Problem 6 - Thomas Calculus

Question

Find the average rate of change of the function over the given interval or intervals. \(P(\theta)=\theta^{3}-4 \theta^{2}+5 \theta ; \quad[1,2]\)

Step-by-Step Solution

Answer

The average rate of change is 0.

Step 1: Understand the Function and Interval

We are given the function \( P(\theta) = \theta^3 - 4\theta^2 + 5\theta \) and the interval \([1, 2]\). The goal is to find the average rate of change of the function over this interval.

Step 2: Calculate Function Values

Calculate \( P(1) \) and \( P(2) \).For \( \theta = 1 \):\[P(1) = 1^3 - 4(1)^2 + 5(1) = 1 - 4 + 5 = 2\]For \( \theta = 2 \):\[P(2) = 2^3 - 4(2)^2 + 5(2) = 8 - 16 + 10 = 2\]

Step 3: Use the Average Rate of Change Formula

The average rate of change of a function \( f(x) \) over the interval \([a, b]\) is given by the formula:\[\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}\]In this case, \( f(x) = P(\theta) \), \( a = 1 \), and \( b = 2 \). Substituting the values, we have:\[\frac{P(2) - P(1)}{2 - 1} = \frac{2 - 2}{2 - 1} = 0\]

Step 4: Interpret the Result

The average rate of change of \( P(\theta) \) from \( \theta = 1 \) to \( \theta = 2 \) is 0. This means that the value of the function didn't change over this interval.

Key Concepts

Function Values

When analyzing a function like \(P(\theta) = \theta^3 - 4\theta^2 + 5\theta\), one key concept to grasp is the idea of function values. A function value represents the output we get when we feed a particular input into the function. For example, to calculate the function value at any point, such as \(\theta = 1\), we substitute \(1\) into the equation \(P(\theta)\).

• At \(\theta = 1\): \(P(1) = 1^3 - 4(1)^2 + 5(1) = 1 - 4 + 5 = 2\)
• At \(\theta = 2\): \(P(2) = 2^3 - 4(2)^2 + 5(2) = 8 - 16 + 10 = 2\)

Here, both P(1) and P(2) result in the same answer of 2, highlighting that the function has the same output for different inputs in this context.
This concept is fundamental in calculating other mathematical properties like rates of change over different intervals.

Interval

An interval is a range between two points on a number line, often used in mathematics to define a segment over which we can analyze a function. In this exercise, the interval is \([1, 2]\), which means we are looking at the behavior of the function \(P(\theta)\) between these two points. Intervals are essential in calculus for determining how functions behave across specific ranges.

• Closed intervals like \([1, 2]\) include the endpoints 1 and 2.
• They help in establishing boundaries for calculating changes, such as the average rate over a particular domain.

Interpreting intervals correctly allows us to understand where and how a function begins and ends within the specified segment, like the change from \(\theta = 1\) to \(\theta = 2\) in this problem.

Average Rate of Change Formula

The average rate of change formula is a crucial tool in calculus. It helps us measure how a function changes on average over a specific interval. For a function \(f(x)\), the formula is: \[ \text{Average rate of change} = \frac{f(b) - f(a)}{b-a} \] where \([a, b]\) is the interval.
In our example, \(f(x) = P(\theta)\), with \(a = 1\) and \(b = 2\).
Using this formula:

• Substitute the function values: \(\frac{P(2) - P(1)}{2 - 1} = \frac{2 - 2}{1} = 0\)

The result is \(0\), meaning that the function doesn't change between \(\theta = 1\) and \(\theta = 2\).
This outcome tells us that for the given interval, the function \(P(\theta)\) maintains a constant value, demonstrating how the formula provides insights into the behavior of functions in specific ranges.