| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (tan/sec/cot/cosec identities) |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: chain rule for dy/dx, finding a tangent line, and converting to Cartesian form using trigonometric identities. Part (c) requires recognizing that sin²t = tan²t/(1+tan²t), which is a standard identity manipulation. All parts are routine C4-level exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = 2(\tan t)\sec^2 t\), \(\frac{dy}{dt} = \cos t\) | B1 | Correct \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) |
| \(\frac{dy}{dx} = \frac{\cos t}{2\tan t \sec^2 t}\) \(\left(= \frac{\cos^4 t}{2\sin t}\right)\) | M1 | \(\frac{\pm\cos t}{\text{their } \frac{dx}{dt}}\) |
| Correct simplified expression | A1 \(\checkmark\) | \(\frac{+\cos t}{\text{their } \frac{dx}{dt}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(t = \frac{\pi}{4}\): \(x = 1\), \(y = \frac{1}{\sqrt{2}}\) | B1, B1 | The point \(\left(1, \frac{1}{\sqrt{2}}\right)\) or \((1, \text{ awrt } 0.71)\). These coordinates can be implied. (\(y = \sin\left(\frac{\pi}{4}\right)\) is not sufficient for B1) |
| \(m(T) = \frac{dy}{dx} = \frac{\cos\frac{\pi}{4}}{2\tan\frac{\pi}{4}\sec^2\frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{2(1)\left(\frac{1}{\frac{1}{\sqrt{2}}}\right)^2} = \frac{\frac{1}{\sqrt{2}}}{2(1)\left(\frac{1}{\frac{1}{2}}\right)} = \frac{\frac{1}{\sqrt{2}}}{2(1)(2)} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}\) | B1 aef | Any of the five underlined expressions or awrt 0.18 |
| \(y - \frac{1}{\sqrt{2}} = \frac{1}{4\sqrt{2}}(x-1)\) | M1 \(\checkmark\) aef | Finding equation of tangent with their point and their tangent gradient, or finds \(c\) using \(y = (\text{their gradient})x + c\) |
| \(y = \frac{1}{4\sqrt{2}}x + \frac{3}{4\sqrt{2}}\) or \(y = \frac{\sqrt{2}}{8}x + \frac{3\sqrt{2}}{8}\) | A1 aef cso | Correct simplified EXACT equation of tangent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \frac{\sin^2 t}{\cos^2 t}\), uses \(\cos^2 t = 1 - \sin^2 t\) so \(x = \frac{\sin^2 t}{1-\sin^2 t}\) | M1 | Uses \(\cos^2 t = 1 - \sin^2 t\) |
| \(x = \frac{y^2}{1-y^2}\) | M1 | Eliminates \(t\) to write equation involving \(x\) and \(y\) |
| \(x(1-y^2) = y^2 \Rightarrow x - xy^2 = y^2 \Rightarrow x = y^2 + xy^2 \Rightarrow x = y^2(1+x)\) | ddM1 | Rearranging and factorising with attempt to make \(y^2\) the subject |
| \(y^2 = \dfrac{x}{1+x}\) | A1 | \(\dfrac{x}{1+x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 + \cot^2 t = \csc^2 t\) | M1 | Uses \(1 + \cot^2 t = \csc^2 t\) |
| \(= \dfrac{1}{\sin^2 t}\) | M1 implied | Uses \(\csc^2 t = \dfrac{1}{\sin^2 t}\) |
| \(1 + \dfrac{1}{x} = \dfrac{1}{y^2}\) | ddM1 | Eliminates \(t\) to write equation involving \(x\) and \(y\) |
| \(y^2 = 1 - \dfrac{1}{(1+x)}\) or \(\dfrac{x}{1+x}\) | A1 | \(1 - \dfrac{1}{(1+x)}\) or \(\dfrac{x}{1+x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 + \tan^2 t = \sec^2 t\) | M1 | Uses \(1 + \tan^2 t = \sec^2 t\) |
| \(= \dfrac{1}{\cos^2 t}\) | M1 | Uses \(\sec^2 t = \dfrac{1}{\cos^2 t}\) |
| \(1 + x = \dfrac{1}{1 - \sin^2 t}\), hence \(1 + x = \dfrac{1}{1-y^2}\) | ddM1 | Eliminates \(t\) to write equation involving \(x\) and \(y\) |
| \(y^2 = 1 - \dfrac{1}{(1+x)}\) or \(\dfrac{x}{1+x}\) | A1 | \(1 - \dfrac{1}{(1+x)}\) or \(\dfrac{x}{1+x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y^2 = \sin^2 t = 1 - \cos^2 t\) | M1 | Uses \(\sin^2 t = 1 - \cos^2 t\) |
| \(= 1 - \dfrac{1}{\sec^2 t}\) | M1 | Uses \(\cos^2 t = \dfrac{1}{\sec^2 t}\) |
| \(= 1 - \dfrac{1}{(1+\tan^2 t)}\) | ddM1 | Then uses \(\sec^2 t = 1 + \tan^2 t\) |
| \(y^2 = 1 - \dfrac{1}{(1+x)}\) or \(\dfrac{x}{1+x}\) | A1 | \(1 - \dfrac{1}{(1+x)}\) or \(\dfrac{x}{1+x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \tan^2 t \Rightarrow \tan t = \sqrt{x}\); draws right-angled triangle with \(\sqrt{x}\) and \(1\) on triangle | M1 | Draws right-angled triangle and places both \(\sqrt{x}\) and \(1\) on the triangle |
| Uses Pythagoras to deduce hypotenuse \(= \sqrt{1+x}\) | M1 | Uses Pythagoras to deduce hypotenuse |
| \(y = \sin t = \dfrac{\sqrt{x}}{\sqrt{1+x}}\) | ddM1 | Eliminates \(t\) to write equation involving \(x\) and \(y\) |
| \(y^2 = \dfrac{x}{1+x}\) | A1 | \(\dfrac{x}{1+x}\) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 2(\tan t)\sec^2 t$, $\frac{dy}{dt} = \cos t$ | B1 | Correct $\frac{dx}{dt}$ and $\frac{dy}{dt}$ |
| $\frac{dy}{dx} = \frac{\cos t}{2\tan t \sec^2 t}$ $\left(= \frac{\cos^4 t}{2\sin t}\right)$ | M1 | $\frac{\pm\cos t}{\text{their } \frac{dx}{dt}}$ |
| Correct simplified expression | A1 $\checkmark$ | $\frac{+\cos t}{\text{their } \frac{dx}{dt}}$ |
**[3 marks]**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $t = \frac{\pi}{4}$: $x = 1$, $y = \frac{1}{\sqrt{2}}$ | B1, B1 | The point $\left(1, \frac{1}{\sqrt{2}}\right)$ or $(1, \text{ awrt } 0.71)$. These coordinates can be implied. ($y = \sin\left(\frac{\pi}{4}\right)$ is not sufficient for B1) |
| $m(T) = \frac{dy}{dx} = \frac{\cos\frac{\pi}{4}}{2\tan\frac{\pi}{4}\sec^2\frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{2(1)\left(\frac{1}{\frac{1}{\sqrt{2}}}\right)^2} = \frac{\frac{1}{\sqrt{2}}}{2(1)\left(\frac{1}{\frac{1}{2}}\right)} = \frac{\frac{1}{\sqrt{2}}}{2(1)(2)} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}$ | B1 aef | Any of the five underlined expressions or awrt 0.18 |
| $y - \frac{1}{\sqrt{2}} = \frac{1}{4\sqrt{2}}(x-1)$ | M1 $\checkmark$ aef | Finding equation of tangent with their point and their tangent gradient, or finds $c$ using $y = (\text{their gradient})x + c$ |
| $y = \frac{1}{4\sqrt{2}}x + \frac{3}{4\sqrt{2}}$ or $y = \frac{\sqrt{2}}{8}x + \frac{3\sqrt{2}}{8}$ | A1 aef cso | Correct simplified EXACT equation of tangent |
**[5 marks]**
---
### Part (c) — Way 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{\sin^2 t}{\cos^2 t}$, uses $\cos^2 t = 1 - \sin^2 t$ so $x = \frac{\sin^2 t}{1-\sin^2 t}$ | M1 | Uses $\cos^2 t = 1 - \sin^2 t$ |
| $x = \frac{y^2}{1-y^2}$ | M1 | Eliminates $t$ to write equation involving $x$ and $y$ |
| $x(1-y^2) = y^2 \Rightarrow x - xy^2 = y^2 \Rightarrow x = y^2 + xy^2 \Rightarrow x = y^2(1+x)$ | ddM1 | Rearranging and factorising with attempt to make $y^2$ the subject |
| $y^2 = \dfrac{x}{1+x}$ | A1 | $\dfrac{x}{1+x}$ |
**[4 marks]**
---
### Part (c) — Way 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 + \cot^2 t = \csc^2 t$ | M1 | Uses $1 + \cot^2 t = \csc^2 t$ |
| $= \dfrac{1}{\sin^2 t}$ | M1 implied | Uses $\csc^2 t = \dfrac{1}{\sin^2 t}$ |
| $1 + \dfrac{1}{x} = \dfrac{1}{y^2}$ | ddM1 | Eliminates $t$ to write equation involving $x$ and $y$ |
| $y^2 = 1 - \dfrac{1}{(1+x)}$ or $\dfrac{x}{1+x}$ | A1 | $1 - \dfrac{1}{(1+x)}$ or $\dfrac{x}{1+x}$ |
**[4 marks]**
---
### Part (c) — Way 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 + \tan^2 t = \sec^2 t$ | M1 | Uses $1 + \tan^2 t = \sec^2 t$ |
| $= \dfrac{1}{\cos^2 t}$ | M1 | Uses $\sec^2 t = \dfrac{1}{\cos^2 t}$ |
| $1 + x = \dfrac{1}{1 - \sin^2 t}$, hence $1 + x = \dfrac{1}{1-y^2}$ | ddM1 | Eliminates $t$ to write equation involving $x$ and $y$ |
| $y^2 = 1 - \dfrac{1}{(1+x)}$ or $\dfrac{x}{1+x}$ | A1 | $1 - \dfrac{1}{(1+x)}$ or $\dfrac{x}{1+x}$ |
**[4 marks]**
---
### Part (c) — Way 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^2 = \sin^2 t = 1 - \cos^2 t$ | M1 | Uses $\sin^2 t = 1 - \cos^2 t$ |
| $= 1 - \dfrac{1}{\sec^2 t}$ | M1 | Uses $\cos^2 t = \dfrac{1}{\sec^2 t}$ |
| $= 1 - \dfrac{1}{(1+\tan^2 t)}$ | ddM1 | Then uses $\sec^2 t = 1 + \tan^2 t$ |
| $y^2 = 1 - \dfrac{1}{(1+x)}$ or $\dfrac{x}{1+x}$ | A1 | $1 - \dfrac{1}{(1+x)}$ or $\dfrac{x}{1+x}$ |
**[4 marks]**
---
### Part (c) — Way 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \tan^2 t \Rightarrow \tan t = \sqrt{x}$; draws right-angled triangle with $\sqrt{x}$ and $1$ on triangle | M1 | Draws right-angled triangle and places both $\sqrt{x}$ and $1$ on the triangle |
| Uses Pythagoras to deduce hypotenuse $= \sqrt{1+x}$ | M1 | Uses Pythagoras to deduce hypotenuse |
| $y = \sin t = \dfrac{\sqrt{x}}{\sqrt{1+x}}$ | ddM1 | Eliminates $t$ to write equation involving $x$ and $y$ |
| $y^2 = \dfrac{x}{1+x}$ | A1 | $\dfrac{x}{1+x}$ |
**[4 marks]**
> **Note:** $\dfrac{1}{1+\frac{1}{x}}$ is an acceptable response for the final accuracy A1 mark.
**Total: 12 marks**6. A curve has parametric equations
$$x = \tan ^ { 2 } t , \quad y = \sin t , \quad 0 < t < \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. You need not simplify your answer.
\item Find an equation of the tangent to the curve at the point where $t = \frac { \pi } { 4 }$.
Give your answer in the form $y = a x + b$, where $a$ and $b$ are constants to be determined.
\item Find a cartesian equation of the curve in the form $y ^ { 2 } = \mathrm { f } ( x )$.\\
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q6 [12]}}