Problem 66 - Thomas Calculus

Question

The derivative \(d t / d x\) in Example 4 \begin{equation} \begin{array}{c}{\text { a. Show that }} \\\ {f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}} \\ {\text { is an increasing function of } x \text { . }}\\\\{\text { b. Show that }} \\\ {g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}} \\ {\text { is a decreasing function of } x .}\\\\{\text { c. Show that }} \\ {\frac{d t}{d x}=\frac{x}{c_{1} \sqrt{a^{2}+x^{2}}}-\frac{d-x}{c_{2} \sqrt{b^{2}+(d-x)^{2}}}} \\ {\text { is an increasing function of } x .}\end{array} \end{equation}

Step-by-Step Solution

Answer

f(x) is increasing, g(x) is decreasing, and the derivative \( \frac{dt}{dx} \) is increasing.

Step 1: Differentiate f(x)

To show that a function is increasing, we need to show that its derivative is positive. Start by differentiating the function \( f(x) = \frac{x}{\sqrt{a^2 + x^2}} \) with respect to \( x \).Using the quotient rule, where \( u = x \) and \( v = \sqrt{a^2 + x^2} \), gives us:\[ f'(x) = \frac{(v)(du/dx) - (u)(dv/dx)}{v^2} = \frac{\sqrt{a^2 + x^2} - \frac{x^2}{\sqrt{a^2 + x^2}}}{a^2 + x^2}\]Simplifying the numerator, we get:\[f'(x) = \frac{a^2}{(a^2 + x^2)^{3/2}}\]Since \( a^2 > 0 \), \( f'(x) > 0 \) for all \( x \), showing that \( f(x) \) is an increasing function.

Step 2: Differentiate g(x)

Now let's differentiate \( g(x) = \frac{d-x}{\sqrt{b^2 + (d-x)^2}} \) with respect to \( x \).Using the same quotient rule, let \( u = d-x \) and \( v = \sqrt{b^2 + (d-x)^2} \):\[g'(x) = \frac{(\sqrt{b^2 + (d-x)^2})(-1) - (d-x)(\frac{-(d-x)}{\sqrt{b^2 + (d-x)^2}})}{b^2 + (d-x)^2}\]Simplifying gives:\[g'(x) = \frac{-b^2}{(b^2 + (d-x)^2)^{3/2}}\]Since \( b^2 > 0 \), \( g'(x) < 0 \) for all \( x \), proving that \( g(x) \) is a decreasing function.

Step 3: Differentiate the combined function and check

Next, we show that \( \frac{dt}{dx} = \frac{x}{c_1 \sqrt{a^2 + x^2}} - \frac{d-x}{c_2 \sqrt{b^2 + (d-x)^2}} \) is an increasing function.First, differentiate the left part \( \frac{x}{c_1 \sqrt{a^2 + x^2}} \):\[\text{Using a similar method as before, the derivative is } f'(x) = \frac{a^2}{c_1(a^2 + x^2)^{3/2}}\]For the right part \( \frac{d-x}{c_2 \sqrt{b^2 + (d-x)^2}} \):\[g'(x) = -\frac{b^2}{c_2(b^2 + (d-x)^2)^{3/2}}\]Thus, \[\frac{d}{dx}\left(\frac{dt}{dx}\right) = f'(x) - g'(x) = \frac{a^2}{c_1(a^2 + x^2)^{3/2}} + \frac{b^2}{c_2(b^2 + (d-x)^2)^{3/2}}\]Both derivatives are positive, making the whole expression positive, so \( \frac{dt}{dx} \) is increasing.

Key Concepts

Derivative

The concept of a derivative is one of the cornerstones of calculus. In simple terms, a derivative measures how a function changes as its input changes. It provides the rate at which one quantity changes with respect to another. Mathematically, if you have a function \( f(x) \), its derivative is usually denoted as \( f'(x) \) or \( \frac{df}{dx} \). This tells us the slope of the tangent line to the curve of \( f(x) \) at any given point.

When working with derivatives, you are essentially looking at the "instantaneous rate of change". It's like a snapshot of how the function is behaving—whether it's going up, going down, or staying steady, which leads us into our next concept: increasing and decreasing functions.

Increasing and Decreasing Functions

In calculus, we often want to determine whether a function is increasing or decreasing within a particular interval. An increasing function is one where, as \( x \) becomes larger, \( f(x) \) becomes larger as well. Mathematically speaking, if the derivative \( f'(x) > 0 \) for all \( x \) in some interval, then \( f(x) \) is increasing on that interval.

On the other hand, a decreasing function is one where, as \( x \) increases, \( f(x) \) decreases. This occurs when \( f'(x) < 0 \) over the interval you're considering.

To identify these trends, we calculate the derivative and examine its sign across the desired intervals. For example, showing that a derivative \( f'(x) \) is positive across an interval proves the function is increasing in that region.

Quotient Rule

The quotient rule is a technique used to differentiate functions that are presented as one function divided by another. In calculus, it's presented as a particular formula:

\[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]

Here, \( u \) and \( v \) are functions of \( x \), with \( u' \) representing the derivative of \( u \) and \( v' \) the derivative of \( v \). This rule is particularly useful because it gives a clear way to find derivatives of complex fractions by breaking them down systematically. Utilizing this rule, we are able to handle functions like \( f(x) = \frac{x}{\sqrt{a^2 + x^2}} \).

The quotient rule makes differentiation manageable and is critical when analyzing functions described with division.

Differentiation

Differentiation is the process of calculating a derivative. It is a fundamental tool in calculus with vast applications. By performing differentiation, you can find out how functions behave, which can lead to more abstract conclusions about them.

There are numerous rules of differentiation, such as the power rule, product rule, chain rule, and the previously discussed quotient rule. Each of these rules provides a roadmap for deriving new functions from old ones, quickly and accurately.

Differentiation contributes greatly to understanding real-world dynamics. For instance, the accelerated speed of a car or the diminishing returns in economics can be analyzed through this powerful mathematical technique.

Mathematical Proof

To establish the claims made in mathematics, rigorous mathematical proofs are used. A proof shows that a statement is true based on logical reasoning and established knowledge.

In our example, mathematical proof involves demonstrating that a function is increasing or decreasing, based on its derivative. We derived the derivative using established rules, then analyzed its behavior over specific intervals. This serves as a proof of the nature of functions \( f(x) \) and \( g(x) \).

Ensuring the proof is sound, we consider the inequality of the derivative: if the derivative is positive across an interval, it supports the claim that the function is increasing; if negative, then the function is decreasing. A mathematical proof thus relies on precise and logical progression from known principles to our target conclusion.