Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose \(P(t)=t^{4}\). a. Write an expression for \((P(b)-P(a)) /(b-a)\), the difference quotient of \(P\) on the interval, \([a, b] .\) b. Simplify your expression. c. Use your simplified expression to show that the rate of change of \(P\) at \(a\) is \(4 a^{3}\).

Step-by-Step Solution

Answer

Rate of change of \(P\) at \(a\) is \(4a^3\).

Step 1: Understand the Difference Quotient

The difference quotient of a function \(P(t)\) on the interval \([a, b]\) is given by \((P(b) - P(a)) / (b - a)\). In our problem, \(P(t) = t^4\). So, we'll substitute \(P(b)\) and \(P(a)\) into the formula.

Step 2: Substitute Function Values

Compute \(P(b) = b^4\) and \(P(a) = a^4\). Thus, the difference quotient becomes \((b^4 - a^4) / (b - a)\).

Step 3: Simplify the Expression

Notice that \(b^4 - a^4\) is a difference of two fourth powers. This can be factored as \((b - a)(b^3 + ab^2 + a^2b + a^3)\). So, the expression simplifies to:\[\frac{b^4 - a^4}{b - a} = b^3 + ab^2 + a^2b + a^3\]

Step 4: Determine the Rate of Change at a Point

The rate of change of \(P\) at \(a\) is found by taking the limit as \(b\) approaches \(a\) of the simplified difference quotient \(b^3 + ab^2 + a^2b + a^3\). Substitute \(b = a\) to find the rate of change:\[3a^3 + 3a^3 + a^3 = 4a^3\]Thus, the rate of change of \(P\) at \(a\) is \(4a^3\).

Key Concepts

Rate of Change

Understanding the rate of change is fundamental in calculus and mathematics in general. It describes how a quantity changes in relation to another quantity, often time. In the context of our exercise, we examine the rate of change of a polynomial function, specifically when time changes from one point to another.

The difference quotient is a tool we use to estimate the average rate of change over an interval, \([a, b]\). It is expressed as:

• \((P(b) - P(a)) / (b - a)\)

In this exercise, the function \(P(t) = t^4\) is used. By evaluating this difference quotient, we gather important insights into how the function behaves over specific intervals. When the interval \([a, b]\) becomes extremely small, this average rate of change gives us the instantaneous rate at a point, otherwise known as the derivative.

This transition from average to instantaneous rate of change forms a key foundation of calculus, enabling us to better understand how functions behave over tiny increments.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers of variables that are multiplied by constants. They can take many forms, such as linear, quadratic, cubic, etc. The function we are discussing, \(P(t) = t^4\), is a quartic polynomial, as it involves a degree of four.

A key property of polynomial functions is their smooth and continuous nature. They are differentiable, meaning we can calculate a derivative for them at any given point. With \(P(t) = t^4\), the exercise leads us to use algebraic manipulation to explore its rate of change. By simplifying expressions like \((b^4 - a^4)\), we unlock deeper insights into how polynomials can be analyzed and understood through their algebraic structure.

Factoring polynomials, understanding their powers, and recognizing their behavior over intervals are essential skills in calculus. They lay the groundwork for more advanced mathematical concepts and help students appreciate the diverse behaviors of different functions.

Calculus Concepts

Calculus provides us with powerful tools for analyzing changes and patterns. The difference quotient is a fundamental concept that bridges algebra and calculus. While it initially gives us an average change, the magic of calculus allows us to find the precise rate when intervals shrink to zero.

In our exercise, we delve into the derivative, which is central to calculus. As demonstrated, the difference quotient was simplified to \(b^3 + ab^2 + a^2b + a^3\), and taking the limit as \(b\) approaches \(a\) revealed the instantaneous rate of change, specifically \(4a^3\). This process illustrates how calculus captures the exact behavior of a function at any point.

Key takeaways from calculus include:

• Understanding limits and how they facilitate the transition from average to instantaneous change
• Recognizing the power of derivatives to predict and analyze the behavior of functions
• Realizing that calculus concepts like differentiation and integration are interconnected in describing and solving real-world problems

These foundational concepts make calculus a critical tool for scientists, engineers, economists, and many other professions dealing with change and motion.