| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - Nā)) |
| Difficulty | Moderate -0.3 This is a standard first-order linear differential equation with straightforward setup and solution. Part (a) requires basic interpretation of rates, part (b) uses routine separation of variables or integrating factor method with given initial condition, and part (c) involves substituting known values to find constants. The question is slightly easier than average because it's highly scaffolded with the form of the answer provided and follows a predictable tank problem template common in C4. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| (a) \(\frac{dV}{dt}\) is the rate of increase of volume (with respect to time) | B1 |
| \(-kV\): \(k\) is constant of proportionality and the negative shows decrease (or loss) | B1 |
| giving \(\frac{dV}{dt} = 20 - kV\) ā | These B's are to be awarded independently |
| Answer | Marks |
|---|---|
| (b) \(\int\frac{1}{20-kV}dV = \int 1\,dt\) | separating variables M1 |
| \(-\frac{1}{k}\ln(20-kV) = t\) \((+C)\) | M1 A1 |
| Using \(V = 0, t = 0\) to evaluate the constant of integration | M1 |
| \(c = -\frac{1}{k}\ln 20\) | |
| \(t = \frac{1}{k}\ln\left(\frac{20}{20-kV}\right)\) | |
| Obtaining answer in the form \(V = A + Be^{-kt}\) | M1 |
| \(V = \frac{20}{k} - \frac{20}{k}e^{-kt}\) | Accept \(\frac{20}{k}(1-e^{-kt})\) M1 A1 |
| Answer | Marks |
|---|---|
| (c) \(\frac{dV}{dt} = 20e^{-kt}\) | Can be implied M1 |
| \(\frac{dV}{dt} = 10, t = 5 \Rightarrow 10 = 20e^{-kt} \Rightarrow k = \frac{1}{5}\ln 2 \approx 0.139\) | M1 A1 |
| At \(t = 10, V = \frac{75}{\ln 2}\) | awrt 108 M1 A1 |
| Answer | Marks |
|---|---|
| Using printed answer and differentiating: \(\frac{dV}{dt} = -kB e^{-kt}\) | M1 |
| Substituting into differential equation | M1 |
| \(-kB e^{-kt} = 20 - kA - kB e^{-kt}\) | M1 |
| \(A = \frac{20}{k}\) | M1 A1 |
| Using \(V = 0, t = 0\) in printed answer to obtain \(A + B = 0\) | M1 |
| \(B = -\frac{20}{k}\) | A1 |
**(a)** $\frac{dV}{dt}$ is the rate of increase of volume (with respect to time) | B1 |
$-kV$: $k$ is constant of proportionality and the negative shows decrease (or loss) | B1 |
giving $\frac{dV}{dt} = 20 - kV$ ā
| These B's are to be awarded independently |
**Total for (a): [2]**
**(b)** $\int\frac{1}{20-kV}dV = \int 1\,dt$ | separating variables M1 |
$-\frac{1}{k}\ln(20-kV) = t$ $(+C)$ | M1 A1 |
Using $V = 0, t = 0$ to evaluate the constant of integration | M1 |
$c = -\frac{1}{k}\ln 20$ | |
$t = \frac{1}{k}\ln\left(\frac{20}{20-kV}\right)$ | |
Obtaining answer in the form $V = A + Be^{-kt}$ | M1 |
$V = \frac{20}{k} - \frac{20}{k}e^{-kt}$ | Accept $\frac{20}{k}(1-e^{-kt})$ M1 A1 |
**Total for (b): [6]**
**(c)** $\frac{dV}{dt} = 20e^{-kt}$ | Can be implied M1 |
$\frac{dV}{dt} = 10, t = 5 \Rightarrow 10 = 20e^{-kt} \Rightarrow k = \frac{1}{5}\ln 2 \approx 0.139$ | M1 A1 |
At $t = 10, V = \frac{75}{\ln 2}$ | awrt 108 M1 A1 |
**Total for (c): [5]**
**Alternative to (b):**
Using printed answer and differentiating: $\frac{dV}{dt} = -kB e^{-kt}$ | M1 |
Substituting into differential equation | M1 |
$-kB e^{-kt} = 20 - kA - kB e^{-kt}$ | M1 |
$A = \frac{20}{k}$ | M1 A1 |
Using $V = 0, t = 0$ in printed answer to obtain $A + B = 0$ | M1 |
$B = -\frac{20}{k}$ | A1 |
**Total: [13]**\begin{enumerate}
\item Liquid is pouring into a container at a constant rate of $20 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ and is leaking out at a rate proportional to the volume of liquid already in the container.\\
(a) Explain why, at time $t$ seconds, the volume, $V \mathrm {~cm} ^ { 3 }$, of liquid in the container satisfies the differential equation
\end{enumerate}
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = 20 - k V$$
where $k$ is a positive constant.
The container is initially empty.\\
(b) By solving the differential equation, show that
$$V = A + B \mathrm { e } ^ { - k t }$$
giving the values of $A$ and $B$ in terms of $k$.
Given also that $\frac { \mathrm { d } V } { \mathrm {~d} t } = 10$ when $t = 5$,\\
(c) find the volume of liquid in the container at 10 s after the start.\\
\hfill \mbox{\textit{Edexcel C4 2005 Q8 [13]}}