Problem 13 - College Algebra and Calculus: An Applied Approach

Question

Find the function values. \(f(x, y)=x^{2}-2 y\) (a) \(f(x+\Delta x, y)\) (b) \(\frac{f(x, y+\Delta y)-f(x, y)}{\Delta y}\)

Step-by-Step Solution

Answer

(a) The function value is \(f(x + \Delta x, y) = x^{2} + 2x\Delta x + (\Delta x)^{2} - 2y\), (b) The approximate change rate of the function w. r. t. y is \(-2\).

Step 1: Compute Function Value with Input (x + Δx, y)

Substitute \(x\) with \((x + \Delta x)\) in the function \(f(x, y) = x^{2} - 2y\). The function then becomes \(f(x + \Delta x, y) = (x + \Delta x)^{2} - 2y = x^{2} + 2x\Delta x + (\Delta x)^{2} - 2y\).

Step 2: Compute the Approximate Change Rate of the Function w. r. t. y

Substitute \(y\) with \((y + \Delta y)\) in the function \(f(x, y)\). This yields \(f(x, y + \Delta y) = x^{2} - 2(y + \Delta y)\). Subsequently, compute the expression \(\frac{f(x, y+\Delta y)-f(x, y)}{\Delta y}\) which results in \(\frac{x^{2} - 2y - 2\Delta y - (x^{2} - 2y)}{\Delta y} = -2\).

Key Concepts

Function Evaluation

When working with functions, it's essential to understand the process of Function Evaluation. This involves substituting specific values into the function's expression and simplifying to compute the output. In this exercise, we evaluated the function \( f(x, y) = x^2 - 2y \) for \( f(x + \Delta x, y) \).

Here's how it works:

• Substitute \( x \) with \( x + \Delta x \).
• Replace \( x^2 \) in the original function with \( (x + \Delta x)^2 \).
• Use binomial expansion to simplify: \( (x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2 \).
• Thus, \( f(x + \Delta x, y) = x^2 + 2x\Delta x + (\Delta x)^2 - 2y \).

Function evaluation is crucial as it helps determine how a function behaves as its inputs change.
Understanding this concept is key to working with variables and calculating precise function values.

Partial Derivatives

Partial Derivatives are an essential concept in calculus, representing the rate of change of a function with respect to one variable while keeping the others constant. This exercise required understanding the effect of small changes in \( y \) on \( f(x, y) \).

To achieve this:

• Consider the expression \( f(x, y + \Delta y) \).
• Substitute \( y \) with \( y + \Delta y \) in \( f(x, y) = x^2 - 2y \).
• This yields: \( f(x, y + \Delta y) = x^2 - 2(y + \Delta y) = x^2 - 2y - 2\Delta y \).

The key part is finding the derivative by simplifying:

• Calculate: \( \frac{f(x, y+\Delta y)-f(x, y)}{\Delta y} = \frac{x^2 - 2y - 2\Delta y - (x^2 - 2y)}{\Delta y} \).
• Simplify to get \( -2 \), which is the partial derivative of \( f \) with respect to \( y \).

Partial derivatives allow us to examine how each individual variable influences the function's value.

Change Rate

The Change Rate tells us how a quantity changes concerning another. Here, we looked at how \( f(x, y) \) changes as \( y \) changes, known as the rate of change with respect to \( y \). This was done by calculating the difference quotient:

In this problem, to find the rate of change of the function when \( y \) changes, we computed:

• \( f(x, y + \Delta y) \) as \( x^2 - 2y - 2\Delta y \).
• Next, determine the difference: \( f(x, y + \Delta y) - f(x, y) = -2\Delta y \).
• Finally, divide by \( \Delta y \) to find the rate: \( \frac{-2 \Delta y}{\Delta y} = -2 \).

This rate, \( -2 \), tells us that for every unit increase in \( y \), the function \( f \) decreases by 2 units. Change rates are vital for predicting behaviors and trends in many fields, including physics, engineering, and economics.