| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard technique (product rule and chain rule) followed by solving a simple equation. Part (a) is routine verification, and part (b) requires recognizing that vertical tangents occur when the denominator equals zero, leading to a quadratic equation. The multi-step nature and 4-mark allocation place it slightly above average, but no novel insight is required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(2xy + x^2\frac{dy}{dx}\) as derivative of \(x^2y\) | B1 | Accept partial: \(\frac{\partial}{\partial x} \to 2xy\) |
| State or imply \(2ay\frac{dy}{dx}\) as derivative of \(ay^2\) | B1 | Accept partial: \(\frac{\partial}{\partial y} \to x^2 - 2ay\) |
| Equate attempted derivative to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain answer \(\frac{dy}{dx} = \frac{2xy}{2ay - x^2}\) from correct working | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(2ay - x^2 = 0\) | *M1 | |
| Substitute into equation of curve to obtain equation in \(x\) and \(a\) or in \(y\) and \(a\) | DM1 | e.g. \(2ay^2 - ay^2 = 4a^3\) or \(\frac{x^4}{2a} - \frac{x^4}{4a} = 4a^3\) |
| Obtain one correct point | A1 | e.g. \((2a, 2a)\) |
| Obtain second correct point and no others | A1 | e.g. \((-2a, 2a)\) |
| 4 | SC: Allow A1 A0 for \(x = \pm 2a\) or for \(y = 2a\) |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $2xy + x^2\frac{dy}{dx}$ as derivative of $x^2y$ | B1 | Accept partial: $\frac{\partial}{\partial x} \to 2xy$ |
| State or imply $2ay\frac{dy}{dx}$ as derivative of $ay^2$ | B1 | Accept partial: $\frac{\partial}{\partial y} \to x^2 - 2ay$ |
| Equate attempted derivative to zero and solve for $\frac{dy}{dx}$ | M1 | |
| Obtain answer $\frac{dy}{dx} = \frac{2xy}{2ay - x^2}$ from correct working | A1 | AG |
| | **4** | |
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## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $2ay - x^2 = 0$ | *M1 | |
| Substitute into equation of curve to obtain equation in $x$ and $a$ or in $y$ and $a$ | DM1 | e.g. $2ay^2 - ay^2 = 4a^3$ or $\frac{x^4}{2a} - \frac{x^4}{4a} = 4a^3$ |
| Obtain one correct point | A1 | e.g. $(2a, 2a)$ |
| Obtain second correct point and no others | A1 | e.g. $(-2a, 2a)$ |
| | **4** | SC: Allow A1 A0 for $x = \pm 2a$ or for $y = 2a$ |
---5 The equation of a curve is $x ^ { 2 } y - a y ^ { 2 } = 4 a ^ { 3 }$, where $a$ is a non-zero constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x y } { 2 a y - x ^ { 2 } }$.
\item Hence find the coordinates of the points where the tangent to the curve is parallel to the $y$-axis. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q5 [8]}}