| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (tan/sec/cot/cosec identities) |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine techniques: finding dy/dx using the chain rule, finding a tangent equation at a specific point, and converting to Cartesian form using trigonometric identities. While it involves multiple parts and requires knowledge of cot t and sin²t identities, these are well-practiced techniques with no novel problem-solving required. Slightly easier than average due to the straightforward nature of each part. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| (a) \(\frac{dx}{dt} = -2\cos ec^2 t, \quad \frac{dy}{dt} = 4\sin t\cos t\) | both M1 A1 |
| \(\frac{dy}{dx} = \frac{-2\sin t\cos t}{\cos ec^2 t} \left(= -2\sin^3 t\cos t\right)\) | M1 A1 |
| Answer | Marks |
|---|---|
| (b) At \(t = \frac{\pi}{4}, x = 2, y = 1\) | both B1 |
| Substitutes \(t = \frac{\pi}{4}\) into an attempt at \(\frac{dy}{dx}\) to obtain gradient \(\left(-\frac{1}{2}\right)\) | M1 |
| Equation of tangent is \(y - 1 = -\frac{1}{2}(x-2)\) | M1 A1 |
| Accept \(x + 2y = 4\) or any correct equivalent |
| Answer | Marks |
|---|---|
| (c) Uses \(1 + \cot^2 t = \cos ec^2 t\), or equivalent, to eliminate \(t\) | M1 |
| \(1 + \left(\frac{x}{2}\right)^2 = \frac{2}{y}\) | correctly eliminates \(t\) A1 |
| \(y = \frac{8}{4+x^2}\) | cao A1 |
| The domain is \(x > 0\) | B1 |
| Answer | Marks |
|---|---|
| \(\sin t = \left(\frac{y}{2}\right)^{\frac{1}{2}}, \quad \cos t = \frac{x}{2}\sin t = \frac{x}{2}\left(\frac{y}{2}\right)^{\frac{1}{2}}\) | |
| \(\sin^2 t + \cos^2 t = 1 \Rightarrow \frac{y}{2} + \frac{x^2}{4} \times \frac{y}{2} = 1\) | M1 A1 |
| Leading to \(y = \frac{8}{4+x^2}\) | A1 |
**(a)** $\frac{dx}{dt} = -2\cos ec^2 t, \quad \frac{dy}{dt} = 4\sin t\cos t$ | both M1 A1 |
$\frac{dy}{dx} = \frac{-2\sin t\cos t}{\cos ec^2 t} \left(= -2\sin^3 t\cos t\right)$ | M1 A1 |
**Total for (a): [4]**
**(b)** At $t = \frac{\pi}{4}, x = 2, y = 1$ | both B1 |
Substitutes $t = \frac{\pi}{4}$ into an attempt at $\frac{dy}{dx}$ to obtain gradient $\left(-\frac{1}{2}\right)$ | M1 |
Equation of tangent is $y - 1 = -\frac{1}{2}(x-2)$ | M1 A1 |
Accept $x + 2y = 4$ or any correct equivalent | |
**Total for (b): [4]**
**(c)** Uses $1 + \cot^2 t = \cos ec^2 t$, or equivalent, to eliminate $t$ | M1 |
$1 + \left(\frac{x}{2}\right)^2 = \frac{2}{y}$ | correctly eliminates $t$ A1 |
$y = \frac{8}{4+x^2}$ | cao A1 |
The domain is $x > 0$ | B1 |
**Total for (c): [4]**
**An alternative in (c):**
$\sin t = \left(\frac{y}{2}\right)^{\frac{1}{2}}, \quad \cos t = \frac{x}{2}\sin t = \frac{x}{2}\left(\frac{y}{2}\right)^{\frac{1}{2}}$ | |
$\sin^2 t + \cos^2 t = 1 \Rightarrow \frac{y}{2} + \frac{x^2}{4} \times \frac{y}{2} = 1$ | M1 A1 |
Leading to $y = \frac{8}{4+x^2}$ | A1 |
**Total: [12]**
---\begin{enumerate}
\item A curve has parametric equations
\end{enumerate}
$$x = 2 \cot t , \quad y = 2 \sin ^ { 2 } t , \quad 0 < t \leqslant \frac { \pi } { 2 }$$
(a) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of the parameter $t$.\\
(b) Find an equation of the tangent to the curve at the point where $t = \frac { \pi } { 4 }$.\\
(c) Find a cartesian equation of the curve in the form $y = \mathrm { f } ( x )$. State the domain on which the curve is defined.\\
\hfill \mbox{\textit{Edexcel C4 2005 Q6 [12]}}