| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard application of the product rule and chain rule, followed by routine normal line calculation. The algebra is manageable and the methods are textbook exercises, making it slightly easier than average for A-level. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(3\frac{dy}{dx}\) as derivative of \(3y\), or equivalent | B1 | |
| State \(4xy + 2x^2\frac{dy}{dx}\) as a derivative of \(2x^2y\), or equivalent | B1 | |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain given answer correctly | A1 | [4] |
| (ii) Substitute \(x = 2\) into given equation and solve for \(y\) | M1 | |
| Obtain gradient \(= \frac{12}{5}\) correctly | A1 | |
| Form equation of the normal at their point, using negative recip of their \(\frac{dy}{dx}\) | M1 | |
| State correct equation of normal \(5x + 12y + 2 = 0\) or equivalent | A1 | [4] |
**(i)** State $3\frac{dy}{dx}$ as derivative of $3y$, or equivalent | B1 |
State $4xy + 2x^2\frac{dy}{dx}$ as a derivative of $2x^2y$, or equivalent | B1 |
Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 |
Obtain given answer correctly | A1 | [4]
**(ii)** Substitute $x = 2$ into given equation and solve for $y$ | M1 |
Obtain gradient $= \frac{12}{5}$ correctly | A1 |
Form equation of the normal at their point, using negative recip of their $\frac{dy}{dx}$ | M1 |
State correct equation of normal $5x + 12y + 2 = 0$ or equivalent | A1 | [4]5 The equation of a curve is
$$x ^ { 2 } - 2 x ^ { 2 } y + 3 y = 9$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 4 x y } { 2 x ^ { 2 } - 3 }$.\\
(ii) Find the equation of the normal to the curve at the point where $x = 2$, giving your answer in the form $a x + b y + c = 0$.
\hfill \mbox{\textit{CAIE P2 2013 Q5 [8]}}