Problem 12 - 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 \(f(c)\) $$ f(x)=\cot (x), c=\pi / 3, x=1 $$

Step-by-Step Solution

Answer

Estimate \( f(1) \approx \frac{1}{\sqrt{3}} - \frac{4}{3}(1 - \frac{\pi}{3}) \).

Step 1: Understand the Problem

We need to estimate the value of \( f(x) = \cot(x) \) at \( x = 1 \) using increments and the known value of \( f(\frac{\pi}{3}) \). We are given \( c = \frac{\pi}{3} \) and \( x = 1 \).

Step 2: Calculate Known Value

Find \( f(c) \) where \( c = \frac{\pi}{3} \). We have \( f(c) = \cot(\frac{\pi}{3}) \). Since \( \cot(x) = \frac{1}{\tan(x)} \) and \( \tan(\frac{\pi}{3}) = \sqrt{3} \), therefore \( f(c) = \frac{1}{\sqrt{3}} \).

Step 3: Compute Increment

Calculate the increment \( \Delta x = x - c \). Given \( x = 1 \) and \( c = \frac{\pi}{3} \), the increment \( \Delta x = 1 - \frac{\pi}{3} \).

Step 4: Find Derivative

Calculate the derivative \( f'(x) \) of \( f(x) = \cot(x) \). The derivative is \( f'(x) = -\csc^2(x) \). Evaluate at \( c = \frac{\pi}{3} \): \( f'(\frac{\pi}{3}) = -\csc^2(\frac{\pi}{3}) = -\frac{4}{3} \).

Step 5: Apply Linear Approximation

Estimate \( f(x) \) using the linear approximation formula: \( f(x) \approx f(c) + f'(c) \Delta x \). Use \( f(c) = \frac{1}{\sqrt{3}} \), \( f'(c) = -\frac{4}{3} \), and \( \Delta x = 1 - \frac{\pi}{3} \).

Step 6: Calculate the Approximation

Substitute the values into the approximation formula: \( \frac{1}{\sqrt{3}} + \left(-\frac{4}{3}\right)\left(1 - \frac{\pi}{3}\right) \). Simplify this expression to find the estimated value of \( f(x) \).

Key Concepts

Linear Approximation

Linear approximation is a technique used to estimate the value of a function at a particular point, using the tangent line at a nearby point. This method can be very useful when you need quick estimates and don't have access to a calculator for exact values.

The basic idea is to take the function value and its derivative at a point you're familiar with (let's call it 'c') to approximate an unknown value at another point 'x.' Here's what you need:

• The function value at 'c', written as \( f(c) \).
• The derivative of the function at 'c', known as \( f'(c) \).
• The difference between 'x' and 'c', which we label as \( \Delta x = x - c \).

The approximation formula is:\[ f(x) \approx f(c) + f'(c) \Delta x \]This formula gives you an estimate of \( f(x) \) that's easy to calculate, especially for functions that are smooth and well-behaved around 'c'. With linear approximation, we essentially assume that the function behaves like its tangent line over small intervals, providing insightful but rapid estimates.

Trigonometric Functions

Trigonometric functions are often involved in situations like this because they help describe periodic phenomena, among other things. The cotangent function, denoted as \( \cot(x) \), is one of these and is defined as the reciprocal of the tangent function: \[\cot(x) = \frac{1}{\tan(x)}\]This function is particularly interesting because it behaves like a wave, repeating its pattern over intervals.

For the angles we often use, like \( \frac{\pi}{3} \), the values of common trigonometric functions are well-established. For example:

• \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \)
• Thus, \( \cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}} \)

Recognizing these standard values can help solve problems quickly by understanding how these functions behave at key points, thus aiding in processes like the linear approximation through clear, descriptive increments.

Increments Method

The increments method relies on small changes (increments) in the independent variable to estimate change in the dependent variable. By expressing change in terms of \( \Delta x \), the increments method connects closely with the concept of derivatives and linear approximation.

In practice, it involves:

• Determining \( \Delta x \) as the difference between your point of interest, 'x,' and a base point 'c' (\( \Delta x = x - c \)).
• Calculating the function's derivative to understand how changes in 'x' affect changes in 'f(x)'.
• Using these two factors in the linear approximation formula to estimate the function at 'x'.

For our problem, with \( f(x) = \cot(x) \), this method helps simplify finding the approximate value at \( x = 1 \). Given the derivative, \( f'(x) = -\csc^2(x) \), calculated at a known point, incrementing it smoothly extends the understanding of \( f(x) \) from \( \pi/3 \) to any nearby point, like 1. This approach reduces complex calculations into manageable, smaller problems by focusing on how small stretches change the larger picture of the function.