| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Standard +0.3 This is a well-structured, guided differential equations question with clear steps: simple arithmetic in (i), standard separation of variables in (ii), straightforward substitution to find constants in (iii), routine exponential equation solving in (iv), and basic comparison in (v). While it covers multiple techniques, each part is scaffolded and uses standard C4 methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((A)\ 9/1.5 = 6\) hours | B1 | |
| \((B)\ 18/1.5 = 12\) hours | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d\theta}{dt} = -k(\theta-\theta_0)\) | ||
| \(\Rightarrow \int\frac{d\theta}{\theta-\theta_0} = \int -k\,dt\) | M1 | separating variables |
| \(\Rightarrow \ln(\theta-\theta_0) = -kt+c\) | A1 | \(\ln(\theta-\theta_0)\) |
| A1 | \(-kt+c\) | |
| \(\theta-\theta_0 = e^{-kt+c}\) | M1 | anti-logging correctly (with \(c\)) |
| \(\theta = \theta_0 + Ae^{-kt}\) * | E1 [5] | \(A=e^c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(98 = 50 + Ae^0 \Rightarrow A=48\) | M1, A1 | |
| Initially \(\frac{d\theta}{dt} = -k(98-50) = -48k = -1.5\) | M1 | |
| \(\Rightarrow k = 0.03125\)* | E1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((A)\ 89 = 50+48e^{-0.03125t}\) | M1 | equating |
| \(\Rightarrow 39/48 = e^{-0.03125t}\) | M1 | taking lns correctly for either |
| \(\Rightarrow t = \ln(39/48)/(-0.03125) = 6.64\) hours | A1 | |
| \((B)\ 80 = 50+48e^{-0.03125t}\) | M1 | |
| \(\Rightarrow 30/48 = e^{-0.03125t}\) | ||
| \(\Rightarrow t = \ln(30/48)/(-0.03125) = 15\) hours | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Models disagree more for greater temperature loss | B1 [1] |
# Question 7:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(A)\ 9/1.5 = 6$ hours | B1 | |
| $(B)\ 18/1.5 = 12$ hours | B1 [2] | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d\theta}{dt} = -k(\theta-\theta_0)$ | | |
| $\Rightarrow \int\frac{d\theta}{\theta-\theta_0} = \int -k\,dt$ | M1 | separating variables |
| $\Rightarrow \ln(\theta-\theta_0) = -kt+c$ | A1 | $\ln(\theta-\theta_0)$ |
| | A1 | $-kt+c$ |
| $\theta-\theta_0 = e^{-kt+c}$ | M1 | anti-logging correctly (with $c$) |
| $\theta = \theta_0 + Ae^{-kt}$ * | E1 [5] | $A=e^c$ |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $98 = 50 + Ae^0 \Rightarrow A=48$ | M1, A1 | |
| Initially $\frac{d\theta}{dt} = -k(98-50) = -48k = -1.5$ | M1 | |
| $\Rightarrow k = 0.03125$* | E1 [4] | |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(A)\ 89 = 50+48e^{-0.03125t}$ | M1 | equating |
| $\Rightarrow 39/48 = e^{-0.03125t}$ | M1 | taking lns correctly for either |
| $\Rightarrow t = \ln(39/48)/(-0.03125) = 6.64$ hours | A1 | |
| $(B)\ 80 = 50+48e^{-0.03125t}$ | M1 | |
| $\Rightarrow 30/48 = e^{-0.03125t}$ | | |
| $\Rightarrow t = \ln(30/48)/(-0.03125) = 15$ hours | A1 [5] | |
## Part (v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Models disagree more for greater temperature loss | B1 [1] | |
---7 Scientists can estimate the time elapsed since an animal died by measuring its body temperature.
\begin{enumerate}[label=(\roman*)]
\item Assuming the temperature goes down at a constant rate of 1.5 degrees Fahrenheit per hour, estimate how long it will take for the temperature to drop\\
(A) from $98 ^ { \circ } \mathrm { F }$ to $89 ^ { \circ } \mathrm { F }$,\\
(B) from $98 ^ { \circ } \mathrm { F }$ to $80 ^ { \circ } \mathrm { F }$.
In practice, rate of temperature loss is not likely to be constant. A better model is provided by Newton's law of cooling, which states that the temperature $\theta$ in degrees Fahrenheit $t$ hours after death is given by the differential equation
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k \left( \theta - \theta _ { 0 } \right)$$
where $\theta _ { 0 } { } ^ { \circ } \mathrm { F }$ is the air temperature and $k$ is a constant.
\item Show by integration that the solution of this equation is $\theta = \theta _ { 0 } + A \mathrm { e } ^ { - k t }$, where $A$ is a constant.
The value of $\theta _ { 0 }$ is 50 , and the initial value of $\theta$ is 98 . The initial rate of temperature loss is $1.5 ^ { \circ } \mathrm { F }$ per hour.
\item Find $A$, and show that $k = 0.03125$.
\item Use this model to calculate how long it will take for the temperature to drop\\
(A) from $98 ^ { \circ } \mathrm { F }$ to $89 ^ { \circ } \mathrm { F }$,\\
(B) from $98 ^ { \circ } \mathrm { F }$ to $80 ^ { \circ } \mathrm { F }$.
\item Comment on the results obtained in parts (i) and (iv).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 2009 Q7 [17]}}