Find the domain of the vector function. (Enter your answer in interval notation.) $r(t) = \frac{t - 2}{t + 2} \mathbf{i} + \sin(t) \mathbf{j} + \ln(49 - t^2) \mathbf{k}$

Neetesh Kumar | December 14, 2024

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This is the solution to Math 1C
Assignment: 13.1 Question Number 15
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Step-by-step solution:

To find the domain of the vector function, we need to consider the domain of each component individually.

1. First component: $\frac{t - 2}{t + 2}$

The first component involves a rational function. The denominator $t + 2$ cannot be zero because division by zero is undefined. Therefore, we require:

$t + 2 \neq 0$

Solving for $t$:

$t \neq -2$

Thus, the domain of the first component is all real numbers except $t = -2$.

2. Second component: $\sin(t)$

The sine function, $\sin(t)$, is defined for all real numbers. Therefore, the domain of the second component is $(-\infty, \infty)$.

3. Third component: $\ln(49 - t^2)$

The natural logarithm function, $\ln(x)$, is defined for positive arguments, i.e., $x > 0$. Therefore, for $\ln(49 - t^2)$ to be defined, we require:

$49 - t^2 > 0$

Rearranging:

$t^2 < 49$

Taking the square root of both sides:

$-7 < t < 7$

Thus, the domain of the third component is $(-7, 7)$.

4. Combine the domains

To find the domain of the entire vector function, we take the intersection of the individual domains:

• First component: $(-\infty, \infty)$ except $t = -2$
• Second component: $(-\infty, \infty)$
• Third component: $(-7, 7)$

The intersection of these three domains is:

$t \in (-7, 7) \quad \text{with} \quad t \neq -2$

Thus, the domain of the vector function is:

$(-7, -2) \cup (-2, 7)$

Final Answer:

The domain of the vector function is:

$\boxed{(-7, -2) \cup (-2, 7)}$

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