first M1 note)
| Scheme | Marks | AO |
|---|---|---|
| \(300\sin^2 t + 0.687\sin t = 0\) \(\sin t(300\sin t + 0.687) = 0\) \(\Rightarrow \sin t = \ldots\) | M1 | 3.4 |
| \(t = \pi\) or \(t = \pi - \sin^{-1}\left(-\dfrac{0.687}{300}\right)\) or \(\pi + \sin^{-1}\left(\dfrac{0.687}{300}\right)\) or awrt \(t = 3.14\) | M1 | 1.1b |
| 13:08 or 13:09 or 1.08 (pm) or 1.09 (pm) | A1 | 3.2a |
| (3) |
Notes
M1: Sets \(x = 0\) and solves a quadratic equation of the form \(\lambda\sin^2 t + \mu\sin t = 0\), where \(\lambda, \mu \neq 0\) finding at least one value for \(\sin t\)
M1: Solves a quadratic equation of the form \(2A\sin^2 t + c\sin t = 0\) where \(A,\ c \gt 0\) to find a correct value of \(t\). May be implied by a value of \(t\) awrt 3.14
Award for \(t = \pi\) but accept the use of \(\arcsin\left(\dfrac{-\mu}{\lambda}\right) + \pi\) oe, where they must add \(\pi\) to a positive value of \(t\). They must also be working in radians.
A1: Correct time awrt 13:08 or 13:09 or 1.08 (pm), 1.09 (pm), which comes from a correct equation
| Scheme | Marks | AO |
|---|---|---|
| \(300\sin^2 t + 0.687\sin t = 300.68\) \(300\sin^2 t + 0.687\sin t - 300.68 = 0\) \(\Rightarrow \sin t = \ldots\) | M1 | 3.4 |
| \(t =\) awrt 1.57 | A1 | 1.1b |
| (2) |
Notes
M1: Solves their 3TQ equation of the form \(2A\sin^2 t + c\sin t = 300.68\) where \(A,\ c \gt 0\) to find an answer for \(\sin t\). You may need to check the value for \(\sin t\) or \(t\) obtained from their quadratic. A correct equation followed by \(t =\) awrt 1.57 is sufficient.
A1: \(t =\) awrt 1.57 isw. Answer must come from a correct equation. Do not allow an answer of \(\dfrac{\pi}{2}\) which may have been obtained from \(\sin t = 1\) even if it proceeds to 1.57
Here it asks for the value of \(t\), so do not accept a time, for example 11.42am
Alternatively uses
\(300.0872\sin^2 t + 0.613\sin t - 300.68 = 0 \Rightarrow \sin t = \ldots\ t =\) awrt 1.56
| Scheme | Marks | AO |
|---|---|---|
| The model suggests that the time to go up the hill and the time to go down will be the similar. In reality the times will be different. | B1ft | 3.5b |
| (1) | ||
| (12 marks) |
Notes
B1ft: They must have stated times or values for \(t\) in (c) and (d) for this mark.
We need a comparison consistent with their times from (c) and (d) and reality.
Their values may not be correct, but their statement should be consistent with their times.
You may ignore incorrect statements once a correct statement is given, provided their incorrect statement does not contradict their correct statement.
e.g.
Accept
- The time to go up and down is the same, which is not realistic
- The times are similar which is unlikely
- It should take longer to go up the hill
- In reality it will take longer to go up the hill than down the hill (because of the slope)
Condone
- It takes longer to go up the hill than down the hill
- It takes the same amount of time to go up and down
- In the model it takes them longer to go down the hill
- The rambler may not walk up and down at the same speed, or rate
Reject
- the rambler may need to take a rest
- the model is inaccurate or invalid for certain values of \(t\)
- the rambler may walk down a different path or route
- the rambler is modelled as a particle
- the rambler walks at a constant velocity
- the rambler may not walk at the same speed/pace for the entire journey
If you are uncertain please send to review