| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Particle motion - velocity/displacement (v dv/dx = f(v,x)) |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard C4 techniques: separating variables in a differential equation, verifying initial conditions, using R-cos(θ-α) form, and finding maxima. Each part follows directly from the previous with clear signposting and no novel problem-solving required. Slightly easier than average due to the guided structure and routine methods. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v\frac{dv}{dx} + 4x = 0\) | ||
| \(\int v\, dv = -\int 4x\, dx\) | M1 | Separating variables and intending to integrate |
| \(\frac{1}{2}v^2 = -2x^2 + c\) | A1 | oe condone absence of \(c\). [Not immediate \(v^2 = -4x^2 (+c)\)] |
| When \(x = 1\), \(v = 4\), so \(c = 10\) | B1 | Finding \(c\), must be convinced as AG, need to see at least the statement given here oe (condone change of \(c\)) |
| \(v^2 = 20 - 4x^2\) | A1 | AG following finding \(c\) convincingly. Alternatively SC \(v^2=20-4x^2\), by differentiation \(2v\frac{dv}{dx}=-8x\), so \(v\frac{dv}{dx}+4x=0\) scores B2; if initial conditions checked a further B1 scored. Total possible 3/4. |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \cos 2t + 2\sin 2t\) | ||
| When \(t=0\), \(x = \cos 0 + 2\sin 0 = 1\) | B1 | AG need some justification |
| \(v = \frac{dx}{dt} = -2\sin 2t + 4\cos 2t\) | M1 | Differentiating, accept \(\pm 2, \pm 4\) as coefficients but not \(\pm 1, \pm 2\) and not \(\pm\frac{1}{2}, \pm 1\) from integrating |
| A1 | cao | |
| \(v = 4\cos 0 - 2\sin 0 = 4\) | A1 | ww AG |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos 2t + 2\sin 2t = R\cos(2t - \alpha) = R(\cos 2t \cos\alpha + \sin 2t \sin\alpha)\) | SEE APPENDIX 1 for further guidance | |
| \(R = \sqrt{5}\) | B1 | or 2.24 or better (not \(\pm\) unless negative rejected) |
| \(R\cos\alpha = 1\), \(R\sin\alpha = 2\) | M1 | Correct pairs soi |
| \(\tan\alpha = 2\) | M1 | Correct method |
| \(\alpha = 1.107\) | A1 | cao radians only, 1.11 or better (or multiples of \(\pi\) that round to 1.11) |
| \(x = \sqrt{5}\cos(2t - 1.107)\) | ||
| \(v = -2\sqrt{5}\sin(2t - 1.107)\) | A1 | Differentiating or otherwise, ft their numerical \(R\), \(\alpha\) (not degrees). Required form. SC B1 for \(v = \sqrt{20}\cos(2t + 0.464)\) oe |
| EITHER \(v^2 = 20\sin^2(2t-\alpha)\) | ||
| \(20 - 4x^2 = 20 - 20\cos^2(2t-\alpha)\) | M1 | Squaring their \(v\) (if of required form with same \(\alpha\) as \(x\)), and \(x\), and attempting to show \(v^2 = 20-4x^2\) ft their \(R\), \(\alpha\) (incl. degrees) [\(\alpha\) may not be specified] |
| \(= 20(1-\cos^2(2t-\alpha)) = 20\sin^2(2t-\alpha)\) | A1 | cao www (condone the use of over-rounded \(\alpha\) (radians) or degrees) |
| so \(v^2 = 20 - 4x^2\) | ||
| OR multiplying out \(v^2 = (-2\sin 2t + 4\cos 2t)^2 = 4\sin^2 2t - 16\sin 2t\cos 2t + 16\cos^2 2t\) and \(4x^2 = 4(\cos^2 2t + 4\sin 2t\cos 2t + 4\sin^2 2t) = 4\cos^2 2t + 16\sin 2t\cos 2t + 16\sin^2 2t\) (need middle term) and attempting to show \(v^2 + 4x^2 = 4(\sin^2 2t+\cos^2 2t) + 16(\cos^2 2t+\sin^2 2t) = 4+16 = 20\) | M1 | Differentiating to find \(v\) (condone coefficient errors), squaring \(v\) and \(x\) and multiplying out (need attempt at middle terms) and attempting to show \(v^2 = 20-4x^2\) |
| so \(v^2 = 20 - 4x^2\) | A1 | cao www |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \sqrt{5}\cos(2t-\alpha)\) or otherwise | ||
| \(x_{\max} = \sqrt{5}\) | B1 | ft their \(R\) |
| When \(\cos(2t-\alpha)=1\), \(2t - 1.107 = 0\), \(2t = 1.107\), \(t = 0.55\) | M1 | oe (say by differentiation) ft their \(\alpha\) in radians or degrees for method only |
| A1 | cao (or answers that round to 0.554) | |
| [3] |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v\frac{dv}{dx} + 4x = 0$ | | |
| $\int v\, dv = -\int 4x\, dx$ | M1 | Separating variables and intending to integrate |
| $\frac{1}{2}v^2 = -2x^2 + c$ | A1 | oe condone absence of $c$. [Not immediate $v^2 = -4x^2 (+c)$] |
| When $x = 1$, $v = 4$, so $c = 10$ | B1 | Finding $c$, must be convinced as AG, need to see at least the statement given here oe (condone change of $c$) |
| $v^2 = 20 - 4x^2$ | A1 | **AG** following finding $c$ convincingly. Alternatively **SC** $v^2=20-4x^2$, by differentiation $2v\frac{dv}{dx}=-8x$, so $v\frac{dv}{dx}+4x=0$ scores B2; if initial conditions checked a further B1 scored. Total possible 3/4. |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \cos 2t + 2\sin 2t$ | | |
| When $t=0$, $x = \cos 0 + 2\sin 0 = 1$ | B1 | **AG** need some justification |
| $v = \frac{dx}{dt} = -2\sin 2t + 4\cos 2t$ | M1 | Differentiating, accept $\pm 2, \pm 4$ as coefficients but not $\pm 1, \pm 2$ and not $\pm\frac{1}{2}, \pm 1$ from integrating |
| | A1 | cao |
| $v = 4\cos 0 - 2\sin 0 = 4$ | A1 | ww **AG** |
| **[4]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 2t + 2\sin 2t = R\cos(2t - \alpha) = R(\cos 2t \cos\alpha + \sin 2t \sin\alpha)$ | | **SEE APPENDIX 1 for further guidance** |
| $R = \sqrt{5}$ | B1 | or 2.24 or better (not $\pm$ unless negative rejected) |
| $R\cos\alpha = 1$, $R\sin\alpha = 2$ | M1 | Correct pairs soi |
| $\tan\alpha = 2$ | M1 | Correct method |
| $\alpha = 1.107$ | A1 | cao radians only, 1.11 or better (or multiples of $\pi$ that round to 1.11) |
| $x = \sqrt{5}\cos(2t - 1.107)$ | | |
| $v = -2\sqrt{5}\sin(2t - 1.107)$ | A1 | Differentiating or otherwise, ft their numerical $R$, $\alpha$ (not degrees). Required form. **SC** B1 for $v = \sqrt{20}\cos(2t + 0.464)$ oe |
| **EITHER** $v^2 = 20\sin^2(2t-\alpha)$ | | |
| $20 - 4x^2 = 20 - 20\cos^2(2t-\alpha)$ | M1 | Squaring their $v$ (if of required form with same $\alpha$ as $x$), and $x$, and attempting to show $v^2 = 20-4x^2$ ft their $R$, $\alpha$ (incl. degrees) [$\alpha$ may not be specified] |
| $= 20(1-\cos^2(2t-\alpha)) = 20\sin^2(2t-\alpha)$ | A1 | cao www (condone the use of over-rounded $\alpha$ (radians) or degrees) |
| so $v^2 = 20 - 4x^2$ | | |
| **OR** multiplying out $v^2 = (-2\sin 2t + 4\cos 2t)^2 = 4\sin^2 2t - 16\sin 2t\cos 2t + 16\cos^2 2t$ and $4x^2 = 4(\cos^2 2t + 4\sin 2t\cos 2t + 4\sin^2 2t) = 4\cos^2 2t + 16\sin 2t\cos 2t + 16\sin^2 2t$ (need middle term) and attempting to show $v^2 + 4x^2 = 4(\sin^2 2t+\cos^2 2t) + 16(\cos^2 2t+\sin^2 2t) = 4+16 = 20$ | M1 | Differentiating to find $v$ (condone coefficient errors), squaring $v$ and $x$ and multiplying out (need attempt at middle terms) and attempting to show $v^2 = 20-4x^2$ |
| so $v^2 = 20 - 4x^2$ | A1 | cao www |
| **[7]** | | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \sqrt{5}\cos(2t-\alpha)$ or otherwise | | |
| $x_{\max} = \sqrt{5}$ | B1 | ft their $R$ |
| When $\cos(2t-\alpha)=1$, $2t - 1.107 = 0$, $2t = 1.107$, $t = 0.55$ | M1 | oe (say by differentiation) ft their $\alpha$ in radians or degrees for method only |
| | A1 | cao (or answers that round to 0.554) |
| **[3]** | | |
---4 The motion of a particle is modelled by the differential equation
$$v \frac { \mathrm {~d} v } { \mathrm {~d} x } + 4 x = 0$$
where $x$ is its displacement from a fixed point, and $v$ is its velocity.
Initially $x = 1$ and $v = 4$.\\
(i) Solve the differential equation to show that $v ^ { 2 } = 20 - 4 x ^ { 2 }$.
Now consider motion for which $x = \cos 2 t + 2 \sin 2 t$, where $x$ is the displacement from a fixed point at time $t$.\\
(ii) Verify that, when $t = 0 , x = 1$. Use the fact that $v = \frac { \mathrm { d } x } { \mathrm {~d} t }$ to verify that when $t = 0 , v = 4$.\\
(iii) Express $x$ in the form $R \cos ( 2 t - \alpha )$, where $R$ and $\alpha$ are constants to be determined, and obtain the corresponding expression for $v$. Hence or otherwise verify that, for this motion too, $v ^ { 2 } = 20 - 4 x ^ { 2 }$.\\
(iv) Use your answers to part (iii) to find the maximum value of $x$, and the earliest time at which $x$ reaches this maximum value.
\hfill \mbox{\textit{OCR MEI C4 Q4 [18]}}