CAIE FP1 2011 June — Question 7 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeParametric surface area of revolution
DifficultyChallenging +1.2 This is a Further Maths question requiring arc length and surface of revolution formulas for parametric curves. While the calculus is straightforward (derivatives of e^t cos t and e^t sin t are standard, and the resulting integral simplifies nicely to √2 e^t), it requires knowledge of specialized formulas and careful algebraic manipulation. The two-part structure and Further Maths context place it above average difficulty, but the integrals are tractable without advanced techniques.
Spec1.03g Parametric equations: of curves and conversion to cartesian8.06b Arc length and surface area: of revolution, cartesian or parametric
7 A curve \(C\) has parametric equations \(x = \mathrm { e } ^ { t } \cos t , y = \mathrm { e } ^ { t } \sin t\), for \(0 \leqslant t \leqslant \pi\). Find the arc length of \(C\). Find the area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
Question 7:
Arc length section:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\dot{x} = e^t(\cos t - \sin t)\), \(\dot{y} = e^t(\sin t + \cos t)\)B1
\(\dot{s} = \sqrt{e^{2t}(1 - 2\sin t\cos t + 1 + 2\sin t\cos t)} = \sqrt{2}e^t\)B1
\(s = \sqrt{2}\int_0^{\pi} e^t \, dt\)M1 Uses arc length formula
\(= \sqrt{2}(e^{\pi} - 1) \approx 31.3\)A1 Part total: 4
Surface area section:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S = 2\pi\int_0^{\pi} e^t \sin t\sqrt{2}e^t \, dt = 2\sqrt{2}\pi\int_0^{\pi} e^{2t}\sin t \, dt\)M1A1 Uses surface area formula and obtains correct integral
Let \(I = \int e^{2t}\sin t \, dt\); \(= -e^{2t}\cos t + \int 2e^{2t}\cos t \, dt\)M1 Integrates by parts twice
\(-e^{2t}\cos t + 2e^{2t}\sin t - \int 4e^{2t}\sin t \, dt\)A1
\(5I = 2e^{2t}\sin t - e^{2t}\cos t\)M1 Sees original integral again
\(\Rightarrow I = \frac{e^{2t}}{5}(2\sin t - \cos t)\)A1
\(S = 2\sqrt{2}\pi\left[\frac{e^{2t}}{5}(2\sin t - \cos t)\right]_0^{\pi} = \frac{2\sqrt{2}\pi}{5}(e^{2\pi}+1)\)A1 \(\approx 953\); if 953 written with no working award B1 in place of final 5 marks. Part total: 7
Alternative method for \(\int e^{2t}\sin t \, dt\):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\text{Im}\left\{\int e^{2t}\cdot e^{it}\,dt\right\} = \text{Im}\left\{\int e^{(2+i)t}\,dt\right\}\)M1
\(= \text{Im}\left[\frac{e^{(2+i)t}}{2+i}\right] = \text{Im}\left[\frac{e^{2t}}{5}(\cos t + i\sin t)(2-i)\right]\)A1M1
\(= \frac{1}{5}e^{2t}(2\sin t - \cos t)\)A1 Total: [11] 
## Question 7:

**Arc length section:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dot{x} = e^t(\cos t - \sin t)$, $\dot{y} = e^t(\sin t + \cos t)$ | B1 | |
| $\dot{s} = \sqrt{e^{2t}(1 - 2\sin t\cos t + 1 + 2\sin t\cos t)} = \sqrt{2}e^t$ | B1 | |
| $s = \sqrt{2}\int_0^{\pi} e^t \, dt$ | M1 | Uses arc length formula |
| $= \sqrt{2}(e^{\pi} - 1) \approx 31.3$ | A1 | **Part total: 4** |

**Surface area section:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S = 2\pi\int_0^{\pi} e^t \sin t\sqrt{2}e^t \, dt = 2\sqrt{2}\pi\int_0^{\pi} e^{2t}\sin t \, dt$ | M1A1 | Uses surface area formula and obtains correct integral |
| Let $I = \int e^{2t}\sin t \, dt$; $= -e^{2t}\cos t + \int 2e^{2t}\cos t \, dt$ | M1 | Integrates by parts twice |
| $-e^{2t}\cos t + 2e^{2t}\sin t - \int 4e^{2t}\sin t \, dt$ | A1 | |
| $5I = 2e^{2t}\sin t - e^{2t}\cos t$ | M1 | Sees original integral again |
| $\Rightarrow I = \frac{e^{2t}}{5}(2\sin t - \cos t)$ | A1 | |
| $S = 2\sqrt{2}\pi\left[\frac{e^{2t}}{5}(2\sin t - \cos t)\right]_0^{\pi} = \frac{2\sqrt{2}\pi}{5}(e^{2\pi}+1)$ | A1 | $\approx 953$; if 953 written with no working award B1 in place of final 5 marks. **Part total: 7** |

**Alternative method for $\int e^{2t}\sin t \, dt$:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Im}\left\{\int e^{2t}\cdot e^{it}\,dt\right\} = \text{Im}\left\{\int e^{(2+i)t}\,dt\right\}$ | M1 | |
| $= \text{Im}\left[\frac{e^{(2+i)t}}{2+i}\right] = \text{Im}\left[\frac{e^{2t}}{5}(\cos t + i\sin t)(2-i)\right]$ | A1M1 | |
| $= \frac{1}{5}e^{2t}(2\sin t - \cos t)$ | A1 | **Total: [11]** |

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7 A curve $C$ has parametric equations $x = \mathrm { e } ^ { t } \cos t , y = \mathrm { e } ^ { t } \sin t$, for $0 \leqslant t \leqslant \pi$. Find the arc length of $C$.

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

\hfill \mbox{\textit{CAIE FP1 2011 Q7 [11]}}
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