Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

If \(b\) approaches 3 a. \(b\) approaches _________. b. \(2 \div b\) approaches _________. c. \(\pi\) approaches _________. d. \(\frac{2}{\sqrt{b}+\sqrt{3}}\) approaches _________. e. \(b^{3}+b^{2}+b\) approaches _________. f. \(\frac{b}{1+b}\) approaches _________. g. \(2^{b}\) approaches _________. h. \(\log _{3} b\) approaches _________.

Step-by-Step Solution

Answer

a. 3; b. \(\frac{2}{3}\); c. \(\pi \); d. \(\frac{\sqrt{3}}{3}\); e. 39; f. \(\frac{3}{4}\); g. 8; h. 1.

Step 1: Calculating Part a

Since \( b \) approaches 3, the answer to part a simply states that \( b \) approaches 3.

Step 2: Calculating Part b

Substituting \( b = 3 \) into the expression \( 2 \div b = \frac{2}{b} \). So, \( \frac{2}{3} \) is the solution. Therefore, as \( b \) approaches 3, \( \frac{2}{b} \) approaches \( \frac{2}{3} \).

Step 3: Calculating Part c

Since \( \pi \) is a constant and not affected by \( b \), \( \pi \) approaches \( \pi \).

Step 4: Calculating Part d

Consider \( \frac{2}{\sqrt{b} + \sqrt{3}} \). As \( b \to 3 \), \( \sqrt{b} \to \sqrt{3} \). Therefore, the expression becomes \( \frac{2}{\sqrt{3} + \sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \).

Step 5: Calculating Part e

Substitute \( b = 3 \) into \( b^3 + b^2 + b \). We get \( 3^3 + 3^2 + 3 = 27 + 9 + 3 = 39 \). Thus, as \( b \to 3 \), \( b^3 + b^2 + b \to 39 \).

Step 6: Calculating Part f

For \( \frac{b}{1+b} \), substituting \( b = 3 \), we get \( \frac{3}{1+3} = \frac{3}{4} \). Therefore, as \( b \to 3 \), \( \frac{b}{1+b} \to \frac{3}{4} \).

Step 7: Calculating Part g

Substitute \( b = 3 \) into \( 2^b \). This gives us \( 2^3 = 8 \). Hence, as \( b \to 3 \), \( 2^b \to 8 \).

Step 8: Calculating Part h

Applying \( b = 3 \) in \( \log_3 b \), we have \( \log_3 3 = 1 \). Consequently, as \( b \to 3 \), \( \log_3 b \to 1 \).

Key Concepts

Calculus

Calculus is an essential branch of mathematics dealing with the study of change. Its two main branches, differential and integral calculus, provide tools to understand a variety of mathematical models. Limits are a foundational building block in calculus. They help specify the behavior of a function as it approaches a particular point or value. For instance, if we say that "as \( b \) approaches 3, \( f(b) \) approaches \( L \)," we describe the limit of \( f(b) \) as \( b \) tends toward 3. This allows us to analyze and solve more complex problems in calculus, beyond simple substitutions, by predicting function behavior close to specific values.

Approaching Values

Understanding the concept of approaching values is crucial when dealing with limits. When we say a value approaches a number, we imagine getting infinitely close to a specific point.

• Consider \( b \) approaching 3. This means that \( b \) gets nearer and nearer to 3, but is not necessarily equal to 3.
• Let's take an example: \( \frac{2}{b} \) approaches \( \frac{2}{3} \) as \( b \) nears 3. Here, the expression simplifies by substituting \( b = 3 \) directly, resulting in \( \frac{2}{3} \).

This idea is instrumental in evaluating expressions where direct substitution might not give an actual value due to division by zero or undefined forms, but approaching techniques help us to understand the limit's approach.

Mathematical Constants

Mathematical constants are numbers that have fixed values, unlike variables which can change. Common constants include \( \pi \), \( e \), and \( i \). They remain unchanged regardless of variations in other elements of an equation. For example, in the expression \( \pi \) given in part c of the exercise, regardless of how \( b \) changes, \( \pi \) will always approach \( \pi \). Constants provide stable points in mathematical equations and formulas, offering a crucial aspect of predictability in calculations. Their unchanging nature makes them valuable references when evaluating limits and functions.

Algebraic Expressions

Algebraic expressions involve variables, numbers, and operations (addition, subtraction, multiplication, division) all grouped together to represent a mathematical concept. Evaluating these expressions often requires substituting numerical values and performing operations according to algebraic rules.

• Let's examine \( b^3 + b^2 + b \). By substituting \( b = 3 \), our expression becomes \( 39 \), demonstrating how substitution helps determine the outcome when variables approach certain values.
• Another example, \( \frac{b}{1+b} \), approaches \( \frac{3}{4} \) as \( b \) nears 3.

Algebraic manipulations and substitutions ease calculations and solutions, providing clarity in expression analysis and evaluation.