Problem 14 - Calculus Early Transcendentals

Question

A function \(y=f(x)\) and values of \(x_{0}\) and \(x_{1}\) are given. (a) Find the average rate of change of \(y\) with respect to \(x\) over the interval \(\left[x_{0}, x_{1}\right] .\) (b) Find the instantaneous rate of change of \(y\) with respect to \(x\) at the specified value of \(x_{0}\) (c) Find the instantaneous rate of change of \(y\) with respect to \(x\) a) The arbitrary value of \(x_{0} .\) (d) The average rate of change in part (a) is the slope of a certain secant line, and the instantaneous rate of change in part (b) is the slope of a certain tangent line. Sketch the graph of \(y=f(x)\) together with those two lines. $$ y=1 / x^{2} ; x_{0}=1, x_{1}=2 $$

Step-by-Step Solution

Answer

(a) Average rate of change is \(-\frac{3}{4}\); (b) Instantaneous rate at \(x_0=1\) is \(-2\); (d) Graph shows secant and tangent slopes.

Step 1: Compute the Average Rate of Change

The average rate of change of a function \( y = f(x) \) over the interval \( [x_0, x_1] \) can be calculated using the formula: \[ \frac{f(x_1) - f(x_0)}{x_1 - x_0} \]. For the function \( y = \frac{1}{x^2} \), and the interval \( [1, 2] \), substitute the values: \( f(1) = \frac{1}{1^2} = 1 \) and \( f(2) = \frac{1}{2^2} = \frac{1}{4} \). Thus, the average rate of change is: \[ \frac{\frac{1}{4} - 1}{2 - 1} = \frac{-\frac{3}{4}}{1} = -\frac{3}{4} \].

Step 2: Compute the Instantaneous Rate of Change at \( x_0 = 1 \)

The instantaneous rate of change of the function at a specific point \( x_0 \) is its derivative evaluated at that point. First, find the derivative of \( y = \frac{1}{x^2} \). Using the power rule, \( y' = -\frac{2}{x^3} \). Evaluate this derivative at \( x_0 = 1 \): \( y'(1) = -\frac{2}{1^3} = -2 \).

Step 3: Compute the Instantaneous Rate of Change at Arbitrary \( x_0 \)

The derivative function represents the instantaneous rate of change at any point \( x = x_0 \). We have already calculated it as \( y' = -\frac{2}{x^3} \). This formula provides the instantaneous rate of change for any chosen \( x_0 \).

Step 4: Graph the Function and Lines

To illustrate the secant and tangent lines, plot \( y = \frac{1}{x^2} \). The secant line is drawn between points \( (1, 1) \) and \( (2, \frac{1}{4}) \) with a slope of \( -\frac{3}{4} \). The tangent line at \( x_0 = 1 \) has a slope of \( -2 \) and passes through point \( (1, 1) \). This secant line approximates the average rate of change, while the tangent line represents the instantaneous rate of change at \( x = 1 \).

Step 5: Interpret the Secant and Tangent Slopes

The slope of the secant line \( -\frac{3}{4} \) shows the average rate of change of the function from \( x = 1 \) to \( x = 2 \). The tangent line slope, \( -2 \), quantifies how quickly \( y \) changes at exactly \( x = 1 \). Graphically, the secant line goes through two points on the curve, whereas the tangent line just "touches" at \( x_0 \) providing the precise slope at that point.

Key Concepts

Average Rate of Change

The average rate of change is a fundamental concept in calculus that helps you understand how a function behaves over a specific interval. Imagine you're driving a car and want to find out your average speed over a certain distance. Similarly, in mathematics, we use the average rate of change to determine how a function changes from one point to another along the x-axis. When you compute the average rate of change of a function \( y = f(x) \) over the interval \([x_0, x_1]\), you are essentially finding the slope of the line that connects the points \( (x_0, f(x_0)) \) and \( (x_1, f(x_1)) \). This line is known as a secant line. The formula is:

• \( \frac{f(x_1) - f(x_0)}{x_1 - x_0} \)

In the given example, the function is \( y = \frac{1}{x^2} \) and the interval is \([1, 2]\). By applying the formula, you calculate:

• \( f(1) = 1 \)
• \( f(2) = \frac{1}{4} \)
• Average rate of change = \( \frac{\frac{1}{4} - 1}{2 - 1} = -\frac{3}{4} \)

This result tells us that on average, the function \( y = \frac{1}{x^2} \) decreases by \( \frac{3}{4} \) as you move from \( x = 1 \) to \( x = 2 \).

Instantaneous Rate of Change

The instantaneous rate of change differs from the average rate of change as it represents how fast a function is changing at a specific point. Think about a car's speedometer showing how fast you're going at an exact moment. In calculus, this concept is captured by finding the derivative of a function. The derivative tells us the slope of the tangent line at any given point on the curve.
To find the instantaneous rate of change for a function \( y = f(x) \) at a point \( x_0 \), you calculate its derivative and then evaluate it at \( x_0 \). For example, the derivative of \( y = \frac{1}{x^2} \) is calculated using the power rule:

• \( y' = -\frac{2}{x^3} \)

You then plug in \( x = 1 \) to get:

• \( y'(1) = -2 \)

This tells you that at \( x = 1 \), the function is decreasing at a rate of -2. The negative sign shows that the function is going downwards at this point.

Secant Line and Tangent Line

In calculus, the secant line and tangent line help us visualize the rate of change in different contexts. The secant line is a straight line connecting two points on a curve, expressing the average rate of change over a given interval. In contrast, the tangent line "touches" the curve at just one point, displaying the instantaneous rate of change.
For the function \( y = \frac{1}{x^2} \), we already calculated:

• The average rate of change over \([1, 2]\) as \(-\frac{3}{4}\)
• The instantaneous rate of change at \( x = 1 \) as \(-2\)

Graphically:

• The secant line joins the points \( (1, 1) \) and \( (2, \frac{1}{4}) \).
• The tangent line at \( x = 1 \) just touches the curve at \( (1, 1) \). This tangent line provides a precise snapshot of how steep the curve is at that exact point.

Understanding these two lines helps you get a clearer picture of how functions change over different scales, whether it's across intervals or at a single point.