CAIE FP1 2015 June — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArc length of polar curve
DifficultyChallenging +1.2 This is a straightforward application of the polar arc length formula with an exponential function. The integration is routine (exponential remains exponential), and the question only requires substituting into the formula, integrating, and solving a simple logarithmic equation. While it's a Further Maths topic, the execution is mechanical with no conceptual challenges beyond knowing the formula.
Spec4.09c Area enclosed: by polar curve
2 The curve \(C\) has polar equation \(r = \mathrm { e } ^ { 4 \theta }\) for \(0 \leqslant \theta \leqslant \alpha\), where \(\alpha\) is measured in radians. The length of \(C\) is 2015 . Find the value of \(\alpha\).
Question 2:
AnswerMarks Guidance
Working/AnswerMarks Notes
\(s = \int_0^\alpha \sqrt{e^{8\theta} + 16e^{8\theta}}\, d\theta = \int_0^\alpha \sqrt{17}e^{4\theta}\, d\theta\)M1A1 AEF, LNR
\(= \left[\frac{\sqrt{17}}{4}e^{4\theta}\right]_0^\alpha \Rightarrow \frac{\sqrt{17}}{4}e^{4\alpha} - \frac{\sqrt{17}}{4} = 2015\)dM1A1
\(e^{4\alpha} = 1955.837\ldots \Rightarrow \alpha = 1.89\)dM1A1 CAO 
## Question 2:

| Working/Answer | Marks | Notes |
|---|---|---|
| $s = \int_0^\alpha \sqrt{e^{8\theta} + 16e^{8\theta}}\, d\theta = \int_0^\alpha \sqrt{17}e^{4\theta}\, d\theta$ | M1A1 | AEF, LNR |
| $= \left[\frac{\sqrt{17}}{4}e^{4\theta}\right]_0^\alpha \Rightarrow \frac{\sqrt{17}}{4}e^{4\alpha} - \frac{\sqrt{17}}{4} = 2015$ | dM1A1 | |
| $e^{4\alpha} = 1955.837\ldots \Rightarrow \alpha = 1.89$ | dM1A1 | CAO |

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2 The curve $C$ has polar equation $r = \mathrm { e } ^ { 4 \theta }$ for $0 \leqslant \theta \leqslant \alpha$, where $\alpha$ is measured in radians. The length of $C$ is 2015 . Find the value of $\alpha$.

\hfill \mbox{\textit{CAIE FP1 2015 Q2 [6]}}
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Q1 6 Q2 6 Q3 7 Q4 7 Q5 9 Q6 9 Q7 10 Q8 10 Q9 11 Q10 11 Q11 EITHER Q11 OR