Edexcel P4 2024 June — Question 4 6 marks

Exam BoardEdexcel
ModuleP4 (Pure Mathematics 4)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeSegment or sector area rate
DifficultyStandard +0.3 This is a standard connected rates of change question requiring the area formula for a circular segment (A = ½r²(2θ - sin 2θ)), differentiation, and chain rule application. Part (a) is routine calculus with a given answer to show, and part (b) is straightforward substitution. Slightly easier than average due to the structured guidance and standard techniques.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
\includegraphics{figure_1} Figure 1 shows a sketch of a segment \(PQRP\) of a circle with centre \(O\) and radius \(5\) cm. Given that • angle \(PQR\) is \(\theta\) radians • \(\theta\) is increasing, from \(0\) to \(\pi\), at a constant rate of \(0.1\) radians per second • the area of the segment \(PQRP\) is \(A\) cm²
  1. show that $$\frac{dA}{d\theta} = K(1 - \cos \theta)$$ where \(K\) is a constant to be found. [2]
  2. Find, in cm²s⁻¹, the rate of increase of the area of the segment when \(\theta = \frac{\pi}{3}\) [4]
Question 4:
AnswerMarks
4(a)1 ( ) d A 2 5 ( )
A 5 2 s i n   = − 1 c o s   = 
AnswerMarks
2 d 2 M1
d A 2 5 ( )
1 c o s  = −
AnswerMarks
d 2 A1
(2)
AnswerMarks
(b)d 
0 . 1 =
AnswerMarks
d tB1
d A d A d 
=  (oe)
AnswerMarks
d t d d t B1
dA 25 
= 1−cos 0.1
AnswerMarks
dt 2  3M1
5
= (cm2 per second)
AnswerMarks
8A1
(4)
(6 marks)
(a)
M1: Attempts formula for area of segment with r = 5 and differentiates with respect to θ.
Look for 5 2
(
− s i n
)
  
d
d
A (
1 c o s
)
 
 =  Allow with either
1
2
 = or 1
You may see A=52−52sin 
d
d
A
=  c o s    again with
1
2
 = or 1
A1: Reaches correct result with no errors seen. Allow with 12.5.
Must follow M1 (the form is given). It cannot just appear without sight of some correct working
(b)
B1: States or uses
d
d t
0 . 1
=
This may be seen anywhere in the question, even in part (a) or next to the Figure
B1: States or uses a correct version of the chain rule for the given problem.
May be seen in (a) or in the question itself.
E.g
d
d
A
t
d
d
A d
d t
=  or
d
d A
d
d
A
t
d
d t
 
 = are two correct versions
This may be implied by an equation involving their
d
d
A
or
d
d
A
= K
(
1 − c o s
)
 and their
d
d t
M1: Attempts the chain rule with their (a), a correct
d
d t
0 . 1
= and
3
 = to find a value for
d
d
A
t
If they don’t attempt (a), or use a made up K, condone
d
d
A
t
= 0 .1 
1
2
' K ' =
1
2 0
' K ' , but
d
d
A
must be of
the form K
(
1 − c o s
)
 via this route.
A1: CSO. units not required. Decimal answer is 0.625.
Cannot be scored following an incorrect part (a) but allow if they have
(a) left as
d
d
A r
2
2 (
1 c o s
)
= − followed by (b)
d
d
A
t
2
2
5
1 c o s
3
0 . 1
5
8
=

−

 =
PMT
which scores 00 1111.
AnswerMarks Guidance
QuestionScheme Marks 
Question 4:
--- 4(a) ---
4(a) | 1 ( ) d A 2 5 ( )
A 5 2 s i n   = − 1 c o s   = 
2 d 2  | M1
d A 2 5 ( )
1 c o s  = −
d 2  | A1
(2)
(b) | d 
0 . 1 =
d t | B1
d A d A d 
=  (oe)
d t d d t  | B1
dA 25 
= 1−cos 0.1
dt 2  3 | M1
5
= (cm2 per second)
8 | A1
(4)
(6 marks)
(a)
M1: Attempts formula for area of segment with r = 5 and differentiates with respect to θ.
Look for 5 2
(
− s i n
)
  
d
d
A (
1 c o s
)
 

 =  Allow with either
1
2
 = or 1
You may see A=52−52sin 
d
d
A
=  c o s    again with

1
2
 = or 1
A1: Reaches correct result with no errors seen. Allow with 12.5.
Must follow M1 (the form is given). It cannot just appear without sight of some correct working
(b)
B1: States or uses
d
d t
0 . 1

=
This may be seen anywhere in the question, even in part (a) or next to the Figure
B1: States or uses a correct version of the chain rule for the given problem.
May be seen in (a) or in the question itself.
E.g
d
d
A
t
d
d
A d
d t


=  or
d
d A
d
d
A
t
d
d t
 
 = are two correct versions
This may be implied by an equation involving their
d
d
A
or

d
d
A
= K
(
1 − c o s
)
 and their

d
d t

M1: Attempts the chain rule with their (a), a correct
d
d t
0 . 1

= and
3

 = to find a value for
d
d
A
t
If they don’t attempt (a), or use a made up K, condone
d
d
A
t
= 0 .1 
1
2
' K ' =
1
2 0
' K ' , but
d
d
A
must be of

the form K
(
1 − c o s
)
 via this route.
A1: CSO. units not required. Decimal answer is 0.625.
Cannot be scored following an incorrect part (a) but allow if they have
(a) left as
d
d
A r
2
2 (
1 c o s
)


= − followed by (b)
d
d
A
t
2
2
5
1 c o s
3
0 . 1
5
8

=

−

 =
PMT
which scores 00 1111.
Question | Scheme | Marks
\includegraphics{figure_1}

Figure 1 shows a sketch of a segment $PQRP$ of a circle with centre $O$ and radius $5$ cm.

Given that
• angle $PQR$ is $\theta$ radians
• $\theta$ is increasing, from $0$ to $\pi$, at a constant rate of $0.1$ radians per second
• the area of the segment $PQRP$ is $A$ cm²

\begin{enumerate}[label=(\alph*)]
\item show that

$$\frac{dA}{d\theta} = K(1 - \cos \theta)$$

where $K$ is a constant to be found.
[2]

\item Find, in cm²s⁻¹, the rate of increase of the area of the segment when $\theta = \frac{\pi}{3}$
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel P4 2024 Q4 [6]}}
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