| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find parameter from gradient condition |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring routine substitution and solving simple equations. Part (a) involves direct substitution at t=0, solving a quadratic for v=0, and another substitution. Part (b) requires finding the maximum of a quadratic (completing the square or differentiation) and substitution. Part (c) is a standard modelling comment. All techniques are standard A-level fare 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 differentiation |
| Answer | Marks |
|---|---|
| (a)(iii) | Obtains correct h for their positive t |
| Answer | Marks | Guidance |
|---|---|---|
| FT their t allow negative h | AO3.4 | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| 15(c) | Identifies t=3 (for maximum | |
| velocity) | AO3.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI their answer | 3.4 | M1 |
| Finds correct height with units | 3.2a | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| model is limited by time. | 3.5b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| times when v=0 | 3.5a | B1 |
| Total | 10 | |
| Q | Marking instructions | AO |
Question 15:
--- 15
(a)(iii) ---
15
(a)(iii) | Obtains correct h for their positive t
provided h < 5.88
Accept 0.12
If given to more decimal places
AWFW 0.115 to 0.116
FT their t allow negative h | AO3.4 | B1F | 0.12m
15(b)
15(c) | Identifies t=3 (for maximum
velocity) | AO3.1b | B1 | t = 3
h=3−23 3−3
h = 3 metres
Substitutes their t into model for h
PI their answer | 3.4 | M1
Finds correct height with units | 3.2a | A1
Explains that the validity of the
model is limited by time. | 3.5b | B1 | The model breaks down after one
cycle of the tide.
After 6 hours the model shows
the height continues to decrease.
Explains that the height continues
to decrease after 6 hours (or after
the first cycle or first low tide).
Or explains there are no other
times when v=0 | 3.5a | B1
Total | 10
Q | Marking instructions | AO | Mark | Typical solutionAt time $t$ hours after a high tide, the height, $h$ metres, of the tide and the velocity, $v$ knots, of the tidal flow can be modelled using the parametric equations
$$v = 4 - \left(\frac{2t}{3} - 2\right)^2$$
$$h = 3 - 2\sqrt[3]{t - 3}$$
High tides and low tides occur alternately when the velocity of the tidal flow is zero.
A high tide occurs at 2am.
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Use the model to find the height of this high tide. [1 mark]
\item Find the time of the first low tide after 2am. [3 marks]
\item Find the height of this low tide. [1 mark]
\end{enumerate}
\item Use the model to find the height of the tide when it is flowing with maximum velocity. [3 marks]
\item Comment on the validity of the model. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2019 Q15 [13]}}