| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (polynomial/exponential x-side) |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation requiring standard integration techniques (integrating cos to get sin), followed by applying initial conditions and interpreting the exponential function. Part (b) requires recognizing that max occurs when sin=1, and part (c) involves solving a simple trigonometric equation. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.06a Exponential function: a^x and e^x graphs and properties1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dH}{dt} = \frac{H\cos 0.25t}{40} \Rightarrow \int\frac{1}{H}\,dH = \int\frac{\cos 0.25t}{40}\,dt\) | M1 | Separates variables; integral signs on both sides, \(dH\) and \(dt\) in correct positions |
| \(\ln H = A\sin 0.25t\) | M1 | Integrates both sides; with or without \(+c\) |
| \(\ln H = \frac{1}{10}\sin 0.25t\,(+c)\) | A1 | Allow two constants one either side; if 40 on lhs: \(40\ln H = 4\sin 0.25t + c\) |
| Substitutes \(t=0,\, H=5 \Rightarrow c = \ln 5\) | dM1 | Dependent on previous M; needs single \(+c\) to find |
| \(\ln\!\left(\frac{H}{5}\right) = \frac{1}{10}\sin 0.25t \Rightarrow H = 5e^{0.1\sin 0.25t}\) | A1* | Must proceed via \(\ln H = \frac{1}{10}\sin 0.25t + \ln 5\) with at least one correct intermediate line, no incorrect work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max height \(= 5e^{0.1} = 5.53\text{ m}\) | B1 | Accept \(5e^{0.1}\); condone lack of units; penalise incorrect units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(0.25t = \frac{5\pi}{2}\) | M1 | Identifies maximum height reached 2nd time when \(0.25t = \frac{5\pi}{2}\) or 450 |
| \(31.4\) | A1 | Accept awrt \(31.4\) or \(10\pi\); allow if units seen |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dH}{dt} = \frac{H\cos 0.25t}{40} \Rightarrow \int\frac{1}{H}\,dH = \int\frac{\cos 0.25t}{40}\,dt$ | M1 | Separates variables; integral signs on both sides, $dH$ and $dt$ in correct positions |
| $\ln H = A\sin 0.25t$ | M1 | Integrates both sides; with or without $+c$ |
| $\ln H = \frac{1}{10}\sin 0.25t\,(+c)$ | A1 | Allow two constants one either side; if 40 on lhs: $40\ln H = 4\sin 0.25t + c$ |
| Substitutes $t=0,\, H=5 \Rightarrow c = \ln 5$ | dM1 | Dependent on previous M; needs single $+c$ to find |
| $\ln\!\left(\frac{H}{5}\right) = \frac{1}{10}\sin 0.25t \Rightarrow H = 5e^{0.1\sin 0.25t}$ | A1* | Must proceed via $\ln H = \frac{1}{10}\sin 0.25t + \ln 5$ with at least one correct intermediate line, no incorrect work |
**Total: (5)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max height $= 5e^{0.1} = 5.53\text{ m}$ | B1 | Accept $5e^{0.1}$; condone lack of units; penalise incorrect units |
**Total: (1)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $0.25t = \frac{5\pi}{2}$ | M1 | Identifies maximum height reached 2nd time when $0.25t = \frac{5\pi}{2}$ or 450 |
| $31.4$ | A1 | Accept awrt $31.4$ or $10\pi$; allow if units seen |
**Total: (2)**
---\begin{enumerate}
\item The height above ground, $H$ metres, of a passenger on a roller coaster can be modelled by the differential equation
\end{enumerate}
$$\frac { \mathrm { d } H } { \mathrm {~d} t } = \frac { H \cos ( 0.25 t ) } { 40 }$$
where $t$ is the time, in seconds, from the start of the ride.
Given that the passenger is 5 m above the ground at the start of the ride,\\
(a) show that $H = 5 \mathrm { e } ^ { 0.1 \sin ( 0.25 t ) }$\\
(b) State the maximum height of the passenger above the ground.
The passenger reaches the maximum height, for the second time, $T$ seconds after the start of the ride.\\
(c) Find the value of $T$.
\hfill \mbox{\textit{Edexcel Paper 1 2018 Q10 [8]}}