Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

y=\cosh x, \text { between } x=0 \text { and } x=4

Step-by-Step Solution

Answer

\( y = \cosh x \) increases from 1 to about 27.309 as \( x \) goes from 0 to 4.

Step 1: Understand Cosh Function

The function \( y = \cosh x \) is known as the hyperbolic cosine function. It is defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \). Unlike trigonometric cosine, it takes all real numbers and always gives values equal to or greater than 1.

Step 2: Evaluate Function at x=0

Substitute \( x = 0 \) into the equation: \( y = \cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1 \). So, \( y = 1 \) when \( x = 0 \).

Step 3: Evaluate Function at x=4

Substitute \( x = 4 \) into the equation: \( y = \cosh(4) = \frac{e^4 + e^{-4}}{2} \). Using approximations, \( e^4 \approx 54.6 \) and \( e^{-4} \approx 0.018 \), so \( \cosh(4) \approx \frac{54.6 + 0.018}{2} \approx 27.309 \).

Step 4: Analyzing Behavior Between x=0 and x=4

The function \( y = \cosh x \) is always increasing because its derivative, \( \sinh x \), is positive for \( x > 0 \). Thus, as \( x \) ranges from 0 to 4, \( \cosh x \) increases from 1 to approximately 27.309.

Key Concepts

Understanding Calculus in Hyperbolic Functions

Calculus plays a vital role in the study and evaluation of functions, including hyperbolic functions. It provides tools to describe and understand the changing behavior of functions. By using calculus, we can not only find the values of functions at particular points but also analyze their progression over intervals.
For the hyperbolic cosine function, or \( y = \cosh x \), a calculus approach involves understanding its derivative and considering its behavior. The derivative of \( \cosh x \) is \( \sinh x \), which helps us determine where \( \cosh x \) is increasing or decreasing.
In this context, between \( x = 0 \) and \( x = 4 \), \( \sinh x \) is positive, indicating that \( \cosh x \) is indeed increasing throughout this range. Knowing how to apply derivatives informs you about the shape and direction of the function, enhancing your understanding of calculus in hyperbolic functions.

Hyperbolic Functions Overview

Hyperbolic functions, such as \( \cosh x \), are analogs to trigonometric functions but are based on hyperbolas rather than circles. They are defined using exponential functions. The hyperbolic cosine is given by \( \cosh x = \frac{e^x + e^{-x}}{2} \).
Unlike its trigonometric counterpart, \( \cos \theta \), which takes values between -1 and 1, \( \cosh x \) always returns a value equal to or greater than 1 for real numbers. This means \( \cosh x \) never dips below 1 on the Cartesian plane, providing unique properties to explore.
Hyperbolic functions appear frequently in different branches of mathematics and applications in physics due to their interesting properties such as rapid growth and real-valued nature, which are leveraged in solving certain types of differential equations and modeling hyperbolic geometry.

Evaluating Functions at Specific Points

Evaluating a function at specific points is a foundational concept in mathematics, allowing us to determine the function's output at designated inputs. This process is key when dealing with hyperbolic functions like \( y = \cosh x \).
To evaluate \( \cosh x \) at \( x = 0 \), substitute the value into the function. \( \cosh(0) = \frac{e^0 + e^{-0}}{2} \) simplifies to 1, as both \( e^0 \) and \( e^0 \) equal 1. Similarly, at \( x = 4 \), \( \cosh(4) = \frac{e^4 + e^{-4}}{2} \), which calculates approximately to 27.309 using typical exponential approximations.
By evaluating functions at specific points, you gain precise snapshots of the function's behavior, which is particularly helpful for plotting graphs and understanding how a function behaves over intervals.