Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Use the definition of rate of change to find the rate of change of \(P(t)=\frac{1}{t}\) at \(a=5\). Repeat for \(a\) unspecified. Complete the formula $$ P(t)=\frac{1}{t} \quad \Rightarrow \quad P^{\prime}(t)= $$ The functions \(F\) and \(G\) of the next two exercises present interesting challenges.

Step-by-Step Solution

Answer

The rate of change is \(-\frac{1}{25}\) at \(t = 5\) and \(-\frac{1}{t^2}\) for any \(t\).

Step 1: Understand the Problem

We need to find the rate of change (first derivative) of the function \(P(t) = \frac{1}{t}\) at \(a = 5\) and generally for any \(a\). This involves using the definition of the derivative.

Step 2: Apply the Definition of Rate of Change

The rate of change (derivative) of \(P(t)\) is given by \(P'(t)\). According to the definition, \(P'(t) = \lim_{h \to 0} \frac{P(t + h) - P(t)}{h}\). We substitute \(P(t) = \frac{1}{t}\) into this expression.

Step 3: Find the Derivative Expression

Substitute \(P(t) = \frac{1}{t}\) into the derivative formula: \[P'(t) = \lim_{h \to 0} \frac{\frac{1}{t+h} - \frac{1}{t}}{h}\]. Simplify this expression: \[= \lim_{h \to 0} \frac{t - (t + h)}{h(t+h)t} = \lim_{h \to 0} \frac{-h}{h(t+h)t}\]. This simplifies to \(-\frac{1}{t^2}\).

Step 4: Substitute for Specific Value \(a = 5\)

Now substitute \(t = 5\) into the derivative: \(P'(5) = -\frac{1}{5^2} = -\frac{1}{25}\).

Step 5: General Expression for Any \(a\)

The rate of change for \(P(t) = \frac{1}{t}\) in general is \(P'(a) = -\frac{1}{a^2}\). Thus, for any \(a\), the derivative remains in the same form: \(P'(t) = -\frac{1}{t^2}\).

Step 6: Solution Completion

So, the completed part of the problem is: \[P(t) = \frac{1}{t} \quad \Rightarrow \quad P^{\prime}(t) = -\frac{1}{t^2}\]. This is the general derivative formula.

Key Concepts

calculus

Calculus is a branch of mathematics that focuses on the concepts of change and motion. It provides the tools to analyze and understand rates of change and accumulation. The two main branches of calculus are differential calculus and integral calculus.

• Differential Calculus: Concerns itself with the concept of a derivative, which represents the rate at which a quantity changes.
• Integral Calculus: Deals with the concept of the integral, focusing on the accumulation of quantities, such as areas under curves.

By applying calculus, we can solve complex problems related to changing systems, such as finding the slope of a curve at any given point. In this exercise, we specifically utilize differential calculus to determine how a function behaves as its variables change, demonstrating the real-world application and importance of calculus for forecasting future behavior.

rate of change

The rate of change is a fundamental concept in calculus that measures how one quantity changes in relation to another. It is often considered synonymous with the derivative in mathematical terms. In simpler words, it shows how one variable affects another when there's a slight change.
For instance, if we want to know how fast a car is moving, we use the rate of change—the speed—to determine its motion over time. In the exercise provided, you are finding the rate of change of the function \( P(t) = \frac{1}{t} \) at a specific point \(a = 5\) and generally for any \(a\). This function demonstrates how the values change as the variable \(t\) changes.

differentiation

Differentiation is the process of calculating a derivative. In mathematics, it's a technique used to find the rate at which a function is changing at any given point. By using differentiation, we are equipped with the derivative, a powerful tool that helps us understand and predict how functions evolve.
In this particular case, differentiating \( P(t) = \frac{1}{t}\) gives us the expression \( P'(t) = -\frac{1}{t^2} \), which tells us that the rate of change of \(P\) at any value \(t\) is \(-\frac{1}{t^2}\). The negative sign indicates that the function is decreasing at an accelerating rate as \(t\) increases. Differentiation is essential in many fields such as physics, where it can determine how velocities change over time, and economics, where it helps in analyzing cost functions.

limit definition of derivative

The limit definition of derivative is the cornerstone for understanding how derivatives are derived. Formally, the derivative of a function \( f(t) \) at a point \( t \) is defined as:\[ f'(t) = \lim_{h \to 0} \frac{f(t + h) - f(t)}{h} \]This formula calculates the derivative by observing the function's behavior as the change in \(t\), represented by \(h\), becomes infinitesimally small. In simpler terms, it's like zooming in on a curve to find its slope at a precise point.
In the example, you used this definition to find the derivative of \( P(t) = \frac{1}{t} \). By substituting \( P(t + h) \) and \( P(t) \) into the limit equation and simplifying, you arrive at \( P'(t) = -\frac{1}{t^2} \). This method is crucial because it's the foundational concept that enables the calculation of derivatives, allowing us to glean insights into the behavior of functions as they change over time.