CAIE FP1 2010 November — Question 1 4 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionNovember
Marks4
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TopicHyperbolic functions
TypeArc length with hyperbolic curves
DifficultyChallenging +1.2 This is a standard arc length calculation with hyperbolic functions from Further Maths. While it requires recognizing that y = cosh(2x)/2, computing dy/dx, and integrating √(1 + (dy/dx)²), the algebra simplifies nicely using hyperbolic identities (cosh² - sinh² = 1). It's a routine Further Maths exercise with a predetermined answer to verify, requiring more steps than typical A-level but following a standard template without novel insight.
Spec1.06j 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials8.06b Arc length and surface area: of revolution, cartesian or parametric
1 The curve \(C\) has equation \(y = \frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 x } + \mathrm { e } ^ { - 2 x } \right)\). Show that the length of the \(\operatorname { arc }\) of \(C\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 2 }\) is \(\frac { \mathrm { e } ^ { 2 } - 1 } { 4 \mathrm { e } }\).
AnswerMarks Guidance
\(1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4}(e^{2x} - e^{-2x})^2 = \frac{1}{4}(e^{2x} + e^{-2x})^2\)M1A1 expression simplified
\(\text{Length} = \int_0^1 \frac{1}{2}(e^{2x} + e^{-2x})dx = \frac{1}{4}[e^{2x} - e^{-2x}]_0^1\)M1 integrate
\(= \frac{1}{4}(e^1 - e^{-1}) - \frac{1}{4}(e^0 - e^0) = \frac{e^2 - 1}{4e}\)A1 cao 
$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4}(e^{2x} - e^{-2x})^2 = \frac{1}{4}(e^{2x} + e^{-2x})^2$ | M1A1 | expression simplified

$\text{Length} = \int_0^1 \frac{1}{2}(e^{2x} + e^{-2x})dx = \frac{1}{4}[e^{2x} - e^{-2x}]_0^1$ | M1 | integrate

$= \frac{1}{4}(e^1 - e^{-1}) - \frac{1}{4}(e^0 - e^0) = \frac{e^2 - 1}{4e}$ | A1 | cao | [4]

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1 The curve $C$ has equation $y = \frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 x } + \mathrm { e } ^ { - 2 x } \right)$. Show that the length of the $\operatorname { arc }$ of $C$ from the point where $x = 0$ to the point where $x = \frac { 1 } { 2 }$ is $\frac { \mathrm { e } ^ { 2 } - 1 } { 4 \mathrm { e } }$.

\hfill \mbox{\textit{CAIE FP1 2010 Q1 [4]}}
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