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Sagot :
To find the domain of the vector function
[tex]\[ r(t) = \cos(t) \mathbf{i} + \ln(t) \mathbf{j} + \frac{1}{t-4} \mathbf{k}, \][/tex]
we need to consider the domain of each individual component of the vector function and then determine their intersection.
1. Component [tex]\(\cos(t)\)[/tex] [tex]\(\mathbf{i}\)[/tex]:
- The function [tex]\(\cos(t)\)[/tex] is defined for all real numbers [tex]\(t\)[/tex]. Therefore, its domain is:
[tex]\[ (-\infty, \infty) \][/tex]
2. Component [tex]\(\ln(t)\)[/tex] [tex]\(\mathbf{j}\)[/tex]:
- The function [tex]\(\ln(t)\)[/tex] is defined for [tex]\(t > 0\)[/tex]. Therefore, its domain is:
[tex]\[ (0, \infty) \][/tex]
3. Component [tex]\(\frac{1}{t-4}\)[/tex] [tex]\(\mathbf{k}\)[/tex]:
- The function [tex]\(\frac{1}{t-4}\)[/tex] is defined for all [tex]\(t\)[/tex] except [tex]\(t = 4\)[/tex]. Therefore, its domain is:
[tex]\[ (-\infty, 4) \cup (4, \infty) \][/tex]
Next, we find the intersection of these domains:
- The domain [tex]\(\cos(t)\)[/tex] is [tex]\((-∞, ∞)\)[/tex].
- The domain [tex]\(\ln(t)\)[/tex] is [tex]\((0, ∞)\)[/tex].
- The domain [tex]\(\frac{1}{t-4}\)[/tex] is [tex]\((-\infty, 4) \cup (4, ∞)\)[/tex].
The intersection of the domains [tex]\((-\infty, \infty)\)[/tex], [tex]\((0, \infty)\)[/tex], and [tex]\((-\infty, 4) \cup (4, \infty)\)[/tex] is where all three conditions are satisfied simultaneously. This gives:
[tex]\[ (0, 4) \cup (4, \infty) \][/tex]
Therefore, the domain of the vector function [tex]\( r(t) = \cos(t) \mathbf{i} + \ln(t) \mathbf{j} + \frac{1}{t-4} \mathbf{k} \)[/tex] is
[tex]\[ (0, 4) \cup (4, \infty). \][/tex]
[tex]\[ r(t) = \cos(t) \mathbf{i} + \ln(t) \mathbf{j} + \frac{1}{t-4} \mathbf{k}, \][/tex]
we need to consider the domain of each individual component of the vector function and then determine their intersection.
1. Component [tex]\(\cos(t)\)[/tex] [tex]\(\mathbf{i}\)[/tex]:
- The function [tex]\(\cos(t)\)[/tex] is defined for all real numbers [tex]\(t\)[/tex]. Therefore, its domain is:
[tex]\[ (-\infty, \infty) \][/tex]
2. Component [tex]\(\ln(t)\)[/tex] [tex]\(\mathbf{j}\)[/tex]:
- The function [tex]\(\ln(t)\)[/tex] is defined for [tex]\(t > 0\)[/tex]. Therefore, its domain is:
[tex]\[ (0, \infty) \][/tex]
3. Component [tex]\(\frac{1}{t-4}\)[/tex] [tex]\(\mathbf{k}\)[/tex]:
- The function [tex]\(\frac{1}{t-4}\)[/tex] is defined for all [tex]\(t\)[/tex] except [tex]\(t = 4\)[/tex]. Therefore, its domain is:
[tex]\[ (-\infty, 4) \cup (4, \infty) \][/tex]
Next, we find the intersection of these domains:
- The domain [tex]\(\cos(t)\)[/tex] is [tex]\((-∞, ∞)\)[/tex].
- The domain [tex]\(\ln(t)\)[/tex] is [tex]\((0, ∞)\)[/tex].
- The domain [tex]\(\frac{1}{t-4}\)[/tex] is [tex]\((-\infty, 4) \cup (4, ∞)\)[/tex].
The intersection of the domains [tex]\((-\infty, \infty)\)[/tex], [tex]\((0, \infty)\)[/tex], and [tex]\((-\infty, 4) \cup (4, \infty)\)[/tex] is where all three conditions are satisfied simultaneously. This gives:
[tex]\[ (0, 4) \cup (4, \infty) \][/tex]
Therefore, the domain of the vector function [tex]\( r(t) = \cos(t) \mathbf{i} + \ln(t) \mathbf{j} + \frac{1}{t-4} \mathbf{k} \)[/tex] is
[tex]\[ (0, 4) \cup (4, \infty). \][/tex]
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