Chapter 6: Problem 434
For the following exercises, solve each problem. prove the expression for \(\cosh ^{-1}(x)\) . Multiply \(x=\cosh (y)=(1 / 2)\left(e^{y}-e^{-y}\right)\) by 2\(e^{y}\) and solve for \(y\) Does your expression match the textbook?
Short Answer
Expert verified
The derived expression matches the textbook: \( \cosh ^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \).
Step by step solution
01
Define Hyperbolic Cosine
The hyperbolic cosine function is defined as \( \cosh(y) = \frac{1}{2} \left(e^{y} + e^{-y}\right) \). To find \( y \) in terms of \( x \), we start with this equation and set \( x = \cosh(y) \), so \( x = \frac{1}{2} \left(e^{y} + e^{-y}\right) \).
02
Isolate Exponential Terms
Multiply both sides by 2 to eliminate the fraction: \( 2x = e^{y} + e^{-y} \). Then, multiply the entire equation by \( e^{y} \) to clear the denominator of \( e^{-y} \): \( 2xe^{y} = e^{2y} + 1 \).
03
Rearrange to Solve for Exponential
Rearrange the equation to form a quadratic in terms of \( e^{y} \): \( e^{2y} - 2xe^{y} + 1 = 0 \). This expression is a quadratic equation in \( e^{y} \).
04
Apply Quadratic Formula
Use the quadratic formula \( e^{y} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2x, c = 1 \). Substituting these into the quadratic formula gives: \( e^{y} = \frac{2x \pm \sqrt{4x^2 - 4}}{2} \).
05
Simplify the Expression
Simplify the solutions: \( e^{y} = x \pm \sqrt{x^2 - 1} \). Choose the positive branch because \( e^{y} \) is always positive, thus \( e^{y} = x + \sqrt{x^2 - 1} \).
06
Solve for y
Take the natural logarithm on both sides to solve for \( y \): \( y = \ln(x + \sqrt{x^2 - 1}) \). This is the standard expression for the inverse hyperbolic cosine: \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic Cosine
The hyperbolic cosine function, often denoted as \( \cosh(x) \), is a close cousin to the regular cosine function, but it involves exponential growth. It is defined by the formula: \[\cosh(y) = \frac{1}{2} (e^{y} + e^{-y})\]Unlike its trigonometric counterpart, hyperbolic cosine is crucial in areas involving hyperbolic geometry and has properties that make it applicable in hyperbolic equations.
- Hyperbolic functions relate to the shape and properties of a hyperbola, similar to how trigonometric functions relate to a circle.
- In the definition, \( e^{y} \) represents the exponential function, a key player in various fields of mathematical study.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). When dealing with equations that can't be readily factored, this formula comes to the rescue:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the context of our hyperbolic expression, the equation \( e^{2y} - 2xe^{y} + 1 = 0 \) is formed. Here, we identify:
- \(a = 1\),
- \(b = -2x\),
- \(c = 1\)
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). In mathematics, it plays a crucial role in solving equations where an exponent is unknown. It is incredibly useful for inverse operations involving exponential functions.
Once we have \( e^{y} \) from the quadratic equation solved previously, the next step is to determine \( y \). By taking the natural logarithm on both sides:\[y = \ln(x + \sqrt{x^2 - 1})\]This operation effectively 'undoes' the exponent and isolates \( y \), which aligns perfectly with our hyperbolic inverse formula. Often, natural logarithms are used in fields such as calculus, physics, and complex system modeling.
Once we have \( e^{y} \) from the quadratic equation solved previously, the next step is to determine \( y \). By taking the natural logarithm on both sides:\[y = \ln(x + \sqrt{x^2 - 1})\]This operation effectively 'undoes' the exponent and isolates \( y \), which aligns perfectly with our hyperbolic inverse formula. Often, natural logarithms are used in fields such as calculus, physics, and complex system modeling.
- They simplify multiplication into addition, which is useful in many mathematical transformations.
- The natural logarithm is instrumental in solving growth and decay problems, especially where change rates are proportional to the current state.