Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

State the domain of each of the following vector-valued functions: (a) \(\mathbf{r}(t)=\ln (t-1) \mathbf{i}+\sqrt{20-t} \mathbf{j}\) (b) \(\mathbf{r}(t)=\ln \left(t^{-1}\right) \mathbf{i}+\tan ^{-1} t \mathbf{j}+t \mathbf{k}\) (c) \(\mathbf{r}(t)=\frac{1}{\sqrt{1-t^{2}}} \mathbf{j}+\frac{1}{\sqrt{9-t^{2}}} \mathbf{k}\)

Step-by-Step Solution

Answer

(a) Domain is (1, 20]; (b) Domain is (-∞, 0) ∪ (0, ∞); (c) Domain is (-1, 1) ∩ (-3, 3).

Step 1: Understanding the Function

The given function is \(\mathbf{r}(t) = \ln(t-1) \mathbf{i} + \sqrt{20-t} \mathbf{j}\). To find the domain, we need to find where each component is defined.

Step 2: Analyzing Each Component

1. For \(\ln(t-1)\), the argument must be greater than 0. Thus, \(t-1 > 0\) or \(t > 1\). 2. For \(\sqrt{20-t}\), the expression under the square root must be non-negative. Therefore, \(20-t \geq 0\) or \(t \leq 20\).

Step 3: Deriving the Domain for (a)

Combining both conditions, \(t > 1\) and \(t \leq 20\), the domain of the function \(\mathbf{r}(t)\) is \((1, 20]\).

Key Concepts

Domain of Functions

In mathematical analysis and calculus, the domain of a function is crucial for understanding where the function can be evaluated. For any function, the domain represents all the possible input values (usually represented as "x" or, in the case of vector-valued functions, "t") that can be used in the function to produce a valid result.

For vector-valued functions, each component can impose its own restrictions on the domain. Let's look at an example: the vector-valued function \( \mathbf{r}(t) = \ln(t-1) \mathbf{i} + \sqrt{20-t} \mathbf{j} \).The domain is found by considering the restrictions of each component of the function:

• \( \ln(t-1) \): The natural logarithm requires that its input, \(t-1\), must be greater than zero. Therefore, \(t > 1\).
• \( \sqrt{20-t} \): The square root function requires that its argument be non-negative, i.e., \(20-t \geq 0\) or \(t \leq 20\).

Combining these conditions, the domain of \( \mathbf{r}(t) \) is the interval \((1, 20] \). Understanding the domain involves analyzing these constraints and applying set theory to the conditions dictated by each function component.

Logarithmic Functions

Logarithmic functions are a cornerstone in mathematics, often denoted as \( \ln(x) \) for the natural logarithm with base \(e\), where \(e\approx 2.71828\). The domain of logarithmic functions must always be comprised of positive numbers, as the logarithm of non-positive numbers is undefined in real analysis.

For example, consider the logarithmic component in the function \( \ln(t-1) \). The condition for the logarithm to be defined is that \( t-1 > 0 \). This implies \( t > 1 \). The significance of this constraint is clear: it defines one boundary for the domain of any function it is part of. For any point inside this boundary, you can calculate the logarithm.

Key properties to remember about logarithms include:

• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b\ln(a) \)

These properties help manipulate and simplify expressions involving logarithms, assisting in solving equations and understanding function behavior.

Square Root Functions

Square root functions are one of the basic types of functions involving the radical sign. These functions are often written in the form \( \sqrt{x} \) and their domain typically includes only non-negative values of \(x\). This is because the square root of a negative number is not defined in the set of real numbers.

In our example with \( \sqrt{20-t} \), the expression under the square root must be non-negative, meaning \( 20-t \geq 0 \), leading to the condition \( t \leq 20 \). This condition ensures that the result of the square root is a real number.

Here’s what to remember:

• The square root function will only output real numbers for non-negative inputs.
• This property defines an upper or lower boundary for the domain, depending on the context.
• In the vector-valued function \( \mathbf{r}(t) \), it introduced the restriction \( t \leq 20 \).

Understanding these conditions is pivotal when determining the domain of functions that include square root expressions.

Mathematical Analysis

Mathematical analysis involves a wide variety of concepts and techniques used to understand the behavior of functions, their limits, continuity, and derivatives. The study of vector-valued functions is a part of this broader discipline, dealing with functions whose values are vectors.

In examining the domain of vector-valued functions, mathematical analysis uses a combination of understanding individual function components and their constraints to determine the overall function's domain.

Consider the interplay of functions in \( \mathbf{r}(t) = \ln(t-1) \mathbf{i} + \sqrt{20-t} \mathbf{j} \):

• The logarithmic function provides a lower bound \( t > 1 \).
• The square root function provides an upper bound \( t \leq 20 \).

With analysis, we are able to logically combine these constraints to find the largest set of input values for which all components are defined. This logical approach facilitates deeper understanding and solutions to complex problems, embodying the essence of mathematical analysis in practice.