| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Derivative involving harmonic form |
| Difficulty | Standard +0.3 Part (i) is straightforward substitution. Part (ii) requires applying the chain rule (dx/dt = dx/dθ × dθ/dt) with a simple derivative of a sum of trig functions. This is a standard C3 related rates question with clear guidance ('show that'), requiring only routine differentiation and multiplication—slightly easier than average due to its structured nature and minimal problem-solving demand. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 5\sin\frac{\pi}{2} - 4\cos\frac{\pi}{2} = 5\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dx}{d\theta} = 5\cos\theta + 4\sin\theta\): When \(\theta = \frac{\pi}{2}\), \(\frac{dx}{d\theta} = 4\) | M1, A1, A1 | |
| \(\frac{dx}{dt} = \frac{dx}{d\theta} \times \frac{d\theta}{dt} = 4 \times 0.1 = 0.4\) | M1, E1 | |
| Total: 5 |
## Question 7:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 5\sin\frac{\pi}{2} - 4\cos\frac{\pi}{2} = 5$ | B1 | |
| **Total: 1** | | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dx}{d\theta} = 5\cos\theta + 4\sin\theta$: When $\theta = \frac{\pi}{2}$, $\frac{dx}{d\theta} = 4$ | M1, A1, A1 | |
| $\frac{dx}{dt} = \frac{dx}{d\theta} \times \frac{d\theta}{dt} = 4 \times 0.1 = 0.4$ | M1, E1 | |
| **Total: 5** | | |
---7 Two quantities, $x$ and $\theta$, vary with time and are related by the equation $x = 5 \sin \theta - 4 \cos \theta$.\\
(i) Find the value of $x$ when $\theta = \frac { \pi } { 2 }$.\\
(ii) When $\theta = \frac { \pi } { 2 }$, its rate of increase (in suitable units) is given by $\frac { \mathrm { d } \theta } { \mathrm { d } t } = 0.1$. Show that at that moment $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4$.
\hfill \mbox{\textit{OCR MEI C3 Q7 [6]}}