CAIE FP1 2013 June — Question 8 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks11
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of lamina by integration
DifficultyChallenging +1.2 This is a two-part Further Maths question requiring parametric arc length integration and centroid calculation. While it involves multiple techniques (parametric differentiation, integration with substitution), the setup is standard and the parametric forms are designed to integrate cleanly. The arc length formula application is routine for FP1, and the centroid calculation follows textbook methods. More challenging than average A-level due to being Further Maths content, but straightforward execution within that context.
Spec4.08e Mean value of function: using integral8.06b Arc length and surface area: of revolution, cartesian or parametric
8 The curve \(C\) has parametric equations \(x = \frac { 3 } { 2 } t ^ { 2 } , y = t ^ { 3 }\), for \(0 \leqslant t \leqslant 2\). Find the arc length of \(C\). Find the coordinates of the centroid of the region enclosed by \(C\), the \(x\)-axis and the line \(x = 6\).
Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\dot{x} = 3t\), \(\dot{y} = 3t^2 \Rightarrow \frac{ds}{dt} = \sqrt{9t^2 + 9t^4} = 3t\sqrt{1+t^2}\)M1A1
\(s = \int_0^2 3t(1+t^2)^{\frac{1}{2}}\,dt = \left[(1+t^2)^{\frac{3}{2}}\right]_0^2\)M1
\(\Rightarrow s = 5\sqrt{5} - 1\)A1 4 marks (Allow 10.2)
\(\bar{x} = \dfrac{\int_0^6 xy\,dx}{\int_0^6 y\,dx} = \dfrac{\int_0^2 3\frac{t^2}{2} \cdot t^3 \cdot 3t\,dt}{\int_0^2 t^3 \cdot 3t\,dt}\)M1A1
\(= \dfrac{\frac{3}{2}\int_0^2 t^6\,dt}{\int_0^2 t^4\,dt} = \dfrac{\frac{3}{2}\left[\frac{1}{7}t^7\right]_0^2}{\left[\frac{1}{5}t^5\right]_0^2} = \frac{3}{2} \times \frac{5}{7} \times 4 = \frac{30}{7}\)M1A1 (Or 4.29)
\(\bar{y} = \dfrac{\frac{1}{2}\int_0^6 y^2\,dx}{\int_0^6 y\,dx} = \dfrac{\frac{1}{2}\int_0^2 t^6 \cdot 3t\,dt}{\int_0^2 t^3 \cdot 3t\,dt}\)M1
\(= \dfrac{\frac{1}{2}\int_0^2 t^7\,dt}{\int_0^2 t^4\,dt} = \dfrac{\frac{1}{2}\left[\frac{1}{8}t^8\right]_0^2}{\left[\frac{1}{5}t^5\right]_0^2} = \frac{1}{2} \times \frac{5}{8} \times 8 = \frac{5}{2}\)M1A1 7 marks (Or 2.5), [11]
Alternative layout: eliminating \(t\) \(\int y\,dx\) (B1), \(\int xy\,dx\) (B1), \(\int \frac{1}{2}y^2\,dx\) (B1) in terms of \(t\). Then award M1A1 for each of \(\bar{x}\) and \(\bar{y}\) 
# Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dot{x} = 3t$, $\dot{y} = 3t^2 \Rightarrow \frac{ds}{dt} = \sqrt{9t^2 + 9t^4} = 3t\sqrt{1+t^2}$ | M1A1 | |
| $s = \int_0^2 3t(1+t^2)^{\frac{1}{2}}\,dt = \left[(1+t^2)^{\frac{3}{2}}\right]_0^2$ | M1 | |
| $\Rightarrow s = 5\sqrt{5} - 1$ | A1 | 4 marks (Allow 10.2) |
| $\bar{x} = \dfrac{\int_0^6 xy\,dx}{\int_0^6 y\,dx} = \dfrac{\int_0^2 3\frac{t^2}{2} \cdot t^3 \cdot 3t\,dt}{\int_0^2 t^3 \cdot 3t\,dt}$ | M1A1 | |
| $= \dfrac{\frac{3}{2}\int_0^2 t^6\,dt}{\int_0^2 t^4\,dt} = \dfrac{\frac{3}{2}\left[\frac{1}{7}t^7\right]_0^2}{\left[\frac{1}{5}t^5\right]_0^2} = \frac{3}{2} \times \frac{5}{7} \times 4 = \frac{30}{7}$ | M1A1 | (Or 4.29) |
| $\bar{y} = \dfrac{\frac{1}{2}\int_0^6 y^2\,dx}{\int_0^6 y\,dx} = \dfrac{\frac{1}{2}\int_0^2 t^6 \cdot 3t\,dt}{\int_0^2 t^3 \cdot 3t\,dt}$ | M1 | |
| $= \dfrac{\frac{1}{2}\int_0^2 t^7\,dt}{\int_0^2 t^4\,dt} = \dfrac{\frac{1}{2}\left[\frac{1}{8}t^8\right]_0^2}{\left[\frac{1}{5}t^5\right]_0^2} = \frac{1}{2} \times \frac{5}{8} \times 8 = \frac{5}{2}$ | M1A1 | 7 marks (Or 2.5), **[11]** |
| Alternative layout: eliminating $t$ | | $\int y\,dx$ (B1), $\int xy\,dx$ (B1), $\int \frac{1}{2}y^2\,dx$ (B1) in terms of $t$. Then award M1A1 for each of $\bar{x}$ and $\bar{y}$ |

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8 The curve $C$ has parametric equations $x = \frac { 3 } { 2 } t ^ { 2 } , y = t ^ { 3 }$, for $0 \leqslant t \leqslant 2$. Find the arc length of $C$.

Find the coordinates of the centroid of the region enclosed by $C$, the $x$-axis and the line $x = 6$.

\hfill \mbox{\textit{CAIE FP1 2013 Q8 [11]}}
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