| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Easy -1.2 This is a straightforward C1 differentiation question requiring only routine application of the power rule for polynomials. All parts involve standard procedures: differentiating twice, substituting a value, and using the second derivative test. No problem-solving or novel insight required—purely mechanical application of basic calculus rules. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dt} = 2t^3 - 8t + 6t\) | M1 | — |
| A1 | — | One term correct unsimplified |
| A1 | 3 | Further term correct unsimplified; All correct unsimplified (no + c etc) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2V}{dt^2} = 10t^4 - 24t^2 + 6\) | M1 | One term FT correct unsimplified |
| A1 | 2 | CSO. All correct simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(t = 2\) into their \(\frac{dV}{dt}\) | M1 | — |
| \((= 64 - 64 + 12) = 12\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 1 \Rightarrow \frac{dV}{dt} = 2 - 8 + 6 = 0 \Rightarrow\) Stationary value | M1 | Or putting their \(\frac{dV}{dt} = 0\) |
| A1 | 2 | CSO. Shown to = 0 AND statement (If solving equation must obtain t = 1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 1 \Rightarrow \frac{d^2V}{dt^2} = -8\) | M1 | Sub t = 1 into their second derivative or equivalent full test. |
| Maximum value | A1√ | 2 |
## 7(a)(i)
$\frac{dV}{dt} = 2t^3 - 8t + 6t$ | M1 | —
| A1 | — | One term correct unsimplified
| A1 | 3 | Further term correct unsimplified; All correct unsimplified (no + c etc)
## 7(a)(ii)
$\frac{d^2V}{dt^2} = 10t^4 - 24t^2 + 6$ | M1 | One term FT correct unsimplified
| A1 | 2 | CSO. All correct simplified
## 7(b)
Substitute $t = 2$ into their $\frac{dV}{dt}$ | M1 | —
$(= 64 - 64 + 12) = 12$ | A1 | 2 | CSO. Rate of change of volume is $12m^3 s^{-1}$
## 7(c)(i)
$t = 1 \Rightarrow \frac{dV}{dt} = 2 - 8 + 6 = 0 \Rightarrow$ Stationary value | M1 | Or putting their $\frac{dV}{dt} = 0$
| A1 | 2 | CSO. Shown to = 0 **AND** statement (If solving equation must obtain t = 1)
## 7(c)(ii)
$t = 1 \Rightarrow \frac{d^2V}{dt^2} = -8$ | M1 | Sub t = 1 into their second derivative or equivalent full test.
Maximum value | A1√ | 2 | Ft if their test implies minimum
---7 The volume, $V \mathrm {~m} ^ { 3 }$, of water in a tank at time $t$ seconds is given by
$$V = \frac { 1 } { 3 } t ^ { 6 } - 2 t ^ { 4 } + 3 t ^ { 2 } , \quad \text { for } t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } V } { \mathrm {~d} t }$;\\
(3 marks)
\item $\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }$.\\
(2 marks)
\end{enumerate}\item Find the rate of change of the volume of water in the tank, in $\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }$, when $t = 2$.\\
(2 marks)
\item \begin{enumerate}[label=(\roman*)]
\item Verify that $V$ has a stationary value when $t = 1$.\\
(2 marks)
\item Determine whether this is a maximum or minimum value.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2006 Q7 [11]}}