Problem 65 - 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. $$ \operatorname{sech}^{-1}(3 / 5) $$

Step-by-Step Solution

Answer

\( \operatorname{sech}^{-1}(\frac{3}{5}) = \ln(3) \).

Step 1: Identify the Appropriate Formula

We need to find the expression for the inverse hyperbolic secant of \( \frac{3}{5} \). The formula provided for \( \operatorname{sech}^{-1}(x) \) is \( \operatorname{sech}^{-1} x = \ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right) \) for \( 0 < x \leq 1 \).

Step 2: Substitute the Given Value

Substitute \( x = \frac{3}{5} \) into the formula: \[ \operatorname{sech}^{-1} \left(\frac{3}{5}\right) = \ln \left(\frac{1+\sqrt{1-(\frac{3}{5})^{2}}}{\frac{3}{5}}\right) \]

Step 3: Compute the Square of x

Calculate \( (\frac{3}{5})^{2} = \frac{9}{25} \).

Step 4: Calculate the Square Root Term

Compute \( \sqrt{1 - \frac{9}{25}} \). First, calculate the inside expression: \( 1 - \frac{9}{25} = \frac{16}{25} \). Then, find the square root: \( \sqrt{\frac{16}{25}} = \frac{4}{5} \).

Step 5: Complete the Numerator of the Logarithm

Add \( 1 \) to the square root term: \( 1 + \frac{4}{5} = \frac{9}{5} \).

Step 6: Form the Entire Logarithm Argument

The argument of the logarithm is: \[ \operatorname{sech}^{-1} \left(\frac{3}{5}\right) = \ln \left(\frac{\frac{9}{5}}{\frac{3}{5}}\right) \]

Step 7: Simplify the Expression Inside the Logarithm

Simplify \( \frac{\frac{9}{5}}{\frac{3}{5}} = \frac{9}{5} \times \frac{5}{3} = 3 \).

Step 8: Finalize the Solution

Thus, the expression for \( \operatorname{sech}^{-1}(\frac{3}{5}) \) in terms of natural logarithms is \( \ln(3) \).

Key Concepts

Natural Logarithms

The natural logarithm is denoted by the symbol \( \ln \) and is the logarithm to the base \( e \), where \( e \) is approximately 2.718. It is a crucial concept in mathematics, particularly when dealing with exponential growth or decay.
Natural logarithms have properties that make them useful in calculus and algebra, such as the ability to transform multiplication into addition, which simplifies many mathematical operations.

• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b\ln(a) \)

Natural logarithms are particularly useful for solving equations involving exponentials, and they appear frequently in mathematical descriptions of processes like population growth and radioactive decay.

Hyperbolic Secant

The hyperbolic secant (\( \operatorname{sech} \)) is a term from hyperbolic trigonometry. It is similar to trigonometric functions but applied to hyperbolas, much like their circular counterparts apply to circles. The hyperbolic secant is defined as the reciprocal of the hyperbolic cosine function:
\[ \operatorname{sech}(x) = \frac{1}{\cosh(x)} = \frac{2}{e^x + e^{-x}} \]
Understanding hyperbolic functions is essential in fields like calculus and complex analysis. They frequently appear in problems related to hyperbolas and exponential decay - especially within the context of physics and engineering. The inverse hyperbolic secant, \( \operatorname{sech}^{-1}(x) \), is similarly defined and can be expressed through natural logarithms, as shown by the formula:
\[ \operatorname{sech}^{-1} x = \ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right) \]. It allows one to convert the operation into a logarithmic equation, which is easier to solve manually.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging or simplifying expressions using rules from algebra. In the given exercise, it was necessary to engage in algebraic manipulation to solve the problem.

• First, substitution is used to replace \( x \) with \( \frac{3}{5} \), which is the given value.
• Then, the square of \( x \) was calculated: \( \left(\frac{3}{5}\right)^2 = \frac{9}{25} \). This helped determine the terms inside the square root.
• Following this, the square root was calculated, which required subtraction from 1, then taking the square root of the resultant value.
• Finally, the expression inside the logarithm was simplified through multiplication and division to reach a final answer.

These steps help ensure that the expression remains mathematically correct while transitioning from the initial value to the simplified form, \( \ln(3) \). Being proficient in these operations is critical in higher mathematics.

Mathematical Identities

Mathematical identities are equations that hold true for all values of their variables. They are essential tools in simplifying expressions and solving equations.
In the context of inverse hyperbolic functions, these identities allow us to transform the expressions into forms that are easier to handle. The equations provided in the exercise are examples of identities, showing how inverse hyperbolic functions can be represented using natural logarithms.
Using identities such as\( 1 - x^2 = (1-x)(1+x) \) can simplify the square root calculations or help in factorization to make further simplifications. Some well-known mathematical identities include:

• The Pythagorean identity: \( \sin^2(x) + \cos^2(x) = 1 \)
• Exponential identities like \( e^{a+b} = e^a \cdot e^b \)

Understanding mathematical identities can greatly facilitate the manipulation and simplification of complex mathematical expressions, as illustrated by the conversion of \( \operatorname{sech}^{-1}(\frac{3}{5}) \) to \( \ln(3) \). They are foundational in solving complex calculus and algebra problems.