| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question requiring dy/dx = (dy/dt)/(dx/dt), finding a parameter value from coordinates, and locating a stationary point. The techniques are routine for this module, though the trigonometric manipulation (double angle formulas) and algebraic simplification require care. Slightly easier than average due to straightforward application of standard methods. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = 4\sec^2 t\), \(\frac{dy}{dt} = 10\sqrt{3}\cos 2t\); \(\frac{dy}{dx} = \frac{10\sqrt{3}\cos 2t}{4\sec^2 t}\) | M1, A1 oe | M1: both \(x\) and \(y\) differentiated correctly w.r.t. \(t\), or \(\frac{dy}{dt}\) divided by \(\frac{dx}{dt}\); A1: correct \(\frac{dy}{dx}\) (can be implied) |
| At \(P\!\left(4\sqrt{3}, \frac{15}{2}\right)\), \(t = \frac{\pi}{3}\): \(\frac{dy}{dx} = \frac{10\sqrt{3}\cos\!\left(\frac{2\pi}{3}\right)}{4\sec^2\!\left(\frac{\pi}{3}\right)}\) | dM1 | Dependent on previous M; some evidence of substituting \(t=\frac{\pi}{3}\) or \(t=60°\) into their \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} = -\frac{5}{16}\sqrt{3}\) or \(-\frac{15}{16\sqrt{3}}\) | A1 cso | From correct solution only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(10\sqrt{3}\cos 2t = 0 \Rightarrow t = \frac{\pi}{4}\); \(x = 4\tan\!\left(\frac{\pi}{4}\right) = 4\), \(y = 5\sqrt{3}\sin\!\left(\frac{\pi}{2}\right) = 5\sqrt{3}\) | M1 | At least one of \(x = 4\tan\!\left(\frac{\pi}{4}\right)\) or \(y = 5\sqrt{3}\sin\!\left(2\cdot\frac{\pi}{4}\right)\), or \(x=4\) or \(y=5\sqrt{3}\) or \(y \approx 8.7\) |
| Coordinates are \(\left(4,\, 5\sqrt{3}\right)\) | A1 | \((4, 5\sqrt{3})\) or \(x=4,\ y=5\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan t = \frac{x}{4} \Rightarrow \sin t = \frac{x}{\sqrt{x^2+16}}\), \(\cos t = \frac{4}{\sqrt{x^2+16}} \Rightarrow y = \frac{40\sqrt{3}x}{x^2+16}\) | ||
| \(\frac{dy}{dx} = \frac{40\sqrt{3}(x^2+16) - 2x(40\sqrt{3}x)}{(x^2+16)^2} = \frac{40\sqrt{3}(16-x^2)}{(x^2+16)^2}\) | M1 | \(\frac{\pm A(x^2+16) \pm Bx^2}{(x^2+16)^2}\) |
| Correct \(\frac{dy}{dx}\); simplified or un-simplified | A1 | |
| \(\frac{dy}{dx} = \frac{40\sqrt{3}(48+16) - 80\sqrt{3}(48)}{(48+16)^2}\) | dM1 | Dependent on previous M mark. Some evidence of substituting \(x = 4\sqrt{3}\) into their \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} = -\frac{5}{16}\sqrt{3}\) or \(-\frac{15}{16\sqrt{3}}\) | A1 cso | \(-\frac{5}{16}\sqrt{3}\) or \(-\frac{15}{16\sqrt{3}}\) from a correct solution only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 5\sqrt{3}\sin\!\left(2\tan^{-1}\!\left(\frac{x}{4}\right)\right)\) | ||
| \(\frac{dy}{dx} = 5\sqrt{3}\cos\!\left(2\tan^{-1}\!\left(\frac{x}{4}\right)\right)\!\left(\frac{2}{1+\left(\frac{x}{4}\right)^2}\right)\!\left(\frac{1}{4}\right)\) | M1 | \(\frac{dy}{dx} = \pm A\cos\!\left(2\tan^{-1}\!\left(\frac{x}{4}\right)\right)\!\left(\frac{1}{1+x^2}\right)\) |
| Correct \(\frac{dy}{dx}\); simplified or un-simplified | A1 | |
| \(\frac{dy}{dx} = 5\sqrt{3}\cos\!\left(2\tan^{-1}(\sqrt{3})\right)\!\left(\frac{2}{1+3}\right)\!\left(\frac{1}{4}\right) = 5\sqrt{3}\!\left(-\frac{1}{2}\right)\!\left(\frac{1}{2}\right)\!\left(\frac{1}{4}\right)\) | dM1 | Dependent on previous M mark. Some evidence of substituting \(x = 4\sqrt{3}\) into their \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} = -\frac{5}{16}\sqrt{3}\) or \(-\frac{15}{16\sqrt{3}}\) | A1 cso | \(-\frac{5}{16}\sqrt{3}\) or \(-\frac{15}{16\sqrt{3}}\) from a correct solution only |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 4\sec^2 t$, $\frac{dy}{dt} = 10\sqrt{3}\cos 2t$; $\frac{dy}{dx} = \frac{10\sqrt{3}\cos 2t}{4\sec^2 t}$ | M1, A1 oe | M1: both $x$ and $y$ differentiated correctly w.r.t. $t$, or $\frac{dy}{dt}$ divided by $\frac{dx}{dt}$; A1: correct $\frac{dy}{dx}$ (can be implied) |
| At $P\!\left(4\sqrt{3}, \frac{15}{2}\right)$, $t = \frac{\pi}{3}$: $\frac{dy}{dx} = \frac{10\sqrt{3}\cos\!\left(\frac{2\pi}{3}\right)}{4\sec^2\!\left(\frac{\pi}{3}\right)}$ | dM1 | Dependent on previous M; some evidence of substituting $t=\frac{\pi}{3}$ or $t=60°$ into their $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = -\frac{5}{16}\sqrt{3}$ or $-\frac{15}{16\sqrt{3}}$ | A1 cso | From correct solution only |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $10\sqrt{3}\cos 2t = 0 \Rightarrow t = \frac{\pi}{4}$; $x = 4\tan\!\left(\frac{\pi}{4}\right) = 4$, $y = 5\sqrt{3}\sin\!\left(\frac{\pi}{2}\right) = 5\sqrt{3}$ | M1 | At least one of $x = 4\tan\!\left(\frac{\pi}{4}\right)$ or $y = 5\sqrt{3}\sin\!\left(2\cdot\frac{\pi}{4}\right)$, or $x=4$ or $y=5\sqrt{3}$ or $y \approx 8.7$ |
| Coordinates are $\left(4,\, 5\sqrt{3}\right)$ | A1 | $(4, 5\sqrt{3})$ or $x=4,\ y=5\sqrt{3}$ |
# Question 5:
## Part (a) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan t = \frac{x}{4} \Rightarrow \sin t = \frac{x}{\sqrt{x^2+16}}$, $\cos t = \frac{4}{\sqrt{x^2+16}} \Rightarrow y = \frac{40\sqrt{3}x}{x^2+16}$ | | |
| $\frac{dy}{dx} = \frac{40\sqrt{3}(x^2+16) - 2x(40\sqrt{3}x)}{(x^2+16)^2} = \frac{40\sqrt{3}(16-x^2)}{(x^2+16)^2}$ | M1 | $\frac{\pm A(x^2+16) \pm Bx^2}{(x^2+16)^2}$ |
| Correct $\frac{dy}{dx}$; simplified or un-simplified | A1 | |
| $\frac{dy}{dx} = \frac{40\sqrt{3}(48+16) - 80\sqrt{3}(48)}{(48+16)^2}$ | dM1 | Dependent on previous M mark. Some evidence of substituting $x = 4\sqrt{3}$ into their $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = -\frac{5}{16}\sqrt{3}$ or $-\frac{15}{16\sqrt{3}}$ | A1 cso | $-\frac{5}{16}\sqrt{3}$ or $-\frac{15}{16\sqrt{3}}$ from a correct solution only |
**Total: [4]**
## Part (a) — Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 5\sqrt{3}\sin\!\left(2\tan^{-1}\!\left(\frac{x}{4}\right)\right)$ | | |
| $\frac{dy}{dx} = 5\sqrt{3}\cos\!\left(2\tan^{-1}\!\left(\frac{x}{4}\right)\right)\!\left(\frac{2}{1+\left(\frac{x}{4}\right)^2}\right)\!\left(\frac{1}{4}\right)$ | M1 | $\frac{dy}{dx} = \pm A\cos\!\left(2\tan^{-1}\!\left(\frac{x}{4}\right)\right)\!\left(\frac{1}{1+x^2}\right)$ |
| Correct $\frac{dy}{dx}$; simplified or un-simplified | A1 | |
| $\frac{dy}{dx} = 5\sqrt{3}\cos\!\left(2\tan^{-1}(\sqrt{3})\right)\!\left(\frac{2}{1+3}\right)\!\left(\frac{1}{4}\right) = 5\sqrt{3}\!\left(-\frac{1}{2}\right)\!\left(\frac{1}{2}\right)\!\left(\frac{1}{4}\right)$ | dM1 | Dependent on previous M mark. Some evidence of substituting $x = 4\sqrt{3}$ into their $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = -\frac{5}{16}\sqrt{3}$ or $-\frac{15}{16\sqrt{3}}$ | A1 cso | $-\frac{5}{16}\sqrt{3}$ or $-\frac{15}{16\sqrt{3}}$ from a correct solution only |
**Total: [4]**
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cbfbb690-bc85-46e5-a97f-35df4b6f1c84-09_605_1131_248_466}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with parametric equations
$$x = 4 \tan t , \quad y = 5 \sqrt { 3 } \sin 2 t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$
The point $P$ lies on $C$ and has coordinates $\left( 4 \sqrt { 3 } , \frac { 15 } { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point $P$.
Give your answer as a simplified surd.
The point $Q$ lies on the curve $C$, where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$
\item Find the exact coordinates of the point $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2016 Q5 [6]}}