5. Find all values of t for which the vector-valued function is continuous: r(t) = Inti + ₁ + √t-2k 1 t-3 Express your answer in interval notation.
5. Find all values of t for which the vector-valued function is continuous: r(t) = Inti + ₁ + √t-2k 1 t-3 Express your answer in interval notation.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem 5
**Objective:**
Find all values of \( t \) for which the vector-valued function is continuous and express your answer in interval notation.
**Given Function:**
\[ \mathbf{r}(t) = \ln(t) \mathbf{i} + \frac{1}{t-3} \mathbf{j} + \sqrt{t-2} \mathbf{k} \]
**Method:**
To determine the values of \( t \) for which the function \( \mathbf{r}(t) \) is continuous, we need to examine each component of the vector function separately and identify any points of discontinuity:
1. **First Component - \( \ln(t) \mathbf{i} \):**
- The natural logarithm function \( \ln(t) \) is defined and continuous for \( t > 0 \).
2. **Second Component - \( \frac{1}{t-3} \mathbf{j} \):**
- The function \( \frac{1}{t-3} \) has a discontinuity at \( t = 3 \) because it becomes undefined.
3. **Third Component - \( \sqrt{t-2} \mathbf{k} \):**
- The square root function \( \sqrt{t-2} \) is defined and continuous for \( t \geq 2 \).
**Solution:**
Considering the domains of each component, the vector-valued function \( \mathbf{r}(t) \) is continuous where all three components are simultaneously continuous. This occurs where:
- \( t > 0 \)
- \( t \neq 3 \)
- \( t \geq 2 \)
Combining these conditions in interval notation:
- Since \( t \geq 2 \) encompasses \( t > 0 \)
- Excluding \( t = 3 \),
The set of values for which \( \mathbf{r}(t) \) is continuous is:
\[ [2, 3) \cup (3, \infty) \]
**Conclusion:**
The values of \( t \) for which the vector-valued function \( \mathbf{r}(t) \) is continuous are \( \boxed{[2, 3) \cup (3, \infty)} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F55815b3c-2575-4201-a9dc-5a4b41a14c1d%2Fd164893d-85c2-454e-aad4-9d21542d6c70%2F7ij1rt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 5
**Objective:**
Find all values of \( t \) for which the vector-valued function is continuous and express your answer in interval notation.
**Given Function:**
\[ \mathbf{r}(t) = \ln(t) \mathbf{i} + \frac{1}{t-3} \mathbf{j} + \sqrt{t-2} \mathbf{k} \]
**Method:**
To determine the values of \( t \) for which the function \( \mathbf{r}(t) \) is continuous, we need to examine each component of the vector function separately and identify any points of discontinuity:
1. **First Component - \( \ln(t) \mathbf{i} \):**
- The natural logarithm function \( \ln(t) \) is defined and continuous for \( t > 0 \).
2. **Second Component - \( \frac{1}{t-3} \mathbf{j} \):**
- The function \( \frac{1}{t-3} \) has a discontinuity at \( t = 3 \) because it becomes undefined.
3. **Third Component - \( \sqrt{t-2} \mathbf{k} \):**
- The square root function \( \sqrt{t-2} \) is defined and continuous for \( t \geq 2 \).
**Solution:**
Considering the domains of each component, the vector-valued function \( \mathbf{r}(t) \) is continuous where all three components are simultaneously continuous. This occurs where:
- \( t > 0 \)
- \( t \neq 3 \)
- \( t \geq 2 \)
Combining these conditions in interval notation:
- Since \( t \geq 2 \) encompasses \( t > 0 \)
- Excluding \( t = 3 \),
The set of values for which \( \mathbf{r}(t) \) is continuous is:
\[ [2, 3) \cup (3, \infty) \]
**Conclusion:**
The values of \( t \) for which the vector-valued function \( \mathbf{r}(t) \) is continuous are \( \boxed{[2, 3) \cup (3, \infty)} \).
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