4.07e Inverse hyperbolic: definitions, domains, ranges

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1. It is given that \(\mathrm { y } = \operatorname { sech } ^ { - 1 } \left( \mathrm { x } + \frac { 1 } { 2 } \right)\).
   Express cosh \(y\) in terms of \(x\) and hence show that \(\sinh y \frac { d y } { d x } = - \frac { 1 } { \left( x + \frac { 1 } { 2 } \right) ^ { 2 } }\).
2. Find the first three terms in the Maclaurin's series for \(\operatorname { sech } ^ { - 1 } \left( x + \frac { 1 } { 2 } \right)\) in the form $$\ln a + b x + c x ^ { 2 }$$ where \(a\), \(b\) and \(c\) are constants to be determined.

4 It is given that $$x = - t + \tan ^ { - 1 } t \quad \text { and } \quad y = t + \sinh ^ { - 1 } t$$

1. Show that \(\frac { d y } { d x } = - \frac { t ^ { 2 } + 1 + \sqrt { t ^ { 2 } + 1 } } { t ^ { 2 } }\).
2. Find the value of \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) when \(t = \frac { 3 } { 4 }\).

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1. Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\). \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-12_778_1548_1007_296} The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
2. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
3. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).

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1. Starting from the definitions of sech and tanh in terms of exponentials, prove that $$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$ \includegraphics[max width=\textwidth, alt={}]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ ....................................................................................................................................... \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_72_1573_911_324} \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_67_1570_1005_324} The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
2. Show that \(\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t\).
3. Find the coordinates of the point on \(C\) with \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }\), giving your answer in the form \(( a + \ln b , c )\) where \(a , b\) and \(c\) are rational numbers.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

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1. Use the substitution \(\mathrm { u } = \mathrm { x } ^ { 2 } - 1\) to find \(\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x\). \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-12_778_1548_1007_296} The diagram shows the curve with equation \(\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }\) together with a set of \(( N - 1 )\) rectangles of unit width.
2. By considering the sum of the areas of these rectangles, show that $$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
3. Use a similar method to find, in terms of \(N\), an upper bound for \(\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)\).

5 It is given that $$x = \sinh ^ { - 1 } t , \quad y = \cos ^ { - 1 } t$$ where \(- 1 < t < 1\).

1. By differentiating \(\cos y\) with respect to \(t\), show that \(\frac { d y } { d t } = - \frac { 1 } { \sqrt { 1 - t ^ { 2 } } }\).
2. Find \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) in terms of \(t\), simplifying your answer.

3 Find the first three terms in the Maclaurin's series for \(\tanh ^ { - 1 } \left( \frac { 1 } { 2 } e ^ { x } \right)\) in the form \(\frac { 1 } { 2 } \ln a + b x + c x ^ { 2 }\), giving the exact values of the constants \(a , b\) and \(c\).

3 A curve has equation \(y = \mathrm { e } ^ { x }\) for \(\ln \frac { 4 } { 3 } \leqslant x \leqslant \ln \frac { 12 } { 5 }\). The area of the surface generated when the curve is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(A\).

1. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that $$A = 2 \pi \int _ { \frac { 4 } { 3 } } ^ { \frac { 12 } { 5 } } \sqrt { 1 + u ^ { 2 } } \mathrm {~d} u$$
2. Use the substitution \(u = \sinh v\) to show that $$A = \pi \left( \frac { 904 } { 225 } + \ln \frac { 5 } { 3 } \right) .$$ \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-06_2716_38_109_2012} \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-07_2726_35_97_20}

3 A curve has equation \(y = \mathrm { e } ^ { x }\) for \(\ln \frac { 4 } { 3 } \leqslant x \leqslant \ln \frac { 12 } { 5 }\). The area of the surface generated when the curve is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(A\).

1. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that $$A = 2 \pi \int _ { \frac { 4 } { 3 } } ^ { \frac { 12 } { 5 } } \sqrt { 1 + u ^ { 2 } } \mathrm {~d} u$$
2. Use the substitution \(u = \sinh v\) to show that $$A = \pi \left( \frac { 904 } { 225 } + \ln \frac { 5 } { 3 } \right) .$$ \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-06_2716_38_109_2012} \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-07_2726_35_97_20}

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1. Starting from the definition of tanh in terms of exponentials, prove that \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\). [3]
2. Given that \(y = \tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)\), show that \(( 2 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 1 = 0\).
3. Hence find the first three terms in the Maclaurin's series for \(\tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)\) in the form $$a \ln 3 + b x + c x ^ { 2 } ,$$ where \(a , b\) and \(c\) are constants to be determined.

3. (a) Given that \(y = \operatorname { arsech } \left( \frac { x } { 2 } \right)\), where \(0 < x \leqslant 2\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { x \sqrt { q - x ^ { 2 } } }$$ where \(p\) and \(q\) are constants to be determined. In part (b) solutions based entirely on calculator technology are not acceptable. $$\mathrm { f } ( x ) = \operatorname { artanh } ( x ) + \operatorname { arsech } \left( \frac { x } { 2 } \right) \quad 0 < x \leqslant 1$$ (b) Determine, in simplest form, the exact value of \(x\) for which \(\mathrm { f } ^ { \prime } ( x ) = 0\)

3. Without using a calculator, find

1. \(\int _ { - 2 } ^ { 1 } \frac { 1 } { x ^ { 2 } + 4 x + 13 } \mathrm {~d} x\), giving your answer as a multiple of \(\pi\),
2. \(\int _ { - 1 } ^ { 4 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 12 x + 34 } } \mathrm {~d} x\), giving your answer in the form \(p \ln ( q + r \sqrt { 2 } )\),
   where \(p , q\) and \(r\) are rational numbers to be found.
   

5. Given that \(y = \operatorname { artanh } ( \cos x )\)

1. show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } x$$
2. Hence find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \cos x \operatorname { artanh } ( \cos x ) d x$$ giving your answer in the form \(a \ln ( b + c \sqrt { 3 } ) + d \pi\), where \(a , b , c\) and \(d\) are rational numbers to be found.
   (5)
   

3. Given that $$y = x - \operatorname { artanh } \left( \frac { 2 x } { 1 + x ^ { 2 } } \right)$$

1. show that $$1 - \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { 1 - x ^ { 2 } }$$ where \(k\) is a constant to be found.
2. Hence, or otherwise, show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \left( 1 - \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 0$$

3. Using the substitution \(\mathrm { x } = \frac { 3 } { \sinh \theta }\), or otherwise, find the exact value of $$\int _ { 4 } ^ { 3 \sqrt { } 3 } \frac { 1 } { x \sqrt { } \left( x ^ { 2 } + 9 \right) } d x$$ giving your answer in the form a ln b , where a and b are rational numbers.
(Total 8 marks)

4. $$y = \operatorname { artanh } \left( \frac { \cos x + a } { \cos x - a } \right)$$ where \(a\) is a non-zero constant.
Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \tan x$$ where \(k\) is a constant to be determined.

1. (a) Differentiate \(x \operatorname { arcosh } 5 x\) with respect to \(x\) (b) Hence, or otherwise, show that

$$\int _ { \frac { 1 } { 4 } } ^ { \frac { 3 } { 5 } } \operatorname { arcosh } 5 x \mathrm {~d} x = \frac { 3 } { 20 } - \frac { 2 \sqrt { 2 } } { 5 } + \ln ( p + q \sqrt { 2 } ) ^ { k } - \frac { 1 } { 4 } \ln r$$ where \(p , q , r\) and \(k\) are rational numbers to be determined.

1. The curve \(C\) has equation

$$y = \frac { 1 } { 2 } \operatorname { arcosh } ( 2 x ) \quad \frac { 7 } { 2 } \leqslant x \leqslant 13$$ Using calculus, determine the exact length of the curve \(C\).
Give your answer in the form \(p \sqrt { q }\), where \(p\) and \(q\) are constants to be found.

2. Given that $$\cosh y = x \quad \text { and } \quad y < 0$$ use the definition of coshy in terms of exponential functions to prove that $$y = \ln \left( x - \sqrt { x ^ { 2 } - 1 } \right)$$

4. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation $$y = 40 \operatorname { arcosh } x - 9 x , \quad x \geqslant 1$$ Use calculus to find the exact coordinates of the turning point of the curve, giving your answer in the form \(\left( \frac { p } { q } , r \ln 3 + s \right)\), where \(p , q , r\) and \(s\) are integers.

3. (a) Prove that $$\frac { \mathrm { d } ( \operatorname { arcoth } x ) } { \mathrm { d } x } = \frac { 1 } { 1 - x ^ { 2 } }$$ Given that \(y = ( \operatorname { arcoth } x ) ^ { 2 }\),
(b) show that $$\left( 1 - x ^ { 2 } \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 x \frac { d y } { d x } = \frac { k } { 1 - x ^ { 2 } }$$ where \(k\) is a constant to be determined.

1. Given that \(y = \operatorname { arsinh } ( \tanh x )\), show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \operatorname { sech } ^ { 2 } x } { \sqrt { 1 + \tanh ^ { 2 } x } }$$ \section*{-} \includegraphics[max width=\textwidth, alt={}, center]{64dc962a-1788-49ac-a4db-af1241b552a0-03_51_51_276_2012}
N

1. (a) Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, show that, for \(x \in \mathbb { R }\)

$$\tanh x = \frac { \mathrm { e } ^ { 2 x } - 1 } { \mathrm { e } ^ { 2 x } + 1 }$$ (b) Hence, given that \(- 1 < \theta < 1\), prove that $$\operatorname { artanh } \theta = \frac { 1 } { 2 } \ln \left( \frac { 1 + \theta } { 1 - \theta } \right)$$ uestion 1 continued \(\_\_\_\_\) 7