4.07d Differentiate/integrate: hyperbolic functions

8

1. Using the definitions of \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that \(\sinh 2 x = 2 \sinh x \cosh x\). You are given the function \(\mathrm { f } ( x ) = a \cosh x - \cosh 2 x\), where \(a\) is a positive constant.
2. Verify that, for any value of \(a\), the curve \(y = \mathrm { f } ( x )\) has a stationary point on the \(y\)-axis.
3. Find the coordinates of the stationary point found in part (ii).
4. Determine the maximum value of \(a\) for which the stationary point found in part (ii) is the only stationary point on the curve \(y = \mathrm { f } ( x )\). You are given that for any value of \(a\) greater than the value found in part (iv) there are three stationary points, the one found in part (ii) and two others, one of which satisfies \(x > 0\).
5. Find the coordinates of this point when \(a = 6\). Give your answer in the form \(\left( \cosh ^ { - 1 } p , q \right)\).

5. Given that \(y = ( \operatorname { arcosh } 3 x ) ^ { 2 }\), where \(3 x > 1\), show that

1. \(\left( 9 x ^ { 2 } - 1 \right) \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 36 y\),
2. \(\left( 9 x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 9 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 18\).

7

1. Use the definitions $$\sinh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } \right) \quad \text { and } \quad \cosh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } \right)$$ to show that:
   1. \(2 \sinh \theta \cosh \theta = \sinh 2 \theta\);
   2. \(\cosh ^ { 2 } \theta + \sinh ^ { 2 } \theta = \cosh 2 \theta\).
2. A curve is given parametrically by $$x = \cosh ^ { 3 } \theta , \quad y = \sinh ^ { 3 } \theta$$
   1. Show that $$\left( \frac { \mathrm { d } x } { \mathrm {~d} \theta } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} \theta } \right) ^ { 2 } = \frac { 9 } { 4 } \sinh ^ { 2 } 2 \theta \cosh 2 \theta$$
   2. Show that the length of the arc of the curve from the point where \(\theta = 0\) to the point where \(\theta = 1\) is $$\frac { 1 } { 2 } \left[ ( \cosh 2 ) ^ { \frac { 3 } { 2 } } - 1 \right]$$

4

1. Given that \(y = \operatorname { sech } t\), show that:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} t } = - \operatorname { sech } t \tanh t\);
   2. \(\left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \operatorname { sech } ^ { 2 } t - \operatorname { sech } ^ { 4 } t\).
2. The diagram shows a sketch of part of the curve given parametrically by $$x = t - \tanh t \quad y = \operatorname { sech } t$$
   \includegraphics[max width=\textwidth, alt={}]{1891766e-7744-49ac-82b6-7e51cb63b381-3_424_625_863_703}
   The curve meets the \(y\)-axis at the point \(K\), and \(P ( x , y )\) is a general point on the curve. The arc length \(K P\) is denoted by \(s\). Show that:
   1. \(\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \tanh ^ { 2 } t\);
   2. \(s = \ln \cosh t\);
   3. \(y = \mathrm { e } ^ { - s }\).
3. The arc \(K P\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that the surface area generated is $$2 \pi \left( 1 - \mathrm { e } ^ { - S } \right)$$ (4 marks)

7

1. Given that \(y = \ln \tanh \frac { x } { 2 }\), where \(x > 0\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosech } x$$
2. A curve has equation \(y = \ln \tanh \frac { x } { 2 }\), where \(x > 0\). The length of the arc of the curve between the points where \(x = 1\) and \(x = 2\) is denoted by \(s\).
   1. Show that $$s = \int _ { 1 } ^ { 2 } \operatorname { coth } x \mathrm {~d} x$$
   2. Hence show that \(s = \ln ( 2 \cosh 1 )\).

5

1. Given that \(u = \cosh ^ { 2 } x\), show that \(\frac { \mathrm { d } u } { \mathrm {~d} x } = \sinh 2 x\).
2. Hence show that $$\int _ { 0 } ^ { 1 } \frac { \sinh 2 x } { 1 + \cosh ^ { 4 } x } \mathrm {~d} x = \tan ^ { - 1 } \left( \cosh ^ { 2 } 1 \right) - \frac { \pi } { 4 }$$

3 The curve \(C\) has equation $$y = \cosh x - 3 \sinh x$$

1. 1. The line \(y = - 1\) meets \(C\) at the point \(( k , - 1 )\). Show that $$\mathrm { e } ^ { 2 k } - \mathrm { e } ^ { k } - 2 = 0$$
   2. Hence find \(k\), giving your answer in the form \(\ln a\).
2. 1. Find the \(x\)-coordinate of the point where the curve \(C\) intersects the \(x\)-axis, giving your answer in the form \(p \ln a\).
   2. Show that \(C\) has no stationary points.
   3. Show that there is exactly one point on \(C\) for which \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0\).

7 A curve has equation \(y = 4 \sqrt { x }\).

1. Show that the length of arc \(s\) of the curve between the points where \(x = 0\) and \(x = 1\) is given by $$s = \int _ { 0 } ^ { 1 } \sqrt { \frac { x + 4 } { x } } \mathrm {~d} x$$
2. 1. Use the substitution \(x = 4 \sinh ^ { 2 } \theta\) to show that $$\int \sqrt { \frac { x + 4 } { x } } \mathrm {~d} x = \int 8 \cosh ^ { 2 } \theta \mathrm {~d} \theta$$
   2. Hence show that $$s = 4 \sinh ^ { - 1 } 0.5 + \sqrt { 5 }$$

7 The diagram shows a curve which starts from the point \(A\) with coordinates ( 0,2 ). The curve is such that, at every point \(P\) on the curve, $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } s$$ where \(s\) is the length of the \(\operatorname { arc } A P\). \includegraphics[max width=\textwidth, alt={}, center]{587aac5c-fbc2-41d2-b1b3-16f3f7851d9d-4_399_764_1324_605}

1. 1. Show that $$\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \sqrt { 4 + s ^ { 2 } }$$ (3 marks)
   2. Hence show that $$s = 2 \sinh \frac { x } { 2 }$$
   3. Hence find the cartesian equation of the curve.
2. Show that $$y ^ { 2 } = 4 + s ^ { 2 }$$

9

1. Using the definitions of \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), prove that $$\sinh 2 x = 2 \sinh x \cosh x$$
2. Show that the curve with equation $$y = \cosh 2 x - 6 \sinh x$$ has just one stationary point, and find its \(x\)-coordinate in logarithmic form. Determine the nature of the stationary point.

12 The equation \(x ^ { 3 } - 2 x ^ { 2 } - x + 2 = 0\) has three roots. One of the roots is 2 12

1. Find the other two roots of the equation. 12
2. Hence, or otherwise, solve $$\cosh ^ { 3 } \theta - 2 \cosh ^ { 2 } \theta - \cosh \theta + 2 = 0$$ giving your answers in an exact form.

12 The integral \(S _ { n }\) is defined by $$S _ { n } = \int _ { 0 } ^ { a } x ^ { n } \sinh x \mathrm {~d} x \quad ( n \geq 0 )$$ 12

1. Show that for \(n \geq 2\) $$S _ { n } = n ( n - 1 ) S _ { n - 2 } + a ^ { n } \cosh a - n a ^ { n - 1 } \sinh a$$

   12	Hence show that \(\int _ { 0 } ^ { 1 } x ^ { 4 } \sinh x d x = \frac { 9 } { 2 } e + \frac { 65 } { 2 } e ^ { - 1 } - 24\)

3 The diagram shows part of the curve \(y = 5 \cosh x + 3 \sinh x\). \includegraphics[max width=\textwidth, alt={}, center]{ef967953-70b5-4dd1-a342-ad488b5fa79f-02_426_661_906_260}

1. Solve the equation \(5 \cosh x + 3 \sinh x = 4\) giving your solution in exact form.
2. In this question you must show detailed reasoning. Find \(\int _ { - 1 } ^ { 1 } ( 5 \cosh x + 3 \sinh x ) \mathrm { d } x\) giving your answer in the form \(a \mathrm { e } + \frac { b } { \mathrm { e } }\) where \(a\) and \(b\) are integers to be determined.

5 The function \(\operatorname { sech } x\) is defined by \(\operatorname { sech } x = \frac { 1 } { \cosh x }\).

1. Show that \(\operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 }\).
2. Using a suitable substitution, find \(\int \operatorname { sech } x \mathrm {~d} x\).

2 In this question you must show detailed reasoning.
Solve the equation \(2 \cosh ^ { 2 } x + 5 \sinh x - 5 = 0\) giving each answer in the form \(\ln ( p + q \sqrt { r } )\) where \(p\) and \(q\) are rational numbers, and \(r\) is an integer, whose values are to be determined. You are given that the matrix \(\mathbf { A } = \left( \begin{array} { c c c } 1 & 0 & 0 \\ 0 & \frac { 2 a - a ^ { 2 } } { 3 } & 0 \\ 0 & 0 & 1 \end{array} \right)\), where \(a\) is a positive constant, represents the transformation R which is a reflection in 3-D.

1. State the plane of reflection of \(R\).
2. Determine the value of \(a\).
3. With reference to R explain why \(\mathbf { A } ^ { 2 } = \mathbf { I }\), the \(3 \times 3\) identity matrix.
   1. By using Euler's formula show that \(\cosh ( \mathrm { iz } ) = \cos z\).
   2. Hence, find, in logarithmic form, a root of the equation \(\cos z = 2\). [You may assume that \(\cos z = 2\) has complex roots.] A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it. The extent to which the door is open at any time, \(t\) seconds, is measured by the angle at the hinge, \(\theta\), which the plane of the door makes with the plane of the equilibrium position. See the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-03_317_954_497_255} In an initial model of the motion of a certain swing door it is suggested that \(\theta\) satisfies the following differential equation. $$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 25 \theta = 0$$
   3. 1. Write down the general solution to (\textit{).
      2. With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (}) is unlikely to be realistic. In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes $$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + \lambda \frac { \mathrm { d } \theta } { \mathrm {~d} t } + 25 \theta = 0$$ where \(\lambda\) is a positive constant.
   4. In the case where \(\lambda = 16\) the door is held open at an angle of 0.9 radians and then released from rest at time \(t = 0\).
      1. Find, in a real form, the general solution of ( \(\dagger\) ).
      2. Find the particular solution of ( \(\dagger\) ).
      3. With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in ( \(\dagger\) ) improves the model.
      4. Find the value of \(\lambda\) for which the door is critically damped.

1

1. 1. Given that \(\mathrm { f } ( x ) = \arctan x\), write down an expression for \(\mathrm { f } ^ { \prime } ( x )\). Assuming that \(x\) is small, use a binomial expansion to express \(\mathrm { f } ^ { \prime } ( x )\) in ascending powers of \(x\) as far as the term in \(x ^ { 4 }\).
   2. Hence express \(\arctan x\) in ascending powers of \(x\) as far as the term in \(x ^ { 5 }\).
2. Find, in exact form, the value of the following integral. $$\int _ { 0 } ^ { \frac { 3 } { 4 } } \frac { 1 } { \sqrt { 3 - 4 x ^ { 2 } } } \mathrm {~d} x$$
3. A curve has polar equation \(r = \frac { a } { \sqrt { \theta } }\) where \(a > 0\).
   1. Sketch the curve for \(\frac { \pi } { 4 } \leqslant \theta \leqslant 2 \pi\).
   2. State what happens to \(r\) as \(\theta\) tends to zero.
   3. Find the area of the region enclosed by the part of the curve sketched in part (i) and the lines \(\theta = \frac { \pi } { 4 }\) and \(\theta = 2 \pi\). Give your answer in an exact simplified form.
      1. (i) Express \(2 \sin \frac { 1 } { 2 } \theta \left( \sin \frac { 1 } { 2 } \theta - \mathrm { j } \cos \frac { 1 } { 2 } \theta \right)\) in terms of \(z\) where \(z = \cos \theta + \mathrm { j } \sin \theta\).
         (ii) The series \(C\) and \(S\) are defined as follows. $$\begin{aligned} C & = 1 - \binom { n } { 1 } \cos \theta + \binom { n } { 2 } \cos 2 \theta - \ldots + ( - 1 ) ^ { n } \binom { n } { n } \cos n \theta \\ S & = - \binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta - \ldots + ( - 1 ) ^ { n } \binom { n } { n } \sin n \theta \end{aligned}$$ Show that $$C + \mathrm { j } S = \left\{ - 2 \mathrm { j } \sin \frac { 1 } { 2 } \theta \left( \cos \frac { 1 } { 2 } \theta + \mathrm { j } \sin \frac { 1 } { 2 } \theta \right) \right\} ^ { n } .$$ Hence show that, for even values of \(n\), $$\frac { C } { S } = \cot \left( \frac { 1 } { 2 } n \theta \right)$$
      2. Write the complex number \(z = \sqrt { 6 } + \mathrm { j } \sqrt { 2 }\) in the form \(r \mathrm { e } ^ { \mathrm { j } \theta }\), expressing \(r\) and \(\theta\) as simply as possible. Hence find the cube roots of \(z\) in the form \(r \mathrm { e } ^ { \mathrm { j } \theta }\). Show the points representing \(z\) and its cube roots on an Argand diagram.
         1. Find the eigenvalues and eigenvectors of the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { l l } \frac { 1 } { 2 } & \frac { 1 } { 2 } \\ \frac { 2 } { 3 } & \frac { 1 } { 3 } \end{array} \right)$$ Hence express \(\mathbf { M }\) in the form \(\mathbf { P D P } ^ { - 1 }\) where \(\mathbf { D }\) is a diagonal matrix.
         2. Write down an equation for \(\mathbf { M } ^ { n }\) in terms of the matrices \(\mathbf { P }\) and \(\mathbf { D }\). Hence obtain expressions for the elements of \(\mathbf { M } ^ { n }\).
            Show that \(\mathbf { M } ^ { n }\) tends to a limit as \(n\) tends to infinity. Find that limit.
         3. Express \(\mathbf { M } ^ { - 1 }\) in terms of the matrices \(\mathbf { P }\) and \(\mathbf { D }\). Hence determine whether or not \(\left( \mathbf { M } ^ { - 1 } \right) ^ { n }\) tends to a limit as \(n\) tends to infinity. Section B (18 marks)
            1. Given that \(y = \cosh x\), use the definition of \(\cosh x\) in terms of exponential functions to prove that $$x = \pm \ln \left( y + \sqrt { y ^ { 2 } - 1 } \right) .$$
            2. Solve the equation $$\cosh x + \cosh 2 x = 5$$ giving the roots in an exact logarithmic form.
            3. Sketch the curve with equation \(y = \cosh x + \cosh 2 x\). Show on your sketch the line \(y = 5\). Find the area of the finite region bounded by the curve and the line \(y = 5\). Give your answer in an exact form that does not involve hyperbolic functions. \section*{END OF QUESTION PAPER}

12

1. Let \(I _ { n } = \int \frac { x ^ { n } } { \sqrt { x ^ { 2 } + 1 } } \mathrm {~d} x\), for integers \(n \geqslant 0\).
   By writing \(\frac { x ^ { n } } { \sqrt { x ^ { 2 } + 1 } }\) as \(x ^ { n - 1 } \times \frac { x } { \sqrt { x ^ { 2 } + 1 } }\), or otherwise, show that, for \(n \geqslant 2\), $$n I _ { n } = x ^ { n - 1 } \sqrt { x ^ { 2 } + 1 } - ( n - 1 ) I _ { n - 2 } .$$
2. The diagram shows a sketch of the hyperbola \(H\) with equation \(\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1\). \includegraphics[max width=\textwidth, alt={}, center]{32ed7cc8-3456-4cf0-952a-ee04eada1298-6_593_666_776_776}
   1. Find the coordinates of the points where \(H\) crosses the \(x\)-axis.
   2. The curve \(J\) has parametric equations \(x = 2 \cosh \theta , y = 4 \sinh \theta\), for \(\theta \geqslant 0\). Show that these parametric equations satisfy the cartesian equation of \(H\), and indicate on a copy of the above diagram which part of \(H\) is \(J\).
   3. The arc of the curve \(J\) between the points where \(x = 2\) and \(x = 34\) is rotated once completely about the \(x\)-axis to form a surface of revolution with area \(S\). Show that $$S = 16 \pi \int _ { \alpha } ^ { \beta } \sinh \theta \sqrt { 5 \cosh ^ { 2 } \theta - 1 } \mathrm {~d} \theta$$ for suitable constants \(\alpha\) and \(\beta\).
   4. Use the substitution \(u ^ { 2 } = 5 \cosh ^ { 2 } \theta - 1\) to show that $$S = \frac { 8 \pi } { \sqrt { 5 } } ( 644 \sqrt { 5 } - \ln ( 9 + 4 \sqrt { 5 } ) )$$

3

1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
2. Find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).

13

1. Use the definitions \(\tanh \theta = \frac { \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } } { \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } }\) and \(\operatorname { sech } \theta = \frac { 2 } { \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } }\) to prove the results
   1. \(\tanh ^ { 2 } \theta \equiv 1 - \operatorname { sech } ^ { 2 } \theta\),
   2. \(\frac { \mathrm { d } } { \mathrm { d } \theta } ( \tanh \theta ) = \operatorname { sech } ^ { 2 } \theta\).
   3. Let \(I _ { n } = \int _ { 0 } ^ { \alpha } \tanh ^ { 2 n } \theta \mathrm {~d} \theta\) for \(n \geqslant 0\), where \(\alpha > 0\).
      (a) Show that \(I _ { n - 1 } - I _ { n } = \frac { \tanh ^ { 2 n - 1 } \alpha } { 2 n - 1 }\) for \(n \geqslant 1\). Given that \(\alpha = \frac { 1 } { 2 } \ln 3\),
      (b) evaluate \(I _ { 0 }\),
   4. use the method of differences to show that \(I _ { n } = \frac { 1 } { 2 } \ln 3 - \sum _ { r = 1 } ^ { n } \frac { \left( \frac { 1 } { 2 } \right) ^ { 2 r - 1 } } { 2 r - 1 }\) and deduce the sum of the infinite series \(\sum _ { r = 0 } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) 4 ^ { r } }\).

12 The curve \(C\) has equation \(y = \ln \left( \tanh \frac { 1 } { 2 } x \right)\), for \(x > 0\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosech } x\).
2. For positive integers \(n\), the length of the arc of \(C\) between \(x = n\) and \(x = 2 n\) is \(L _ { n }\).
   1. Show by calculus that, when \(n\) is large, \(L _ { n } \approx n\).
   2. Explain how this result corresponds to the shape of \(C\).

4

1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
2. Hence or otherwise find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).

4

1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
2. Hence or otherwise find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).

12 The curve \(C\) is defined parametrically by $$x = t + \ln ( \cosh t ) , \quad y = \sinh t$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \cosh ^ { 2 } t\).
2. Hence show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \cosh ^ { 2 } t ( 2 \sinh t - \cosh t )\).
3. Find the exact value of \(t\) at the point on \(C\) where \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0\).

The curve \(C\) is defined parametrically by $$x = 2\cos^3 t \quad \text{and} \quad y = 2\sin^3 t, \quad \text{for } 0 < t < \frac{1}{2}\pi.$$ Show that, at the point with parameter \(t\), $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{1}{6}\sec^4 t \cosec t.$$ [4]

1. Starting from the definitions of \(\tanh\) and \(\sech\) in terms of exponentials, prove that $$1 - \tanh^2 \theta = \sech^2 \theta.$$ [3]

The variables \(x\) and \(y\) are such that \(\tanh y = \cos\left(x + \frac{1}{4}\pi\right)\), for \(-\frac{1}{4}\pi < x < \frac{3}{4}\pi\).

1. By differentiating the equation \(\tanh y = \cos\left(x + \frac{1}{4}\pi\right)\) with respect to \(x\), show that $$\frac{dy}{dx} = -\operatorname{cosec}\left(x + \frac{1}{4}\pi\right).$$ [4]
2. Hence find the first three terms in the Maclaurin's series for \(\tanh^{-1}\left(\cos\left(x + \frac{1}{4}\pi\right)\right)\) in the form \(\frac{1}{2}\ln a + bx + cx^2\), giving the exact values of the constants \(a\), \(b\) and \(c\). [5]