Find the domain of the vector function 7(t) < vt + 4, ,log(11 – t) > Domain: {t

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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**Finding the Domain of the Vector Function**

Given the vector function:

\[
\vec{r}(t) = \left< \sqrt{t + 4}, \frac{1}{\sqrt{t + 4}}, \log(11 - t) \right>
\]

We need to determine the domain of \(\vec{r}(t)\).

**Considerations for the Domain:**

1. **Square Root:**
   - The expression \(\sqrt{t + 4}\) requires that \(t + 4 \geq 0\).
   - Therefore, \(t \geq -4\).

2. **Fraction with Square Root:**
   - The expression \(\frac{1}{\sqrt{t + 4}}\) requires that \(\sqrt{t + 4} \neq 0\).
   - Therefore, \(t + 4 > 0\) which means \(t > -4\).

3. **Logarithm:**
   - The expression \(\log(11 - t)\) requires that \(11 - t > 0\).
   - Therefore, \(t < 11\).

**Combining All Conditions:**

Taking into account all these conditions, the domain of \(\vec{r}(t)\) is:

\[
\text{Domain: } \{ t \mid -4 < t < 11 \}
\]
Transcribed Image Text:**Finding the Domain of the Vector Function** Given the vector function: \[ \vec{r}(t) = \left< \sqrt{t + 4}, \frac{1}{\sqrt{t + 4}}, \log(11 - t) \right> \] We need to determine the domain of \(\vec{r}(t)\). **Considerations for the Domain:** 1. **Square Root:** - The expression \(\sqrt{t + 4}\) requires that \(t + 4 \geq 0\). - Therefore, \(t \geq -4\). 2. **Fraction with Square Root:** - The expression \(\frac{1}{\sqrt{t + 4}}\) requires that \(\sqrt{t + 4} \neq 0\). - Therefore, \(t + 4 > 0\) which means \(t > -4\). 3. **Logarithm:** - The expression \(\log(11 - t)\) requires that \(11 - t > 0\). - Therefore, \(t < 11\). **Combining All Conditions:** Taking into account all these conditions, the domain of \(\vec{r}(t)\) is: \[ \text{Domain: } \{ t \mid -4 < t < 11 \} \]
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