| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Velocity and acceleration problems |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of the power rule, finding stationary points using standard second derivative test, and interpreting derivatives. All parts follow textbook procedures with no problem-solving or novel insight required, making it easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t^3 - 52t + 96\) | M1, A1, A1 | one term correct; another term correct; all correct (no \(+ c\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3t^2 - 52\) | M1 | ft one term correct |
| A1\(\checkmark\) | ft all "correct" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dt} = 8 - 104 + 96\) | M1 | substitute \(t = 2\) into their \(\frac{dy}{dt}\) |
| \(= 0 \Rightarrow\) stationary value | A1 | CSO; shown \(= 0 +\) statement |
| Substitute \(t = 2\) into \(\frac{d^2y}{dt^2}\) \((= -40)\) | M1 | any appropriate test, e.g. \(y'(1)\) and \(y'(3)\) |
| \(\frac{d^2y}{dt^2} < 0 \Rightarrow\) max value | A1 | all values (if stated) must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(t = 1\) into their \(\frac{dy}{dt}\) | M1 | must be their \(\frac{dy}{dt}\) NOT \(\frac{d^2y}{dt^2}\) |
| Rate of change \(= 45 \; (\text{cms}^{-1})\) | A1\(\checkmark\) | ft their \(y'(1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(t = 3\) into their \(\frac{dy}{dt}\) | M1 | interpreting their value of \(\frac{dy}{dt}\) |
| \((27 - 156 + 96 = -33 < 0)\) | ||
| \(\Rightarrow\) decreasing when \(t = 3\) | E1\(\checkmark\) | allow increasing if their \(\frac{dy}{dt} > 0\) |
## Question 4:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t^3 - 52t + 96$ | M1, A1, A1 | one term correct; another term correct; all correct (no $+ c$ etc) |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3t^2 - 52$ | M1 | ft one term correct |
| | A1$\checkmark$ | ft all "correct" |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dt} = 8 - 104 + 96$ | M1 | substitute $t = 2$ into their $\frac{dy}{dt}$ |
| $= 0 \Rightarrow$ stationary value | A1 | CSO; shown $= 0 +$ statement |
| Substitute $t = 2$ into $\frac{d^2y}{dt^2}$ $(= -40)$ | M1 | any appropriate test, e.g. $y'(1)$ and $y'(3)$ |
| $\frac{d^2y}{dt^2} < 0 \Rightarrow$ max value | A1 | all values (if stated) must be correct |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $t = 1$ into their $\frac{dy}{dt}$ | M1 | must be their $\frac{dy}{dt}$ NOT $\frac{d^2y}{dt^2}$ |
| Rate of change $= 45 \; (\text{cms}^{-1})$ | A1$\checkmark$ | ft their $y'(1)$ |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $t = 3$ into their $\frac{dy}{dt}$ | M1 | interpreting their value of $\frac{dy}{dt}$ |
| $(27 - 156 + 96 = -33 < 0)$ | | |
| $\Rightarrow$ decreasing when $t = 3$ | E1$\checkmark$ | allow increasing if their $\frac{dy}{dt} > 0$ |
---4 A model helicopter takes off from a point $O$ at time $t = 0$ and moves vertically so that its height, $y \mathrm {~cm}$, above $O$ after time $t$ seconds is given by
$$y = \frac { 1 } { 4 } t ^ { 4 } - 26 t ^ { 2 } + 96 t , \quad 0 \leqslant t \leqslant 4$$
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} t }$;\\
(3 marks)
\item $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }$.\\
(2 marks)
\end{enumerate}\item Verify that $y$ has a stationary value when $t = 2$ and determine whether this stationary value is a maximum value or a minimum value.\\
(4 marks)
\item Find the rate of change of $y$ with respect to $t$ when $t = 1$.
\item Determine whether the height of the helicopter above $O$ is increasing or decreasing at the instant when $t = 3$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2007 Q4 [13]}}