CAIE FP1 2012 November — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeParametric surface area of revolution
DifficultyChallenging +1.2 This is a standard Further Maths parametric calculus question requiring arc length and surface area of revolution formulas. While it involves multiple steps and algebraic manipulation (finding dx/dt and dy/dt, then simplifying the square root expression), the techniques are routine for FP1 students and the algebra simplifies cleanly. It's harder than typical A-level Pure questions due to the Further Maths content, but straightforward application of memorized formulas without requiring novel insight.
Spec4.08d Volumes of revolution: about x and y axes4.08f Integrate using partial fractions
8 The curve \(C\) has parametric equations $$x = \frac { 1 } { 3 } t ^ { 3 } - \ln t , \quad y = \frac { 4 } { 3 } t ^ { \frac { 3 } { 2 } }$$ for \(1 \leqslant t \leqslant 3\). Find the arc length of \(C\). Find also the area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
Question 8:
Arc Length Part (6 marks):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\dot{x} = t^2 - \frac{1}{t}\), \(\dot{y} = 2t^{\frac{1}{2}}\)B1 Differentiates
\(\frac{ds}{dt} = \sqrt{\left(t^2-\frac{1}{t}\right)^2+4t} = \sqrt{\left(t^2+\frac{1}{t}\right)^2}\)M1A1 Squares and adds
\(s = \int_1^3 \left(t^2+\frac{1}{t}\right)dt\)M1 Uses arc length formula
\(= \left[\frac{t^3}{3}+\ln t\right]_1^3\)A1 Integrates
\(= 9+\ln 3 - \frac{1}{3} = \frac{26}{3}+\ln 3 \quad (=9.77)\)A1 6
Surface Area Part (4 marks):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S = 2\pi\int_1^3\left(\frac{4}{3}t^{\frac{3}{2}}\right)\left(t^2+\frac{1}{t}\right)dt = \frac{8\pi}{3}\int_1^3\left(t^{\frac{7}{2}}+t^{\frac{1}{2}}\right)dt\)M1 Uses surface area formula
\(= \frac{8\pi}{3}\left[\frac{2}{9}t^{\frac{9}{2}}+\frac{2}{3}t^{\frac{3}{2}}\right]_1^3\)A1 Integrates
\(= \frac{8\pi}{3}\left\{\left[18\sqrt{3}+2\sqrt{3}\right]-\left[\frac{2}{9}+\frac{2}{3}\right]\right\}\)M1 Inserts limits
\(= \pi\left(\frac{160\sqrt{3}}{3}-\frac{64}{27}\right) \quad (=283 \text{ or } 90.0\pi)\)A1 4 
# Question 8:

## Arc Length Part (6 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dot{x} = t^2 - \frac{1}{t}$, $\dot{y} = 2t^{\frac{1}{2}}$ | B1 | Differentiates |
| $\frac{ds}{dt} = \sqrt{\left(t^2-\frac{1}{t}\right)^2+4t} = \sqrt{\left(t^2+\frac{1}{t}\right)^2}$ | M1A1 | Squares and adds |
| $s = \int_1^3 \left(t^2+\frac{1}{t}\right)dt$ | M1 | Uses arc length formula |
| $= \left[\frac{t^3}{3}+\ln t\right]_1^3$ | A1 | Integrates |
| $= 9+\ln 3 - \frac{1}{3} = \frac{26}{3}+\ln 3 \quad (=9.77)$ | A1 | 6 | Obtains result |

## Surface Area Part (4 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S = 2\pi\int_1^3\left(\frac{4}{3}t^{\frac{3}{2}}\right)\left(t^2+\frac{1}{t}\right)dt = \frac{8\pi}{3}\int_1^3\left(t^{\frac{7}{2}}+t^{\frac{1}{2}}\right)dt$ | M1 | Uses surface area formula |
| $= \frac{8\pi}{3}\left[\frac{2}{9}t^{\frac{9}{2}}+\frac{2}{3}t^{\frac{3}{2}}\right]_1^3$ | A1 | Integrates |
| $= \frac{8\pi}{3}\left\{\left[18\sqrt{3}+2\sqrt{3}\right]-\left[\frac{2}{9}+\frac{2}{3}\right]\right\}$ | M1 | Inserts limits |
| $= \pi\left(\frac{160\sqrt{3}}{3}-\frac{64}{27}\right) \quad (=283 \text{ or } 90.0\pi)$ | A1 | 4 | **[10]** |

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8 The curve $C$ has parametric equations

$$x = \frac { 1 } { 3 } t ^ { 3 } - \ln t , \quad y = \frac { 4 } { 3 } t ^ { \frac { 3 } { 2 } }$$

for $1 \leqslant t \leqslant 3$. Find the arc length of $C$.

Find also the area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis.

\hfill \mbox{\textit{CAIE FP1 2012 Q8 [10]}}
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