Problem 35 - Calculus Early Transcendentals

Question

(a) Show that if \(f\) and \(g\) are functions for which $$ f^{\prime}(x)=g(x) \text { and } g^{\prime}(x)=-f(x) $$ for all \(x,\) then \(f^{2}(x)+g^{2}(x)\) is a constant. (b) Give an example of functions \(f\) and \(g\) with this property.

Step-by-Step Solution

Answer

(a) \( f^2(x) + g^2(x) \) is constant because \( h'(x) = 0 \). (b) Example: \( f(x) = \sin(x), g(x) = \cos(x) \).

Step 1: Define the function

First, consider the function \( h(x) = f^2(x) + g^2(x) \). Our goal is to determine whether this function is constant.

Step 2: Differentiate the function

Differentiate \( h(x) = f^2(x) + g^2(x) \) with respect to \( x \). Using the chain rule, we get:\[ h'(x) = 2f(x)f'(x) + 2g(x)g'(x) \]

Step 3: Substitute given relationships

Replace \( f'(x) \) with \( g(x) \) and \( g'(x) \) with \( -f(x) \) as given in the problem statement. Substituting these into the differentiated function gives:\[ h'(x) = 2f(x)g(x) + 2g(x)(-f(x)) \]

Step 4: Simplify the derivative

Simplify \( h'(x) = 2f(x)g(x) - 2g(x)f(x) \). Notice that this results in:\[ h'(x) = 0 \].

Step 5: Interpret the result

Since \( h'(x) = 0 \) for all \( x \), this means that \( h(x) = f^2(x) + g^2(x) \) is constant. This proves part (a) of the problem.

Step 6: Example of functions \( f \) and \( g \)

Consider \( f(x) = \sin(x) \) and \( g(x) = \cos(x) \). Their derivatives are \( f'(x) = \cos(x) \) and \( g'(x) = -\sin(x) \), which satisfy the conditions \( f'(x) = g(x) \) and \( g'(x) = -f(x) \). Therefore, these functions are an example for part (b).

Key Concepts

Differentiation

In calculus, differentiation is one of the fundamental concepts. It allows us to find the rate at which a function is changing at any point. In simpler terms, differentiation provides us the derivative of a function. The derivative shows how much a function's output changes as its input changes slightly.

In our original exercise, we have functions:

• \( f'(x) = g(x) \)
• and \( g'(x) = -f(x) \)

These derivatives show the relationship between the functions \( f(x) \) and \( g(x) \). When we differentiated the function \( h(x) = f^2(x) + g^2(x) \), it helped us to determine whether this function, \( h(x) \), is constant.

When you differentiate a function, you use various rules and techniques to make it easier. For example, in this exercise, we utilized both chain rule and knowledge of trigonometric functions to simplify and find the result.

Chain Rule

The Chain Rule is a valuable technique in calculus, especially when dealing with composite functions. It allows one to differentiate a function that is nested within another without unwieldy expansions of expressions.

Think of the chain rule as a way to 'unravel' a function within another function. It's particularly useful when you have functions of the form \( f(g(x)) \). The Chain Rule states:

• The derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).

In this problem, when we differentiated \( h(x) = f^2(x) + g^2(x) \), the Chain Rule allows us to differentiate each part of the function like this:

• \( 2f(x)f'(x) \)
• \( 2g(x)g'(x) \)

This way, we could easily substitute the given relationships between \( f'(x) = g(x) \) and \( g'(x) = -f(x) \) to find that the derivative of \( h(x) \) was \( 0 \). Hence, \( h(x) \) is constant.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are extremely important in calculus and beyond, serving as fundamental building blocks for a variety of mathematical applications.

In the example provided in the exercise, \( f(x) = \sin(x) \) and \( g(x) = \cos(x) \) were chosen to satisfy the conditions laid out in the problem:

• \( f'(x) = \cos(x) = g(x) \)
• \( g'(x) = -\sin(x) = -f(x) \)

This illustrates a beautiful property of these functions: their derivatives are also trigonometric functions with cyclical relationships. For instance, if you differentiate \( \sin(x) \), you get \( \cos(x) \), and differentiating \( \cos(x) \) gives \( -\sin(x) \).

Trigonometric functions often help in simplifying complex calculus problems, especially when they appear in integrals, derivatives, or when modeling cyclic behavior.