Problem 16 - Advanced Calculus

Question

Suppose that \(f: \mathbb{R} \rightarrow \mathbb{R}\) and \(g: \mathbb{R} \rightarrow \mathbb{R}\) are each differentiable and that $$ \left\\{\begin{array}{l} f^{\prime}(x)=g(x) \text { and } \quad g^{\prime}(x)=-f(x) \quad \text { for all } x \\ f(0)=0 \quad \text { and } \quad g(0)=1 \end{array}\right. $$ Prove that $$ [f(x)]^{2}+[g(x)]^{2}=1 $$ for all \(x\). (Hint: Define \(h(x) \equiv[f(x)]^{2}+[g(x)]^{2}\) for all \(x\). Show that \(h: \mathbb{R} \rightarrow \mathbb{R}\) is a constant function.)

Step-by-Step Solution

Answer

Through defining the function \(h(x) = [f(x)]^{2} + [g(x)]^{2}\), differentiating it, using the given initial conditions, and demonstrating that its derivative is 0, we showed that \(h(x)\) is constant for all x, specifically \(h(x)=1\). Hence, \([f(x)]^{2} + [g(x)]^{2} = 1\) for all x.

Step 1: Define the function h(x)

Let's define a new function as suggested by the hint: $$h(x)=f(x)^2+g(x)^2$$We will show that this function is constant by proving that its derivative, h'(x), is 0 for all x.

Step 2: Find the derivative of h(x)

We differentiate h(x) with respect to x using the power rule and the chain rule. Since h(x) is a sum of functions, we differentiate each term separately:$$ h'(x) = \frac{d}{dx}[f(x)^2] + \frac{d}{dx}[g(x)^2] $$Applying the chain rule, this yields:$$ h'(x) = 2f(x)f'(x) + 2g(x)g'(x) $$Substituting the given information that \(f'(x) = g(x)\) and \(g'(x) = -f(x)\), we get:$$ h'(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 0 $$

Step 3: Conclude that h(x) is a constant function

Since the derivative of h(x) is zero, we conclude that h(x) is a constant function. That means for all x in the domain, h(x) has the same value. We will use the given initial conditions to find the value of this constant.

Step 4: Use the initial conditions to find the constant value of h(x)

Given that \(f(0) = 0\) and \(g(0) = 1\), we evaluate h(x) at x = 0 to find the constant value. Plugging the initial conditions into the definition of h(x) gives us:$$ h(0) = f(0)^2 + g(0)^2 = 0^2 + 1^2 = 1 $$Since h(x) is constant for all x, we must have \(h(x) = 1\) for all x.

Step 5: Conclude the proof

Thus, we've shown that the function \(h(x) = f(x)^2 + g(x)^2\) is equal to 1 for all x, which completes the proof that:$$ [f(x)]^{2} + [g(x)]^{2} = 1 $$

Key Concepts

Derivative

The derivative of a function represents the rate at which the function's value changes with respect to a change in its input value. Essentially, it measures how a function reacts to small changes in its input and is a foundational concept in differential calculus.

When we write the derivative of a function, say, \( f(x) \), as \( f'(x) \), we are referring to its instantaneous rate of change at any point \( x \). Derivatives can be found using various rules of differentiation, such as the power rule, which is handy when dealing with monomials like \( x^n \), where the derivative is \( nx^{n-1} \).

In the context of our given exercise, derivatives help us analyze the behavior of the functions \( f(x) \) and \( g(x) \) and, consequently, the function \( h(x) \), by allowing us to express changes in \( h(x) \) quantitatively. It is through the derivative that we can find the connection between these functions and solve the problem at hand.

Chain Rule

Differential calculus is not just about functions in their simplest forms; oftentimes, we deal with composite functions. The chain rule is a method we use to differentiate composite functions – functions made by combining several other functions.

The beauty of the chain rule is that we can break down the complex function into its constituent parts, differentiate each part, and then piece them together to find the derivative of the whole. The chain rule formula is:\[ \left( f(g(x)) \right)' = f'(g(x))g'(x) \].

Let's apply this knowledge to the function \( h(x) = f(x)^2 + g(x)^2 \) from our exercise. To find the derivative of \( h(x) \), we consider \( f(x)^2 \) and \( g(x)^2 \) as composite functions where an outer function \( u^2 \) is combined with inner functions \( u = f(x) \) and \( u = g(x) \), respectively. By utilizing the chain rule, we can find the derivative of \( h(x) \) in a stepwise manner, as demonstrated in the solution. Understanding and applying the chain rule is vital for such problems where functions are intertwined.

Constant Function

A constant function is a type of function that always returns the same value, no matter the input. In mathematical terms, a function \( c(x) \) is constant if the derivative \( c'(x) \) is zero for all values of \( x \) in the domain of the function.

The exercise shows us that if the derivative of a function is zero across its entire domain, the function must be constant. This is why after differentiating \( h(x) \) and finding that its derivative is zero, we determine \( h(x) \) to be constant. A constant function, quite literally, does not change; it's a straight horizontal line on a graph.

In solving the original problem, we hinge on this property of a constant function. By finding out that the derivative of \( h(x) \) is zero and knowing two initial conditions, we can assert the constant value of \( h(x) \), which is necessary to prove our final result. The concept of the constant function is key to unlocking complex problems like this where we are trying to prove constancy across an entire domain.