| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2009 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard application of the chain rule and product rule, followed by finding a tangent equation at a given point. While it involves multiple steps, the techniques are routine for P3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(2xy + x^2\frac{dy}{dx}\) as derivative of \(x^2y\) | B1 | |
| State \(3y^2\frac{dy}{dx}\) as derivative of \(y^3\) | B1 | |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain answer \(\frac{3x^2 - 2xy}{x^2 + 3y^2}\), or equivalent | A1 | [4] |
| (ii) Find gradient of tangent at (2, 1) and form equation of tangent | M1 | |
| Obtain answer \(8x - 7y - 9 = 0\), or equivalent | A1√ | [2] |
**(i)** State $2xy + x^2\frac{dy}{dx}$ as derivative of $x^2y$ | B1 |
State $3y^2\frac{dy}{dx}$ as derivative of $y^3$ | B1 |
Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 |
Obtain answer $\frac{3x^2 - 2xy}{x^2 + 3y^2}$, or equivalent | A1 | [4]
**(ii)** Find gradient of tangent at (2, 1) and form equation of tangent | M1 |
Obtain answer $8x - 7y - 9 = 0$, or equivalent | A1√ | [2]3 The equation of a curve is $x ^ { 3 } - x ^ { 2 } y - y ^ { 3 } = 3$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(ii) Find the equation of the tangent to the curve at the point $( 2,1 )$, giving your answer in the form $a x + b y + c = 0$.
\hfill \mbox{\textit{CAIE P3 2009 Q3 [6]}}