Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Problem 2: Determine the Domains of the Vector-Valued Function**
Given the vector-valued function:
\[
\mathbf{r}_2(t) = \left\langle \sqrt{8 - t^3}, \ln t, e^{\sqrt{t}} \right\rangle
\]
**Explanation:**
To find the domain of the vector-valued function \(\mathbf{r}_2(t)\), we must consider the domain restrictions for each component of the vector:
1. **Square Root Component \(\sqrt{8 - t^3}\):**
- The expression under the square root, \(8 - t^3\), must be non-negative.
- Therefore, \(8 - t^3 \geq 0\).
- Solving for \(t\), we get \(t^3 \leq 8\), which implies \(t \leq 2\).
2. **Logarithmic Component \(\ln t\):**
- The argument of the logarithm, \(t\), must be positive.
- Therefore, \(t > 0\).
3. **Exponential Component \(e^{\sqrt{t}}\):**
- The base of the exponent is valid for all real numbers, but the square root requires \(t\) to be non-negative.
- Thus, \(t \geq 0\).
**Overall Domain:**
By combining these conditions, the domain for \(\mathbf{r}_2(t)\) is \(0 < t \leq 2\).
The entire function is defined for the intersection of the domains of its components, where \(t\) is strictly greater than zero and less than or equal to two.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93c425ca-5be8-4a48-b5f2-917fc035445e%2F7b5721a0-4246-435a-bb25-8d1ae80ad702%2F8m523l8n_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2: Determine the Domains of the Vector-Valued Function**
Given the vector-valued function:
\[
\mathbf{r}_2(t) = \left\langle \sqrt{8 - t^3}, \ln t, e^{\sqrt{t}} \right\rangle
\]
**Explanation:**
To find the domain of the vector-valued function \(\mathbf{r}_2(t)\), we must consider the domain restrictions for each component of the vector:
1. **Square Root Component \(\sqrt{8 - t^3}\):**
- The expression under the square root, \(8 - t^3\), must be non-negative.
- Therefore, \(8 - t^3 \geq 0\).
- Solving for \(t\), we get \(t^3 \leq 8\), which implies \(t \leq 2\).
2. **Logarithmic Component \(\ln t\):**
- The argument of the logarithm, \(t\), must be positive.
- Therefore, \(t > 0\).
3. **Exponential Component \(e^{\sqrt{t}}\):**
- The base of the exponent is valid for all real numbers, but the square root requires \(t\) to be non-negative.
- Thus, \(t \geq 0\).
**Overall Domain:**
By combining these conditions, the domain for \(\mathbf{r}_2(t)\) is \(0 < t \leq 2\).
The entire function is defined for the intersection of the domains of its components, where \(t\) is strictly greater than zero and less than or equal to two.
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