C4 June 2005 Q8
8. Liquid is pouring into a container at a constant rate of \(20\ \text{cm}^3\,\text{s}^{-1}\) 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, \(V\ \text{cm}^3\), of liquid in the container satisfies the differential equation
\[\frac{\mathrm{d}V}{\mathrm{d}t} = 20 - kV,\]
where \(k\) is a positive constant. (2)The container is initially empty.
(b) By solving the differential equation, show that
\[V = A + B\mathrm{e}^{-kt},\]
giving the values of \(A\) and \(B\) in terms of \(k\). (6)Given also that \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 10\) when \(t = 5\),
(c) find the volume of liquid in the container at 10 s after the start. (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) 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) giving \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 20 - kV\) \(\ast\) | B1 |
| (2) |
Notes
These Bs are to be awarded independently
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{20 - kV}\,\mathrm{d}V = \int 1\,\mathrm{d}t\) separating variables | M1 |
| \(-\dfrac{1}{k}\ln(20 - kV) = t \quad (+C)\) | M1 A1 |
| Using \(V = 0, t = 0\) to evaluate the constant of integration \(c = -\dfrac{1}{k}\ln 20\) \(t = \dfrac{1}{k}\ln\left(\dfrac{20}{20 - kV}\right)\) | M1 |
| Obtaining answer in the form \(V = A + B\mathrm{e}^{-kt}\) | M1 |
| \(V = \dfrac{20}{k} - \dfrac{20}{k}\mathrm{e}^{-kt}\) Accept \(\dfrac{20}{k}\left(1 - \mathrm{e}^{-kt}\right)\) | A1 |
| (6) |
Alternative to (b)
| Using printed answer and differentiating \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -kB\mathrm{e}^{-kt}\) | M1 |
| Substituting into differential equation \(-kB\mathrm{e}^{-kt} = 20 - kA - kB\mathrm{e}^{-kt}\) | M1 |
| \(A = \dfrac{20}{k}\) | M1 A1 |
| Using \(V = 0, t = 0\) in printed answer to obtain \(A + B = 0\) | M1 |
| \(B = -\dfrac{20}{k}\) | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 20\mathrm{e}^{-kt}\) Can be implied | M1 |
| \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 10, t = 5 \Rightarrow 10 = 20\mathrm{e}^{-kt} \Rightarrow k = \dfrac{1}{5}\ln 2 \approx 0.139\) | M1 A1 |
| At \(t = 10,\ V = \dfrac{75}{\ln 2}\) awrt 108 | M1 A1 |
| (5) | |
| (13 marks) |