Question 3 - A-Level Maths

AQA C1 2011 June — Question 3 11 marks

Exam Board	AQA
Module	C1 (Core Mathematics 1)
Year	2011
Session	June
Marks	11
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 only routine application of power rule, evaluation at a point, and basic second derivative test. All steps are standard textbook exercises 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 notation 1.07i Differentiate x^n: for rational n and sums 1.07n Stationary points: find maxima, minima using derivatives 1.07p Points of inflection: using second derivative

3 The volume, $V \mathrm {~m} ^ { 3 }$, of water in a tank after time $t$ seconds is given by $$V = \frac { t ^ { 3 } } { 4 } - 3 t + 5$$

Find $\frac { \mathrm { d } V } { \mathrm {~d} t }$.
1. Find the rate of change of volume, in $\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }$, when $t = 1$.
2. Hence determine, with a reason, whether the volume is increasing or decreasing when $t = 1$.
1. Find the positive value of $t$ for which $V$ has a stationary value.
2. Find $\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }$, and hence determine whether this stationary value is a maximum value or a minimum value.
  (3 marks)

Show mark scheme Show mark scheme source

3(a)

Answer	Marks	Guidance
$\frac{dV}{dr} = \frac{3r^2}{4} - 3$	M1	one of these terms correct
A1	all correct (no $+c$ etc)

3(b)(i)

Answer	Marks	Guidance
$t = 1 \Rightarrow \frac{dV}{dr} = \frac{3}{4} - 3 = -2\frac{1}{4}$	M1	substituting $t = 1$ into their $\frac{dV}{dr}$
A1 also	$(-2.25$ OE$)$ BUT must have $\frac{dV}{dr}$ correct

3(b)(ii)

Answer	Marks	Guidance
Volume is decreasing when $t = 1$ because $\frac{dV}{dr} < 0$	E1 $\checkmark$	must state that $\frac{dV}{dr} < 0$ (or $-2\frac{1}{4} < 0$ etc)
ft increasing plus explanation
if their $\frac{dV}{dr} > 0$

3(c)(i)

Answer	Marks	Guidance
$\frac{dV}{dr} = 0 \Rightarrow \frac{3t^2}{4} - 3 = 0$	M1	PI by "correct" equation being solved
$\Rightarrow t^2 = 4$	A1 $\checkmark$	obtaining $t^n = k$ correctly from their $\frac{dV}{dr}$
$t = 2$	A1 also	withhold if answer left as $t = \pm 2$

3(c)(ii)

Answer	Marks	Guidance
$\frac{d^2V}{dr^2} = \frac{3t}{2}$	B1 $\checkmark$	(condone unsimplified) ft their $\frac{dV}{dr}$
When $t = 2$, $\frac{d^2V}{dr^2} = 3$ or $\frac{d^2V}{dr^2} > 0$ $\Rightarrow$ minimum	M1	ft their $\frac{d^2V}{dr^2}$ and value of $t$ from (c)(i)
A1 also

**3(a)**

$\frac{dV}{dr} = \frac{3r^2}{4} - 3$ | M1 | one of these terms correct

| A1 | all correct (no $+c$ etc)

**3(b)(i)**

$t = 1 \Rightarrow \frac{dV}{dr} = \frac{3}{4} - 3 = -2\frac{1}{4}$ | M1 | substituting $t = 1$ into their $\frac{dV}{dr}$

| A1 also | $(-2.25$ OE$)$ BUT must have $\frac{dV}{dr}$ correct

**3(b)(ii)**

Volume is decreasing when $t = 1$ because $\frac{dV}{dr} < 0$ | E1 $\checkmark$ | must state that $\frac{dV}{dr} < 0$ (or $-2\frac{1}{4} < 0$ etc)

| | ft increasing plus explanation

| | if their $\frac{dV}{dr} > 0$

**3(c)(i)**

$\frac{dV}{dr} = 0 \Rightarrow \frac{3t^2}{4} - 3 = 0$ | M1 | PI by "correct" equation being solved

$\Rightarrow t^2 = 4$ | A1 $\checkmark$ | obtaining $t^n = k$ correctly from their $\frac{dV}{dr}$

$t = 2$ | A1 also | withhold if answer left as $t = \pm 2$

**3(c)(ii)**

$\frac{d^2V}{dr^2} = \frac{3t}{2}$ | B1 $\checkmark$ | (condone unsimplified) ft their $\frac{dV}{dr}$

When $t = 2$, $\frac{d^2V}{dr^2} = 3$ or $\frac{d^2V}{dr^2} > 0$ $\Rightarrow$ minimum | M1 | ft their $\frac{d^2V}{dr^2}$ and value of $t$ from (c)(i)

| A1 also |

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Show LaTeX source

3 The volume, $V \mathrm {~m} ^ { 3 }$, of water in a tank after time $t$ seconds is given by

$$V = \frac { t ^ { 3 } } { 4 } - 3 t + 5$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } V } { \mathrm {~d} t }$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the rate of change of volume, in $\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }$, when $t = 1$.
\item Hence determine, with a reason, whether the volume is increasing or decreasing when $t = 1$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the positive value of $t$ for which $V$ has a stationary value.
\item Find $\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }$, and hence determine whether this stationary value is a maximum value or a minimum value.\\
(3 marks)
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2011 Q3 [11]}}

This paper (8 questions)

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Q1 9 Q2 8 Q3 11 Q4 12 Q5 8 Q6 8 Q7 6 Q8 13

Answer	Marks	Guidance
\(\frac{dV}{dr} = \frac{3r^2}{4} - 3\)	M1	one of these terms correct
A1	all correct (no \(+c\) etc)

Answer	Marks	Guidance
\(t = 1 \Rightarrow \frac{dV}{dr} = \frac{3}{4} - 3 = -2\frac{1}{4}\)	M1	substituting \(t = 1\) into their \(\frac{dV}{dr}\)
A1 also	\((-2.25\) OE\()\) BUT must have \(\frac{dV}{dr}\) correct

Answer	Marks	Guidance
Volume is decreasing when \(t = 1\) because \(\frac{dV}{dr} < 0\)	E1 \(\checkmark\)	must state that \(\frac{dV}{dr} < 0\) (or \(-2\frac{1}{4} < 0\) etc)
ft increasing plus explanation
if their \(\frac{dV}{dr} > 0\)

Answer	Marks	Guidance
\(\frac{dV}{dr} = 0 \Rightarrow \frac{3t^2}{4} - 3 = 0\)	M1	PI by "correct" equation being solved
\(\Rightarrow t^2 = 4\)	A1 \(\checkmark\)	obtaining \(t^n = k\) correctly from their \(\frac{dV}{dr}\)
\(t = 2\)	A1 also	withhold if answer left as \(t = \pm 2\)

Answer	Marks	Guidance
\(\frac{d^2V}{dr^2} = \frac{3t}{2}\)	B1 \(\checkmark\)	(condone unsimplified) ft their \(\frac{dV}{dr}\)
When \(t = 2\), \(\frac{d^2V}{dr^2} = 3\) or \(\frac{d^2V}{dr^2} > 0\) \(\Rightarrow\) minimum	M1	ft their \(\frac{d^2V}{dr^2}\) and value of \(t\) from (c)(i)
A1 also