Problem 7 - Calculus: Early Transcendentals

Question

Let \(f(x)=\sin (2 x) .\) If \(a=\pi\) and \(d x=\Delta x=\pi / 100,\) what are \(\Delta y\) and \(d y ?\)

Step-by-Step Solution

Answer

\(\Delta y = \sin\left(\frac{\pi}{50}\right)\) and \(dy = 0\).

Step 1: Understanding the Problem

The problem requires us to find the changes in the function value, \(\Delta y\), and the differential, \(dy\), for the function \(f(x) = \sin(2x)\) at \(a = \pi\) with \(dx = \Delta x = \frac{\pi}{100}\).

Step 2: Calculate \(\Delta y\)

To find \(\Delta y\), we use \(f(x + \Delta x) - f(x)\).First, calculate \(f(a)\):\[f(\pi) = \sin(2 \times \pi) = \sin(2\pi) = 0.\]Next, calculate \(f(a + \Delta x)\):\[f\left(\pi + \frac{\pi}{100}\right) = \sin\left(2\left(\pi + \frac{\pi}{100}\right)\right) = \sin\left(2\pi + \frac{2\pi}{100}\right) = \sin\left(2\pi + \frac{\pi}{50}\right).\]Knowing the periodicity of sine, \(\sin(\theta + 2\pi) = \sin(\theta)\), so:\[\sin\left(2\pi + \frac{\pi}{50}\right) = \sin\left(\frac{\pi}{50}\right).\]Thus, \(\Delta y = \sin\left(\frac{\pi}{50}\right) - 0 = \sin\left(\frac{\pi}{50}\right).\)

Key Concepts

Sine Function

The sine function is a fundamental concept in both trigonometry and calculus. It is represented as \( \sin(\theta) \), where \( \theta \) is an angle measured in radians. The sine function oscillates between -1 and 1, creating a smooth, wave-like pattern. This periodicity is what makes sine such an important function in modeling waves and cyclical patterns in the natural world.

• The sine function is periodic with a period of \( 2\pi \).
• This means that \( \sin(\theta + 2\pi) = \sin(\theta) \).
• Sine has zeros at multiples of \( \pi \), where \( \sin(n\pi) = 0 \) for integer \( n \).

In the exercise provided, the sine function is modified to \( f(x) = \sin(2x) \), which means the function has a period of \( \pi \). This is because multiplying \( x \) by 2 compresses the standard sine wave horizontally. Thus, its periodicity is affected, making it crucial to understand the domain transformations that occur.

Differential Approximation

Differential approximation is a method in calculus used to estimate small changes in function values. By using the derivative, we can approximate how much a function's value changes as its input changes by a small amount. This is particularly useful because calculating exact values can be complex.

• For a given function \( f(x) \), the differential \( dy \) can be calculated as \( f'(a) \cdot dx \).
• The derivative \( f'(x) \) of a function at a point \( a \) represents the rate at which the function value changes.

In our problem, for \( f(x) = \sin(2x) \), the derivative \( f'(x) \) is \( 2\cos(2x) \). At \( a = \pi \), substitute to find the specific derivative: \( f'(\pi) = 2\cos(2\pi) = 2 \).
The differential \( dy \) for \( dx = \frac{\pi}{100} \) is calculated as: \[ dy = 2 \cdot \frac{\pi}{100} = \frac{2\pi}{100} = \frac{\pi}{50}. \] This small \( dy \) value is used to approximate changes swiftly and effectively.

Change in Function Value

The change in function value, \( \Delta y \), represents how much a function's output changes when its input changes. Unlike differential approximation, \( \Delta y \) doesn't approximate but instead calculates the actual change by evaluating the function at different inputs.
To compute \( \Delta y \), we use the formula:\[\Delta y = f(x + \Delta x) - f(x).\]
For the function \( f(x) = \sin(2x) \) at \( a = \pi \), we have:\[f\left(\pi + \frac{\pi}{100}\right) - f(\pi) = \sin\left(\frac{\pi}{50}\right) - 0 = \sin\left(\frac{\pi}{50}\right).\]

• \( \Delta y \) is exact and shows how the sine function responds to small input changes.
• The exact value expression represents the true nature of the function's behavior over small intervals.

This highlights the behavior of \( \sin \), especially its reaction to small deviations from established nodes, like \( n\pi \), where the function consistently transitions between decreasing and increasing values.