Chapter 11: Problem 10
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) \(\mathbf{r}(t)=\frac{1}{\sqrt{1-t^{2}}} \mathbf{j}+\frac{1}{\sqrt{9-t^{2}}} \mathbf{k}\)
Short Answer
Step by step solution
Determine the domain of \( \\mathbf{r}(t)=\ln (t-1) \\mathbf{i}+\sqrt{20-t} \\mathbf{j} \)
Determine the domain of \( \\mathbf{r}(t)=\ln \left(t^{-1}\right) \\mathbf{i}+\tan ^{-1} t \\mathbf{j}+t \\mathbf{k} \)
Determine the domain of \( \\mathbf{r}(t)=\frac{1}{\sqrt{1-t^{2}}} \\mathbf{j}+\frac{1}{\sqrt{9-t^{2}}} \\mathbf{k} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of Functions
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.
Logarithmic Functions
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.
Trigonometric Functions
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.
Square Root Function
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.