Edexcel CP AS 2018 June — Question 9 11 marks

Exam BoardEdexcel
ModuleCP AS (Core Pure AS)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeApplied context: real-world solid
DifficultyStandard +0.3 This is a straightforward volumes of revolution question with a real-world context. Part (a) requires solving simultaneous equations from boundary conditions (routine algebra), part (b) is a standard π∫y² dx integration, part (c) asks for a model limitation (accessible discussion), and part (d) requires percentage comparison. All techniques are standard Core Pure AS level with no novel insights required, making it slightly easier than average.
Spec1.02m Graphs of functions: difference between plotting and sketching4.08d Volumes of revolution: about x and y axes

9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e48fac26-15a2-4a5e-9204-9d49db8a998a-32_789_452_331_497} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e48fac26-15a2-4a5e-9204-9d49db8a998a-32_681_523_424_1248} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A mathematics student is modelling the profile of a glass bottle of water. Figure 1 shows a sketch of a central vertical cross-section \(A B C D E F G H A\) of the bottle with the measurements taken by the student. The horizontal cross-section between \(C F\) and \(D E\) is a circle of diameter 8 cm and the horizontal cross-section between \(B G\) and \(A H\) is a circle of diameter 2 cm . The student thinks that the curve \(G F\) could be modelled as a curve with equation $$y = a x ^ { 2 } + b \quad 1 \leqslant x \leqslant 4$$ where \(a\) and \(b\) are constants and \(O\) is the fixed origin, as shown in Figure 2.
  1. Find the value of \(a\) and the value of \(b\) according to the model.
  2. Use the model to find the volume of water that the bottle can contain.
  3. State a limitation of the model. The label on the bottle states that the bottle holds approximately \(750 \mathrm {~cm} ^ { 3 }\) of water.
  4. Use this information and your answer to part (b) to evaluate the model, explaining your reasoning.

Question 9(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((4,14),(1,18)\Rightarrow 14=a(4)^2+b,\ 18=a(1)^2+b \Rightarrow a=\ldots, b=\ldots\)M1
\(a=-\dfrac{4}{15},\ b=\dfrac{274}{15}\)A1
Question 9(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\pi\times4^2\times14\) and \(\pi\times1^2\times10\)B1
\(\pi\int x^2\,dy=\dfrac{\pi}{4}\int(274-15y)\,dy\)B1ft
\(=\dfrac{\pi}{4}\displaystyle\int_{14}^{18}(274-15y)\,dy\)M1
\(=\dfrac{\pi}{4}\left[274y-\dfrac{15y^2}{2}\right]_{14}^{18}\)M1, A1
\(V=234\pi+\dfrac{\pi}{4}\left[274(18)-\dfrac{15(18)^2}{2}-\left(274(14)-\dfrac{15(14)^2}{2}\right)\right]\)ddM1
\(V=268\pi\approx842\text{ cm}^3\)A1
Question 9(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Any one of: measurements may not be accurate; equation of curve may not be suitable model; bottom of bottle may not be flat; thickness of glass not considered; glass may not be smoothB1 Must refer to limitation of the model, e.g. measuring of dimensions, model used for curve, simplified model (thickness of glass, shape, smoothness)
Question 9(d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Comparison of their value to 750 (e.g. \(810-750=\ldots\)) AND conclusion consistent with their value (e.g. not a good model / this is a good model)B1ft Must demonstrate comparison, not just state percentage error or difference with no calculation. If answer \(<750\) conclude not good model; if \(>750\) give sensible consistent comment
Question (from mark scheme notes):
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Substitutes \((4, 14)\) and \((1, 18)\) into the curve equation to obtain at least one correct equationM1 Attempts to find values of \(a\) and \(b\)
Correct values of \(a\) and \(b\) inferred from the data in the modelA1
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct expressions for the 2 cylindrical partsB1 May be seen as a sum or as separate cylinders
Uses the model to obtain \(\pi\int\left(\dfrac{y - b}{a}\right)dy\)B1ft Note that the \(\pi\) may be recovered later
Chooses limits appropriate to the model: \(14\) and \(18\)M1
Integrates to obtain an expression of the form \(\alpha y + \beta y^2\)M1
Uses their model correctly to give \(274y - \dfrac{15y^2}{2}\)A1
Uses the model to find the sum of their cylinders + their integrated volumeddM1 Must be fully correct method; dependent on both previous M marks; must have attempted volumes of cylinders "AHBG" and "CFED" and adds these to the magnitude of their integrated volume
\(268\pi\) or awrt \(842\)A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
States an acceptable limitation of the model with no contradictory statementsB1 Independent of part (b)
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Compares the actual volume to their answer to part (b) and makes an assessment of the model with a reasonB1ft No contradictory statements
## Question 9(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(4,14),(1,18)\Rightarrow 14=a(4)^2+b,\ 18=a(1)^2+b \Rightarrow a=\ldots, b=\ldots$ | M1 | |
| $a=-\dfrac{4}{15},\ b=\dfrac{274}{15}$ | A1 | |

---

## Question 9(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pi\times4^2\times14$ and $\pi\times1^2\times10$ | B1 | |
| $\pi\int x^2\,dy=\dfrac{\pi}{4}\int(274-15y)\,dy$ | B1ft | |
| $=\dfrac{\pi}{4}\displaystyle\int_{14}^{18}(274-15y)\,dy$ | M1 | |
| $=\dfrac{\pi}{4}\left[274y-\dfrac{15y^2}{2}\right]_{14}^{18}$ | M1, A1 | |
| $V=234\pi+\dfrac{\pi}{4}\left[274(18)-\dfrac{15(18)^2}{2}-\left(274(14)-\dfrac{15(14)^2}{2}\right)\right]$ | ddM1 | |
| $V=268\pi\approx842\text{ cm}^3$ | A1 | |

---

## Question 9(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any one of: measurements may not be accurate; equation of curve may not be suitable model; bottom of bottle may not be flat; thickness of glass not considered; glass may not be smooth | B1 | Must refer to limitation of the model, e.g. measuring of dimensions, model used for curve, simplified model (thickness of glass, shape, smoothness) |

---

## Question 9(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Comparison of their value to 750 (e.g. $810-750=\ldots$) AND conclusion consistent with their value (e.g. not a good model / this is a good model) | B1ft | Must demonstrate comparison, not just state percentage error or difference with no calculation. If answer $<750$ conclude not good model; if $>750$ give sensible consistent comment |

# Question (from mark scheme notes):

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $(4, 14)$ and $(1, 18)$ into the curve equation to obtain at least one correct equation | M1 | Attempts to find values of $a$ and $b$ |
| Correct values of $a$ and $b$ inferred from the data in the model | A1 | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct expressions for the 2 cylindrical parts | B1 | May be seen as a sum or as separate cylinders |
| Uses the model to obtain $\pi\int\left(\dfrac{y - b}{a}\right)dy$ | B1ft | Note that the $\pi$ may be recovered later |
| Chooses limits appropriate to the model: $14$ and $18$ | M1 | |
| Integrates to obtain an expression of the form $\alpha y + \beta y^2$ | M1 | |
| Uses their model correctly to give $274y - \dfrac{15y^2}{2}$ | A1 | |
| Uses the model to find the sum of their cylinders + their integrated volume | ddM1 | Must be fully correct method; dependent on both previous M marks; must have attempted volumes of cylinders "AHBG" and "CFED" and adds these to the magnitude of their integrated volume |
| $268\pi$ or awrt $842$ | A1 | |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States an acceptable limitation of the model with no contradictory statements | B1 | Independent of part (b) |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Compares the actual volume to their answer to part (b) and makes an assessment of the model with a reason | B1ft | No contradictory statements |
9.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e48fac26-15a2-4a5e-9204-9d49db8a998a-32_789_452_331_497}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e48fac26-15a2-4a5e-9204-9d49db8a998a-32_681_523_424_1248}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

A mathematics student is modelling the profile of a glass bottle of water. Figure 1 shows a sketch of a central vertical cross-section $A B C D E F G H A$ of the bottle with the measurements taken by the student.

The horizontal cross-section between $C F$ and $D E$ is a circle of diameter 8 cm and the horizontal cross-section between $B G$ and $A H$ is a circle of diameter 2 cm .

The student thinks that the curve $G F$ could be modelled as a curve with equation

$$y = a x ^ { 2 } + b \quad 1 \leqslant x \leqslant 4$$

where $a$ and $b$ are constants and $O$ is the fixed origin, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$ according to the model.
\item Use the model to find the volume of water that the bottle can contain.
\item State a limitation of the model.

The label on the bottle states that the bottle holds approximately $750 \mathrm {~cm} ^ { 3 }$ of water.
\item Use this information and your answer to part (b) to evaluate the model, explaining your reasoning.
\end{enumerate}

\hfill \mbox{\textit{Edexcel CP AS 2018 Q9 [11]}}