| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a standard parametric equations question covering routine techniques: finding intersection points by setting y=0, finding parameter values, computing dy/dx using the chain rule, and converting to Cartesian form using trigonometric identities. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.03h Parametric equations: in modelling contexts1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \((3,0)\) and \((-3,0)\) | B1, B1 | One B1 for one correct value of \(x\) or for seeing \((0,3)\) or \((0,-3)\); second B1 for both coordinates correct; ignore any reference to \(A\), \(B\) or \(O\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\) | M1 A1 | M1 for one correct in degrees or radians; A1 for both correct in radians and no others inside the range |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{18\cos 2t}{-3\sin t}\) | M1 A1 | M1: attempts to differentiate both \(x\) and \(y\) wrt \(t\) and uses \(\frac{dy/dt}{dx/dt}\); A1: correct result with no errors e.g. \(\frac{dy}{dx} = \frac{18\cos^2 t - 18\sin^2 t}{-3\sin t}\) |
| \(= \frac{18 \times \frac{1}{2}}{-3 \times \frac{1}{2}} = -6\) | dM1 A1 | dM1: substitute \(t = \frac{\pi}{6}\); A1: cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(y^2 = 81 \times 4\sin^2 t\cos^2 t\) | M1 | Attempts to use double angle formula for \(\sin 2t\) to reach \(y = 18\sin t\cos t\) or equivalent |
| Use \(\cos^2 t = \frac{x^2}{9}\) and \(\sin^2 t = 1 - \frac{x^2}{9}\); or \(\cos t = \frac{x}{3}\) and \(\sin t = \sqrt{1-\frac{x^2}{9}}\) | M1 | Uses correct trig identities; condone \(x = \cos t\) for this mark |
| Correct equation \(y^2 = 81\times4\times\left(1-\frac{x^2}{9}\right)\times\frac{x^2}{9}\); or \(y = 9\times2\times\sqrt{1-\frac{x^2}{9}}\times\frac{x}{3}\) | A1 | |
| \(y^2 = 4x^2(9-x^2)\) | A1 | cso |
# Question 11:
## Part (a):
| $(3,0)$ and $(-3,0)$ | B1, B1 | One B1 for one correct value of $x$ or for seeing $(0,3)$ or $(0,-3)$; second B1 for both coordinates correct; ignore any reference to $A$, $B$ or $O$ | **(2)** |
## Part (b):
| $t = \frac{\pi}{2}$ and $t = \frac{3\pi}{2}$ | M1 A1 | M1 for one correct in degrees or radians; A1 for both correct in radians and no others inside the range | **(2)** |
## Part (c):
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{18\cos 2t}{-3\sin t}$ | M1 A1 | M1: attempts to differentiate both $x$ and $y$ wrt $t$ and uses $\frac{dy/dt}{dx/dt}$; A1: correct result with no errors e.g. $\frac{dy}{dx} = \frac{18\cos^2 t - 18\sin^2 t}{-3\sin t}$ |
| $= \frac{18 \times \frac{1}{2}}{-3 \times \frac{1}{2}} = -6$ | dM1 A1 | dM1: substitute $t = \frac{\pi}{6}$; A1: cso | **(4)** |
## Part (d):
| $y^2 = 81 \times 4\sin^2 t\cos^2 t$ | M1 | Attempts to use double angle formula for $\sin 2t$ to reach $y = 18\sin t\cos t$ or equivalent |
| Use $\cos^2 t = \frac{x^2}{9}$ and $\sin^2 t = 1 - \frac{x^2}{9}$; or $\cos t = \frac{x}{3}$ and $\sin t = \sqrt{1-\frac{x^2}{9}}$ | M1 | Uses correct trig identities; condone $x = \cos t$ for this mark |
| Correct equation $y^2 = 81\times4\times\left(1-\frac{x^2}{9}\right)\times\frac{x^2}{9}$; or $y = 9\times2\times\sqrt{1-\frac{x^2}{9}}\times\frac{x}{3}$ | A1 | |
| $y^2 = 4x^2(9-x^2)$ | A1 | cso | **(4)** |
**(12 marks)**
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7d07e1ad-d87a-4eb5-a15e-05b927892915-32_858_743_118_603}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The curve $C$ shown in Figure 3 has parametric equations
$$x = 3 \cos t , \quad y = 9 \sin 2 t , \quad 0 \leqslant t \leqslant 2 \pi$$
The curve $C$ meets the $x$-axis at the origin and at the points $A$ and $B$, as shown in Figure 3 .
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of $A$ and $B$.
\item Find the values of $t$ at which the curve passes through the origin.
\item Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, and hence find the gradient of the curve when $t = \frac { \pi } { 6 }$
\item Show that the cartesian equation for the curve $C$ can be written in the form
$$y ^ { 2 } = a x ^ { 2 } \left( b - x ^ { 2 } \right)$$
where $a$ and $b$ are integers to be determined.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q11 [12]}}