Problem 8 - Calculus Single Variable

Question

Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=\sin (x), c=0, x=0.02 $$

Step-by-Step Solution

Answer

The estimated value of \( f(0.02) \) is approximately 0.02.

Step 1: Understand the Increment Formula

The increment formula is a way to approximate the change in a function's value for a small change in the input. For a function \( f(x) \), the increment (or change) is given by \( \Delta y \approx f'(c) \Delta x \), where \( c \) is the known value, \( \Delta x = x - c \), and \( f'(c) \) is the derivative of \( f(x) \) evaluated at \( c \).

Step 2: Find the Derivative of f(x)

The function is \( f(x) = \sin(x) \). The derivative, \( f'(x) \), of the sine function is \( \cos(x) \). So, \( f'(x) = \cos(x) \).

Step 3: Evaluate the Derivative at c

Substitute \( c = 0 \) into the derivative we found. This gives \( f'(0) = \cos(0) = 1 \).

Step 4: Calculate the Increment \( \Delta x \)

Find \( \Delta x = x - c = 0.02 - 0 = 0.02 \).

Step 5: Calculate \( \Delta y \) Using the Increment Formula

Use the formula \( \Delta y \approx f'(c) \Delta x \), substitute the values we found to get \( \Delta y \approx 1 \times 0.02 = 0.02 \).

Step 6: Estimate \( f(x) \)

The initial known value is \( f(c) = f(0) = \sin(0) = 0 \). Thus, \( f(x) \approx f(c) + \Delta y \). Substitute the known values: \( f(0.02) \approx 0 + 0.02 = 0.02 \).

Key Concepts

Increment Formula

The Increment Formula is a powerful mathematical tool used to estimate how much a function's value changes as the input changes by a small amount. This method is particularly useful when you want to estimate the function's value near a specific point, especially when finding an exact solution is challenging.
Simply put, the increment of a function, represented as \( \Delta y \), is approximately equal to the product of its derivative at a known point \( c \) and the small change in \( x \), denoted as \( \Delta x \). More formally, we can write it as:

• \( \Delta y \approx f'(c) \Delta x \)

This approach assumes that the change in the function is linear over very small intervals, making it a handy approximation method when precision over small changes is acceptable.
To use the Increment Formula effectively, it's essential to know both the derivative of the function and the exact point around which you're estimating the change.

Derivative of Trigonometric Functions

Understanding the derivative of trigonometric functions, like sine, is crucial when using the Increment Formula. The derivative tells us the rate at which the function's value is changing at any given point. For the sine function, \( f(x) = \sin(x) \), its derivative is the cosine function:

• \( f'(x) = \cos(x) \)

This means that at any point \( x \), the rate of change of \( \sin(x) \) is given by \( \cos(x) \). In the context of the exercise, you need to evaluate this derivative at the known point \( c \).
If \( c = 0 \), then we substitute into the derivative equation:

• \( f'(0) = \cos(0) = 1 \)

This result indicates that at \( x = 0 \), the sine function is increasing at a rate of 1. Such details are pivotal for accurately employing the Increment Formula.

Estimating Function Values

Estimating the value of a function using the Increment Formula involves a series of calculated steps, ensuring precision as much as possible with available information. Let's break down the process of estimating \( f(x) \) at a new point \( x \), given the value at \( c \): Firstly, compute \( \Delta x \), the small change in your input, which is given by \( x - c \). In our exercise, \( x = 0.02 \) and \( c = 0 \), so \( \Delta x = 0.02 \). This tells us how far we're estimating from the known point.
Knowing \( \Delta x \), use the Increment Formula to calculate \( \Delta y \), which approximates how much \( f(x) \) changes from \( f(c) \). The calculation, as shown, is:

• \( \Delta y \approx f'(c) \Delta x = 1 \times 0.02 = 0.02 \)

Finally, find \( f(x) \) by adding \( \Delta y \) to \( f(c) \):

• \( f(0.02) \approx f(0) + 0.02 = 0 + 0.02 = 0.02 \)

These estimations provide rapid and efficient ways to calculate values of functions that would otherwise require intricate calculations. By understanding these steps, you will not only improve your estimative skills but also enhance your comfort with calculus-oriented problems.