| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Newton's law of cooling |
| Difficulty | Moderate -0.3 Part (a) involves straightforward substitution into a given formula (routine evaluation at t=0, t=30, and solving a simple logarithmic equation). Part (b) requires separating variables and integrating a standard differential equation, then rearranging—this is a textbook application of Newton's law of cooling with no novel insight required. Slightly easier than average due to the scaffolded structure and standard techniques. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(t = 0\), \(x = 85\) | B1 | |
| (ii) \(t = 30\), \(x = 15 + 70e^{-\frac{30}{40}}\) | M1 | |
| \(= 48.06 \approx 48°C\) | A1 | |
| (iii) \(x = 60\): \(\frac{60 - 15}{70} = e^{-\frac{t}{40}}\) | M1 | |
| \(\ln\left(\frac{45}{70}\right) = -\frac{t}{40}\) | m1 | Their 60-15 |
| \(t = 17.67 \approx 18\) minutes | A1 | |
| (b)(i) \(\int\frac{dx}{x-15} = -\int\frac{dt}{40}\) | M1, A1 | Attempt to separate and integrate; Correct expression and used |
| \(\ln(x - 15) = -\frac{t}{40}(+c)\) | A1A1 | |
| \((0, 85): c = \ln 70\) | m1 | Use (0,85) to find c, which must now appear in expression |
| \(\frac{t}{40} = \ln 70 - \ln(x - 15)\), \(t = 40\ln\left(\frac{70}{x-15}\right)\) | A1 | Manipulate to \(t = \ldots\) |
| (ii) \(e^{\frac{t}{40}} = \frac{70}{x - 15}\) | M1 | Manipulate expression including a ln towards \(x = \ldots\) |
| \(x - 15 = 70e^{-\frac{t}{40}}\) | M1A1 | AG; convincingly obtained |
**(a)(i)** $t = 0$, $x = 85$ | B1 |
**(ii)** $t = 30$, $x = 15 + 70e^{-\frac{30}{40}}$ | M1 |
$= 48.06 \approx 48°C$ | A1 |
**(iii)** $x = 60$: $\frac{60 - 15}{70} = e^{-\frac{t}{40}}$ | M1 |
$\ln\left(\frac{45}{70}\right) = -\frac{t}{40}$ | m1 | Their 60-15
$t = 17.67 \approx 18$ minutes | A1 |
**(b)(i)** $\int\frac{dx}{x-15} = -\int\frac{dt}{40}$ | M1, A1 | Attempt to separate and integrate; Correct expression and used
$\ln(x - 15) = -\frac{t}{40}(+c)$ | A1A1 |
$(0, 85): c = \ln 70$ | m1 | Use (0,85) to find c, which must now appear in expression
$\frac{t}{40} = \ln 70 - \ln(x - 15)$, $t = 40\ln\left(\frac{70}{x-15}\right)$ | A1 | Manipulate to $t = \ldots$
**(ii)** $e^{\frac{t}{40}} = \frac{70}{x - 15}$ | M1 | Manipulate expression including a ln towards $x = \ldots$
$x - 15 = 70e^{-\frac{t}{40}}$ | M1A1 | AG; convincingly obtained
**Total: 14 marks**
**GRAND TOTAL: 75 marks**8
\begin{enumerate}[label=(\alph*)]
\item A cup of coffee is cooling down in a room. At time $t$ minutes after the coffee is made, its temperature is $x ^ { \circ } \mathrm { C }$, where
$$x = 15 + 70 \mathrm { e } ^ { - \frac { t } { 40 } }$$
\begin{enumerate}[label=(\roman*)]
\item Find the temperature of the coffee when it is made.
\item Find the temperature of the coffee 30 minutes after it is made.
\item Find how long it will take for the coffee to cool down to $60 ^ { \circ } \mathrm { C }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use integration to solve the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 1 } { 40 } ( x - 15 ) , \quad x > 15$$
given that $x = 85$ when $t = 0$, expressing $t$ in terms of $x$.
\item Hence show that $x = 15 + 70 \mathrm { e } ^ { - \frac { t } { 40 } }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2005 Q8 [14]}}