| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Moderate -0.3 This is a straightforward separable differential equation with standard integration and parameter fitting. Part (i) requires writing dV/dt = k√V and verifying a given solution by differentiation (not solving from scratch). Part (ii) involves simple substitution of boundary conditions to find constants. The verification approach makes it easier than typical C4 differential equations questions. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dV}{dt} = k\sqrt{V}\) | B1 | cao condone different \(k\) (allow MR B1 for \(= kV^2\)) |
| \(V = (\frac{1}{2}kt + c)^2\) | ||
| \(\Rightarrow \frac{dV}{dt} = 2(\frac{1}{2}kt+c)\cdot\frac{1}{2}k\) | M1 | \(2(\frac{1}{2}kt+c) \times\) constant multiple of \(k\) (or from multiplying out oe; or implicit differentiation) |
| \(= k(\frac{1}{2}kt + c)\) | A1 | cao www any equivalent form (including unsimplified) |
| \(= k\sqrt{V}\) | A1 | Allow SCB2 if \(V=(\frac{1}{2}kt+c)^2\) fully obtained by integration including convincing change of constant if used. Can score B1 M0 SCB2 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\frac{1}{2}k+c)^2 = 10000 \Rightarrow \frac{1}{2}k + c = 100\) | B1 | Substituting any one from \(t=1\), \(V=10000\) or \(t=0\), \(V=0\) or \(t=2\), \(V=40000\) into squared form or rooted form of equation (Allow \(-/\pm100\) or \(-/\pm200\)) |
| \((k+c)^2 = 40000 \Rightarrow k+c = 200\) | B1 | Substituting any other from above |
| \(\Rightarrow \frac{1}{2}k = 100 \Rightarrow k = 200,\ c = 0\) | M1 | Solving correct equations for both www (possible solutions are \((200,0)\), \((-200,0)\), \((600,-400)\), \((-600,400)\) (some from \(-\)ve root)) |
| \(\Rightarrow V = (100t)^2 = 10000t^2\) | A1 | either form www. SC B2 for \(V=(100t)^2\) oe stated without justification. SCB4 if justification eg showing substitution. SC those working with \((k+c)^2 = 30000\) can score a maximum of B1B0 M1A0 (leads to \(k\approx146\), \(c\approx26.8\)) |
| [4] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dV}{dt} = k\sqrt{V}$ | B1 | cao condone different $k$ (allow MR B1 for $= kV^2$) |
| $V = (\frac{1}{2}kt + c)^2$ | | |
| $\Rightarrow \frac{dV}{dt} = 2(\frac{1}{2}kt+c)\cdot\frac{1}{2}k$ | M1 | $2(\frac{1}{2}kt+c) \times$ constant multiple of $k$ (or from multiplying out oe; or implicit differentiation) |
| $= k(\frac{1}{2}kt + c)$ | A1 | cao www any equivalent form (including unsimplified) |
| $= k\sqrt{V}$ | A1 | Allow SCB2 if $V=(\frac{1}{2}kt+c)^2$ fully obtained by integration including convincing change of constant if used. Can score B1 M0 SCB2 |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\frac{1}{2}k+c)^2 = 10000 \Rightarrow \frac{1}{2}k + c = 100$ | B1 | Substituting any one from $t=1$, $V=10000$ or $t=0$, $V=0$ or $t=2$, $V=40000$ into squared form or rooted form of equation (Allow $-/\pm100$ or $-/\pm200$) |
| $(k+c)^2 = 40000 \Rightarrow k+c = 200$ | B1 | Substituting any other from above |
| $\Rightarrow \frac{1}{2}k = 100 \Rightarrow k = 200,\ c = 0$ | M1 | Solving correct equations for both www (possible solutions are $(200,0)$, $(-200,0)$, $(600,-400)$, $(-600,400)$ (some from $-$ve root)) |
| $\Rightarrow V = (100t)^2 = 10000t^2$ | A1 | either form www. **SC** B2 for $V=(100t)^2$ oe stated without justification. SCB4 if justification eg showing substitution. **SC** those working with $(k+c)^2 = 30000$ can score a maximum of B1B0 M1A0 (leads to $k\approx146$, $c\approx26.8$) |
| **[4]** | | |5 The total value of the sales made by a new company in the first $t$ years of its existence is denoted by $\pounds V$. A model is proposed in which the rate of increase of $V$ is proportional to the square root of $V$. The constant of proportionality is $k$.\\
(i) Express the model as a differential equation.
Verify by differentiation that $V = \left( \frac { 1 } { 2 } k t + c \right) ^ { 2 }$, where $c$ is an arbitrary constant, satisfies this differential equation.\\
(ii) The value of the company's sales in its first year is $\pounds 10000$, and the total value of the sales in the first two years is $\pounds 40000$. Find $V$ in terms of $t$.
\hfill \mbox{\textit{OCR MEI C4 Q5 [8]}}