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Prove: (a) \(\cosh ^{-1} x=\ln (x+\sqrt{x^{2}-1}), x \geq 1\) (b) \(\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right),-1

Short Answer

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(a) Set \(y = \cosh^{-1}(x)\), solve for \(e^y = x + \sqrt{x^2 - 1}\) and take the natural log. (b) Set \(y = \tanh^{-1}(x)\), solve \((1-x)e^{2y} = (1+x)\), and take the log.

Step by step solution

01

Understand Hyperbolic Functions

Hyperbolic functions, such as \(\cosh\) and \(\tanh\), are analogs of trigonometric functions but for hyperbolas instead of circles. The hyperbolic cosine function is defined as \(\cosh(y) = \frac{e^y + e^{-y}}{2}\) and the hyperbolic tangent function as \(\tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}\). The inverse functions \(\cosh^{-1}(x)\) and \(\tanh^{-1}(x)\) are used to find the value that, when applied to the hyperbolic cosine or tangent, yields \(x\).
02

Prove \(\cosh^{-1}(x)\) Formula

To prove \(\cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})\), start by setting \(y = \cosh^{-1}(x)\). Therefore, \(x = \cosh(y) = \frac{e^y + e^{-y}}{2}\). Multiply both sides by 2 to get \(2x = e^y + e^{-y}\). Rearrange this to form a quadratic expression in terms of \(e^y\): \((e^y)^2 - 2xe^y + 1 = 0\). Solve this quadratic equation using the quadratic formula \(e^y = \frac{2x \pm \sqrt{(2x)^2 - 4}}{2}\). Simplifying gives \(e^y = x \pm \sqrt{x^2 - 1}\). Since \(e^y > 0\), we take the positive root: \(e^y = x + \sqrt{x^2 - 1}\). Taking natural logarithms on both sides yields \(y = \ln(x + \sqrt{x^2 - 1})\).
03

Prove \(\tanh^{-1}(x)\) Formula

For \(\tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\), begin by setting \(y = \tanh^{-1}(x)\), which implies \(x = \tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}\). Cross-multiply to obtain \(x(e^y + e^{-y}) = e^y - e^{-y}\). Rearrange this to \(xe^y + xe^{-y} = e^y - e^{-y}\). Group terms with \(e^y\) and \(e^{-y}\) and factor to get \((1 - x)e^y = (1 + x)e^{-y}\). Multiply by \(e^y\) to eliminate \(e^{-y}\), yielding \((1-x)e^{2y} = (1+x)\). Solve for \(e^{2y}\) to get \(e^{2y} = \frac{1+x}{1-x}\). Take the natural logarithm of both sides: \(2y = \ln\left(\frac{1+x}{1-x}\right)\), and finally, solving for \(y\) gives \(y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbolic Functions
Hyperbolic functions are counterparts to trigonometric functions but instead relate to hyperbolas. They play a crucial role in many mathematical, physics, and engineering applications. The most common hyperbolic functions are the hyperbolic sine, cosine, and tangent, denoted as \(\sinh(x)\), \(\cosh(x)\), and \(\tanh(x)\), respectively. The hyperbolic cosine \(\cosh(y)\) is defined as:- \(\cosh(y) = \frac{e^y + e^{-y}}{2}\) Whereas, the hyperbolic tangent \(\tanh(y)\) is defined as:- \(\tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}\)Inverse hyperbolic functions, such as \(\cosh^{-1}(x)\) and \(\tanh^{-1}(x)\), are used to determine the inputs that result in a given value from the original hyperbolic functions. Essentially, they 'undo' the hyperbolic function, making them highly useful in solving equations involving these functions.
Logarithms
Logarithms are the inverse operations of exponentiation, meaning they answer the question: "To what exponent must a base be raised to obtain a given number?". In simpler terms, they help us work backward from the result of an exponentiation to find the original exponent. Consider the basic logarithm equation:- If \(b^y = x\), then \(\log_b(x) = y\), where \(b\) is the base.Logarithms have widespread uses in solving exponential equations, compressing data scales, and even in financial growth models. They appear frequently in calculus and algebra, allowing for complex problems to be simplified and solved.A specific type of logarithm relevant to hyperbolic functions is the natural logarithm, often denoted as \(\ln(x)\), with the base \(e\). The definition is:- \(\ln(x) = y\), if \(e^y = x\).
Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The solutions to quadratic equations can be found using the well-known quadratic formula:- \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows us to find the roots of the quadratic equation, which are the values of \(x\) that satisfy the equation. In the context of proving hyperbolic function identities, quadratic equations frequently arise. For instance, when rearranging equations involving \(e^y\) terms, we can often reach a quadratic form, requiring us to solve for \(e^y\) using the quadratic formula.
Natural Logarithm
The natural logarithm is a special logarithm with a base of \(e\), where \(e \approx 2.718\). It is a fundamental concept in mathematics, commonly denoted as \(\ln(x)\). It allows us to solve equations of the form \(e^y = x\) by "undoing" the exponential function.Natural logarithms have unique properties such as:- \(\ln(e) = 1\)- \(\ln(1) = 0\)In problems involving hyperbolic functions, the natural logarithm helps simplify and solve identities due to the exponential nature of hyperbolic functions themselves. For instance, when showing that \(\cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})\), we use the natural logarithm to transition from an equation involving \(e^y\) to one that can be expressed in terms of \(y\) itself, demonstrating how integral this tool is in connecting exponential and logarithmic forms.

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Most popular questions from this chapter

Use cylindrical shells to find the volume of the torus obtained by revolving the circle \(x^{2}+y^{2}=a^{2}\) about the line \(x=b,\) where \(b>a>0 .\) IHint: It may help in the integration to think of an integral as an area.]

(a) Show that the total arc length of the ellipse $$x=2 \cos t, \quad y=\sin t \quad(0 \leq t \leq 2 \pi)$$ is given by $$4 \int_{0}^{\pi / 2} \sqrt{1+3 \sin ^{2} t} d t$$ (b) Use a CAS or a scientific calculator with numerical integration capabilities to approximate the arc length in part (a). Round your answer to two decimal places. (c) Suppose that the parametric equations in part (a) describe the path of a particle moving in the \(x y\) -plane, where \(t\) is time in seconds and \(x\) and \(y\) are in centimeters. Use a CAS or a scientific calculator with numerical integration capabilities to approximate the distance traveled by the particle from \(t=1.5\) s to \(t=4.8 \mathrm{s}\) Round your answer to two decimal places.

Assume that a force of \(6 \mathrm{N}\) is required to compress a spring from a natural length of \(4 \mathrm{m}\) to a length of \(3 \frac{1}{2} \mathrm{m} .\) Find the work required to compress the spring from its natural length to a length of \(2 \mathrm{m}\). (Hooke's law applies to compression as well as extension.)

Prove the identities. (a) \(\cosh x+\sinh x=e^{x}\) (b) \(\cosh x-\sinh x=e^{-x}\) (c) \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\) (d) \(\sinh 2 x=2 \sinh x \cosh x\) (c) \(\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y\) (f) \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\) (g) \(\cosh 2 x=2 \sinh ^{2} x+1\) (h) \(\cosh 2 x=2 \cosh ^{2} x-1\)

On August 10.1972 a meteorite with an estimated mass of \(4 \times 10^{6} \mathrm{kg}\) and an estimated speed of \(15 \mathrm{km} / \mathrm{s}\) skipped across the atmosphere above the western United States and Canada but fortunately did not hit the Earth. (a) Assuming that the meteorite had hit the Earth with a speed of \(15 \mathrm{km} / \mathrm{s}\), what would have been its change in kinetic energy in joules (J)? (b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is \(4.2 \times 10^{15} \mathrm{J}\) (c) The energy associated with the Hiroshima atomic bomb was 13 kilotons of TNT. To how many such bombs would the meteorite impact have been equivalent?

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