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) \( 1 < t \leq 20 \); (b) \( t > 0 \); (c) \( -1 < t < 1 \).

Step 1: Determine the domain of \( \\mathbf{r}(t)=\ln (t-1) \\mathbf{i}+\sqrt{20-t} \\mathbf{j} \)

The function contains two expressions: \( \ln(t-1) \) and \( \sqrt{20-t} \).- The domain of \( \ln(t-1) \) requires \( t-1 > 0 \), thus \( t > 1 \).- The domain of \( \sqrt{20-t} \) requires \( 20-t \geq 0 \), thus \( t \leq 20 \).Combine these conditions, the domain is \( 1 < t \leq 20 \).

Step 2: Determine the domain of \( \\mathbf{r}(t)=\ln \left(t^{-1}\right) \\mathbf{i}+\tan ^{-1} t \\mathbf{j}+t \\mathbf{k} \)

The function contains three expressions: \( \ln(t^{-1}) \), \( \tan^{-1} t \), and \( t \).- The domain of \( \ln(t^{-1}) \) requires \( t^{-1} > 0 \), which implies \( t > 0 \).- \( \tan^{-1} t \) and \( t \) are defined for all real numbers.Thus, the only restriction comes from \( \ln(t^{-1}) \), making the domain \( t > 0 \).

Step 3: Determine the domain of \( \\mathbf{r}(t)=\frac{1}{\sqrt{1-t^{2}}} \\mathbf{j}+\frac{1}{\sqrt{9-t^{2}}} \\mathbf{k} \)

The function has two expressions: \( \frac{1}{\sqrt{1-t^2}} \) and \( \frac{1}{\sqrt{9-t^2}} \).- The expression \( \frac{1}{\sqrt{1-t^2}} \) requires \( 1-t^2 > 0 \), resulting in \( -1 < t < 1 \).- The expression \( \frac{1}{\sqrt{9-t^2}} \) requires \( 9-t^2 > 0 \), resulting in \( -3 < t < 3 \).The domain is the intersection of \( -1 < t < 1 \) and \( -3 < t < 3 \), which is \( -1 < t < 1 \).

Key Concepts

Domain of Functions

Understanding the domain of a function is a vital part of analyzing its behavior. The domain refers to all the possible input values, typically represented by the variable, for which the function is defined. This concept is crucial when examining vector-valued functions, as each component function can have its own domain restrictions.
In vector-valued functions like \(\mathbf{r}(t) = \ln(t-1) \mathbf{i} + \sqrt{20-t} \mathbf{j}\), the domain is determined by the intersection of the domains of each individual component. Here, \(\ln(t-1)\) is defined only when \(t-1 > 0\) (i.e., \(t > 1\)), and \(\sqrt{20-t}\) is valid when \(20-t \geq 0\) (i.e., \(t \leq 20\)). Therefore, the domain of \(\mathbf{r}(t)\) is \(1 < t \leq 20\). Simplifying the restrictions for each part ensures you find the valid range where all conditions are satisfied simultaneously.

• The domain is a foundational element in calculus and function analysis.
• Both mathematical operations and function definitions, like logarithms and square roots, contribute to determining domain restrictions.
• Evaluating the domain involves ensuring inputs don't lead to undefined or non-real results.

Always think about how inputs affect each component of your function to find where the function is fully defined.

Logarithmic Functions

Logarithmic functions are a fundamental part of calculus, appearing often in vector-valued functions. The natural logarithm, denoted as \(\ln(x)\), is primarily defined for positive values of \(x\). This stems from the inherent properties of logarithms, where the base raised to some power results in the input value; in the natural logarithm, the base is \(e\), the mathematical constant approximately equal to 2.718.
For instance, in the vector-valued function \(\mathbf{r}(t) = \ln(t-1) \mathbf{i} + \cdots\), the domain restriction for \(\ln(t-1)\) requires \(t-1 > 0\) (hence, \(t > 1\)) to ensure the argument of the logarithm is positive.

• Logarithmic functions are undefined for zero or negative inputs due to their base-exponent nature.
• The properties of logarithms include the ability to transform multiplicative relationships into additive ones, thereby simplifying many calculus problems.

When dealing with logarithms, always check the sign of the input inside the function. This guarantees you stay within the domain where the logarithm is properly defined.

Trigonometric Functions

Trigonometric functions are another staple of mathematics, frequently integrated into vector-valued functions. Functions like sine, cosine, and tangent provide periodic behavior and are defined over a wide range of values. In the context of vector-valued functions, the arctangent or inverse tangent, denoted as \(\tan^{-1}(t)\), is quite common.
The domain for \(\tan^{-1}(t)\) is all real numbers, meaning there are no restrictions on \(t\) from this component of a function. This is true for many inverse trigonometric functions, which serve as the bridge from trigonometric ratios to angles.

• Inverse trigonometric functions such as \(\tan^{-1}(t)\) are defined everywhere on the real line.
• These functions invert the process of finding trigonometric values, allowing determination of angles from ratios.

Understanding the characteristics of trigonometric functions enables you to comprehend their effects within a vector-valued function without needing to impose additional domain constraints.

Square Root Function

Square root functions appear frequently in mathematical expressions and have specific domain constraints. The square root of a non-negative number \(x\) is denoted \(\sqrt{x}\) and is defined only when \(x \geq 0\). This is because there's no real number whose square is negative.
In vector-valued functions like \(\mathbf{r}(t) = \frac{1}{\sqrt{1-t^2}} \mathbf{j} + \cdots\), square roots set domain limits. For example, \(\sqrt{1-t^2}\) is defined only when \(1-t^2 > 0\), which leads to the range \(-1 < t < 1\). This ensures the expression under the square root remains positive.

• Square roots demand non-negative values under the radical sign.
• The positivity of the expression is crucial to avoid imaginary numbers, which are undefined in real-valued functions.

When assessing vector-valued functions' domains, carefully handle components with square roots to pinpoint their permissible range precisely.