Find the domain of the vector function. (Enter your answer using interval notation.) $r(t) = \left\langle \sqrt{36 - t^2}, e^{-5t}, \ln(t + 1) \right\rangle$

Neetesh Kumar | December 14, 2024

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This is the solution to Math 1C
Assignment: 13.1 Question Number 14
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Step-by-step solution:

To find the domain of the vector function, we must determine the domain of each individual component of the vector function.

1. First component: $\sqrt{36 - t^2}$

For the square root function to be real, the argument inside the square root must be non-negative:

$36 - t^2 \geq 0$

Rearranging:

$t^2 \leq 36$

Taking the square root of both sides:

$-6 \leq t \leq 6$

So, the domain of the first component is $[-6, 6]$.

2. Second component: $e^{-5t}$

The exponential function $e^{-5t}$ is defined for all real numbers. Therefore, the domain of the second component is $(-\infty, \infty)$.

3. Third component: $\ln(t + 1)$

For the natural logarithm function to be defined, its argument must be positive:

$t + 1 > 0$

Solving for $t$:

$t > -1$

Thus, the domain of the third component is $(-1, \infty)$.

4. Combine the domains

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

• First component: $[-6, 6]$
• Second component: $(-\infty, \infty)$
• Third component: $(-1, \infty)$

The intersection of these three domains is the set of values of $t$ that satisfy all three conditions:

$t \in (-1, 6]$

Final Answer:

The domain of the vector function is:

$\boxed{(-1, 6]}$

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