Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 41-48, the density of a rod is given. Find c so that the mass from 0 to \(c\) is equal to 1 . Whenever possible find an exact solution. If this is not possible, find an approximation for c. (See Examples 4 and 5 ). $$ \delta(x)=\frac{1}{x+1} $$

Step-by-Step Solution

Answer

The exact value for \( c \) is \( e - 1 \).

Step 1: Understand the problem

We need to find a value for \( c \) such that the mass of the rod from 0 to \( c \) is equal to 1, given the density function \( \delta(x) = \frac{1}{x+1} \). This involves setting up and solving the integral \( \int_0^c \delta(x)\, dx = 1 \).

Step 2: Set up the integral for mass

The mass, \( m \), from 0 to \( c \) is given by the integral \( m = \int_0^c \frac{1}{x+1}\, dx \). We want \( m = 1 \).

Step 3: Solve the integral

Calculate \( \int \frac{1}{x+1} \, dx \). This is a standard integral with the antiderivative \( \ln|x+1| + C \). Therefore, \( \int_0^c \frac{1}{x+1} \, dx = \ln(c+1) - \ln(1) = \ln(c+1) \).

Step 4: Set the integral equal to 1

Since the mass is 1, we equate \( \ln(c+1) \) to 1: \( \ln(c+1) = 1 \).

Step 5: Solve for c

To solve for \( c \), exponentiate both sides: \( c+1 = e^1 \), thus \( c = e - 1 \).

Step 6: Check the solution

Verify the solution by substituting \( c = e-1 \) back into the expression for mass: \( \ln(e) = 1 \), confirming that the mass is indeed 1 for \( c = e-1 \).

Key Concepts

Density Function

The concept of a density function is crucial in understanding mass distribution along an object, such as a rod. A density function, typically denoted by \( \delta(x) \), provides the mass per unit length at each point \(x\) on the object. In this exercise, the density function given is \( \delta(x) = \frac{1}{x+1} \), which indicates that the mass is spread differently along the rod depending on the position \(x\). This function decreases as \(x\) increases, meaning less mass is found per unit length further along the rod.

Understanding the role of a density function can help in visualizing how mass is not always uniform but varies depending on the position, which is key to applying calculus to physical problems.

Integration

Integration is an essential technique in calculus for finding quantities like areas, volumes, and in this case, mass from a density function. It allows us to calculate the total accumulation of a quantity given its rate of change or density. Here, we integrate the density function \( \delta(x) = \frac{1}{x+1} \) over the interval from 0 to \(c\) to find the total mass.

We set up the integral as \( \int_0^c \frac{1}{x+1} \, dx \) and know that the result must equal 1 to solve for \( c \). This involves calculating the area under the curve of the density function, which represents the total mass of the rod from \(x = 0\) to \(x = c\). Integration is a powerful tool that transforms density functions into meaningful physical quantities like mass.

Exponential Functions

Exponential functions are a class of mathematical functions where the variable appears in the exponent. In this problem, they arise when solving the equation \( \ln(c+1) = 1 \). Exponentiating both sides, we take the exponent \( e \), the base of the natural logarithm, to solve for \( c \).

The equation transforms to \( c+1 = e^1 \), demonstrating how exponential functions are inversely related to the natural logarithm. The value \( e \), approximately 2.718, is a fundamental constant in mathematics, representing the base of natural growth processes. Mastering exponential functions is key in understanding their role in calculus and solving a variety of mathematical problems.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the inverse function of the exponential function with base \( e \). This property is critical in the current exercise where the integral of the density function results in a natural logarithm expression: \( \ln(c+1) \).

By setting \( \ln(c+1) = 1 \), we can find \( c \) through exponentiation, where \( e^1 = c+1 \). Natural logarithms allow us to solve for unknowns in equations involving growth or decay, which are represented by exponential functions. Understanding their relationship with exponentials is essential for solving calculus problems that involve growth rates or densities.