| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 Part (a) is standard implicit differentiation with straightforward algebra to find dy/dx at a given point. Part (b) requires finding a normal equation using parametric coordinates and then proving an inequality, but follows routine techniques (differentiate, find gradient, write normal equation, analyze x-intercept). The multi-step nature and inequality proof elevate it slightly above average, but all techniques are standard A-level fare with no novel insights required. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(4x^3 + 2xy + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0\) | B2 | (B2, 4 correct terms) (B1, 3 correct terms) |
| Now, \(x = -1, y = 3\) | ||
| so that \(-4 - 6 + \frac{dy}{dx} + 6\frac{dy}{dx} = 0\) | B1 | |
| \(\frac{dy}{dx} = \frac{10}{7}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy}{dp} \cdot \frac{dp}{dx} = \frac{2}{2p} = \frac{1}{p}\) | M1, A1 | |
| Gradient of normal is \(-p\) | B1 | |
| Equation of normal is \((y - 2p) = -p(x - p^2)\) | m1 | |
| \(y - 2p = -px + p^3\) | ||
| so that \(y + px = 2p + p^3\) | A1 | (convincing) |
| When \(y = 0\), \(x = b\): \(b = 2 + p^2\) | B1 | |
| Since \(p^2 > 0\), \(b > 2\) | E1 |
### Part (a):
$4x^3 + 2xy + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$ | B2 | (B2, 4 correct terms) (B1, 3 correct terms)
Now, $x = -1, y = 3$ | |
so that $-4 - 6 + \frac{dy}{dx} + 6\frac{dy}{dx} = 0$ | B1 |
$\frac{dy}{dx} = \frac{10}{7}$ | B1 |
### Part (b):
$\frac{dy}{dx} = \frac{dy}{dp} \cdot \frac{dp}{dx} = \frac{2}{2p} = \frac{1}{p}$ | M1, A1 |
Gradient of normal is $-p$ | B1 |
Equation of normal is $(y - 2p) = -p(x - p^2)$ | m1 |
$y - 2p = -px + p^3$ | |
so that $y + px = 2p + p^3$ | A1 | (convincing)
When $y = 0$, $x = b$: $b = 2 + p^2$ | B1 |
Since $p^2 > 0$, $b > 2$ | E1 |
**Total: [11]**\begin{enumerate}[label=(\alph*)]
\item The curve $C$ is given by the equation
$$x^4 + x^2 y + y^2 = 13.$$
Find the value of $\frac{dy}{dx}$ at the point $(-1, 3)$. [4]
\item Show that the equation of the normal to the curve $y^2 = 4x$ at the point $P(p^2, 2p)$ is
$$y + px = 2p + p^3.$$
Given that $p \neq 0$ and that the normal at $P$ cuts the $x$-axis at $B(b, 0)$, show that $b > 2$. [7]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q11 [11]}}