| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question covering routine techniques: substituting parameter values, finding dy/dx using the chain rule, locating turning points by setting dy/dx = 0, converting to Cartesian form, and calculating a volume of revolution. All parts follow textbook methods 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 differentiation4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\theta = -\pi/2\): O \((0, 0)\) | B1 | Origin or O, condone omission of \((0,0)\) or O |
| \(\theta = 0\): P \((2, 0)\) | B1 | Or, say at P \(x = 2\), \(y = 0\), need P stated |
| \(\theta = \pi/2\): O \((0, 0)\) | B1 | Origin or O, condone omission of \((0,0)\) or O |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) | M1 | their \(dy/d\theta \div dx/d\theta\) |
| \(= \frac{2\cos 2\theta}{-2\sin\theta} = -\frac{\cos 2\theta}{\sin\theta}\) | A1 | any equivalent form www (not from \(-2\cos 2\theta/2\sin\theta\)) |
| When \(\theta = \pi/2\), \(dy/dx = -\cos\pi/\sin(\pi/2) = 1\) | M1 | subst \(\theta = \pi/2\) in their equation |
| When \(\theta = -\pi/2\), \(dy/dx = -\cos(-\pi)/\sin(-\pi/2) = -1\) | A1 | Obtaining \(dy/dx = 1\) and \(dy/dx = -1\) shown (or explaining using symmetry of curve) www |
| Either \(1 \times -1 = -1\) so perpendicular Or gradient tangent \(= 1 \Rightarrow\) meets axis at \(45°\), similarly gradient \(= -1 \Rightarrow\) meets axis at \(45°\) oe | A1 | justification that tangents are perpendicular www dependent on previous A1 |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At Q, \(\sin 2\theta = 1 \Rightarrow 2\theta = \pi/2\), \(\theta = \pi/4\) | M1 | or, using the derivative, \(\cos 2\theta = 0\) soi or their \(dy/dx = 0\) to find \(\theta\). If the only error is in the sign or the coeff of the derivative in (ii), allow full marks in this part (condone \(\theta = 45°\)) |
| coordinates of Q are \((2\cos\pi/4,\ \sin\pi/2)\) \(= (\sqrt{2}, 1)\) | A1 A1 | www (exact only) accept \(2/\sqrt{2}\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin^2\theta = (1 - \cos^2\theta) = 1 - \frac{1}{4}x^2\) | B1 | oe, eg may be \(x^2 = \ldots\) |
| \(\Rightarrow y = \sin 2\theta = 2\sin\theta\cos\theta\) | M1 | Use of \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(= (\pm)\ x\sqrt{(1 - \frac{1}{4}x^2)}\) | A1 | subst for \(x\) or \(y^2 = 4\sin^2\theta\cos^2\theta\) (squaring), either order oe |
| \(\Rightarrow y^2 = x^2(1 - \frac{1}{4}x^2)\)* | A1 | squaring or subst for \(x\), either order oe |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \int_0^2 \pi x^2(1 - \frac{1}{4}x^2)\,dx\) | M1 | integral including correct limits but ft their '2' from (i); limits may appear later; condone omission of \(dx\) if intention clear |
| \(= \int_0^2 (\pi x^2 - \frac{1}{4}\pi x^4)\,dx\) | ||
| \(= \pi\left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]_0^2\) | B1 | \(\left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]\) ie allow if no \(\pi\) and/or incorrect/no limits (or equivalent by parts) |
| \(= \pi\left[\frac{8}{3} - \frac{32}{20}\right]\) | A1 | substituting limits into correct expression (including \(\pi\)) ft their '2' |
| \(= 16\pi/15\) | A1 | cao oe, 3.35 or better (any multiple of \(\pi\) must round to 3.35 or better) |
| [4] |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta = -\pi/2$: O $(0, 0)$ | B1 | Origin or O, condone omission of $(0,0)$ or O |
| $\theta = 0$: P $(2, 0)$ | B1 | Or, say at P $x = 2$, $y = 0$, need P stated |
| $\theta = \pi/2$: O $(0, 0)$ | B1 | Origin or O, condone omission of $(0,0)$ or O |
| **[3]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ | M1 | their $dy/d\theta \div dx/d\theta$ |
| $= \frac{2\cos 2\theta}{-2\sin\theta} = -\frac{\cos 2\theta}{\sin\theta}$ | A1 | any equivalent form www (not from $-2\cos 2\theta/2\sin\theta$) |
| When $\theta = \pi/2$, $dy/dx = -\cos\pi/\sin(\pi/2) = 1$ | M1 | subst $\theta = \pi/2$ in their equation |
| When $\theta = -\pi/2$, $dy/dx = -\cos(-\pi)/\sin(-\pi/2) = -1$ | A1 | Obtaining $dy/dx = 1$ and $dy/dx = -1$ **shown** (or explaining using symmetry of curve) www |
| **Either** $1 \times -1 = -1$ so perpendicular **Or** gradient tangent $= 1 \Rightarrow$ meets axis at $45°$, similarly gradient $= -1 \Rightarrow$ meets axis at $45°$ oe | A1 | justification that tangents are perpendicular www dependent on previous A1 |
| **[5]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At Q, $\sin 2\theta = 1 \Rightarrow 2\theta = \pi/2$, $\theta = \pi/4$ | M1 | or, using the derivative, $\cos 2\theta = 0$ soi or their $dy/dx = 0$ to find $\theta$. If the only error is in the sign or the coeff of the derivative in (ii), allow full marks in this part (condone $\theta = 45°$) |
| coordinates of Q are $(2\cos\pi/4,\ \sin\pi/2)$ $= (\sqrt{2}, 1)$ | A1 A1 | www (exact only) accept $2/\sqrt{2}$ |
| **[3]** | | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin^2\theta = (1 - \cos^2\theta) = 1 - \frac{1}{4}x^2$ | B1 | oe, eg may be $x^2 = \ldots$ |
| $\Rightarrow y = \sin 2\theta = 2\sin\theta\cos\theta$ | M1 | **Use** of $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $= (\pm)\ x\sqrt{(1 - \frac{1}{4}x^2)}$ | A1 | subst for $x$ **or** $y^2 = 4\sin^2\theta\cos^2\theta$ (squaring), either order oe |
| $\Rightarrow y^2 = x^2(1 - \frac{1}{4}x^2)$* | A1 | squaring **or** subst for $x$, either order oe |
| **[4]** | | |
### Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^2 \pi x^2(1 - \frac{1}{4}x^2)\,dx$ | M1 | integral including correct limits but ft their '2' from (i); limits may appear later; condone omission of $dx$ if intention clear |
| $= \int_0^2 (\pi x^2 - \frac{1}{4}\pi x^4)\,dx$ | | |
| $= \pi\left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]_0^2$ | B1 | $\left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]$ ie allow if no $\pi$ and/or incorrect/no limits (or equivalent by parts) |
| $= \pi\left[\frac{8}{3} - \frac{32}{20}\right]$ | A1 | substituting limits into correct expression (including $\pi$) ft their '2' |
| $= 16\pi/15$ | A1 | cao oe, 3.35 or better (any multiple of $\pi$ must round to 3.35 or better) |
| **[4]** | | |
---2 Fig. 7a shows the curve with the parametric equations
$$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$
The curve meets the $x$-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_509_660_571_714}
\captionsetup{labelformat=empty}
\caption{Fig. 7a}
\end{center}
\end{figure}
(i) State, with their coordinates, the points on the curve for which $\theta = - \frac { \pi } { 2 } , \theta = 0$ and $\theta = \frac { \pi } { 2 }$.\\
(ii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. Hence find the gradient of the curve when $\theta = \frac { \pi } { 2 }$, and verify that the two tangents to the curve at the origin meet at right angles.\\
(iii) Find the exact coordinates of the turning point Q .
When the curve is rotated about the $x$-axis, it forms a paperweight shape, as shown in Fig. 7b.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_324_389_1692_857}
\captionsetup{labelformat=empty}
\caption{Fig. 7b}
\end{center}
\end{figure}
(iv) Express $\sin ^ { 2 } \theta$ in terms of $x$. Hence show that the cartesian equation of the curve is $y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)$.\\
(v) Find the volume of the paperweight shape.
\hfill \mbox{\textit{OCR MEI C4 Q2 [19]}}