| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Rate of change from explicit formula |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of power rule, evaluation at given points, and standard interpretation of first/second derivatives. All parts follow textbook procedures with no problem-solving or novel insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07d Second derivatives: d^2y/dx^2 notation1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{dy}{dt}\right) = \frac{4t^3}{8} - 2t\) | M1 | one of these terms correct |
| A1 | 2 marks | all correct (no \(+c\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 1 \Rightarrow \frac{dy}{dt} = \frac{4}{8} - 2 = -1\frac{1}{2}\) | M1 | Correctly sub \(t = 1\) into their \(\frac{dy}{dt}\) |
| A1 cso | 2 marks | must have \(\frac{dy}{dt}\) correct (watch for \(t^3\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dt} < 0\) | must have used \(\frac{dy}{dt}\) in part (b)(i) | |
| \(\Rightarrow\) (height is) decreasing (when \(t = 1\)) | E1/√ | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{d^2y}{dt^2}\right) = \frac{4}{8} \times 3t^2 - 2\) | M1 | Correctly differentiating their \(\frac{dy}{dt}\) even if wrongly simplified |
| \(\left(t = 2, \frac{d^2y}{dt^2} =\right) 4\) | A1 cso | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow\) minimum | E1/√ | 1 mark |
**2(a)**
$\left(\frac{dy}{dt}\right) = \frac{4t^3}{8} - 2t$ | M1 | one of these terms correct
| A1 | 2 marks | all correct (no $+c$ etc)
**2(b)(i)**
$t = 1 \Rightarrow \frac{dy}{dt} = \frac{4}{8} - 2 = -1\frac{1}{2}$ | M1 | Correctly sub $t = 1$ into their $\frac{dy}{dt}$
| A1 cso | 2 marks | must have $\frac{dy}{dt}$ correct (watch for $t^3$ etc)
**2(b)(ii)**
$\frac{dy}{dt} < 0$ | | must have used $\frac{dy}{dt}$ in part (b)(i)
$\Rightarrow$ (height is) **decreasing** (when $t = 1$) | E1/√ | 1 mark | must state that "$\frac{dy}{dt} < 0$" or "$-1.5 < 0$" or the equivalent in words; FT their value of $\frac{dy}{dt}$ with appropriate explanation and conclusion
**2(c)(i)**
$\left(\frac{d^2y}{dt^2}\right) = \frac{4}{8} \times 3t^2 - 2$ | M1 | Correctly differentiating their $\frac{dy}{dt}$ even if wrongly simplified
$\left(t = 2, \frac{d^2y}{dt^2} =\right) 4$ | A1 cso | 2 marks | Both derivatives correct and simplified to 4
**2(c)(ii)**
$\Rightarrow$ minimum | E1/√ | 1 mark | FT their numerical value of $\frac{d^2y}{dt^2}$ from part (c)(i)
---2 A bird flies from a tree. At time $t$ seconds, the bird's height, $y$ metres, above the horizontal ground is given by
$$y = \frac { 1 } { 8 } t ^ { 4 } - t ^ { 2 } + 5 , \quad 0 \leqslant t \leqslant 4$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} t }$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the rate of change of height of the bird in metres per second when $t = 1$.
\item Determine, with a reason, whether the bird's height above the horizontal ground is increasing or decreasing when $t = 1$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }$ when $t = 2$.
\item Given that $y$ has a stationary value when $t = 2$, state whether this is a maximum value or a minimum value.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q2 [8]}}