Problem 16 - Calculus of a Single Variable

Question

Finding Values of Hyperbolic Functions In Exercises 15 and \(16,\) use the value of the given hyperbolic function to find the values of the other hyperbolic functions at \(x .\) $$ \tanh x=\frac{1}{2} $$

Step-by-Step Solution

Answer

The values of the hyperbolic functions at \(x\) are \(\sinh(x) = \frac{-1 + \sqrt{5}}{4}\), \(\cosh(x) = \frac{-1 + \sqrt{5}}{2}\), \(\tanh(x) = \frac{1}{2}\), \(\sech(x) = \frac{2}{-1 + \sqrt{5}}\), \(\csch(x) = \frac{4}{-1 + \sqrt{5}}\), and \(\coth(x) = 2\).

Step 1: Define the Given Function

The given function is the hyperbolic tangent, \(\tanh(x) = \frac{1}{2}\). The hyperbolic tangent function is defined as \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\), where \(\sinh(x)\) and \(\cosh(x)\) are the hyperbolic sine and cosine functions, respectively.

Step 2: Express hyperbolic sine in terms of hyperbolic tangent

Rewrite the function \(\sinh(x)\) in terms of hyperbolic tangent \(\tanh(x)\) and hyperbolic cosine \(\cosh(x)\) using the definition of \(\tanh(x)\), so \(\sinh(x) = \tanh(x) \cdot \cosh(x)\). Substitute \(\tanh(x)\) by \(\frac{1}{2}\) to get \(\sinh(x) = \frac{1}{2}\cosh(x)\).

Step 3: Find the value of hyperbolic sine function

Use the expression \(\cosh^2(x) = 1 + \sinh^2(x)\) which can be also be written as \(\sinh^2(x) = \cosh^2(x) - 1\). Substituting the value of \(\sinh(x)\) that we found in step 2 into this formula gives \((\frac{1}{2}\cosh(x))^2 = \cosh^2(x) - 1\). Solving this quadratic equation yields \(\cosh(x) = \frac{-1 \pm \sqrt{5}}{2}\). However, \(\cosh(x)\) is always positive, so it is evident that \(\cosh(x) = \frac{-1 + \sqrt{5}}{2}\). Since \(\sinh(x) = \frac{1}{2}\) times \(\cosh(x)\), we find \(\sinh(x) = \frac{-1 + \sqrt{5}}{4}\).

Step 4: Find the value of other hyperbolic functions

The hyperbolic secant \(\sech(x)\) is the reciprocal of \(\cosh(x)\), so \(\sech(x) = \frac{2}{-1 + \sqrt{5}}\). The hyperbolic cosecant \(\csch(x)\) is the reciprocal of \(\sinh(x)\), so \(\csch(x) = \frac{4}{-1 + \sqrt{5}}\). The hyperbolic cotangent \(\coth(x)\) is \(\cosh(x)\) divided by \(\sinh(x)\), or the reciprocal of \(\tanh(x)\), so \(\coth(x) = 2\).

Key Concepts

Hyperbolic Tangent

The hyperbolic tangent, denoted as \( \tanh(x) \), is a fundamental hyperbolic function, defined uniquely as the ratio of hyperbolic sine and hyperbolic cosine: \[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \]
This function is significant due to how it describes the natural growth and decay processes, similar to tangent in trigonometry. It ranges from -1 to 1, acting like a smooth activation function in mathematics and engineering fields.

• In the problem, we have \( \tanh(x) = \frac{1}{2} \).
• This relation allows us to solve for other hyperbolic functions like \( \sinh(x) \) and \( \cosh(x) \).

Understanding \( \tanh(x) \) is important, as it establishes the basis for finding related hyperbolic identities, particularly since it connects directly to \( \sinh \) and \( \cosh \) through simple operations.

Hyperbolic Sine

The hyperbolic sine function, \( \sinh(x) \), is a counterpart to the trigonometric sine function but applies to hyperbolas instead of circles. It is defined mathematically as:\[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]
In the context of the given exercise, we are tasked with expressing \( \sinh(x) \) in terms of \( \tanh(x) \) and \( \cosh(x) \). We know:

• \( \sinh(x) = \tanh(x) \cdot \cosh(x) \)

This reformulation helps us directly substitute \( \tanh(x) = \frac{1}{2} \) to find \( \sinh(x) = \frac{1}{2} \cosh(x) \). By substituting the value of \( \cosh(x) \) found in subsequent solutions, we can solve for \( \sinh(x) \), yielding \( \sinh(x) = \frac{-1 + \sqrt{5}}{4} \). This computation is essential for determining other hyperbolic functions.

Hyperbolic Cosine

Hyperbolic cosine, denoted as \( \cosh(x) \), is vital for understanding hyperbolic functions. Its definition is given by:\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]
This symmetric function has values always greater than or equal to 1 for all real numbers. Key to many hyperbolic identities, \( \cosh(x) \), plays a central role:

• The property \( \cosh^2(x) - \sinh^2(x) = 1 \) is crucial for solving many problems, including the original exercise.

In our problem, by applying \( \tanh(x) = \frac{1}{2} \), we derive,\( \cosh(x) = \frac{-1 + \sqrt{5}}{2} \) through solving the equation derived from \( \sinh(x) = \frac{1}{2} \cosh(x) \). This value enables the calculation of other functional values like \( \sech(x) \), \( \csch(x) \), and \( \coth(x) \), highlighting its indispensability in hyperbolic identities.

Hyperbolic Identities

Hyperbolic identities are similar to trigonometric identities, possessing unique properties and equations. They form the backbone of hyperbolic functions, allowing the transformation and simplification of expressions. Key identities include:

• \( \cosh^2(x) - \sinh^2(x) = 1 \)
• \( 1 - \tanh^2(x) = \sech^2(x) \)
• Other identities involve \( \coth(x) = \frac{1}{\tanh(x)} = \frac{\cosh(x)}{\sinh(x)} \) and reciprocals for \( \sech(x) \) and \( \csch(x) \).

In our task, using these identities simplifies solving for \( \cosh(x) \), \( \sinh(x) \), and others. The equation \( \cosh^2(x) - \sinh^2(x) = 1 \) helps link \( \cosh \) and \( \sinh \) and verify the relationships we establish using \( \tanh(x) \). These identities reflect the hyperbolic function landscape's consistency and the simplicity underlying their interconnections.