/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 State the domain of each of the ... [FREE SOLUTION] | 91影视

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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

Expert verified
(a) \(1 < t \leq 20\), (b) \(t > 0\), (c) \(-1 < t < 1\)

Step by step solution

01

Analyze Function (a)

For the vector-valued function \( \mathbf{r}(t) = \ln(t-1) \mathbf{i} + \sqrt{20-t} \mathbf{j} \):1. The term \( \ln(t-1) \) requires that \( t-1 > 0 \), hence \( t > 1 \).2. The term \( \sqrt{20-t} \) requires that \( 20-t \geq 0 \), hence \( t \leq 20 \).Therefore, the domain for (a) is \( 1 < t \leq 20 \).
02

Analyze Function (b)

For the vector-valued function \( \mathbf{r}(t) = \ln(t^{-1}) \mathbf{i} + \tan^{-1}t \mathbf{j} + t \mathbf{k} \):1. The term \( \ln(t^{-1}) \) can be rewritten as \( \ln(\frac{1}{t}) = -\ln(t) \), which implies \( t > 0 \).2. The term \( \tan^{-1}t \) is defined for all real numbers.3. The term \( t \mathbf{k} \) is defined for all real numbers.Therefore, the domain for (b) is \( t > 0 \).
03

Analyze Function (c)

For the vector-valued function \( \mathbf{r}(t) = \frac{1}{\sqrt{1-t^2}} \mathbf{j} + \frac{1}{\sqrt{9-t^2}} \mathbf{k} \):1. The term \( \frac{1}{\sqrt{1-t^2}} \) is defined when \( 1-t^2 > 0 \), leading to \( -1 < t < 1 \).2. The term \( \frac{1}{\sqrt{9-t^2}} \) is defined when \( 9-t^2 > 0 \), leading to \( -3 < t < 3 \).The intersection of these intervals is obtained by considering both, \( -1 < t < 1 \), which is a stricter constraint within \( -3 < t < 3 \).Therefore, the domain for (c) is \( -1 < t < 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of Functions
Understanding the domain of a function is crucial to analyze any vector-valued function accurately. The domain refers to the set of all possible input values (in this case, the variable 't') that will not lead to undefined or nonsensical outputs in the function. Each component of a vector-valued function can influence its domain:

  • For logarithmic functions, like \( \ln(t-1) \) from function (a), the domain is restricted to values where the argument is positive. This condition ensures that the logarithm is defined.
  • Roots restrict the domain as well. For instance, \( \sqrt{20-t} \) requires that the expression inside the square root be non-negative.
  • Rational expressions can introduce additional constraints, as division by zero is undefined. This is seen in function (c) where expressions such as \( \frac{1}{\sqrt{1-t^2}} \) necessitate that the denominator is non-zero.
By examining each term separately and determining their individual constraints, you can derive the overall domain of the vector-valued function using set intersections of these domains.
Logarithmic Functions
Logarithmic functions play a significant role in determining the domain of vector-valued functions. The logarithm, denoted as \( \ln(x) \), is defined only when the argument \( x \) is greater than zero. Understanding this is essential because:

  • The absence of a positive input results in undefined logarithmic values, leading to invalid elements within the function.
  • As seen in part (a) where the expression \( \ln(t-1) \) implies that \( t \) should be greater than 1 to satisfy the condition that \( t-1 > 0 \).
In function (b), transforming \( \ln(t^{-1}) \) to \( -\ln(t) \) helps to understand that \( t \) must indeed be positive due to the same fundamental principle. By verifying logarithmic components individually, one can accurately determine sections of the domain critical for proper functioning.
Trigonometric Functions
Trigonometric functions such as \( \tan^{-1}(t) \), found in vector-valued functions, have distinct characteristics with regard to their domain. Inverse trigonometric functions like \( \tan^{-1}(t) \) are intriguing as they are defined for all real numbers. The behavior of these functions eliminates the worries of having restrictive conditions tied to their operation:

  • Unlike traditional trigonometric functions which can sometimes be undefined due to vertical asymptotes, \( \tan^{-1}(t) \) gives a valid output for any real number \( t \).
  • This absolute broad domain is reflected well in function (b), where \( \tan^{-1}(t) \) allows the rest of the function to be unbounded in that particular component.
Thus, trigonometric functions can both simplify and extend the domain of vector-valued functions by introducing more inclusive intervals. Recognizing this property is beneficial when analyzing the domain influence of trigonometric components.
Interval Notation
Interval notation is a simplified way to represent the domain of functions. This notation provides a clear illustration of which values belong to the domain without relying solely on inequalities. The format is as follows:
  • Square brackets \([a, b]\) indicate that the interval includes the endpoints \(a\) and \(b\), making them part of the solution set.
  • Parentheses \((a, b)\) are used when the endpoints are not included in the interval.
For example, the domain of function (a) is written as \( (1, 20] \), meaning \( t \) can take any value greater than 1 and up to and including 20. Meanwhile, function (c) is represented as \((-1, 1)\), excluding the endpoints, because \( t \) cannot equal -1 or 1 without making the expression undefined.

By mastering interval notation, you can concisely and effectively convey complex domain information in a straightforward format.

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Most popular questions from this chapter

Find the curvature \(\kappa,\) the unit tangent vector \(\mathbf{T},\) the unit normal vector \(\mathbf{N},\) and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). $$ \mathbf{r}(t)=\cos ^{3} t \mathbf{i}+\sin ^{3} t \mathbf{k} ; t_{1}=\pi / 2 $$

Sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T},\) and \(\kappa\) at the point where \(t=t_{1}\). $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} ; \quad 0 \leq t \leq 2 ; t_{1}=1 $$

Sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T},\) and \(\kappa\) at the point where \(t=t_{1}\). $$ \mathbf{r}(t)=t^{2} \mathbf{i}+(2 t+1) \mathbf{j} ; \quad 0 \leq t \leq 2 ; t_{1}=1 $$

Sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T},\) and \(\kappa\) at the point where \(t=t_{1}\). $$ \mathbf{r}(t)=\frac{t^{2}}{8} \mathbf{i}+5 \cos t \mathbf{j}+5 \sin t \mathbf{k} ; \quad 0 \leq t \leq 4 \pi ; t_{1}=\pi $$

Find the point of the curve at which the curvature is a maximum. $$ y=\sinh x $$
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