| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| 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 C4 parametric equations question requiring routine techniques: differentiation using the chain rule (dy/dx = (dy/dt)/(dx/dt)), and converting to Cartesian form using the double angle identity cos(2t) = 1 - 2sin²(t). The range follows directly from the domain. Slightly above average difficulty due to multiple parts and requiring recall of trigonometric identities, but all techniques are standard textbook exercises. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = 2\cos t\), \(\frac{dy}{dt} = 2\sin 2t\) or \(\frac{dy}{dt} = 4\sin t \cos t\) | B1 | At least one of \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\) correct |
| Both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) correct | B1 | Can be implied from working |
| \(\frac{dy}{dx} = \frac{2\sin 2t}{2\cos t} \left\{= \frac{4\cos t \sin t}{2\cos t} = 2\sin t\right\}\) | M1 | Applies \(\frac{dy}{dt}\) divided by \(\frac{dx}{dt}\) AND substitutes \(t = \frac{\pi}{6}\) into their \(\frac{dy}{dx}\) |
| At \(t = \frac{\pi}{6}\), \(\frac{dy}{dx} = \frac{2\sin\left(\frac{2\pi}{6}\right)}{2\cos\left(\frac{\pi}{6}\right)} = 1\) | A1 cao cso | Correct value of \(\frac{dy}{dx}\) of 1. Note: Don't accept incorrect methods leading to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 1 - \cos 2t = 1-(1-2\sin^2 t) = 2\sin^2 t\) | M1 | Uses correct double angle formula \(\cos 2t = 1-2\sin^2 t\) or \(\cos 2t = 2\cos^2 t -1\) or \(\cos 2t = \cos^2 t - \sin^2 t\) |
| \(y = 2\left(\frac{x}{2}\right)^2\) or \(y = \frac{x^2}{2}\) or \(y = 2-2\left(1-\left(\frac{x}{2}\right)^2\right)\) | A1 cso isw | \(y = \frac{x^2}{2}\) or equivalent. Award A0 if \(+c\) added |
| Either \(k=2\) or \(-2 \leqslant x \leqslant 2\) | B1 | Either \(k=2\) or \(-2 \leqslant k \leqslant 2\); note \(-2 \leqslant k \leqslant 2\) unless \(k\) stated as 2 is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Range: \(0 \leqslant f(x) \leqslant 2\) or \(0 \leqslant y \leqslant 2\) or \(0 \leqslant f \leqslant 2\) | B1 B1 | Must evaluate 0 and/or 2; achieves inclusive upper or lower limit using acceptable notation |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 2\cos t$, $\frac{dy}{dt} = 2\sin 2t$ or $\frac{dy}{dt} = 4\sin t \cos t$ | B1 | At least one of $\frac{dx}{dt}$ or $\frac{dy}{dt}$ correct |
| Both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ correct | B1 | Can be implied from working |
| $\frac{dy}{dx} = \frac{2\sin 2t}{2\cos t} \left\{= \frac{4\cos t \sin t}{2\cos t} = 2\sin t\right\}$ | M1 | Applies $\frac{dy}{dt}$ divided by $\frac{dx}{dt}$ AND substitutes $t = \frac{\pi}{6}$ into their $\frac{dy}{dx}$ |
| At $t = \frac{\pi}{6}$, $\frac{dy}{dx} = \frac{2\sin\left(\frac{2\pi}{6}\right)}{2\cos\left(\frac{\pi}{6}\right)} = 1$ | A1 cao cso | Correct value of $\frac{dy}{dx}$ of 1. Note: Don't accept incorrect methods leading to 1 |
**Special Case:** B0B0M1A1 for $\frac{dx}{dt} = -2\cos t$, $\frac{dy}{dt} = -2\sin 2t$ leading to $\frac{dy}{dx} = \frac{-2\sin 2t}{-2\cos t}$, substituting $t = \frac{\pi}{6}$ to give $\frac{dy}{dx} = 1$
**Note:** Applying $\frac{dx}{dt}$ divided by $\frac{dy}{dt}$ is M0, even if they state $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$
**[4 marks]**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 1 - \cos 2t = 1-(1-2\sin^2 t) = 2\sin^2 t$ | M1 | Uses correct double angle formula $\cos 2t = 1-2\sin^2 t$ or $\cos 2t = 2\cos^2 t -1$ or $\cos 2t = \cos^2 t - \sin^2 t$ |
| $y = 2\left(\frac{x}{2}\right)^2$ or $y = \frac{x^2}{2}$ or $y = 2-2\left(1-\left(\frac{x}{2}\right)^2\right)$ | A1 cso isw | $y = \frac{x^2}{2}$ or equivalent. Award A0 if $+c$ added |
| Either $k=2$ or $-2 \leqslant x \leqslant 2$ | B1 | Either $k=2$ or $-2 \leqslant k \leqslant 2$; note $-2 \leqslant k \leqslant 2$ unless $k$ stated as 2 is B0 |
**[3 marks]**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range: $0 \leqslant f(x) \leqslant 2$ or $0 \leqslant y \leqslant 2$ or $0 \leqslant f \leqslant 2$ | B1 B1 | Must evaluate 0 and/or 2; achieves inclusive upper or lower limit using acceptable notation |
**Special Cases:**
- SC: B1B0 for either $0 < f(x) < 2$ or $0 < f < 2$ or $0 < y < 2$ or $(0,2)$
- SC: B1B0 for $0 \leqslant x \leqslant 2$
- $[0,2]$ is B1B1; $(0,2)$ is SC B1B0
**[2 marks]**
---\begin{enumerate}
\item A curve $C$ has parametric equations
\end{enumerate}
$$x = 2 \sin t , \quad y = 1 - \cos 2 t , \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point where $t = \frac { \pi } { 6 }$\\
(b) Find a cartesian equation for $C$ in the form
$$y = \mathrm { f } ( x ) , \quad - k \leqslant x \leqslant k$$
stating the value of the constant $k$.\\
(c) Write down the range of $\mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel C4 2013 Q4 [9]}}