| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Moderate -0.8 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/du)/(dx/du), followed by finding the parameter value at the given point and substituting. Both parts require only routine calculus techniques with no problem-solving insight needed, making it easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(dx/du = 2u\), \(dy/du = 6u^2\) | B1 | both \(2u\) and \(6u^2\) |
| \(\frac{dy}{dx} = \frac{dy/du}{dx/du} = \frac{6u^2}{2u} = 3u\) | M1 A1 | |
| OR \(y = 2(x-1)^{3/2}\), \(dy/dx = 3(x-1)^{1/2} = 3u\) | B1(\(y=f(x)\)), M1 differentiation, A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \((5, 16)\), \(u = 2\) | M1 | |
| \(dy/dx = 6\) | A1 | cao |
| [2] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $dx/du = 2u$, $dy/du = 6u^2$ | B1 | both $2u$ and $6u^2$ |
| $\frac{dy}{dx} = \frac{dy/du}{dx/du} = \frac{6u^2}{2u} = 3u$ | M1 A1 | |
| **OR** $y = 2(x-1)^{3/2}$, $dy/dx = 3(x-1)^{1/2} = 3u$ | | B1($y=f(x)$), M1 differentiation, A1 |
| **[3]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $(5, 16)$, $u = 2$ | M1 | |
| $dy/dx = 6$ | A1 | cao |
| **[2]** | | |
---7 A curve has parametric equations $x = 1 + u ^ { 2 } , y = 2 u ^ { 3 }$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $u$.\\
(ii) Hence find the gradient of the curve at the point with coordinates $( 5,16 )$.
\hfill \mbox{\textit{OCR MEI C4 Q7 [5]}}