| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2005 |
| Session | November |
| Marks | 7 |
| 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 routine tangent equation finding. The algebra is manageable and the method is well-practiced, making it slightly easier than average for A-level. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(3y^2 = \frac{dy}{dx}\) as derivative of \(y^3\) | B1 | |
| State \(9y + y = \frac{dy}{dx}\) as derivative of \(9xy\) | B1 | |
| Express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) | M1 | |
| Obtain given answer correctly [The M1 is conditional on at least one B mark being obtained.] | A1 | 4 |
| (ii) Obtain gradient at (2, 4) in any correct unsimplified form | B1 | |
| Form the equation of the tangent at (2, 4) | M1 | |
| Obtain answer \(5y - 4x = 12\), or equivalent | A1 | 3 |
(i) State $3y^2 = \frac{dy}{dx}$ as derivative of $y^3$ | B1 |
State $9y + y = \frac{dy}{dx}$ as derivative of $9xy$ | B1 |
Express $\frac{dy}{dx}$ in terms of $x$ and $y$ | M1 |
Obtain given answer correctly [The M1 is conditional on at least one B mark being obtained.] | A1 | 4
(ii) Obtain gradient at (2, 4) in any correct unsimplified form | B1 |
Form the equation of the tangent at (2, 4) | M1 |
Obtain answer $5y - 4x = 12$, or equivalent | A1 | 34 The equation of a curve is $x ^ { 3 } + y ^ { 3 } = 9 x y$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y - x ^ { 2 } } { y ^ { 2 } - 3 x }$.\\
(ii) Find the equation of the tangent to the curve at the point ( 2,4 ), giving your answer in the form $a x + b y = c$.
\hfill \mbox{\textit{CAIE P2 2005 Q4 [7]}}