| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question requiring students to differentiate implicitly, set dy/dx equal to the given gradient, and solve the resulting system. While it involves multiple steps (implicit differentiation, algebraic manipulation, substitution into the original equation, solving a quadratic), these are all routine techniques for C4 students with no novel insight required. The 'show that' in part (i) provides significant scaffolding for part (ii). |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\) | B1 | |
| \(\frac{d}{dx}(-5xy) = (-)(5)\frac{dy}{dx} + (-)(5)y\) | M1 | i.e. reasonably clear use of product rule |
| LHS completely correct \(4x - 5x\frac{dy}{dx} - 5y + 2y\frac{dy}{dx}(= 0)\) | A1 | Accept "\(\frac{dy}{dx} = \)" provided it is not used |
| Substitute \(\frac{dy}{dx} = \frac{3}{8}\) or solve for \(\frac{dy}{dx}\) & then equate to \(\frac{3}{8}\) | M1 | Accuracy not required for "solve for \(\frac{dy}{dx}\)" |
| Produce \(x = 2y\) WWW AG (Converse acceptable) | A1 5 | Expect \(17x = 34y\) and/or \(\frac{dy}{dx} = \frac{5y - 4x}{2y - 5x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(2y\) for \(x\) or \(\frac{1}{2}x\) for \(y\) in curve equation | M1 | |
| Produce either \(x^2 = 36\) or \(y^2 = 9\) | A1 | |
| AEF of \((\pm6, \pm3)\) | A1 3 | ISW Any correct format acceptable |
**(i)**
$\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ | B1 |
$\frac{d}{dx}(-5xy) = (-)(5)\frac{dy}{dx} + (-)(5)y$ | M1 | i.e. reasonably clear use of product rule
LHS completely correct $4x - 5x\frac{dy}{dx} - 5y + 2y\frac{dy}{dx}(= 0)$ | A1 | Accept "$\frac{dy}{dx} = $" provided it is not used
Substitute $\frac{dy}{dx} = \frac{3}{8}$ or solve for $\frac{dy}{dx}$ & then equate to $\frac{3}{8}$ | M1 | Accuracy not required for "solve for $\frac{dy}{dx}$"
Produce $x = 2y$ WWW AG (Converse acceptable) | A1 5 | Expect $17x = 34y$ and/or $\frac{dy}{dx} = \frac{5y - 4x}{2y - 5x}$
**(ii)**
Substitute $2y$ for $x$ or $\frac{1}{2}x$ for $y$ in curve equation | M1 |
Produce either $x^2 = 36$ or $y^2 = 9$ | A1 |
AEF of $(\pm6, \pm3)$ | A1 3 | ISW Any correct format acceptable
---8 The points $P$ and $Q$ lie on the curve with equation
$$2 x ^ { 2 } - 5 x y + y ^ { 2 } + 9 = 0$$
The tangents to the curve at $P$ and $Q$ are parallel, each having gradient $\frac { 3 } { 8 }$.\\
(i) Show that the $x$ - and $y$-coordinates of $P$ and $Q$ are such that $x = 2 y$.\\
(ii) Hence find the coordinates of $P$ and $Q$.
\hfill \mbox{\textit{OCR C4 2011 Q8 [8]}}