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Find the numerical value of each expression. $${ (a) }\cosh 3 \quad \text { (b) } \cosh (\ln 3)$$

Short Answer

Expert verified
(a) \( \cosh 3 \approx 10.06765 \), (b) \( \cosh (\ln 3) \approx 1.6665 \).

Step by step solution

01

Understanding Hyperbolic Cosine

The hyperbolic cosine function is defined as \( \cosh x = \frac{e^x + e^{-x}}{2}\) . Our task for part (a) is to evaluate \(\cosh 3\) using this definition.
02

Calculate \( \cosh 3 \)

We substitute \( x = 3 \) into the formula for hyperbolic cosine:\[ \cosh 3 = \frac{e^3 + e^{-3}}{2} \]Next, calculate \( e^3 \approx 20.0855 \) and \( e^{-3} \approx 0.0498 \). Substituting these values, we get:\[ \cosh 3 = \frac{20.0855 + 0.0498}{2} \approx \frac{20.1353}{2} \approx 10.06765 \]
03

Evaluate \( \cosh (\ln 3) \)

We substitute \( x = \ln 3 \) in the definition of hyperbolic cosine:\[ \cosh (\ln 3) = \frac{e^{\ln 3} + e^{-\ln 3}}{2} \]Use properties of logarithms and exponents, where \( e^{\ln 3} = 3 \) and \( e^{-\ln 3} = \frac{1}{3} \). Hence:\[ \cosh (\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{3.333}{2} = 1.6665 \]

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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, denoted as \( \cosh{x} \), has a pivotal role when working with hyperbolic functions. It shares similarities with trigonometric cosine but is distinct in its definition and application. The hyperbolic cosine of a number \( x \) is calculated using the formula:\[\cosh{x} = \frac{e^x + e^{-x}}{2}\]This mathematical expression involves the use of exponential functions. To find a specific value like \( \cosh{3} \), simply substitute \( x = 3 \) into the formula. The result involves computing two exponential terms: \( e^3 \) and \( e^{-3} \). These values are then summed and divided by 2 for the final answer.
  • \( e^3 \approx 20.0855 \)
  • \( e^{-3} \approx 0.0498 \)
Thus, \( \cosh{3} \approx 10.06765 \). This process illustrates how hyperbolic cosine is connected to exponential growth and decay.
Natural Logarithm
The natural logarithm, represented as \( \ln{x} \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It's a fundamental concept used to reverse the effects of exponential growth. In mathematical expressions, \( \ln{x} \) tells us the power to which \( e \) must be raised to obtain the number \( x \).

When calculating \( \cosh(\ln{3}) \), understanding the properties of natural logarithms is key. These properties simplify the expression by converting exponents and logs effectively.
  • \( e^{\ln{3}} = 3 \) (since the exponential and logarithm functions are inverses)
  • \( e^{-\ln{3}} = \frac{1}{3} \)
By substituting these values, \( \cosh(\ln{3}) \) simplifies to \( \frac{3 + \frac{1}{3}}{2} \), resulting in \( 1.6665 \). This eases the computation and showcases the powerful interplay between exponentials and logarithms.
Exponential Function
Exponential functions are essential in mathematics, especially in the context of calculating hyperbolic functions and logarithms. The exponential function \( e^x \) denotes the constant \( e \) raised to the power of \( x \). It serves as the backbone of continuous growth processes and shapes a wide range of mathematical equations.

The defining characteristic of the exponential function is that its rate of growth is directly proportional to its current value. This intrinsic quality is captured in the way we calculate hyperbolic cosine, where both \( e^x \) and \( e^{-x} \) are used to form the equation:
\[\cosh{x} = \frac{e^x + e^{-x}}{2}\]
  • For \( x = 3 \), we calculate \( e^3 \approx 20.0855 \)
  • For \( x = \ln{3} \), \( e^{\ln{3}} = 3 \) and \( e^{-\ln{3}} = \frac{1}{3} \)
This representation of exponential functions allows not just for computation of hyperbolic cosines, but also provides insights into logarithmic transformations. Understanding these objectives highlights how exponential and logarithmic functions intertwine, offering robust solutions in calculus and algebra.

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Most popular questions from this chapter

If \(R\) denotes the reaction of the body to some stimulus of strength \(x,\) the sensitivity \(S\) is defined to be the rate of change of the reaction with respect to \(x\) . A particular example is that when the brightness \(x\) of a light source is increased, the eye reacts by decreasing the area \(R\) of the pupil. The experimental formula $$R=\frac{40+24 x^{0.4}}{1+4 x^{0.4}}$$ has been used to model the dependence of \(\mathrm{R}\) on \(\mathrm{x}\) when \(\mathrm{R}\) is measured in square millimeters and \(\mathrm{x}\) is measured in appropriate units of brightness. (a) Find the sensitivity. (b) Illustrate part (a) by graphing both \(\mathrm{R}\) at low levels of of \(\mathrm{x} .\) Comment on the values of \(\mathrm{R}\) and \(\mathrm{S}\) at low levels of brightness. Is this what you would expect?

Establish the following rules for working with differentials (where c denotes a constant and u and \(v\) are functions of \(x )\) . (a) \(\mathrm{dc}=0\) $$ \mathrm{d}(\mathrm{u}+v)=\mathrm{du}+\mathrm{d} v \quad \text { (d) } \mathrm{d}(\mathrm{u} v)=\mathrm{u} \mathrm{d} v+v $$ $$ \mathrm{d}\left(\frac{\mathrm{u}}{v}\right)=\frac{v \mathrm{du}-\mathrm{u} \mathrm{d} v}{v^{2}} \quad \text { (f) } \mathrm{d}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \mathrm{d} \mathrm{x} $$

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A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 \(\mathrm{m}\) higher than the bow of the boat. If the rope is pulled in at a rate of \(1 \mathrm{m} / \mathrm{s},\) how fast is the boat approaching the dock when it is 8 \(\mathrm{m}\) from tho dock?

Find the derivative. Simplify where possible. $$y=\operatorname{sech}^{-1} \sqrt{1-x^{2}}, \quad x>0$$

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