| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Standard +0.3 This is a straightforward C3 chain rule question with standard steps: finding an intercept, differentiating using chain rule (though dx/dy rather than dy/dx), and finding a tangent equation. The implicit differentiation aspect and dx/dy orientation add slight complexity beyond routine exercises, but all techniques are standard and the question is well-scaffolded across three parts. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| (a) \(x = 3\) or \((3, 0)\) | B1 |
| (b) \(\frac{dx}{dy} = \frac{1}{2}(9 + 16y - 2y^2)^{-\frac{1}{2}}(16 - 4y)\) or equivalent | M1, M1, A1 |
| (c) Substitute \(y = 0\) into their \(\frac{dy}{dx}\) or \(\frac{dx}{dy}\) | M1 |
| \(\frac{dx}{dy} = \frac{8}{3}\) or \(\frac{dy}{dx} = \frac{3}{8}\) | A1 |
| Uses their numerical value and their \(\frac{dy}{dx}\) from \((3, 0)\) to find equation of tangent | M1 |
| \(\frac{y - 0}{x - 3} = \frac{3}{8}\) or \(y - 0 = \frac{3}{8}(x - 3)\) | A1 |
(a) $x = 3$ or $(3, 0)$ | B1
(b) $\frac{dx}{dy} = \frac{1}{2}(9 + 16y - 2y^2)^{-\frac{1}{2}}(16 - 4y)$ or equivalent | M1, M1, A1
(c) Substitute $y = 0$ into their $\frac{dy}{dx}$ or $\frac{dx}{dy}$ | M1
$\frac{dx}{dy} = \frac{8}{3}$ or $\frac{dy}{dx} = \frac{3}{8}$ | A1
Uses their numerical value and their $\frac{dy}{dx}$ from $(3, 0)$ to find equation of tangent | M1
$\frac{y - 0}{x - 3} = \frac{3}{8}$ or $y - 0 = \frac{3}{8}(x - 3)$ | A19.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0f6fd881-4d4b-4f80-96cc-6da41cc33c60-16_570_903_237_534}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve with equation $x = \left( 9 + 16 y - 2 y ^ { 2 } \right) ^ { \frac { 1 } { 2 } }$.\\
The curve crosses the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $A$.
\item Find an expression for $\frac { \mathrm { d } x } { \mathrm {~d} y }$, in terms of $y$.
\item Find an equation of the tangent to the curve at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q9 [8]}}