| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Verification of solutions |
| Difficulty | Standard +0.8 This is a multi-part differential equations question requiring verification, separation of variables, and solving a non-standard DE with algebraic manipulation. Part (iii) involves integrating terms like h^(1/2)/(1+h)^2 which requires substitution and partial fractions, going beyond routine C4 exercises. The final algebraic verification adds complexity, making this moderately challenging for A-level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(h = (1 - \frac{1}{2}At)^2 \Rightarrow dh/dt = -A(1 - \frac{1}{2}At) = -A\sqrt{h}\) | M1 | Including function of a function; need to see middle step |
| \(= -A\sqrt{h}\) | A1 | AG |
| When \(t=0\), \(h = (1-0)^2 = 1\) as required | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \frac{dh}{\sqrt{h}} = \int -A\,dt\) | M1 | Separating variables correctly and integrating |
| \(2h^{1/2} = -At + c\) | A1 | Including \(c\). [Condone change of \(c\).] |
| \(h = \left(\frac{-At+c}{2}\right)^2\); at \(t=0\), \(h=1\): \(1 = (c/2)^2 \Rightarrow c=2\), \(h=(1-At/2)^2\) | B1 | Using initial conditions; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t = 20\), \(h = 0\) | M1 | Subst and solve for \(A\) |
| \(\Rightarrow 1 - 10A = 0\), \(A = 0.1\) | A1 | cao |
| When depth is 0.5 m: \(0.5 = (1 - 0.05t)^2\) | M1 | Substitute \(h = 0.5\) and their \(A\), solve for \(t\) |
| \(\Rightarrow 1 - 0.05t = \sqrt{0.5}\), \(t = (1-\sqrt{0.5})/0.05 = 5.86\) s | A1 | www cao; accept 5.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dh}{dt} = -B\frac{\sqrt{h}}{(1+h)^2} \Rightarrow \int \frac{(1+h)^2}{\sqrt{h}}\,dh = -\int B\,dt\) | M1 | Separating variables correctly, intending to integrate both sides. NB reading \((1+h)^2\) as \(1+h^2\) is MR — do not mark as MR |
| \(\int \frac{1+2h+h^2}{\sqrt{h}}\,dh\) | M1 | Expanding \((1+h)^2\) and dividing by \(\sqrt{h}\) to form a one-line function of \(h\) with each term as a single power of \(h\) |
| \(= \int (h^{-1/2} + 2h^{1/2} + h^{3/2})\,dh\) | A1 | \(h^{-1/2} + 2h^{1/2} + h^{3/2}\); cao dep on second M only |
| \(2h^{1/2} + \frac{4h^{3/2}}{3} + \frac{2h^{5/2}}{5} = -Bt + c\) | A1 | cao oe; both sides dependent on first M1 |
| A1 | cao; need \(-Bt\) and \(c\) for this A1; constant may be on either side | |
| When \(t=0\), \(h=1 \Rightarrow c = 56/15\) | A1 | From correct work only (accept 3.73 or rounded; not \(c = -56/15\) if constant on opposite side) |
| \(\Rightarrow h^{1/2}(30 + 20h + 6h^2) = 56 - 15Bt\) * | A1 | AG must be from all correct exact work including exact \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(h = 0\) when \(t = 20\) | M1 | Substituting \(h=0\), \(t=20\) |
| \(\Rightarrow B = 56/300 = 0.187\) | A1 | Accept 0.187 |
| When \(h = 0.5\): \(56 - 2.8t = 29.3449\ldots\) | M1 | Subst their \(h=0.5\), their \(B\), and attempt to solve |
| \(\Rightarrow t = 9.52\) s | A1 | Accept answers that round to 9.5s www |
## Question 3:
### Part (i)
**EITHER:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $h = (1 - \frac{1}{2}At)^2 \Rightarrow dh/dt = -A(1 - \frac{1}{2}At) = -A\sqrt{h}$ | M1 | Including function of a function; need to see middle step |
| $= -A\sqrt{h}$ | A1 | AG |
| When $t=0$, $h = (1-0)^2 = 1$ as required | B1 | |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{dh}{\sqrt{h}} = \int -A\,dt$ | M1 | Separating variables correctly and integrating |
| $2h^{1/2} = -At + c$ | A1 | Including $c$. [Condone change of $c$.] |
| $h = \left(\frac{-At+c}{2}\right)^2$; at $t=0$, $h=1$: $1 = (c/2)^2 \Rightarrow c=2$, $h=(1-At/2)^2$ | B1 | Using initial conditions; AG |
**[3 marks]**
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 20$, $h = 0$ | M1 | Subst and solve for $A$ |
| $\Rightarrow 1 - 10A = 0$, $A = 0.1$ | A1 | cao |
| When depth is 0.5 m: $0.5 = (1 - 0.05t)^2$ | M1 | Substitute $h = 0.5$ and their $A$, solve for $t$ |
| $\Rightarrow 1 - 0.05t = \sqrt{0.5}$, $t = (1-\sqrt{0.5})/0.05 = 5.86$ s | A1 | www cao; accept 5.9 |
**[4 marks]**
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dh}{dt} = -B\frac{\sqrt{h}}{(1+h)^2} \Rightarrow \int \frac{(1+h)^2}{\sqrt{h}}\,dh = -\int B\,dt$ | M1 | Separating variables correctly, intending to integrate both sides. NB reading $(1+h)^2$ as $1+h^2$ is MR — do not mark as MR |
| $\int \frac{1+2h+h^2}{\sqrt{h}}\,dh$ | M1 | Expanding $(1+h)^2$ and dividing by $\sqrt{h}$ to form a one-line function of $h$ with each term as a single power of $h$ |
| $= \int (h^{-1/2} + 2h^{1/2} + h^{3/2})\,dh$ | A1 | $h^{-1/2} + 2h^{1/2} + h^{3/2}$; cao dep on second M only |
| $2h^{1/2} + \frac{4h^{3/2}}{3} + \frac{2h^{5/2}}{5} = -Bt + c$ | A1 | cao oe; both sides dependent on first M1 |
| | A1 | cao; need $-Bt$ and $c$ for this A1; constant may be on either side |
| When $t=0$, $h=1 \Rightarrow c = 56/15$ | A1 | From correct work only (accept 3.73 or rounded; not $c = -56/15$ if constant on opposite side) |
| $\Rightarrow h^{1/2}(30 + 20h + 6h^2) = 56 - 15Bt$ * | A1 | **AG** must be from all correct exact work including exact $c$ |
**[7 marks]**
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $h = 0$ when $t = 20$ | M1 | Substituting $h=0$, $t=20$ |
| $\Rightarrow B = 56/300 = 0.187$ | A1 | Accept 0.187 |
| When $h = 0.5$: $56 - 2.8t = 29.3449\ldots$ | M1 | Subst their $h=0.5$, their $B$, and attempt to solve |
| $\Rightarrow t = 9.52$ s | A1 | Accept answers that round to 9.5s www |
**[4 marks]**3 Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_260_447_281_824}
\captionsetup{labelformat=empty}
\caption{Fig. 8.1}
\end{center}
\end{figure}
The height of the water surface above the hole $t$ seconds after opening the hole is $h$ metres, where
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A \sqrt { h }$$
and where $A$ is a positive constant. Initially the water surface is 1 metre above the hole.\\
(i) Verify that the solution to this differential equation is
$$h = \left( 1 - \frac { 1 } { 2 } A t \right) ^ { 2 } .$$
The water stops leaking when $h = 0$. This occurs after 20 seconds.\\
(ii) Find the value of $A$, and the time when the height of the water surface above the hole is 0.5 m .
Fig. 8.2 shows a similar situation with a different barrel; $h$ is in metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_235_455_1425_820}
\captionsetup{labelformat=empty}
\caption{Fig. 8.2}
\end{center}
\end{figure}
For this barrel,
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = - B \frac { \sqrt { h } } { ( 1 + h ) ^ { 2 } } ,$$
where $B$ is a positive constant. When $t = 0 , h = 1$.\\
(iii) Solve this differential equation, and hence show that
$$h ^ { \frac { 1 } { 2 } } \left( 30 + 20 h + 6 h ^ { 2 } \right) = 56 - 15 B t .$$
(iv) Given that $h = 0$ when $t = 20$, find $B$.
Find also the time when the height of the water surface above the hole is 0.5 m .
\hfill \mbox{\textit{OCR MEI C4 Q3 [18]}}