| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Properties of specific curves |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question with routine follow-through parts. Part (i) uses the chain rule formula dx/dt and dy/dt (bookwork), part (ii) is straightforward tangent equation derivation, part (iii) requires finding intercepts and area formula, and part (iv) is basic optimization using the double angle result. All techniques are standard for C4 with no novel insights required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) | M1 | Used, not just quoted |
| \(\frac{dx}{dt} = -4\sin t\) or \(\frac{dy}{dt} = 3\cos t\) | *B1 | |
| \(\frac{dy}{dx} = \frac{3\cos t}{4\sin t}\) or \(\frac{3\cos t}{-4\sin t}\) ISW | dep*A1 | 3 Also \(-\frac{3\cos t}{4\sin t}\) provided B0 not awarded |
| SR: M1 for Cartesian eqn attempt + B1 for \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\) | +A1 as before(must be in terms of t) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 3\sin p = (\text{their}\frac{dy}{dx})(x - 4\cos p)\) | M1 | Accept p or t here |
| or \(y = (\text{their}\frac{dy}{dx})x + c\) & subst cords to find c | Ditto | |
| \(4y\sin p - 12\sin^2 p = -3x\cos p + 12\cos^2 p\) | A1 | Correct equation cleared of fractions |
| or \(c = \frac{12\sin^2 p + 12\cos^2 p}{4\sin p}\) | ||
| \(3x\cos p + 4y\sin p = 12\) | WWW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subst \(x = 0\) and \(y = 0\) separately in tangent eqn | M1 | to find R & S |
| Produce \(\frac{3}{\sin p}\) and \(\frac{4}{\cos p}\) | A1 | Accept \(\frac{12}{4\sin p}\) and/or \(\frac{12}{3\cos p}\) |
| Use \(\Delta = \frac{1}{2}(\frac{3}{\sin p} \cdot \frac{4}{\cos p}) = \frac{12}{\sin 2p}\) | WWW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Least area = 12 | B1 | |
| \(p = \frac{1}{4}\pi\) as final or only answer | B2 | 3 These B marks are independent. |
| S.R. 45° \(\to\) B1; | S.R. [\(-12\) and e.g. \(-\pi/4 \to\) B1] |
### (i)
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 | Used, not just quoted
$\frac{dx}{dt} = -4\sin t$ or $\frac{dy}{dt} = 3\cos t$ | *B1 |
$\frac{dy}{dx} = \frac{3\cos t}{4\sin t}$ or $\frac{3\cos t}{-4\sin t}$ ISW | dep*A1 | 3 Also $-\frac{3\cos t}{4\sin t}$ provided B0 not awarded
SR: M1 for Cartesian eqn attempt + B1 for $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ | | +A1 as before(must be in terms of t)
---
### (ii)
$y - 3\sin p = (\text{their}\frac{dy}{dx})(x - 4\cos p)$ | M1 | Accept p or t here
or $y = (\text{their}\frac{dy}{dx})x + c$ & subst cords to find c | | Ditto
$4y\sin p - 12\sin^2 p = -3x\cos p + 12\cos^2 p$ | A1 | Correct equation cleared of fractions
or $c = \frac{12\sin^2 p + 12\cos^2 p}{4\sin p}$ | |
$3x\cos p + 4y\sin p = 12$ | WWW | A1 | 3 AG Only p here. Mixture earlier $\to$ A0
---
### (iii)
Subst $x = 0$ and $y = 0$ separately in tangent eqn | M1 | to find R & S
Produce $\frac{3}{\sin p}$ and $\frac{4}{\cos p}$ | A1 | Accept $\frac{12}{4\sin p}$ and/or $\frac{12}{3\cos p}$
Use $\Delta = \frac{1}{2}(\frac{3}{\sin p} \cdot \frac{4}{\cos p}) = \frac{12}{\sin 2p}$ | WWW | A1 | 3 AG
---
### (iv)
Least area = 12 | B1 |
$p = \frac{1}{4}\pi$ as final or only answer | B2 | 3 These B marks are independent.
S.R. 45° $\to$ B1; | | S.R. [$-12$ and e.g. $-\pi/4 \to$ B1]
---A curve is given parametrically by the equations
$$x = 4\cos t, \quad y = 3\sin t,$$
where $0 \leq t \leq \frac{1}{2}\pi$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ in terms of $t$. [3]
\item Show that the equation of the tangent at the point $P$, where $t = p$, is
$$3x\cos p + 4y\sin p = 12.$$ [3]
\item The tangent at $P$ meets the $x$-axis at $R$ and the $y$-axis at $S$. $O$ is the origin. Show that the area of triangle $ORS$ is $\frac{6}{\sin 2p}$. [3]
\item Write down the least possible value of the area of triangle $ORS$, and give the corresponding value of $p$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2006 Q9 [12]}}