| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Moderate -0.3 This is a standard Newton's law of cooling problem requiring students to form a differential equation from a verbal description, then solve it using separation of variables with given initial conditions. While it involves multiple steps (forming the DE, finding the constant of proportionality, solving, and applying boundary conditions), each step follows a well-practiced routine with no novel insight required. It's slightly easier than average because Newton's cooling is one of the most common DE applications in A-level Further Maths. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dT}{dt} = k(T-20)\) where \(k\) is a constant | B2 (B1 if no constant) | May be \(-k\) and/or \((20-T)\); if additional terms seen B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = +0.07\) OR \(-0.07\) oe | B1 | |
| \(\int \frac{1}{(T-20)}\, dT = \int k\, dt\) | M1 | Appropriate separation of variables for their DE; condone missing \(dt\) and/or \(dT\) |
| \(\ln(T-20) = kt\ [+c]\) | M1 | o.e. For correct integration to include a \(\ln\) term; or \(T-20 = e^{kt\ [+c]}\) |
| \(\ln 70 = c\) | M1 | To find \(c\) from their equation using \(t=0\), \(T=90\); or \(e^c = 70\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ln(20) = -0.07t + \ln 70\) | M1 | Clear substitution of \(T=40\) into their equation in \(T\) and \(t\) coming from integration; \(-\frac{2}{7} = -0.07t\) |
| \(t \approx 17.9\) [minutes] | A1 | Any accuracy \(\geq\) 2sf; 17.9 nfww gets 6 |
## Question 6:
### Part (a):
$\frac{dT}{dt} = k(T-20)$ where $k$ is a constant | **B2** (B1 if no constant) | May be $-k$ and/or $(20-T)$; if additional terms seen **B0**
### Part (b):
$k = +0.07$ OR $-0.07$ oe | **B1** |
$\int \frac{1}{(T-20)}\, dT = \int k\, dt$ | **M1** | Appropriate separation of variables for their DE; condone missing $dt$ and/or $dT$
$\ln(T-20) = kt\ [+c]$ | **M1** | o.e. For correct integration to include a $\ln$ term; or $T-20 = e^{kt\ [+c]}$
$\ln 70 = c$ | **M1** | To find $c$ from their equation using $t=0$, $T=90$; or $e^c = 70$
$\ln(T-20) = -0.07t + \ln 70$
$\ln(20) = -0.07t + \ln 70$ | **M1** | Clear substitution of $T=40$ into their equation in $T$ and $t$ coming from integration; $-\frac{2}{7} = -0.07t$
$t \approx 17.9$ [minutes] | **A1** | Any accuracy $\geq$ 2sf; 17.9 nfww gets 6
---6 A hot drink is cooling. The temperature of the drink at time $t$ minutes is $T ^ { \circ } \mathrm { C }$.\\
The rate of decrease in temperature of the drink is proportional to $( T - 20 )$.
\begin{enumerate}[label=(\alph*)]
\item Write down a differential equation to describe the temperature of the drink as a function of time.
\item When $t = 0$, the temperature of the drink is $90 ^ { \circ } \mathrm { C }$ and the temperature is decreasing at a rate of $4.9 ^ { \circ } \mathrm { C }$ per minute.
Determine how long it takes for the drink to cool from $90 ^ { \circ } \mathrm { C }$ to $40 ^ { \circ } \mathrm { C }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2022 Q6 [8]}}