CAIE Further Paper 2 2022 November — Question 3 6 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2022
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeParametric arc length calculation
DifficultyChallenging +1.2 This is a standard arc length parametric question requiring computation of √((dx/dt)² + (dy/dt)²) and integration. The derivatives involve straightforward differentiation (exponentials and polynomials), and the resulting integrand simplifies to a perfect square that integrates cleanly. While it requires careful algebraic manipulation and is a Further Maths topic, it follows a completely standard template with no conceptual surprises, making it moderately above average difficulty.
Spec1.03g Parametric equations: of curves and conversion to cartesian8.06b Arc length and surface area: of revolution, cartesian or parametric
3 The curve \(C\) has parametric equations $$\mathrm { x } = \mathrm { e } ^ { \mathrm { t } } - \frac { 1 } { 3 } \mathrm { t } ^ { 3 } , \quad \mathrm { y } = 4 \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { t } } ( \mathrm { t } - 2 ) , \quad \text { for } 0 \leqslant t \leqslant 2$$ Find, in terms of e , the length of \(C\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\dot{x} = e^t - t^2\)B1
\(\dot{y} = 2te^{\frac{1}{2}t}\)B1
\(\sqrt{(e^t - t^2)^2 + 4t^2e^t} = \sqrt{e^{2t} + 2t^2e^t + t^4} = e^t + t^2\)M1 A1 Finds \(\sqrt{\dot{x}^2 + \dot{y}^2}\)
\(\int_0^2 e^t + t^2 \, dt = \left[e^t + \frac{1}{3}t^3\right]_0^2 = e^2 + \frac{5}{3}\)M1 A1 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dot{x} = e^t - t^2$ | B1 | |
| $\dot{y} = 2te^{\frac{1}{2}t}$ | B1 | |
| $\sqrt{(e^t - t^2)^2 + 4t^2e^t} = \sqrt{e^{2t} + 2t^2e^t + t^4} = e^t + t^2$ | M1 A1 | Finds $\sqrt{\dot{x}^2 + \dot{y}^2}$ |
| $\int_0^2 e^t + t^2 \, dt = \left[e^t + \frac{1}{3}t^3\right]_0^2 = e^2 + \frac{5}{3}$ | M1 A1 | |

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3 The curve $C$ has parametric equations

$$\mathrm { x } = \mathrm { e } ^ { \mathrm { t } } - \frac { 1 } { 3 } \mathrm { t } ^ { 3 } , \quad \mathrm { y } = 4 \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { t } } ( \mathrm { t } - 2 ) , \quad \text { for } 0 \leqslant t \leqslant 2$$

Find, in terms of e , the length of $C$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q3 [6]}}
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Q1 5 Q2 7 Q3 6 Q4 12 Q5 10 Q6 11 Q7 10 Q8 14