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Chapter 3: Problem 1

Find the numerical value of each expression. $${ (a) }\sinh 0 \quad \text { (b) } \cosh 0$$

Short Answer

Expert verified
\( \sinh 0 = 0 \) and \( \cosh 0 = 1 \).

Step by step solution

01

Understand the Problem

We need to calculate the hyperbolic sine of 0, which is \( \sinh 0 \), and the hyperbolic cosine of 0, which is \( \cosh 0 \). These functions are similar to the trigonometric sine and cosine but are used in hyperbolic settings.
02

Hyperbolic Sine Formula

The formula for hyperbolic sine is \( \sinh x = \frac{e^x - e^{-x}}{2} \). Let's substitute \( x = 0 \) into this formula.
03

Calculate \( \sinh 0 \)

Substitute \( 0 \) into the formula: \( \sinh 0 = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0 \). Therefore, \( \sinh 0 = 0 \).
04

Hyperbolic Cosine Formula

The formula for hyperbolic cosine is \( \cosh x = \frac{e^x + e^{-x}}{2} \). Let's substitute \( x = 0 \) into this formula.
05

Calculate \( \cosh 0 \)

Substitute \( 0 \) into the formula: \( \cosh 0 = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1 \). Therefore, \( \cosh 0 = 1 \).
06

Conclusion

The hyperbolic sine of 0 is 0, and the hyperbolic cosine of 0 is 1. So, \( \sinh 0 = 0 \) and \( \cosh 0 = 1 \).

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Key Concepts

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

sinh
The hyperbolic sine function, denoted as \( \sinh x \), is a fundamental element in the world of hyperbolic functions. It is defined using exponential functions, specifically as \( \sinh x = \frac{e^x - e^{-x}}{2} \). This formula looks quite similar to the trigonometric sine function, but instead of circular geometry, it applies to hyperbolic geometry.

The hyperbolic sine
  • Has an exponential growth property, rather than a periodic nature like the trigonometric sine.
  • Can take any real number as its input.
  • Returns values that are also real numbers, but they are based on hyperbolic identities.
When calculating \( \sinh 0 \), we substitute 0 into the formula: \( \sinh 0 = \frac{e^0 - e^{-0}}{2} \). This simplifies to \( \frac{1 - 1}{2} = 0 \), illustrating that the hyperbolic sine of 0 is 0.
cosh
The hyperbolic cosine function, or \( \cosh x \), is another key player in hyperbolic functions. The definition also involves exponential functions: \( \cosh x = \frac{e^x + e^{-x}}{2} \). Unlike trigonometric cosine, which oscillates between -1 and 1, the hyperbolic cosine depicts a shape similar to a "U", crucial for modeling hyperbolic paraboloids.

Notable characteristics of \( \cosh x \) include:
  • The minimum value of \( \cosh x \) is 1, occurring at \( x = 0 \).
  • It is always positive, with no negative outputs.
  • It grows exponentially as \( x \) moves away from zero.
To find \( \cosh 0 \), plug 0 into the formula: \( \cosh 0 = \frac{e^0 + e^{-0}}{2} \). This simplifies to \( \frac{1+1}{2} = 1 \), which means the hyperbolic cosine of 0 is 1.
exponential functions
Exponential functions are an essential tool for defining hyperbolic functions like \( \sinh \) and \( \cosh \). Generally expressed as \( e^x \), where \( e \) is approximately 2.71828, these functions exhibit continuous growth and apply to various fields such as biology, finance, and physics.

Here's why exponential functions matter so much:
  • They model real-world phenomena, like population growth and radioactive decay.
  • Provide a basis for solving differential equations.
  • Integral to defining \( \sinh x \) and \( \cosh x \), where \( e^x \) and \( e^{-x} \) portray exponential growth and decay.
Using exponential functions, \( \sinh x \) is calculated as \( \frac{e^x - e^{-x}}{2} \), and \( \cosh x \) as \( \frac{e^x + e^{-x}}{2} \). Understanding how \( e^x \) operates allows us to grasp the mechanics behind More abstract concepts.

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Most popular questions from this chapter

A television camera is positioned 4000 \(\mathrm{ft}\) from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600 \(\mathrm{ft} / \mathrm{s}\) when it has risen 3000 \(\mathrm{ft}\) . (a) How fast is the distance from the television camera to the rocket changing at that moment? (b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

Find the derivative. Simplify where possible. $$y=x^{2} \sinh ^{-1}(2 x)$$

(a) If \(n\) is a positive integer, prove that $$\frac { d } { d x } \left( \sin ^ { n } x \cos n x \right) = n \sin ^ { n - 1 } x \cos ( n + 1 ) x$$ (b) Find a formula for the derivative of \(y = \cos ^ { n } x \cos n x\) that is similar to the one in part (a).

If \(p(x)\) is the total value of the production when there are \(x\) workers in a plant, then the average productivity of the workforce at the plant is $$A(x)=\frac{p(x)}{x}$$ (a) Find \(A^{\prime}(x) .\) Why does the company want to hire more workers if \(A^{\prime}(x)>0 ?\) (b) Show that \(A^{\prime}(x)>0\) if \(p^{\prime}(x)\) is greater than the average productivity.

If \(y = f ( u )\) and \(u = g ( x ) ,\) where \(f\) and \(g\) are twice differen- tiable functions, show that $$\frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { d ^ { 2 } y } { d u ^ { 2 } } \left( \frac { d u } { d x } \right) ^ { 2 } + \frac { d y } { d u } \frac { d ^ { 2 } u } { d x ^ { 2 } }$$
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