| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric loop enclosed area |
| Difficulty | Standard +0.3 This is a standard parametric equations question requiring routine techniques: finding dy/dx using the chain rule, determining limits by solving y=0, converting the integral using dx/dt, and integrating a polynomial. All steps are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(y = 0\) to obtain a non-zero value of \(t\): \(4t^2 - t^3 = 0 \Rightarrow t = 0\) or \(t = 4\) | M1 | AO 3.1a |
| Obtains \(\frac{dy}{dt} = 8t - 3t^2\) or \(\frac{dy}{dt} = -16\) at \(t=4\) | B1 | AO 1.1b |
| Obtains \(\frac{dx}{dt} = 2t + 1\) or \(\frac{dx}{dt} = 9\) at \(t=4\) | B1 | AO 1.1b |
| Uses \(\frac{dy}{dt} \div \frac{dx}{dt} = \frac{dy}{dx}\) with non-zero \(t\) to find numerical value of \(\frac{dy}{dx}\) | M1 | AO 3.1a |
| Obtains \(-\frac{16}{9}\) OE | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(b = 20\) using \(t^2 + t\) for their value of \(t\) | B1F | AO 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(dx = (2t+1)\,dt\) | M1 | AO 3.1a |
| \(A = \int_0^{20} y\,dx = \int_0^4 (4t^2 - t^3)(2t+1)\,dt\) | A1F | AO 1.1b — condone incorrect/omitted limits |
| \(= \int_0^4 4t^2 + 7t^3 - 2t^4\,dt\) | R1 | AO 2.1 — rigorous argument showing \(t=4\) when \(x=20\); may be justified in (b)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = \frac{1856}{15}\) or AWRT 124 | B1 | AO 1.1b |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $y = 0$ to obtain a non-zero value of $t$: $4t^2 - t^3 = 0 \Rightarrow t = 0$ or $t = 4$ | M1 | AO 3.1a |
| Obtains $\frac{dy}{dt} = 8t - 3t^2$ or $\frac{dy}{dt} = -16$ at $t=4$ | B1 | AO 1.1b |
| Obtains $\frac{dx}{dt} = 2t + 1$ or $\frac{dx}{dt} = 9$ at $t=4$ | B1 | AO 1.1b |
| Uses $\frac{dy}{dt} \div \frac{dx}{dt} = \frac{dy}{dx}$ with non-zero $t$ to find numerical value of $\frac{dy}{dx}$ | M1 | AO 3.1a |
| Obtains $-\frac{16}{9}$ OE | A1 | AO 1.1b |
## Question 14(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $b = 20$ using $t^2 + t$ for their value of $t$ | B1F | AO 2.2a |
## Question 14(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $dx = (2t+1)\,dt$ | M1 | AO 3.1a |
| $A = \int_0^{20} y\,dx = \int_0^4 (4t^2 - t^3)(2t+1)\,dt$ | A1F | AO 1.1b — condone incorrect/omitted limits |
| $= \int_0^4 4t^2 + 7t^3 - 2t^4\,dt$ | R1 | AO 2.1 — rigorous argument showing $t=4$ when $x=20$; may be justified in (b)(i) |
## Question 14(b)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \frac{1856}{15}$ or AWRT 124 | B1 | AO 1.1b |
---14 The curve $C$ is defined for $t \geq 0$ by the parametric equations
$$x = t ^ { 2 } + t \quad \text { and } \quad y = 4 t ^ { 2 } - t ^ { 3 }$$
$C$ is shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-26_691_608_541_717}
14
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $C$ at the point where it intersects the positive $x$-axis.\\
14
\item (i) The area $A$ enclosed between $C$ and the $x$-axis is given by
$$A = \int _ { 0 } ^ { b } y \mathrm {~d} x$$
Find the value of $b$.\\
14 (b) (ii) Use the substitution $y = 4 t ^ { 2 } - t ^ { 3 }$ to show that
$$A = \int _ { 0 } ^ { 4 } \left( 4 t ^ { 2 } + 7 t ^ { 3 } - 2 t ^ { 4 } \right) \mathrm { d } t$$
14 (b) (iii) Find the value of $A$.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q14 [10]}}