Problem 63 - Thomas Calculus in SI Units

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. (We give one derivation in Section 8.4.) $$\begin{aligned}&\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1}), \quad-\infty < x<\infty\\\&\cosh ^{-1} x=\ln (x+\sqrt{x^{2}-1}), \quad x \geq 1\\\ &\tanh ^{-1} x=\frac{1}{2} \ln \frac{1+x}{1-x}, \quad|x| < 1\\\&\operatorname{sech}^{-1} x=\ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right), \quad0 < x \leq 1\\\&\operatorname{csch}^{-1} x=\ln \left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right), \quad x \neq 0\\\&\operatorname{coth}^{-1} x=\frac{1}{2} \ln \frac{x+1}{x-1}, \quad |x|>1\end{aligned}$$ Use the formulas in the box here to express the numbers in terms of natural logarithms. $$\tanh ^{-1}(-1 / 2)$$

Step-by-Step Solution

Answer

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

Step 1: Identify the formula for the given inverse hyperbolic function

The problem requires expressing \( \tanh^{-1}(-1/2) \) as a natural logarithm. According to the provided formula, \( \tanh^{-1}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \), applicable for \( |x| < 1 \).

Step 2: Substitute the value into the formula

Substitute \( x = -1/2 \) into the formula: \[ \tanh^{-1}(-1/2) = \frac{1}{2} \ln \left( \frac{1 + (-1/2)}{1 - (-1/2)} \right) \]

Step 3: Simplify the expression inside the logarithm

Simplify the expression in the logarithm: \[ \frac{1 + (-1/2)}{1 - (-1/2)} = \frac{1 - 1/2}{1 + 1/2} = \frac{1/2}{3/2} \]

Step 4: Calculate the fraction

Calculate the fraction: \[ \frac{1/2}{3/2} = \frac{1}{3} \]

Step 5: Complete the natural logarithm expression

Substitute the simplified fraction back into the expression for the inverse hyperbolic function:\[ \tanh^{-1}(-1/2) = \frac{1}{2} \ln \left( \frac{1}{3} \right) \]

Key Concepts

Natural Logarithms

Natural logarithms, denoted by the symbol \( \ln \), are logarithms with a special base: 'e'. The constant 'e' is an irrational number approximately equal to 2.71828. This base is particularly significant in calculus and natural sciences because it offers a unique rate of growth.
This means that natural logarithms help us transform exponential equations into linear ones, making complex calculations more manageable.
For instance, the inverse hyperbolic function \( \tanh^{-1}(x) \) can be expressed in terms of natural logarithms as \( \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \).

The simplistic elegance of \( \ln \) lies in its application, allowing us to solve for unknowns in exponential situations.

• Converts exponential equations into simpler linear forms.
• Used extensively alongside calculus concepts like derivatives and integrals.
• Key in evaluating growth rates and more complex mathematical functions.

Hyperbolic Identities

Hyperbolic identities connect trigonometric-like functions for hyperbolas, much as trigonometric functions relate to circles. These identities define relationships among hyperbolic functions, making them highly useful for calculus operations and transformations.
Some of the essential hyperbolic identities include:

• \( \sinh^{-1} x = \ln (x + \sqrt{x^2 + 1}) \)
• \( \cosh^{-1} x = \ln (x + \sqrt{x^2 - 1}) \)
• \( \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \)

These identities show how inverse hyperbolic functions can be converted into logarithmic forms.
Unlike their circular counterparts, hyperbolic functions don't oscillate. They describe hyperbolic curves and are useful in situations where growth and decay processes are modeled, similar to exponential growth found in nature.
This is why expressions like \( \tanh^{-1}(x) \), an identity linked to natural logarithms, help solve complex calculus problems.

Calculus Applications

Inverse hyperbolic functions hold valuable implications in the field of calculus. They provide the tools necessary for solving integrals and working through differential equations, which are typically challenging to approach with simple functions.
When we use an inverse hyperbolic formula, such as \( \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \), we essentially transform hyperbolic relationships into ones that are solvable using calculus methods.

This transformation lends itself well to solving real-world problems. Here are some fields where inverse hyperbolic functions play an essential role:

• Physics: modeling kinetic energy and wave functions.
• Engineering: analyzing material stresses and electric currents.
• Economics: optimizing cost functions and predicting growth trends.

Thus, whether dealing with exponential decay, growth patterns, or complex motions, these functions have broad applications. They allow us to express complex systems and predict outcomes with higher accuracy and ease.