Problem 63 - Thomas Calculus

Question

When hyperbolic function keys are not available on a calculator, it is still possible to evaluate the inverse hyperbolic functions by expressing them as logarithms, as shown here. $$ \begin{array}{rlrl}{\sinh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {-\infty < x < \infty} \\ {\cosh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {x \geq 1} \\ {\tanh ^{-1} x} & {=\frac{1}{2} \ln \frac{1+x}{1-x},} & {|x| < 1} \\ {\operatorname{sech}^{-1} x} & {=\ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right),} & {0 < x \leq 1} \\\ {\operatorname{csch}^{-1} x} & {=\ln \left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right),} & {x \neq 0} \\\ {\operatorname{coth}^{-1} x} & {=\frac{1}{2} \ln \frac{x+1}{x-1},} & {|x| > 1}\end{array} $$ Use the formulas given above to express the numbers in Exercises \(61-66\) in terms of natural logarithms. $$ \tanh ^{-1}(-1 / 2) $$

Step-by-Step Solution

Answer

\( \tanh^{-1}(-1/2) = -\frac{1}{2} \ln(3) \)

Step 1: Identify the Formula

The problem requires expressing the inverse hyperbolic tangent of a value using logarithms. We use the formula \( \tanh^{-1}x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \) to solve this.

Step 2: Substitute the Given Value

Substitute the given value \(-1/2\) into the formula. This gives us \( \tanh^{-1}(-1/2) = \frac{1}{2} \ln \left(\frac{1+(-1/2)}{1-(-1/2)}\right) \).

Step 3: Simplify Within the Logarithm

Simplify the fraction inside the logarithm: \( \frac{1+(-1/2)}{1-(-1/2)} = \frac{1-1/2}{1+1/2} = \frac{1/2}{3/2} = \frac{1}{3} \).

Step 4: Evaluate the Expression

Substitute back into the formula: \( \tanh^{-1}(-1/2) = \frac{1}{2} \ln \left(\frac{1}{3}\right) = \frac{1}{2} \ln(1) - \frac{1}{2} \ln(3) = -\frac{1}{2} \ln(3) \).

Key Concepts

Natural Logarithms

Natural logarithms are a type of logarithm where the base is the mathematical constant \( e \), approximately 2.718. These logarithms are denoted by \( \ln \). They are used widely in mathematics, especially in calculus and advanced algebra. Natural logarithms have properties that make them very useful for solving a variety of problems.

• They help express relationships in terms of exponential growth or decay, such as in compound interest or population dynamics.
• The simplicity of their derivative \( \frac{d}{dx} \ln(x) = \frac{1}{x} \) makes them integral to calculus operations.
• Logarithms convert multiplication into addition, which simplifies complex calculations.

If you have a function like \( y = b^x \), taking the natural logarithm of both sides allows the equation to be rewritten in a linear format, \( x = \ln(y) / \ln(b) \), which can be easier to handle. This property is particularly helpful in solving expressions involving inverse hyperbolic functions, where these functions are transformed into a form involving \( \ln \).

Hyperbolic Functions

Hyperbolic functions are analogs to the trigonometric functions but for a hyperbola, rather than a circle. They include \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \), among others. These functions are defined in terms of exponential functions and are incredibly useful for describing the geometry of hyperbolas and certain rapid decay models.

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) represents hyperbolic sine and is symmetric about the origin.
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) represents hyperbolic cosine and has a minimum value of 1 at \( x=0 \).
• \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \) represents hyperbolic tangent, which approaches \( -1 \) and \( 1 \) asymptotically as \( x \) becomes large or small respectively.

Using properties of hyperbolic functions, we can translate them into logarithmic terms, facilitating calculation using only basic arithmetic operations when hyperbolic function keys are unavailable on a calculator.

Mathematical Substitution

Mathematical substitution is a method used to simplify expressions and solve equations. By substituting one expression for another, we can make complex calculations more manageable. When working with inverse hyperbolic functions, substitution is employed to express these functions in terms of natural logarithms, which are easier to compute with standard calculators.For instance, when evaluating \( \tanh^{-1}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \), we apply substitution to the value of \( x \). Suppose \( x = -1/2 \). Substituting gives us:- Compute \( \frac{1+(-1/2)}{1-(-1/2)} \).- Simplify to \( \frac{1/3} \).- The expression then becomes \( \frac{1}{2} \ln(1/3) \).Mathematical substitution is a powerful tool not only for simplifying expressions but also for solving complex problems without advanced technology or resources. This technique can be applied in calculus, algebra, and many other branches of mathematics to achieve more efficient problem-solving.