Problem 14 - 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 \(f(c)\) $$ f(x)=\csc (\pi x / 4), c=1, x=0.94 $$

Step-by-Step Solution

Answer

\( f(0.94) \approx \sqrt{2} + \frac{0.06 \pi \sqrt{2}}{4} \).

Step 1: Understand the Concept of Increments

The method of increments is based on the idea that a small change in a function's input near a point can give an estimate of the change in the function's output. Formally, for a small increment \( \Delta x \), the change in the function can be approximated as \( \Delta f \approx f'(c) \cdot \Delta x \). Here, \( \Delta x = x - c = 0.94 - 1 = -0.06 \).

Step 2: Compute the Derivative of the Function

Given \( f(x) = \csc(\pi x / 4) \), the derivative with respect to \(x\) is \( f'(x) = -\csc(\pi x / 4) \cdot \cot(\pi x / 4) \cdot \frac{\pi}{4} \), using the chain rule.

Step 3: Evaluate the Derivative at \( c = 1 \)

Substitute \( c = 1 \) into \( f'(x) \). We have \( \csc(\pi / 4) = \sqrt{2} \) and \( \cot(\pi / 4) = 1 \). Thus, \( f'(1) = -\sqrt{2} \cdot 1 \cdot \frac{\pi}{4} = -\frac{\pi \sqrt{2}}{4} \).

Step 4: Find the Increment \( \Delta f \)

Using the formula \( \Delta f \approx f'(c) \cdot \Delta x \) with \( f'(1) = -\frac{\pi \sqrt{2}}{4} \) and \( \Delta x = -0.06 \), we find \( \Delta f \approx -\frac{\pi \sqrt{2}}{4} \times -0.06 = \frac{0.06 \pi \sqrt{2}}{4} \).

Step 5: Calculate the Estimated Value of \( f(x) \)

Given \( f(c) = f(1) = \sqrt{2} \), estimate \( f(x) = f(1 + \Delta x) \) as \( f(x) \approx f(c) + \Delta f = \sqrt{2} + \frac{0.06 \pi \sqrt{2}}{4} \).

Key Concepts

Method of Increments

The method of increments is a way to predict how a function's value changes when the input changes slightly. It's like using a small step to guess what's around the corner. When you have a function and know its value at a certain point, you can use this method to estimate its value nearby. This is expressed in the formula:

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

where \( \Delta f \) is the change in the function's value, \( f'(c) \) is the derivative at a known point \( c \), and \( \Delta x \) is how much you've changed \( x \).
In our exercise, \( \Delta x = x - c = 0.94 - 1 = -0.06 \). This tells us how much the input \( x \) has moved from \( c \), helping predict the function's new value.

Derivative Calculation

Derivatives tell us how a function changes at a particular point. They're like asking, "How steep is the hill I'm on right now?" Mathematically, for any function \( f(x) \), its derivative \( f'(x) \) gives this rate of change.
In the example, \( f(x) = \csc(\pi x / 4) \), we find its derivative by using standard calculus rules. The derivative is:

• \( f'(x) = -\csc(\pi x / 4) \cdot \cot(\pi x / 4) \cdot \frac{\pi}{4} \)

This expression uses the chain rule, which is explained more fully in a later section. The sign and values indicate how fast and in which direction (up or down) the original function changes.

Function Approximation

Approximating a function means making a close guess about its values without calculating everything exactly. It's useful when precise values are hard to calculate quickly.
Using the method of increments, we adjust by the small changes predicted by the derivative. The process involves:

• Finding the value of the function at a known point, \( f(c) \).
• Using the derivative at that point, \( f'(c) \), and the change in \( x \), \( \Delta x \), to estimate \( \Delta f \).
• Adjusting the known value by \( \Delta f \): \( f(x) \approx f(c) + \Delta f \).

For \( f(x) \) in the exercise, we used \( f(c) = \sqrt{2} \) and \( \Delta f \approx \frac{0.06 \pi \sqrt{2}}{4} \) to approximate the new value of the function.

Chain Rule

The chain rule is a powerful tool in calculus for finding derivatives of complex functions. It answers the question: "If a function is inside another function, how do their rates of change relate?"
The chain rule states:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

This means you multiply the derivative of the outer function at the inner function by the derivative of the inner function itself.
In our example, the function \( f(x) = \csc(\pi x / 4) \) comprises an outer function \( \csc(u) \) with \( u = \pi x / 4 \). Applying the chain rule gets us the detailed expression for the derivative, which combines the rate of change of \( \csc(u) \) and \( \frac{\pi}{4} \), the derivative of \( u \). This method ensures you correctly capture how changes in \( x \) affect the overall function.