4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with polar equation $$r = 2 \sqrt { \sinh \theta + \cosh \theta } \quad 0 \leqslant \theta \leqslant \pi$$ The region \(R\), shown shaded in Figure 1, is bounded by the initial line, the curve and the line with equation \(\theta = \pi\) Use algebraic integration to determine the exact area of \(R\) giving your answer in the form \(p \mathrm { e } ^ { q } - r\) where \(p , q\) and \(r\) are real numbers to be found.

1. Given that

$$y = \mathrm { e } ^ { 2 x } \sinh x$$ prove by induction that for \(n \in \mathbb { N }\) $$\frac { \mathrm { d } ^ { n } y } { \mathrm {~d} x ^ { n } } = \mathrm { e } ^ { 2 x } \left( \frac { 3 ^ { n } + 1 } { 2 } \sinh x + \frac { 3 ^ { n } - 1 } { 2 } \cosh x \right)$$

1. The hyperbola \(H\) has equation

$$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$$ The line \(l _ { 1 }\) is the tangent to \(H\) at the point \(P ( 4 \cosh \theta , 3 \sinh \theta )\).
The line \(l _ { 1 }\) meets the \(x\)-axis at the point \(A\).
The line \(l _ { 2 }\) is the tangent to \(H\) at the point \(( 4,0 )\).
The lines \(l _ { 1 }\) and \(l _ { 2 }\) meet at the point \(B\) and the midpoint of \(A B\) is the point \(M\).

1. Show that, as \(\theta\) varies, a Cartesian equation for the locus of \(M\) is $$y ^ { 2 } = \frac { 9 ( 4 - x ) } { 4 x } \quad p < x < q$$ where \(p\) and \(q\) are values to be determined. Let \(S\) be the focus of \(H\) that lies on the positive \(x\)-axis.
2. Show that the distance from \(M\) to \(S\) is greater than 1

1. A community is concerned about the rising level of pollutant in its local pond and applies a chemical treatment to stop the increase of pollutant.

The concentration, \(x\) parts per million (ppm), of the pollutant in the pond water \(t\) days after the chemical treatment was applied, is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 3 + \cosh t } { 3 x ^ { 2 } \cosh t } - \frac { 1 } { 3 } x \tanh t$$ When the chemical treatment was applied the concentration of pollutant was 3 ppm .

1. Use the iteration formula $$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - y _ { n } \right) } { h }$$ once to estimate the concentration of the pollutant in the pond water 6 hours after the chemical treatment was applied.
2. Show that the transformation \(u = x ^ { 3 }\) transforms the differential equation (I) into the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} t } + u \tanh t = 1 + \frac { 3 } { \cosh t }$$
3. Determine the general solution of equation (II)
4. Hence find an equation for the concentration of pollutant in the pond water \(t\) days after the chemical treatment was applied.
5. Find the percentage error of the estimate found in part (a) compared to the value predicted by the model, stating if it is an overestimate or an underestimate.

6 In this question you must show detailed reasoning. The power output, \(p\) watts, of a machine at time \(t\) hours after it is switched on can be modelled by the equation \(\mathrm { p } = 20 - 20 \tanh ( 1.44 \mathrm { t } )\) for \(t \geqslant 0\). Determine, according to the model, the mean power output of the machine over the first half hour after it is switched on. Give your answer correct to \(\mathbf { 2 }\) decimal places.

1 In this question you must show detailed reasoning.

1. Show that \(\cosh ( 2 \ln 3 ) = \frac { 41 } { 9 }\). The region \(R\) is bounded by the curve with equation \(\mathrm { y } = \sqrt { \operatorname { sinhx } }\), the \(x\)-axis and the line with equation \(x = 2 \ln 3\) (see diagram). The units of the axes are centimetres. \includegraphics[max width=\textwidth, alt={}, center]{23e58e5e-bbaa-4932-aad0-89b3de6647b2-2_652_668_740_242} A manufacturer produces bell-shaped chocolate pieces. Each piece is modelled as being the shape of the solid formed by rotating \(R\) completely about the \(x\)-axis.
2. Determine, according to the model, the exact volume of one chocolate piece.

9 In this question you must show detailed reasoning.

1. Show that \(\operatorname { Re } \left( \mathrm { e } ^ { \mathrm { Ai } \theta } \left( \mathrm { e } ^ { \mathrm { i } \theta } + \mathrm { e } ^ { - \mathrm { i } \theta } \right) ^ { 4 } \right) = a \cos 4 \theta \cos ^ { 4 } \theta\), where \(a\) is an integer to be determined.
2. Hence show that \(\cos \frac { 1 } { 12 } \pi = \frac { 1 } { 2 } \sqrt [ 4 ] { \mathrm { b } + \mathrm { c } \sqrt { 3 } }\), where \(b\) and \(c\) are integers to be determined.

5 In this question you must show detailed reasoning.

1. Using the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, show that \(\sinh 2 x \equiv 2 \sinh x \cosh x\).
2. Solve the equation \(15 \sinh x + 16 \cosh x - 6 \sinh 2 x = 20\), giving all your answers in logarithmic form.

10 In this question you must show detailed reasoning. A region, \(R\), of the floor of an art gallery is to be painted for the purposes of an art installation. A suitable polar coordinate system is set up on the floor of the gallery with units in metres and radians. \(R\) is modelled as being the region enclosed by two curves, \(C _ { 1 }\) and \(C _ { 2 }\). The polar equations of \(C _ { 1 }\) and \(C _ { 2 }\) are $$\begin{array} { l l } C _ { 1 } : r = 5 , & - \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi \\ C _ { 2 } : r = 3 \cosh \theta , & - \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi \end{array}$$ Both curves are shown in the diagram, with \(R\) indicated. \includegraphics[max width=\textwidth, alt={}, center]{7b2bfb4e-524f-4d1c-ae98-075c7fb404f9-6_1481_821_836_251} The gallery must buy tins of paint to paint \(R\). Each tin of paint can cover an area of \(0.5 \mathrm {~m} ^ { 2 }\).
Determine the smallest number of tins of paint that the gallery must buy in order to be able to paint \(R\) completely.

7

1. Using the definition of \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that $$4 \sinh ^ { 3 } x = \sinh 3 x - 3 \sinh x$$ \section*{(ii) In this question you must show detailed reasoning.} By making a suitable substitution, find the real root of the equation $$16 u ^ { 3 } + 12 u = 3 .$$ Give your answer in the form \(\frac { \left( a ^ { \frac { 1 } { b } } - a ^ { - \frac { 1 } { b } } \right) } { c }\) where \(a , b\) and \(c\) are integers.

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 \(\left( \cosh ^ { - 1 } p , q \right)\).

8

1. Given that \(u = \tanh x\), use the definition of \(\tanh x\) in terms of exponentials to show that $$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
2. Solve the equation \(4 \tanh ^ { 2 } x + \tanh x - 3 = 0\), giving the solution in the form \(a \ln b\) where \(a\) and \(b\) are rational numbers to be determined.
3. Explain why the equation in part (b) has only one root.

1. Solve the equation

$$7 \operatorname { sech } x - \tanh x = 5$$ Give your answers in the form \(\ln a\), where \(a\) is a rational number.

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 $$\frac { 1 } { 2 } \left[ ( \cosh 2 ) ^ { \frac { 3 } { 2 } } - 1 \right]$$

1

1. Given that $$4 \cosh ^ { 2 } x = 7 \sinh x + 1$$ find the two possible values of \(\sinh x\).
2. Hence obtain the two possible values of \(x\), giving your answers in the form \(\ln p\).

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 \(s = \ln ( 2 \cosh 1 )\).

1

1. Use the definitions \(\sinh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } \right)\) and \(\cosh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } \right)\) to show that $$1 + 2 \sinh ^ { 2 } \theta = \cosh 2 \theta$$
2. Solve the equation $$3 \cosh 2 \theta = 2 \sinh \theta + 11$$ giving each of your answers in the form \(\ln p\).

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 $$s = 4 \sinh ^ { - 1 } 0.5 + \sqrt { 5 }$$

4

1. Sketch the graph of \(y = \tanh x\).
2. Given that \(u = \tanh x\), use the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\) to show that $$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
3. 1. Show that the equation $$3 \operatorname { sech } ^ { 2 } x + 7 \tanh x = 5$$ can be written as $$3 \tanh ^ { 2 } x - 7 \tanh x + 2 = 0$$
   2. Show that the equation $$3 \tanh ^ { 2 } x - 7 \tanh x + 2 = 0$$ has only one solution for \(x\).
      Find this solution in the form \(\frac { 1 } { 2 } \ln a\), where \(a\) is an integer.

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 $$y ^ { 2 } = 4 + s ^ { 2 }$$

2

1. Sketch the graph of \(y = \tanh x\) and state the equations of its asymptotes.
2. Use the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\) to show that $$\operatorname { sech } ^ { 2 } x + \tanh ^ { 2 } x = 1$$
3. Solve the equation \(6 \operatorname { sech } ^ { 2 } x = 4 + \tanh x\), giving your answers in terms of natural logarithms.
   [0pt] [5 marks] \section*{Answer space for question 2}
   1. \includegraphics[max width=\textwidth, alt={}, center]{bc3aaed2-4aef-4aec-b657-098b1e581e55-04_855_1447_920_324}

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.