| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show dy/dx equals given expression |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question requiring application of the product rule and chain rule, followed by algebraic rearrangement. Part (ii) is trivial (just reciprocate). The 'show that' format makes it easier as students know the target answer. Slightly below average difficulty due to being a standard textbook-style exercise with clear methodology. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(y^3 = 2xy + x^2 \Rightarrow 3y^2\frac{dy}{dx} = 2x\frac{dy}{dx} + 2y + 2x\) | B1 | \(3y^2\frac{dy}{dx} =\) |
| \((3y^2 - 2x)\frac{dy}{dx} = 2y + 2x\) | B1 | \(2x\frac{dy}{dx} + 2y + 2x\) |
| \(\frac{dy}{dx} = \frac{2(x+y)}{3y^2-2x}\) * | M1, E1 | Collecting \(\frac{dy}{dx}\) terms on one side; www |
| (ii) \(\frac{dx}{dy} = \frac{3y^2-2x}{2(x+y)}\) | B1cao |
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $y^3 = 2xy + x^2 \Rightarrow 3y^2\frac{dy}{dx} = 2x\frac{dy}{dx} + 2y + 2x$ | B1 | $3y^2\frac{dy}{dx} =$ |
| $(3y^2 - 2x)\frac{dy}{dx} = 2y + 2x$ | B1 | $2x\frac{dy}{dx} + 2y + 2x$ |
| $\frac{dy}{dx} = \frac{2(x+y)}{3y^2-2x}$ * | M1, E1 | Collecting $\frac{dy}{dx}$ terms on one side; www |
| **(ii)** $\frac{dx}{dy} = \frac{3y^2-2x}{2(x+y)}$ | B1cao | |
---5 A curve is defined implicitly by the equation
$$y ^ { 3 } = 2 x y + x ^ { 2 }$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( x + y ) } { 3 y ^ { 2 } - 2 x }$.\\
(ii) Hence write down $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $x$ and $y$.
\hfill \mbox{\textit{OCR MEI C3 2006 Q5 [5]}}