| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Kinematics: displacement-velocity-acceleration |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of power rule, finding stationary points using standard second derivative test, and interpreting derivatives as rates of change. All parts follow textbook procedures with no problem-solving insight needed, making it easier than average but not trivial due to the polynomial degree and multiple parts. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07d Second derivatives: d^2y/dx^2 notation1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a)(i) | \(\frac{dx}{dt} = 2t^3 - 40t + 66\) | M1 A1 A1 |
| 5(a)(ii) | \(\frac{d^2x}{dt^2} = 6t^2 - 40\) | M1 A1 |
| 5(b) | \(\frac{dx}{dt} = 54 - 120 + 66 = 0 \Rightarrow\) stationary value | M1 A1 |
| Substitute \(t = 3\) into \(\frac{d^2x}{dt^2}\) \((= 14)\) | M1 | |
| \(\frac{d^2x}{dt^2} > 0 \Rightarrow\) minimum value | A1 | CSO; all values (if stated) must be correct |
| 5(c) | Substitute \(t = 1\) into their \(\frac{dx}{dt}\) | M1 |
| \(\frac{dx}{dt} = 28\) | A1 | ft their \(\frac{dx}{dt}\) when \(t = 1\) |
| 5(d) | Substitute \(t = 2\) into their \(\frac{dx}{dt}\) | M1 |
| \(= 16 - 80 + 66 = 2\) \((> 0)\) | Interpreting their value of \(\frac{dx}{dt}\) | |
| \(\Rightarrow\) increasing when \(t = 2\) | E1 | Allow decreasing if their \(\frac{dx}{dt} < 0\) |
**5(a)(i)** | $\frac{dx}{dt} = 2t^3 - 40t + 66$ | M1 A1 A1 | M1: one term correct. A1: another term correct. A1: all correct unsimplified (no + c etc)
**5(a)(ii)** | $\frac{d^2x}{dt^2} = 6t^2 - 40$ | M1 A1 | M1: ft one term correct. A1/: ft all "correct", 2 terms equivalent
**5(b)** | $\frac{dx}{dt} = 54 - 120 + 66 = 0 \Rightarrow$ stationary value | M1 A1 | M1: substitute $t = 3$ into their $\frac{dx}{dt}$. A1: CSO shown $= 0$ (54 or $2 \times 27$ seen) and statement
| Substitute $t = 3$ into $\frac{d^2x}{dt^2}$ $(= 14)$ | M1 | |
| $\frac{d^2x}{dt^2} > 0 \Rightarrow$ minimum value | A1 | CSO; all values (if stated) must be correct
**5(c)** | Substitute $t = 1$ into their $\frac{dx}{dt}$ | M1 | must be their $\frac{dx}{dt}$ NOT $\frac{d^2x}{dt^2}$ etc
| $\frac{dx}{dt} = 28$ | A1 | ft their $\frac{dx}{dt}$ when $t = 1$
**5(d)** | Substitute $t = 2$ into their $\frac{dx}{dt}$ | M1 | must be their $\frac{dx}{dt}$ NOT $\frac{d^2x}{dt^2}$ or $x$
| $= 16 - 80 + 66 = 2$ $(> 0)$ | | Interpreting their value of $\frac{dx}{dt}$
| $\Rightarrow$ increasing when $t = 2$ | E1 | Allow decreasing if their $\frac{dx}{dt} < 0$
**Total: 13 marks**
---5 A model car moves so that its distance, $x$ centimetres, from a fixed point $O$ after time $t$ seconds is given by
$$x = \frac { 1 } { 2 } t ^ { 4 } - 20 t ^ { 2 } + 66 t , \quad 0 \leqslant t \leqslant 4$$
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } x } { \mathrm {~d} t }$;
\item $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } }$.
\end{enumerate}\item Verify that $x$ has a stationary value when $t = 3$, and determine whether this stationary value is a maximum value or a minimum value.
\item Find the rate of change of $x$ with respect to $t$ when $t = 1$.
\item Determine whether the distance of the car from $O$ is increasing or decreasing at the instant when $t = 2$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q5 [13]}}