Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Argue that \(\sin x\) is an increasing function on \(0 \leq x \leq \pi / 2\).

Step-by-Step Solution

Answer

\( \sin x \) is increasing on \([0, \frac{\pi}{2}]\) because its derivative, \( \cos x \), is non-negative and continuous on this interval.

Step 1: Understand the Problem

We need to show that the function \( \sin x \) is increasing on the interval \( 0 \leq x \leq \frac{\pi}{2} \). This means that for any two points \( x_1 \) and \( x_2 \) within \( [0, \frac{\pi}{2}] \), if \( x_1 < x_2 \), then \( \sin x_1 \leq \sin x_2 \).

Step 2: Consider the Derivative of \( \sin x \)

The function \( \sin x \) has a derivative \( \cos x \). If the derivative \( \cos x \) is positive on the interval \( [0, \frac{\pi}{2}] \), it indicates that \( \sin x \) is increasing there.

Step 3: Analyze the Derivative on the Given Interval

On the interval \( [0, \frac{\pi}{2}] \), \( \cos x \) takes values from \( \cos 0 = 1 \) to \( \cos \frac{\pi}{2} = 0 \). Since \( \cos x \) remains non-negative and only decreases from 1 to 0, \( \sin x \) is always increasing on this interval because \( \cos x \) remains positive or zero.

Step 4: Conclusion from the Derivative Analysis

Since \( \cos x > 0 \) for all \( x \) in \( (0, \frac{\pi}{2}) \), the function \( \sin x \) is strictly increasing. At the boundaries, \( x = 0 \) and \( x = \frac{\pi}{2} \), \( \sin x \) remains increasing due to continuity. Therefore, \( \sin x \) is increasing on \( [0, \frac{\pi}{2}] \).

Key Concepts

Trigonometric Functions

Understanding trigonometric functions is key to analyzing various phenomena in calculus. Sine (\( \sin x \)) and cosine (\( \cos x \)) are fundamental trigonometric functions used to describe the relationships in right-angled triangles, circles, and oscillations.

The sine function, particularly in the context of calculus, is often examined across different intervals to understand its behavior. On the interval \([0, \frac{\pi}{2}]\), the sine function has clear properties that make it a focus of study. Generally, sine represents the vertical component of a point on the unit circle. As we start at \(x=0\) and move towards \(x=\frac{\pi}{2}\), the value of \(\sin x\) gradually increases from 0 to 1.

This behavior sets the stage for why one investigates its derivative, cosine, on the same interval. Knowing how \( \cos x \) behaves can tell us a lot about the increasing or decreasing nature of \( \sin x \). This is because the rate of change of the sine—illustrated by its derivative—is directly linked to how the sine function evolves as \( x \) progresses from 0 to \( \frac{\pi}{2} \).

Derivative Analysis

Derivative analysis involves examining how a function's rate of change behaves within a specific interval. The derivative of \( \sin x \), which is \( \cos x \), is a crucial tool for this analysis.

To determine whether a function like \( \sin x \) is increasing or decreasing, we look at the sign of its derivative over the chosen interval, in this case, \([0, \frac{\pi}{2}]\). If the derivative is positive throughout this interval, the function is increasing there.

• At \( x = 0 \), we have \( \cos 0 = 1 \). This indicates a positive initial rate of change, suggesting that \( \sin x \) begins increasing.
• As \( x \) approaches \( \frac{\pi}{2} \), \( cos x \) approaches 0, but remains non-negative, indicating that the rate of change of \( \sin x \) slows but does not reverse.

The fact that \( \cos x \) never becomes negative in the open interval \((0, \frac{\pi}{2})\) assures us that \( \sin x \) is strictly increasing in that portion of the interval.

This is because positive derivative values indicate an upward slope of the sine curve, meaning that for any two points \( x_1 < x_2 \) within the interval, \( \sin x_1 \leq \sin x_2 \).

Monotonic Functions

A monotonic function is one that either never increases or never decreases throughout its domain. Understanding whether a function is monotonic over a certain interval like \([0, \frac{\pi}{2}]\) for \( \sin x \) can simplify analysis significantly.

In the context of the exercise, the sine function is shown to be strictly increasing over its considered interval. This monotonic increasing behavior is characterized by the function's continuous rise without any dips or falls.

Monotonic behavior has several implications in calculus:

• Predictability: If a function is known to be increasing, any movement in its input will result in a predictable movement in its output.
• Uniqueness: An increasing function over a specific interval implies that each input corresponds to a unique output, making it invertible in theory.

Recognizing \( \sin x \) as an increasing function over \([0, \frac{\pi}{2}]\) allows us to leverage these properties in related calculus problems and real-world applications.