CAIE FP1 2012 June — Question 10 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of lamina by integration
DifficultyChallenging +1.2 This is a Further Maths question requiring arc length calculation (with a 'show that' proof) and centroid by integration. The arc length involves differentiating a power function and integrating √(1+y'²), which simplifies nicely. The centroid calculation uses standard formulas with straightforward integration of polynomial terms. While it requires multiple techniques and careful algebra, the methods are direct applications of standard Further Maths formulas without requiring novel insight or particularly complex manipulation.
Spec4.08e Mean value of function: using integral8.06b Arc length and surface area: of revolution, cartesian or parametric
10 The curve \(C\) has equation $$y = 2 \left( \frac { x } { 3 } \right) ^ { \frac { 3 } { 2 } }$$ where \(0 \leqslant x \leqslant 3\). Show that the arc length of \(C\) is \(2 ( 2 \sqrt { 2 } - 1 )\). Find the coordinates of the centroid of the region enclosed by \(C\), the \(x\)-axis and the line \(x = 3\).
Question 10:
Part: Differentiates and squares
AnswerMarks
\(y'=\dfrac{1}{\sqrt{3}}x^{\frac{1}{2}} \Rightarrow (y')^2=\dfrac{x}{3}\)B1
Part: Uses formula for arc length
AnswerMarks
\(s=\int_0^3\sqrt{1+\dfrac{x}{3}}\,dx\)M1
Part: Integrates and obtains value
AnswerMarks Guidance
\(=\left[2\left(1+\dfrac{x}{3}\right)^{\frac{3}{2}}\right]_0^3=4\sqrt{2}-2=2(2\sqrt{2}-1)\)A1A1 (AG)
Part: x-coordinate of centroid
AnswerMarks
\(\bar{x}=\dfrac{\int_0^3\dfrac{2}{3\sqrt{3}}x^{\frac{5}{2}}\,dx}{\int_0^3\dfrac{2}{3\sqrt{3}}x^{\frac{3}{2}}\,dx}\)M1
\(=\dfrac{\left[\dfrac{2}{7}x^{\frac{7}{2}}\right]_0^3}{\left[\dfrac{2}{5}x^{\frac{5}{2}}\right]_0^3}=\dfrac{15}{7}\) \((=2.14)\)A1A1, A1
Part: y-coordinate of centroid
AnswerMarks Guidance
\(\bar{y}=\dfrac{\int_0^3\dfrac{1}{2}\times\dfrac{4}{27}x^3\,dx}{\int_0^3\dfrac{2}{3\sqrt{3}}x^{\frac{3}{2}}\,dx}\)M1
\(=\dfrac{\dfrac{2}{27}\left[\dfrac{x^4}{4}\right]_0^3}{\dfrac{2}{3\sqrt{3}}\left[\dfrac{2}{5}x^{\frac{5}{2}}\right]_0^3}=\dfrac{5}{8}\) \((=0.625)\)A1A1 7 marks 
# Question 10:

## Part: Differentiates and squares
$y'=\dfrac{1}{\sqrt{3}}x^{\frac{1}{2}} \Rightarrow (y')^2=\dfrac{x}{3}$ | B1 |

## Part: Uses formula for arc length
$s=\int_0^3\sqrt{1+\dfrac{x}{3}}\,dx$ | M1 |

## Part: Integrates and obtains value
$=\left[2\left(1+\dfrac{x}{3}\right)^{\frac{3}{2}}\right]_0^3=4\sqrt{2}-2=2(2\sqrt{2}-1)$ | A1A1 | (AG) | 4 marks

## Part: x-coordinate of centroid
$\bar{x}=\dfrac{\int_0^3\dfrac{2}{3\sqrt{3}}x^{\frac{5}{2}}\,dx}{\int_0^3\dfrac{2}{3\sqrt{3}}x^{\frac{3}{2}}\,dx}$ | M1 |

$=\dfrac{\left[\dfrac{2}{7}x^{\frac{7}{2}}\right]_0^3}{\left[\dfrac{2}{5}x^{\frac{5}{2}}\right]_0^3}=\dfrac{15}{7}$ $(=2.14)$ | A1A1, A1 |

## Part: y-coordinate of centroid
$\bar{y}=\dfrac{\int_0^3\dfrac{1}{2}\times\dfrac{4}{27}x^3\,dx}{\int_0^3\dfrac{2}{3\sqrt{3}}x^{\frac{3}{2}}\,dx}$ | M1 |

$=\dfrac{\dfrac{2}{27}\left[\dfrac{x^4}{4}\right]_0^3}{\dfrac{2}{3\sqrt{3}}\left[\dfrac{2}{5}x^{\frac{5}{2}}\right]_0^3}=\dfrac{5}{8}$ $(=0.625)$ | A1A1 | 7 marks

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10 The curve $C$ has equation

$$y = 2 \left( \frac { x } { 3 } \right) ^ { \frac { 3 } { 2 } }$$

where $0 \leqslant x \leqslant 3$. Show that the arc length of $C$ is $2 ( 2 \sqrt { 2 } - 1 )$.

Find the coordinates of the centroid of the region enclosed by $C$, the $x$-axis and the line $x = 3$.

\hfill \mbox{\textit{CAIE FP1 2012 Q10 [11]}}
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