Problem 8 - Calculus

Question

If \(f(t)=1 / t,\) what is the average velocity between \(t=\frac{1}{2}\) and \(t=2 ?\) What is the average between \(t=\frac{1}{2}\) and \(t=1 ?\) What is the average (to one decimal place) between \(t=\frac{1}{2}\) and \(t=101 / 200 ?\)

Step-by-Step Solution

Answer

-1 for \(t = \frac{1}{2}\) to 2; -2 for \(t = \frac{1}{2}\) to 1; -4.0 for \(t = \frac{1}{2}\) to \(\frac{101}{200}\).

Step 1: Calculate Function Values

First, we need to find the function values at the given time intervals. For \(t = \frac{1}{2}\), \(f\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} = 2\). \(f(t)\) at \(t = 2\) is \(f(2) = \frac{1}{2}\). Similarly, find \(f\left(1\right)\) and \(f\left(\frac{101}{200}\right)\). \(f(1) = 1\) and \(f\left(\frac{101}{200}\right) = \frac{200}{101}\).

Step 2: Formula for Average Velocity

The formula for average velocity over an interval from \(t = a\) to \(t = b\) is given by: \[ \text{Average Velocity} = \frac{f(b) - f(a)}{b - a} \] We'll use this formula for each given interval.

Step 3: Calculate Average Velocity Between \(t = \frac{1}{2}\) and \(t = 2\)

Substitute the values into the formula: \(f(b) = f(2) = \frac{1}{2}\), \(f(a) = f\left(\frac{1}{2}\right) = 2\). Then, \[ \text{Average Velocity} = \frac{\frac{1}{2} - 2}{2 - \frac{1}{2}} = \frac{-1.5}{1.5} = -1. \]

Step 4: Calculate Average Velocity Between \(t = \frac{1}{2}\) and \(t = 1\)

Substitute the function values into the formula: \(f(b) = f(1) = 1\), \(f(a) = f\left(\frac{1}{2}\right) = 2\). Then, \[ \text{Average Velocity} = \frac{1 - 2}{1 - \frac{1}{2}} = \frac{-1}{0.5} = -2. \]

Step 5: Calculate Average Velocity Between \(t = \frac{1}{2}\) and \(t = \frac{101}{200}\)

Substitute the values into the formula: \(f(b) = f\left(\frac{101}{200}\right) = \frac{200}{101}\), \(f(a) = 2\). Then, \[ \text{Average Velocity} = \frac{\frac{200}{101} - 2}{\frac{101}{200} - \frac{1}{2}}. \] Calculating: \[ \text{Average Velocity} = \frac{\frac{200}{101} - 2}{\frac{101}{200} - \frac{100}{200}} = \frac{\frac{200}{101} - \frac{202}{101}}{\frac{1}{200}} = \frac{-\frac{2}{101}}{\frac{1}{200}}. \] Which simplifies to: \(-3.9604\) (to one decimal place, it's \(-4.0\)).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with how things change. One of its main branches is differential calculus, which focuses on the concept of the derivative, describing how a function changes over its domain.
In this exercise, we explore the change in the function's values as time progresses, which involves evaluating rates of change and average velocities.
Calculus provides the tools needed to understand these changes by defining derivatives and integrals. When we discuss average velocity, we use the principles of calculus to determine how a particular function behaves over a given interval. This can give insights into concepts like speed and direction within the context of motion or other changing systems.

Function Evaluation

Function evaluation is the process of determining the output of a function for a particular input. This is fundamental in calculus when analyzing the behavior of functions over specific intervals.
In our problem, the function is given by \(f(t) = \frac{1}{t}\). The exercise requires us to calculate the function's output at various points like \(t = \frac{1}{2},\ t = 1,\ t = 2,\) and \(t = \frac{101}{200}\).

• When \(t = \frac{1}{2}\), the function evaluates to 2 because \(f\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} = 2\).
• Similarly, \(f(1) = 1\), as \(\frac{1}{1} = 1\).
• At \(t = 2\), \(f(2) = \frac{1}{2}\).
• For \(t = \frac{101}{200}\), the evaluation \(f\left(\frac{101}{200}\right)\) results in a value of \(\frac{200}{101}\).

These evaluations are crucial for determining how the function behaves over different segments of its domain, which is essential for calculating average velocities.

Interval Analysis

Interval analysis involves examining how a function behaves over a certain range or segment of its domain. In the context of this problem, we are interested in understanding how the function changes between specific time intervals.
The intervals are defined as the differences between two values of \(t\), such as the intervals between \(t = \frac{1}{2}\) and \(t = 2\); \(t = \frac{1}{2}\) and \(t = 1\); and \(t = \frac{1}{2}\) and \(t = \frac{101}{200}\).
Understanding a function's behavior over these intervals allows us to calculate average velocity, a measure of how much change occurs in the function's value over time. This involves analyzing the difference in the function's outputs at the interval's endpoints and dividing it by the length of the interval. Interval analysis is thus an essential part of calculus and function evaluation.

Rate of Change

The rate of change is a fundamental concept in calculus, referring to how fast a quantity changes with respect to another quantity. In terms of functions, it's often described as the derivative, but can also be considered as the average rate of change over an interval.
For the given exercise, the average velocity essentially measures the rate at which the function \(f(t)\) changes between two points in time. The formula used is \( \frac{f(b) - f(a)}{b - a} \), where \(a\) and \(b\) are the time values defining the interval.
This calculation tells us how the function value \(f(t)\) changes per unit of time over that specific interval. Negative results indicate a decrease in the function value, while positive results indicate an increase. Understanding this allows us to interpret how quantities like speed or other variable changes occur over time, thus applying calculus principles in real-world contexts.