CAIE FP1 2018 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeParametric arc length calculation
DifficultyStandard +0.8 This is a standard arc length calculation for parametric equations requiring differentiation of exponential functions, substitution into the arc length formula, and integration. The integral simplifies nicely (the expression under the square root becomes a perfect square), but students must recognize this simplification and execute multiple steps correctly. It's moderately challenging for Further Maths but follows a well-established procedure.
Spec1.03g Parametric equations: of curves and conversion to cartesian8.06b Arc length and surface area: of revolution, cartesian or parametric
1 The curve \(C\) is defined parametrically by $$x = \mathrm { e } ^ { t } - t , \quad y = 4 \mathrm { e } ^ { \frac { 1 } { 2 } t }$$ Find the length of the arc of \(C\) from the point where \(t = 0\) to the point where \(t = 3\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{\mathrm{d}x}{\mathrm{d}t} = e^t - 1\) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = 2e^{\frac{1}{2}t}\)B1
\(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = (e^t+1)^2\)M1 A1 M1 for using \(\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2\)
Arc length is \(\int_0^3 e^t + 1\, \mathrm{d}t = \left[e^t + t\right]_0^3\)M1 M1 for good attempt at correct integral
\(= 2 + e^3\) (or 22.1)A1
Total: 5 
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}x}{\mathrm{d}t} = e^t - 1$ and $\frac{\mathrm{d}y}{\mathrm{d}t} = 2e^{\frac{1}{2}t}$ | B1 | |
| $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = (e^t+1)^2$ | M1 A1 | M1 for using $\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2$ |
| Arc length is $\int_0^3 e^t + 1\, \mathrm{d}t = \left[e^t + t\right]_0^3$ | M1 | M1 for good attempt at correct integral |
| $= 2 + e^3$ (or 22.1) | A1 | |
| **Total: 5** | | |

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1 The curve $C$ is defined parametrically by

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

Find the length of the arc of $C$ from the point where $t = 0$ to the point where $t = 3$.\\

\hfill \mbox{\textit{CAIE FP1 2018 Q1 [5]}}
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Q1 5 Q2 6 Q3 8 Q4 8 Q5 8 Q6 9 Q7 10 Q8 10 Q9 10 Q10 12 Q11 EITHER Q11 OR