| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Moderate -0.3 This is a standard C4 parametric equations question requiring routine differentiation (dy/dx = dy/dt รท dx/dt) and trigonometric manipulation. Part (a) involves straightforward substitution and gradient calculation, while part (b) uses standard trigonometric identities (compound angle formula and Pythagorean identity). Both parts follow well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-05_480_1059_313_438}
\end{center}
\end{figure}
The curve shown in Figure 2 has parametric equations
$$x = \sin t , y = \sin \left( t + \frac { \pi } { 6 } \right) , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the tangent to the curve at the point where $t = \frac { \pi } { 6 }$.
\item Show that a cartesian equation of the curve is
$$y = \frac { \sqrt { } 3 } { 2 } x + \frac { 1 } { 2 } \sqrt { } \left( 1 - x ^ { 2 } \right) , \quad - 1 < x < 1$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2006 Q4 [9]}}