Edexcel PURE 2024 October — Question 5

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicConnected Rates of Change
TypeContainer filling: find depth rate
DifficultyStandard +0.3 This is a standard connected rates of change problem requiring similar triangles to eliminate r, differentiation of the volume formula, and substitution of given values. While it involves multiple steps, each technique is routine for A-level Further Maths students, making it slightly easier than average.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07t Construct differential equations: in context
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa121449-492f-4737-a9eb-a14a62ced47b-14_569_616_242_785} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a container in the shape of a hollow, inverted, right circular cone.
The height of the container is 30 cm and the radius is 12 cm , as shown in Figure 3.
The container is initially empty when water starts flowing into it.
When the height of water is \(h \mathrm {~cm}\), the surface of the water has radius \(r \mathrm {~cm}\) and the volume of water is \(V \mathrm {~cm} ^ { 3 }\)
  1. Show that $$V = \frac { 4 \pi h ^ { 3 } } { 75 }$$ [The volume \(V\) of a right circular cone with vertical height \(h\) and base radius \(r\) is given by the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ] Given that water flows into the container at a constant rate of \(2 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\)
  2. find, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), the rate at which \(h\) is changing, exactly 1.5 minutes after water starts flowing into the container.
Question 5:
AnswerMarks
5(a)2
r 1 2 1 2 1 1 2  
e.g. r h V r 2 h h h   =  =  = =
AnswerMarks
h 3 0 3 0 3 3 5M1
4 h 3 
V = *
AnswerMarks
7 5A1*
(2)
AnswerMarks
(b)4h3 dV 12h2  4h2 
V =  = = 
AnswerMarks
75 dh 75  25 B1
V =1.5602
4 h 3 
1 8 0 h 3 3 3 7 5 h ... ( 1 5 )  =  =  =
AnswerMarks
7 5M1
dh dh dV 75
= = 2=...
AnswerMarks
dt dV dt 12152M1
1
(cm s−1)
AnswerMarks
1 8A1
(4)

Total 6

dh 75
Alt(b) I Solving the differential equation = and then using t =90
dt 6h2
dV 12h2  4h2 
B1: = = 
dh 75  25 
dh dh dV  75 
M1: Attempts =   =  to achieve an expression in h, attempts to solve their
dt dV dt  6h2 
differential equation
6 h 2 d h =
7 5 d t  2 h 3 = 7 5 t + c o.e. and uses t = 0 , h = 0  c = 0
(you may not see the working for c) to find h = f(t) ( h = ( 3 7 .5 t )
13
). Do not be concerned by
the mechanics of the rearrangement.
M1: Attempts
d
d
h
t

=
1
3
3 3 7 .5 t
− 23 
and substitutes t = 9 0 
d
d
h
t
= . . .
A1: As above in main scheme
Alt(b) II Equating volumes, rearranging to h = . . . and then using t = 9 0
B1: V =2t
M1: Equates their 2 t
4h3
 equal to and rearranges to
75
h = f ( t ) ( h = ( 3 7 .5 t )
13
). Do not be
concerned by the mechanics of the rearrangement.
M1: Attempts
d
d
h
t

=
1
3
3 3 7 .5 t
− 23 
and substitutes t = 9 0 
d
d
h
t
= . . .
PMT
A1: As above in main scheme
Question 5:
--- 5(a) ---
5(a) | 2
r 1 2 1 2 1 1 2  
e.g. r h V r 2 h h h   =  =  = =
h 3 0 3 0 3 3 5 | M1
4 h 3 
V = *
7 5 | A1*
(2)
(b) | 4h3 dV 12h2  4h2 
V =  = = 
75 dh 75  25  | B1
V =1.5602
4 h 3 
1 8 0 h 3 3 3 7 5 h ... ( 1 5 )  =  =  =
7 5 | M1
dh dh dV 75
= = 2=...
dt dV dt 12152 | M1
1
(cm s−1)
1 8 | A1
(4)
Total 6
dh 75
Alt(b) I Solving the differential equation = and then using t =90
dt 6h2
dV 12h2  4h2 
B1: = = 
dh 75  25 
dh dh dV  75 
M1: Attempts =   =  to achieve an expression in h, attempts to solve their
dt dV dt  6h2 
differential equation

6 h 2 d h =

7 5 d t  2 h 3 = 7 5 t + c o.e. and uses t = 0 , h = 0  c = 0
(you may not see the working for c) to find h = f(t) ( h = ( 3 7 .5 t )
13
). Do not be concerned by
the mechanics of the rearrangement.
M1: Attempts
d
d
h
t

=
1
3
3 3 7 .5 t
− 23 
and substitutes t = 9 0 
d
d
h
t
= . . .
A1: As above in main scheme
------------------------------------------------------------------------------------------------------------------------
Alt(b) II Equating volumes, rearranging to h = . . . and then using t = 9 0
B1: V =2t
M1: Equates their 2 t
4h3
 equal to and rearranges to
75
h = f ( t ) ( h = ( 3 7 .5 t )
13
). Do not be
concerned by the mechanics of the rearrangement.
M1: Attempts
d
d
h
t

=
1
3
3 3 7 .5 t
− 23 
and substitutes t = 9 0 
d
d
h
t
= . . .
PMT
A1: As above in main scheme
5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa121449-492f-4737-a9eb-a14a62ced47b-14_569_616_242_785}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Figure 3 shows a container in the shape of a hollow, inverted, right circular cone.\\
The height of the container is 30 cm and the radius is 12 cm , as shown in Figure 3.\\
The container is initially empty when water starts flowing into it.\\
When the height of water is $h \mathrm {~cm}$, the surface of the water has radius $r \mathrm {~cm}$ and the volume of water is $V \mathrm {~cm} ^ { 3 }$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$V = \frac { 4 \pi h ^ { 3 } } { 75 }$$

[The volume $V$ of a right circular cone with vertical height $h$ and base radius $r$ is given by the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$ ]

Given that water flows into the container at a constant rate of $2 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$
\item find, in $\mathrm { cm } \mathrm { s } ^ { - 1 }$, the rate at which $h$ is changing, exactly 1.5 minutes after water starts flowing into the container.
\end{enumerate}

\hfill \mbox{\textit{Edexcel PURE 2024 Q5}}
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