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Chapter 10: Problem 27

Two important functions in engineering are \(\sinh x=\frac{e^{x}-e^{-x}}{2} \quad\) hyperbolic sine function \(\cosh x=\frac{e^{x}+e^{-x}}{2} \quad\) hyperbolic cosine function. Show that the following expression can be greatly simplified: $$ (\cosh x)^{2}-(\sinh x)^{2} $$

Short Answer

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Answer: 1

Step by step solution

01

Write down the given expressions

The hyperbolic sine and cosine functions are given by: $$ \sinh x = \frac{e^{x} - e^{-x}}{2} $$ $$ \cosh x = \frac{e^{x} + e^{-x}}{2} $$
02

Square the expressions

Let's square both functions: $$ (\sinh x)^2 = \left(\frac{e^{x} - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4} $$ $$ (\cosh x)^2 = \left(\frac{e^{x} + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} $$
03

Subtract the squared expressions

Now subtract \((\sinh x)^2\) from \((\cosh x)^2\): $$ (\cosh x)^2 - (\sinh x)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} $$
04

Simplify

Combining the fractions and simplifying, we get: $$ (\cosh x)^2 - (\sinh x)^2 = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} $$ Finally, we have: $$ (\cosh x)^2 - (\sinh x)^2 = 1 $$

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Key Concepts

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

Hyperbolic Sine
The hyperbolic sine function, commonly denoted as \( \sinh x \), closely resembles trigonometric sine but relates to the hyperbolic geometry instead. It plays a vital role in both mathematics and engineering. The formula for hyperbolic sine is given by:\[\sinh x = \frac{e^x - e^{-x}}{2}\]This formula utilizes the exponential function, where \( e^x \) is the exponential growth and \( e^{-x} \) represents the exponential decay. These components help to reveal unique properties of the curve represented by \( \sinh x \).

Hyperbolic sine functions can represent scenarios involving dynamic systems such as electrical circuits or mechanical movement, where parameters change over continuous time in a non-linear fashion.
Hyperbolic Cosine
The hyperbolic cosine function, or \( \cosh x \), is a counterpart to the hyperbolic sine and is similarly valued in mathematical applications. Its equation is:\[\cosh x = \frac{e^x + e^{-x}}{2}\]Just like its sine counterpart, it employs exponential terms but adds them. This function is notable for not only its graph resembling the shape of a catenary, found commonly when describing the shape of a hanging chain or cable, but also for its ability to simplify complex equations.
  • Acts as the natural balancing point between growth and decay.
  • Is always positive due to the nature of the exponential terms being added.
The hyperbolic cosine is widely used where smooth transitions are required, such as in computer graphics and the modeling of natural phenomena.
Trigonometric Identities
Trigonometric identities form the backbone of simplifying expressions involving trigonometric and hyperbolic functions. One essential identity involving hyperbolic functions is:\[(\cosh x)^2 - (\sinh x)^2 = 1\]This identity plays a pivotal role in mathematical derivations and solutions, echoing the classic Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \) in traditional trigonometry. It essentially provides a tool to transform or simplify complex forms into workable, manageable pieces. The identity states that no matter the value of \( x \), when you subtract the square of \( \sinh x \) from the square of \( \cosh x \), the result is always 1.
  • Indicative of the intrinsic relationship between hyperbolic sine and cosine.
  • Useful for ensuring correctness in derivations involving hyperbolic terms.
Understanding and using these identities is an essential skill for students dealing with hyperbolic functions in calculus and beyond.

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Most popular questions from this chapter

The value of an investment grows by \(0.06 \%\) every day. By what percent does it increase in a year?

Between 1994 and \(1999,\) the national health expenditures in the United States were rising at an average of 5.3\% per year. The U.S. health expenditures in 1994 were 936.7 billion dollars. (a) Express the national health expenditures, \(P\), in billions of dollars, as a function of the year, \(t,\) with \(t=0\) corresponding to the year \(1994 .\) (b) Use this model to estimate the national health expenditures in the year 1999. Compare this number to the actual 1999 expenditures, which were 1210.7 billion dollars.

A lab receives a 1000 grams of an unknown radioactive substance that decays at a rate of \(7 \%\) per day. (a) Write an expression for \(Q,\) the quantity of substance remaining after \(t\) days. (b) Make a table showing the quantity of the substance remaining at the end of 8,9,10,11,12 days. (c) For what values of \(t\) in the table is the quantity left (i) Less than \(500 \mathrm{gm}\) ? (ii) More than \(500 \mathrm{gm}\) ? (d) A lab worker says that the half-life of the substance is between 11 and 12 days. Is this consistent with your table? If not, how would you correct the estimate?

Solve the equations in Problem using the following approximations: $$ 10^{0.301}=2, \quad 10^{0.477}=3, \quad 10^{0.699}=5 . $$ Example. Solve \(10^{x}=6 .\) Solution. We have $$ \begin{aligned} 10^{x} &=6 \\ &=2 \cdot 3 \\ &=10^{0.301} \cdot 10^{0.477} \\ &=10^{0.301+0.477} \\ &=10^{0.778} \\ \text { so } x=0.778 . \end{aligned} $$ \(10^{x}=32\)

State the starting value \(a\), the growth factor \(b\), and the growth rate \(r\) as a percent correct to 2 decimals for the exponential functions. . \(V=2000(1.0058)^{t}\)
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