Chapter 5: Apply hyperbolic functions

Some shapes and decays in biomedical work are not periodic but still follow a fixed pattern: the sag of a cable between two supports, the fall of a signal along a nerve fibre, the cooling of a fin on a sensor housing. Hyperbolic functions describe these patterns by combining the exponential functions eˣ and e⁻ˣ into sinh, cosh and tanh. This chapter introduces the hyperbolic functions and their graphs, the identities that link them, the solving of hyperbolic equations and the use of the functions in engineering models.

5.1 Introduction to Hyperbolic Functions

Biomedical signals and models often require hyperbolic functions to represent phenomena that cannot be captured by circular trigonometric functions. This section introduces the fundamental hyperbolic functions, sine, cosine, tangent, and their reciprocals, setting the stage for applications such as modeling voltage changes across cell membranes or stress distribution in prosthetic materials.

5.1.1 Definitions of Hyperbolic Sine, Cosine, Tangent, and Their Reciprocals

Hyperbolic functions are defined using exponential functions. The basic hyperbolic functions are hyperbolic sine (\(\sinh\)), hyperbolic cosine (\(\cosh\)), and hyperbolic tangent (\(\tanh\)). Their reciprocals include hyperbolic cosecant (\(\csch\)), hyperbolic secant (\(\sech\)), and hyperbolic cotangent (\(\coth\)).

The definitions are:

$$\sinh x = \frac{e^{x} - e^{-x}}{2}$$

$$\cosh x = \frac{e^{x} + e^{-x}}{2}$$

$$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$$

Reciprocals:

$$\csch x = \frac{1}{\sinh x}$$

$$\sech x = \frac{1}{\cosh x}$$

$$\coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x}$$

Worked Examples

Example 1: Calculate \(\sinh 0.5\) for modeling the displacement in a biomechanical system.

Given: \(x = 0.5\)

$$\sinh 0.5 = \frac{e^{0.5} - e^{-0.5}}{2}$$

Calculate exponentials:

$$e^{0.5} = 1.6487, \quad e^{-0.5} = 0.6065$$

Substitute:

$$\sinh 0.5 = \frac{1.6487 - 0.6065}{2} = \frac{1.0422}{2} = 0.5211$$

Answer: \(\sinh 0.5 = 0.5211\)

Example 2: Find \(\cosh 1.2\) relevant to analyzing voltage response in a nerve fiber.

Given: \(x = 1.2\)

$$\cosh 1.2 = \frac{e^{1.2} + e^{-1.2}}{2}$$

Calculate exponentials:

$$e^{1.2} = 3.3201, \quad e^{-1.2} = 0.3012$$

Substitute:

$$\cosh 1.2 = \frac{3.3201 + 0.3012}{2} = \frac{3.6213}{2} = 1.8107$$

Answer: \(\cosh 1.2 = 1.8107\)

Example 3: Evaluate \(\tanh 0.8\) for blood flow velocity profile analysis.

Given: \(x = 0.8\)

$$\tanh 0.8 = \frac{e^{0.8} - e^{-0.8}}{e^{0.8} + e^{-0.8}}$$

Calculate exponentials:

$$e^{0.8} = 2.2255, \quad e^{-0.8} = 0.4493$$

Substitute:

$$\tanh 0.8 = \frac{2.2255 - 0.4493}{2.2255 + 0.4493} = \frac{1.7762}{2.6748} = 0.6640$$

Answer: \(\tanh 0.8 = 0.6640\)

Example 4: Find \(\sech 0.6\) as part of the analysis of signal attenuation in biological tissues.

Given: \(x = 0.6\)

First, calculate \(\cosh 0.6\):

$$\cosh 0.6 = \frac{e^{0.6} + e^{-0.6}}{2}$$

Calculate exponentials:

$$e^{0.6} = 1.8221, \quad e^{-0.6} = 0.5488$$

Substitute:

$$\cosh 0.6 = \frac{1.8221 + 0.5488}{2} = \frac{2.3709}{2} = 1.1855$$

Then,

$$\sech 0.6 = \frac{1}{\cosh 0.6} = \frac{1}{1.1855} = 0.8433$$

Answer: \(\sech 0.6 = 0.8433\)

Example 5: Compute \(\coth 1.0\), useful in modeling heat transfer in prosthetic limb materials.

Given: \(x = 1.0\)

Calculate \(\sinh 1.0\) and \(\cosh 1.0\):

$$\sinh 1.0 = \frac{e^{1.0} - e^{-1.0}}{2} = \frac{2.7183 - 0.3679}{2} = \frac{2.3504}{2} = 1.1752$$

$$\cosh 1.0 = \frac{e^{1.0} + e^{-1.0}}{2} = \frac{2.7183 + 0.3679}{2} = \frac{3.0862}{2} = 1.5431$$

Then,

$$\coth 1.0 = \frac{\cosh 1.0}{\sinh 1.0} = \frac{1.5431}{1.1752} = 1.3130$$

Answer: \(\coth 1.0 = 1.3130\)

5.1.2 Graphs and Properties of Hyperbolic Functions

Understanding the behavior of hyperbolic functions graphically helps biomedical engineers interpret system responses such as membrane potential changes or tissue deformation profiles. The graphs resemble their trigonometric counterparts but have key differences in range, periodicity, and symmetry.

Graph Characteristics and Properties

• \(\sinh x\) is an odd function: symmetric about the origin, passes through (0,0), and increases exponentially for positive \(x\).
• \(\cosh x\) is an even function: symmetric about the y-axis, minimum value 1 at \(x=0\), and increases exponentially for both positive and negative \(x\).
• \(\tanh x\) is an odd function: asymptotic to \(\pm 1\) as \(x \to \pm \infty\), passing through (0,0).
• None of the hyperbolic functions are periodic.
• Derivatives: \(\frac{d}{dx} \sinh x = \cosh x\), \(\frac{d}{dx} \cosh x = \sinh x\), \(\frac{d}{dx} \tanh x = \sech^2 x\).

Worked Examples

Example 1: Find the derivative of \(\sinh x\) at \(x = 0.7\) to determine rate of change of a biomedical signal.

Given: \(x=0.7\)

Derivative formula:

$$\frac{d}{dx} \sinh x = \cosh x$$

Calculate \(\cosh 0.7\):

$$\cosh 0.7 = \frac{e^{0.7} + e^{-0.7}}{2}$$

Calculate exponentials:

$$e^{0.7} = 2.0138, \quad e^{-0.7} = 0.4966$$

Substitute:

$$\cosh 0.7 = \frac{2.0138 + 0.4966}{2} = \frac{2.5104}{2} = 1.2552$$

Answer: \(\frac{d}{dx} \sinh 0.7 = 1.2552\)

Example 2: Determine the slope of \(\tanh x\) at \(x=1.0\), relevant to neural response modeling.

Given: \(x=1.0\)

Derivative formula:

$$\frac{d}{dx} \tanh x = \sech^2 x = (\frac{1}{\cosh x})^2$$

Calculate \(\cosh 1.0\):

$$\cosh 1.0 = 1.5431$$ (from previous example)

Calculate \(\sech^2 1.0\):

$$\sech 1.0 = \frac{1}{1.5431} = 0.6470$$

$$\sech^2 1.0 = (0.6470)^2 = 0.4186$$

Answer: \(\frac{d}{dx} \tanh 1.0 = 0.4186\)

Example 3: Evaluate \(\cosh x\) and \(\sinh x\) at \(x = -0.9\) to confirm even and odd properties.

Given: \(x = -0.9\)

Calculate \(\cosh (-0.9)\):

$$\cosh (-0.9) = \frac{e^{-0.9} + e^{0.9}}{2}$$

Calculate exponentials:

$$e^{-0.9} = 0.4066, \quad e^{0.9} = 2.4596$$

Substitute:

$$\cosh (-0.9) = \frac{0.4066 + 2.4596}{2} = \frac{2.8662}{2} = 1.4331$$

Calculate \(\sinh (-0.9)\):

$$\sinh (-0.9) = \frac{e^{-0.9} - e^{0.9}}{2} = \frac{0.4066 - 2.4596}{2} = \frac{-2.0530}{2} = -1.0265$$

Check even/odd:

\(\cosh (-0.9) = \cosh 0.9 = 1.4331\) (even function)

\(\sinh (-0.9) = -\sinh 0.9 = -1.0265\) (odd function)

Answer: \(\cosh (-0.9) = 1.4331\), \(\sinh (-0.9) = -1.0265\)

Example 4: Sketch the graph behavior of \(\tanh x\) near \(x=0\) using values at \(x = -0.5, 0, 0.5\).

Calculate \(\tanh -0.5\):

$$\tanh -0.5 = -\tanh 0.5$$ (odd function)

Calculate \(\tanh 0.5\):

$$\tanh 0.5 = \frac{e^{0.5} - e^{-0.5}}{e^{0.5} + e^{-0.5}} = \frac{1.6487 - 0.6065}{1.6487 + 0.6065} = \frac{1.0422}{2.2552} = 0.4621$$

$$\tanh -0.5 = -0.4621$$

At \(x=0\), \(\tanh 0 = 0\).

Answer: Points for graph: (-0.5, -0.4621), (0, 0), (0.5, 0.4621)

Example 5: Calculate the second derivative of \(\sinh x\) at \(x=0.4\) to analyze curvature in a tissue deformation model.

Given: \(x=0.4\)

First derivative:

$$\frac{d}{dx} \sinh x = \cosh x$$

Second derivative:

$$\frac{d^2}{dx^2} \sinh x = \frac{d}{dx} \cosh x = \sinh x$$

Calculate \(\sinh 0.4\):

$$\sinh 0.4 = \frac{e^{0.4} - e^{-0.4}}{2}$$

Calculate exponentials:

$$e^{0.4} = 1.4918, \quad e^{-0.4} = 0.6703$$

Substitute:

$$\sinh 0.4 = \frac{1.4918 - 0.6703}{2} = \frac{0.8215}{2} = 0.4107$$

Answer: \(\frac{d^2}{dx^2} \sinh 0.4 = 0.4107\)

Chapter Summary

This chapter introduced the hyperbolic functions that a biomedical engineering technician meets when a quantity rises or falls exponentially in two directions at once. It began with their definitions from the exponential function: sinh x = (eˣ − e⁻ˣ) ÷ 2, cosh x = (eˣ + e⁻ˣ) ÷ 2 and tanh x = sinh x ÷ cosh x, with the reciprocals csch x, sech x and coth x. Their graphs were compared: sinh x and tanh x are odd functions through the origin, cosh x is an even function whose smallest value is 1, tanh x levels off at −1 and 1, none of them repeats, and the slopes of sinh x, cosh x and tanh x are cosh x, sinh x and sech²x. The next topic set out the identities. The exponential relations cosh x + sinh x = eˣ and cosh x − sinh x = e⁻ˣ lead to the Pythagorean-like identities cosh²x − sinh²x = 1, 1 − tanh²x = sech²x and coth²x − 1 = csch²x, in which the plus of the circular identity has become a minus. The sum and difference formulas, the double-angle formulas and the half-angle formulas followed, with the sign rules that differ from the circular functions. Solving equations came next. A basic equation is solved by writing the function with exponentials, letting y = eˣ and solving the quadratic in y, rejecting any root that is not positive. The equation sinh x = a has one solution for every a, cosh x = b has two solutions when b > 1 and none when b < 1, and tanh x = c has one solution only when −1 < c < 1. Identities reduce equations that mix two functions or contain a double angle. The inverse functions sinh⁻¹ x = ln(x + √(x² + 1)), cosh⁻¹ x = ln(x + √(x² − 1)) and tanh⁻¹ x = ½ ln((1 + x) ÷ (1 − x)) have their own domains and ranges and solve equations directly. The chapter closed with four engineering models: the potential along a nerve fibre, the temperature along a cooling fin, the stress in a cushioning foam that levels off, and the sag and length of a hanging cable.