| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Verify parametric equations |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C4 parametric techniques. Part (i) is direct substitution, (ii) uses the chain rule for parametric differentiation (routine), (iii)-(iv) involve reading/sketching ellipses, (v) applies perpendicular gradient concept, and (vi) is a separable differential equation with boundary condition. All parts are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta = \frac{x}{k}\), \(\sin\theta = \frac{2y}{k}\) | ||
| \(\cos^2\theta + \sin^2\theta = 1\) | M1 | Used |
| \(\left(\frac{x}{k}\right)^2 + \left(\frac{2y}{k}\right)^2 = 1\) | M1 | substitution |
| \(\frac{x^2}{k^2} + \frac{4y^2}{k^2} = 1\) | ||
| \(\Rightarrow x^2 + 4y^2 = k^2\) | E1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{d\theta} = -k\sin\theta\), \(\frac{dy}{d\theta} = \frac{1}{2}k\cos\theta\) | M1 | |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{\frac{1}{2}k\cos\theta}{k\sin\theta}\) | A1 | |
| \(= -\frac{1}{2}\cot\theta\) | ||
| \(-\frac{x}{4y} = -\frac{2k\cos\theta}{4k\sin\theta} = -\frac{1}{2}\cot\theta = \frac{dy}{dx}\) | E1 | oe |
| *or*, differentiating implicitly: \(2x + 8y\frac{dy}{dx} = 0\) | M1 A1 | |
| \(\Rightarrow \frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y}\) | E1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k = 2\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1 correct curve – shape and position | B1 | 1 correct curve – shape and position |
| 2 or more curves correct shape in concentric form | B1 | 2 or more curves correct shape in concentric form |
| all 3 curves correct | B1 [3] | all 3 curves correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| grad of stream path \(= -1/\text{grad of contour}\) | M1 | |
| \(\Rightarrow \frac{dy}{dx} = -\frac{1}{(-x/4y)} = \frac{4y}{x}\) | E1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{4y}{x} \Rightarrow \int\frac{dy}{y} = \int\frac{4dx}{x}\) | M1 | Separating variables |
| \(\ln y = 4\ln x + c = \ln e^c x^4\) | A1, M1 | \(\ln y = 4\ln x\ (+c)\); antilogging correctly (at any stage) |
| \(\Rightarrow y = Ax^4\) where \(A = e^c\) | M1 | substituting \(x=2\), \(y=1\) |
| When \(x=2\), \(y=1 \Rightarrow 1 = 16A \Rightarrow A = \frac{1}{16}\) | A1 | evaluating a correct constant |
| \(\Rightarrow y = \frac{x^4}{16}\) | E1 [6] | www |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = \frac{x}{k}$, $\sin\theta = \frac{2y}{k}$ | | |
| $\cos^2\theta + \sin^2\theta = 1$ | M1 | Used |
| $\left(\frac{x}{k}\right)^2 + \left(\frac{2y}{k}\right)^2 = 1$ | M1 | substitution |
| $\frac{x^2}{k^2} + \frac{4y^2}{k^2} = 1$ | | |
| $\Rightarrow x^2 + 4y^2 = k^2$ | E1 [3] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{d\theta} = -k\sin\theta$, $\frac{dy}{d\theta} = \frac{1}{2}k\cos\theta$ | M1 | |
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{\frac{1}{2}k\cos\theta}{k\sin\theta}$ | A1 | |
| $= -\frac{1}{2}\cot\theta$ | | |
| $-\frac{x}{4y} = -\frac{2k\cos\theta}{4k\sin\theta} = -\frac{1}{2}\cot\theta = \frac{dy}{dx}$ | E1 | oe |
| *or*, differentiating implicitly: $2x + 8y\frac{dy}{dx} = 0$ | M1 A1 | |
| $\Rightarrow \frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y}$ | E1 [3] | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = 2$ | B1 [1] | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1 correct curve – shape and position | B1 | 1 correct curve – shape and position |
| 2 or more curves correct shape in concentric form | B1 | 2 or more curves correct shape in concentric form |
| all 3 curves correct | B1 [3] | all 3 curves correct |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| grad of stream path $= -1/\text{grad of contour}$ | M1 | |
| $\Rightarrow \frac{dy}{dx} = -\frac{1}{(-x/4y)} = \frac{4y}{x}$ | E1 [2] | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{4y}{x} \Rightarrow \int\frac{dy}{y} = \int\frac{4dx}{x}$ | M1 | Separating variables |
| $\ln y = 4\ln x + c = \ln e^c x^4$ | A1, M1 | $\ln y = 4\ln x\ (+c)$; antilogging correctly (at any stage) |
| $\Rightarrow y = Ax^4$ where $A = e^c$ | M1 | substituting $x=2$, $y=1$ |
| When $x=2$, $y=1 \Rightarrow 1 = 16A \Rightarrow A = \frac{1}{16}$ | A1 | evaluating a correct constant |
| $\Rightarrow y = \frac{x^4}{16}$ | E1 [6] | www |
---2 A curve has equation
$$x ^ { 2 } + 4 y ^ { 2 } = k ^ { 2 } ,$$
where $k$ is a positive constant.\\
(i) Verify that
$$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$
are parametric equations for the curve.\\
(ii) Hence or otherwise show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }$.\\
(iii) Fig. 8 illustrates the curve for a particular value of $k$. Write down this value of $k$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-2_658_1070_861_567}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(iv) Copy Fig. 8 and on the same axes sketch the curves for $k = 1 , k = 3$ and $k = 4$.
On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.\\
(v) Explain why the path of the stream is modelled by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
(vi) Solve this differential equation.
Given that the path of the stream passes through the point $( 2,1 )$, show that its equation is $y = \frac { x ^ { 4 } } { 16 }$.
\hfill \mbox{\textit{OCR MEI C4 Q2 [18]}}