Question 5 - A-Level Maths

AQA FP2 2010 June — Question 5 18 marks

Exam Board	AQA
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	June
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Parametric curves with hyperbolic functions
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring hyperbolic function identities, derivatives, parametric arc length formula, and integration by substitution. Part (a) involves routine algebraic manipulation and differentiation using given identities. Part (b)(i) requires applying the arc length formula and simplification using part (a) results. Part (b)(ii) involves a non-trivial substitution to integrate sech t. While systematic, it requires multiple techniques and careful algebra across several steps, placing it moderately above average difficulty.
Spec	1.03g Parametric equations: of curves and conversion to cartesian 1.08h Integration by substitution 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 4.07d Differentiate/integrate: hyperbolic functions 8.06a Reduction formulae: establish, use, and evaluate recursively

Using the identities $$\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1 , \quad \tanh t = \frac { \sinh t } { \cosh t } \quad \text { and } \quad \operatorname { sech } t = \frac { 1 } { \cosh t }$$ show that:
1. $\tanh ^ { 2 } t + \operatorname { sech } ^ { 2 } t = 1$;
2. $\frac { \mathrm { d } } { \mathrm { d } t } ( \tanh t ) = \operatorname { sech } ^ { 2 } t$;
3. $\frac { \mathrm { d } } { \mathrm { d } t } ( \operatorname { sech } t ) = - \operatorname { sech } t \tanh t$.

A curve $C$ is given parametrically by $$x = \operatorname { sech } t , y = 4 - \tanh t$$

Show that the arc length, $s$, of $C$ between the points where $t = 0$ and $t = \frac { 1 } { 2 } \ln 3$ is given by $$s = \int _ { 0 } ^ { \frac { 1 } { 2 } \ln 3 } \operatorname { sech } t \mathrm {~d} t$$

Using the substitution $u = \mathrm { e } ^ { t }$, find the exact value of $s$.

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Show mark scheme Show mark scheme source

Question 5:

Part (a)(i):

Answer	Marks	Guidance
Divide $\cosh^2 t - \sinh^2 t = 1$ by $\cosh^2 t$	M1	Correct division
$1 - \tanh^2 t = \text{sech}^2 t \Rightarrow \tanh^2 t + \text{sech}^2 t = 1$	A1	Correct completion

Part (a)(ii):

Answer	Marks	Guidance
$\tanh t = \frac{\sinh t}{\cosh t}$; using quotient rule: $\frac{d}{dt}(\tanh t) = \frac{\cosh^2 t - \sinh^2 t}{\cosh^2 t}$	M1 A1	Quotient rule applied
$= \frac{1}{\cosh^2 t} = \text{sech}^2 t$	A1	Correct completion

Part (a)(iii):

Answer	Marks	Guidance
$\text{sech}\, t = \frac{1}{\cosh t} = (\cosh t)^{-1}$; $\frac{d}{dt}(\text{sech}\, t) = -(\cosh t)^{-2} \cdot \sinh t$	M1 A1	Chain rule applied
$= -\frac{1}{\cosh t}\cdot\frac{\sinh t}{\cosh t} = -\text{sech}\, t\tanh t$	A1	Correct completion

Part (b)(i):

Answer	Marks	Guidance
$\frac{dx}{dt} = -\text{sech}\, t\tanh t$, $\frac{dy}{dt} = -\text{sech}^2 t$	B1	Correct derivatives
$s = \int_0^{\frac{1}{2}\ln 3}\sqrt{\text{sech}^2 t\tanh^2 t + \text{sech}^4 t}\, dt$	M1	Arc length formula used
$= \int_0^{\frac{1}{2}\ln 3}\text{sech}\, t\sqrt{\tanh^2 t + \text{sech}^2 t}\, dt$	M1	Factoring out $\text{sech}\, t$
$= \int_0^{\frac{1}{2}\ln 3}\text{sech}\, t\, dt$	A1	Using identity from (a)(i)

Part (b)(ii):

Answer	Marks	Guidance
$u = e^t$, $du = e^t dt$, $\text{sech}\, t = \frac{2}{e^t+e^{-t}} = \frac{2u}{u^2+1}$	M1	Correct substitution
Limits: $t=0 \Rightarrow u=1$; $t=\frac{1}{2}\ln 3 \Rightarrow u=\sqrt{3}$	B1	Correct limits
$s = \int_1^{\sqrt{3}} \frac{2u}{u^2+1}\cdot\frac{1}{u}\, du = \int_1^{\sqrt{3}}\frac{2}{u^2+1}\, du$	M1 A1	Correct integral form
$= \left[2\arctan u\right]_1^{\sqrt{3}} = 2\arctan\sqrt{3} - 2\arctan 1$	M1	Integration
$= 2\cdot\frac{\pi}{3} - 2\cdot\frac{\pi}{4} = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}$	A1	Correct exact value

# Question 5:

## Part (a)(i):
| Divide $\cosh^2 t - \sinh^2 t = 1$ by $\cosh^2 t$ | M1 | Correct division |
|---|---|---|
| $1 - \tanh^2 t = \text{sech}^2 t \Rightarrow \tanh^2 t + \text{sech}^2 t = 1$ | A1 | Correct completion |

## Part (a)(ii):
| $\tanh t = \frac{\sinh t}{\cosh t}$; using quotient rule: $\frac{d}{dt}(\tanh t) = \frac{\cosh^2 t - \sinh^2 t}{\cosh^2 t}$ | M1 A1 | Quotient rule applied |
|---|---|---|
| $= \frac{1}{\cosh^2 t} = \text{sech}^2 t$ | A1 | Correct completion |

## Part (a)(iii):
| $\text{sech}\, t = \frac{1}{\cosh t} = (\cosh t)^{-1}$; $\frac{d}{dt}(\text{sech}\, t) = -(\cosh t)^{-2} \cdot \sinh t$ | M1 A1 | Chain rule applied |
|---|---|---|
| $= -\frac{1}{\cosh t}\cdot\frac{\sinh t}{\cosh t} = -\text{sech}\, t\tanh t$ | A1 | Correct completion |

## Part (b)(i):
| $\frac{dx}{dt} = -\text{sech}\, t\tanh t$, $\frac{dy}{dt} = -\text{sech}^2 t$ | B1 | Correct derivatives |
|---|---|---|
| $s = \int_0^{\frac{1}{2}\ln 3}\sqrt{\text{sech}^2 t\tanh^2 t + \text{sech}^4 t}\, dt$ | M1 | Arc length formula used |
| $= \int_0^{\frac{1}{2}\ln 3}\text{sech}\, t\sqrt{\tanh^2 t + \text{sech}^2 t}\, dt$ | M1 | Factoring out $\text{sech}\, t$ |
| $= \int_0^{\frac{1}{2}\ln 3}\text{sech}\, t\, dt$ | A1 | Using identity from (a)(i) |

## Part (b)(ii):
| $u = e^t$, $du = e^t dt$, $\text{sech}\, t = \frac{2}{e^t+e^{-t}} = \frac{2u}{u^2+1}$ | M1 | Correct substitution |
|---|---|---|
| Limits: $t=0 \Rightarrow u=1$; $t=\frac{1}{2}\ln 3 \Rightarrow u=\sqrt{3}$ | B1 | Correct limits |
| $s = \int_1^{\sqrt{3}} \frac{2u}{u^2+1}\cdot\frac{1}{u}\, du = \int_1^{\sqrt{3}}\frac{2}{u^2+1}\, du$ | M1 A1 | Correct integral form |
| $= \left[2\arctan u\right]_1^{\sqrt{3}} = 2\arctan\sqrt{3} - 2\arctan 1$ | M1 | Integration |
| $= 2\cdot\frac{\pi}{3} - 2\cdot\frac{\pi}{4} = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}$ | A1 | Correct exact value |

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5 (a) Using the identities

$$\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1 , \quad \tanh t = \frac { \sinh t } { \cosh t } \quad \text { and } \quad \operatorname { sech } t = \frac { 1 } { \cosh t }$$

show that:\\
(i) $\tanh ^ { 2 } t + \operatorname { sech } ^ { 2 } t = 1$;\\
(ii) $\frac { \mathrm { d } } { \mathrm { d } t } ( \tanh t ) = \operatorname { sech } ^ { 2 } t$;\\
(iii) $\frac { \mathrm { d } } { \mathrm { d } t } ( \operatorname { sech } t ) = - \operatorname { sech } t \tanh t$.\\
(b) A curve $C$ is given parametrically by

$$x = \operatorname { sech } t , y = 4 - \tanh t$$

(i) Show that the arc length, $s$, of $C$ between the points where $t = 0$ and $t = \frac { 1 } { 2 } \ln 3$ is given by

$$s = \int _ { 0 } ^ { \frac { 1 } { 2 } \ln 3 } \operatorname { sech } t \mathrm {~d} t$$

(ii) Using the substitution $u = \mathrm { e } ^ { t }$, find the exact value of $s$.

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\hfill \mbox{\textit{AQA FP2 2010 Q5 [18]}}

This paper (7 questions)

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Q1 9 Q2 8 Q3 9 Q4 13 Q5 18 Q6 8 Q7 10

Answer	Marks	Guidance
Divide \(\cosh^2 t - \sinh^2 t = 1\) by \(\cosh^2 t\)	M1	Correct division
\(1 - \tanh^2 t = \text{sech}^2 t \Rightarrow \tanh^2 t + \text{sech}^2 t = 1\)	A1	Correct completion

Answer	Marks	Guidance
\(\tanh t = \frac{\sinh t}{\cosh t}\); using quotient rule: \(\frac{d}{dt}(\tanh t) = \frac{\cosh^2 t - \sinh^2 t}{\cosh^2 t}\)	M1 A1	Quotient rule applied
\(= \frac{1}{\cosh^2 t} = \text{sech}^2 t\)	A1	Correct completion

Answer	Marks	Guidance
\(\text{sech}\, t = \frac{1}{\cosh t} = (\cosh t)^{-1}\); \(\frac{d}{dt}(\text{sech}\, t) = -(\cosh t)^{-2} \cdot \sinh t\)	M1 A1	Chain rule applied
\(= -\frac{1}{\cosh t}\cdot\frac{\sinh t}{\cosh t} = -\text{sech}\, t\tanh t\)	A1	Correct completion

Answer	Marks	Guidance
\(\frac{dx}{dt} = -\text{sech}\, t\tanh t\), \(\frac{dy}{dt} = -\text{sech}^2 t\)	B1	Correct derivatives
\(s = \int_0^{\frac{1}{2}\ln 3}\sqrt{\text{sech}^2 t\tanh^2 t + \text{sech}^4 t}\, dt\)	M1	Arc length formula used
\(= \int_0^{\frac{1}{2}\ln 3}\text{sech}\, t\sqrt{\tanh^2 t + \text{sech}^2 t}\, dt\)	M1	Factoring out \(\text{sech}\, t\)
\(= \int_0^{\frac{1}{2}\ln 3}\text{sech}\, t\, dt\)	A1	Using identity from (a)(i)

Answer	Marks	Guidance
\(u = e^t\), \(du = e^t dt\), \(\text{sech}\, t = \frac{2}{e^t+e^{-t}} = \frac{2u}{u^2+1}\)	M1	Correct substitution
Limits: \(t=0 \Rightarrow u=1\); \(t=\frac{1}{2}\ln 3 \Rightarrow u=\sqrt{3}\)	B1	Correct limits
\(s = \int_1^{\sqrt{3}} \frac{2u}{u^2+1}\cdot\frac{1}{u}\, du = \int_1^{\sqrt{3}}\frac{2}{u^2+1}\, du\)	M1 A1	Correct integral form
\(= \left[2\arctan u\right]_1^{\sqrt{3}} = 2\arctan\sqrt{3} - 2\arctan 1\)	M1	Integration
\(= 2\cdot\frac{\pi}{3} - 2\cdot\frac{\pi}{4} = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}\)	A1	Correct exact value