Question 6 - A-Level Maths

OCR C4 2010 January — Question 6 6 marks

Exam Board	OCR
Module	C4 (Core Mathematics 4)
Year	2010
Session	January
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric differentiation
Type	Find parameter value given gradient condition
Difficulty	Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt ÷ dx/dt), simplification of logarithms, and solving a simple equation. While it involves multiple steps, each is routine for C4 level—differentiating, simplifying ln(t³) = 3ln(t), and solving a cubic that factors easily. The 'show' command adds minor rigor but the algebra is standard, making this slightly easier than average.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.07s Parametric and implicit differentiation

6 A curve has parametric equations $$x = 9 t - \ln ( 9 t ) , \quad y = t ^ { 3 } - \ln \left( t ^ { 3 } \right)$$ Show that there is only one value of $t$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ and state that value.

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Question 6:

Answer	Marks	Guidance
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$	M1	quoted/implied
$\frac{dx}{dt} = 9 - \frac{9}{9t}$	B1	ISW
$\frac{dy}{dt} = 3t^2 - \frac{3t^2}{t^3}$	B1	ISW
Stating/implying $\frac{3t^2 - \frac{3}{t}}{9 - \frac{1}{t}} = 3 \Rightarrow t^2 = 9$ or $t^3 - 9t = 0$	A1	WWW, totally correct at this stage
$t = 3$ as final answer with clear log indication of invalidity of $-3$; ignore (non) mention of $t=0$	A2	S.R. A1 if $t = \pm 3$ or $t = -3$ or ($t=3$ & wrong/no indication)

# Question 6:

| $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ | M1 | quoted/implied |
|---|---|---|
| $\frac{dx}{dt} = 9 - \frac{9}{9t}$ | B1 | ISW |
| $\frac{dy}{dt} = 3t^2 - \frac{3t^2}{t^3}$ | B1 | ISW |
| Stating/implying $\frac{3t^2 - \frac{3}{t}}{9 - \frac{1}{t}} = 3 \Rightarrow t^2 = 9$ or $t^3 - 9t = 0$ | A1 | WWW, totally correct at this stage |
| $t = 3$ as final answer with clear log indication of invalidity of $-3$; ignore (non) mention of $t=0$ | A2 | **S.R.** A1 if $t = \pm 3$ or $t = -3$ or ($t=3$ & wrong/no indication) |

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6 A curve has parametric equations

$$x = 9 t - \ln ( 9 t ) , \quad y = t ^ { 3 } - \ln \left( t ^ { 3 } \right)$$

Show that there is only one value of $t$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ and state that value.

\hfill \mbox{\textit{OCR C4 2010 Q6 [6]}}

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Q1 4 Q2 6 Q3 5 Q4 6 Q5 7 Q6 6 Q7 8 Q8 7 Q9 10 Q10 13

Answer	Marks	Guidance
\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)	M1	quoted/implied
\(\frac{dx}{dt} = 9 - \frac{9}{9t}\)	B1	ISW
\(\frac{dy}{dt} = 3t^2 - \frac{3t^2}{t^3}\)	B1	ISW
Stating/implying \(\frac{3t^2 - \frac{3}{t}}{9 - \frac{1}{t}} = 3 \Rightarrow t^2 = 9\) or \(t^3 - 9t = 0\)	A1	WWW, totally correct at this stage
\(t = 3\) as final answer with clear log indication of invalidity of \(-3\); ignore (non) mention of \(t=0\)	A2	S.R. A1 if \(t = \pm 3\) or \(t = -3\) or (\(t=3\) & wrong/no indication)