Problem 66 - 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. $$\operatorname{csch}^{-1}(-1 / \sqrt{3})$$

Step-by-Step Solution

Answer

csch ^{-1} (-1 / sqrt {3 } ) ≈ -1.314 .

Step 1: Identify the Formula

The formula for \( csch^{-1} x \) is given as \( ext{csch}^{-1} x = ln \left( \frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|} \right) \), where \(x eq 0\). We'll use this formula to find the inverse hyperbolic cosecant for \(-1 / \sqrt{3}\).

Step 2: Substitute the Value

Substitute \(-1/\sqrt{3}\) into the formula: \( \text{csch}^{-1} \left( - \frac{1}{\sqrt{3}} \right) = \ln \left( \frac{-\sqrt{3}}{1} + \frac{\sqrt{1 + (-1/\sqrt{3})^2}}{|-1/\sqrt{3}|} \right) \).

Step 3: Simplify Inside the Logarithm

Calculate terms inside the logarithm: \( \frac{-\sqrt{3}}{1} = -\sqrt{3} \) and \(\frac{\sqrt{1 + \left(-\frac{1}{\sqrt{3}}\right)^2}}{|\frac{-1}{\sqrt{3}}|} = \frac{\sqrt{1 + \frac{1}{3}}}{\frac{1}{\sqrt{3}}} = \frac{\sqrt{\frac{4}{3}}}{\frac{1}{\sqrt{3}}} = \frac{2}{1} = 2 \).

Step 4: Add Terms in the Logarithm

Add the terms calculated inside the logarithm: \( \ln \left( -\sqrt{3} + 2 \right) \). Now, calculate the result: \( -\sqrt{3} + 2 \approx 0.268 \).

Step 5: Evaluate the Logarithm

Finally, evaluate: \( \ln (0.268) \). Using a calculator or logarithm table, \( \ln (0.268) \approx -1.314 \).

Key Concepts

Logarithmic Expressions

Logarithmic expressions are mathematical expressions that involve the natural logarithm, denoted as \( \ln \). Logarithms are used to solve equations where the variable is an exponent. In calculus and algebra, they help us express multiplicative relationships in an additive form.
To understand logarithmic expressions better, we can consider them as the inverse of exponential functions. For example, if we have \( a = e^b \), then \( b = \ln(a) \). This relationship shows how a logarithm operates: it "undoes" exponentiation to give us back the exponent. This is especially useful when dealing with inverse hyperbolic functions. By expressing an inverse hyperbolic function as a logarithm, you can solve problems even when specific hyperbolic function keys are unavailable.
With natural logarithms, these calculations often require breaking down complex expressions into simpler steps, including substitution or simplification within the logarithmic function itself. Let's see how they come into play in the realm of hyperbolic functions.

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for the hyperbola, not the circle. The common hyperbolic functions are \( \sinh(x) \), \( \cosh(x) \), \( \tanh(x) \), and their respective inverses. These functions are particularly useful in many areas of mathematics, including calculus, where they allow us to describe the shape of a hyperbola.
The expressions for their inverses, such as \( \sinh^{-1}(x) \), are derived from their natural behaviors and are often represented using logarithmic expressions:

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

These equations allow us to calculate inverse hyperbolic functions even when we do not have direct access to their values on a calculator.
In our case of solving for \( \text{csch}^{-1}(-1/\sqrt{3}) \), we use its logarithmic expression, \( \ln \left( \frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|} \right) \). By substituting the value directly into this expression, we can compute even the inverse of more complex hyperbolic functions accurately.

Calculus Problem-Solving

Calculus problem-solving often involves using a strategic combination of functions and expressions to simplify and evaluate complex mathematical problems. With hyperbolic functions, understanding their logarithmic forms becomes essential as it enables evaluating these functions even without specific tools.
Let's break down the steps we use in calculus to solve such a problem using the example of \( \text{csch}^{-1}(-1/\sqrt{3}) \):

• Identify the Formula: Recognize the logarithmic expression that represents the inverse hyperbolic function.
• Substitute the Value: Insert the specified value of the variable into the formula.
• Simplify Inside the Logarithm: Perform any necessary arithmetic to simplify the expression under the natural logarithm.
• Add Terms in the Logarithm: Combine and simplify these terms to get a single expression.
• Evaluate the Logarithm: Use a calculator or logarithm table to find the value of the logarithm.

By following these steps, we create a systematic way to tackle calculus problems involving inverse hyperbolic functions. This approach enables students to confidently work through complexities in calculus, knowing that they can rely on the foundational expressions and operations of logarithms to find solutions.