CAIE FP1 2008 November — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeParametric arc length calculation
DifficultyChallenging +1.2 This is a parametric arc length question requiring the formula $\int \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. While it involves differentiation, algebraic manipulation, and integration, the derivatives are straightforward, the expression under the square root simplifies nicely to a perfect square $(2t^2 - \frac{2}{t})^2$, and the final integration is routine. It's harder than average due to being Further Maths and requiring careful algebra, but the path is standard once the formula is applied.
Spec4.08a Maclaurin series: find series for function

1 The curve \(C\) is defined parametrically by $$x = t ^ { 4 } - 4 \ln t , \quad y = 4 t ^ { 2 }$$ Show that the length of the arc of \(C\) from the point where \(t = 2\) to the point where \(t = 4\) is $$240 + 4 \ln 2 .$$

AnswerMarks
\(x = 4t^3 - 4t, \quad y = 8t\)B1
\(s = \int_0^1 \left[(4t^3 - 4t)^2 + 64t^2\right]^{1/2} dt\)M1A1
\(\left[(4t^3 - 4t)^2 + 64t^2\right]^{1/2} = 4t^3 + 4/t\)B1
\(s = [t^4 + 4\ln t]_2^2 = 240 + 4\ln 2\)A1
$x = 4t^3 - 4t, \quad y = 8t$ | B1 |

$s = \int_0^1 \left[(4t^3 - 4t)^2 + 64t^2\right]^{1/2} dt$ | M1A1 |

$\left[(4t^3 - 4t)^2 + 64t^2\right]^{1/2} = 4t^3 + 4/t$ | B1 |

$s = [t^4 + 4\ln t]_2^2 = 240 + 4\ln 2$ | A1 |
1 The curve $C$ is defined parametrically by

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

Show that the length of the arc of $C$ from the point where $t = 2$ to the point where $t = 4$ is

$$240 + 4 \ln 2 .$$

\hfill \mbox{\textit{CAIE FP1 2008 Q1 [5]}}