Find the domain of 
r(t) = 5e−t i + e−t j + ln(t − 1)k.
 (Enter your answer using interval notation.)

We are tasked with finding the domain of the vector function \( r(t) = 5e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + \ln(t - 1) \mathbf{k} \).

The domain is determined by identifying any restrictions on the input \( t \) that would make any of the terms in the function undefined.

1. **Exponential terms:**
   Both \( 5e^{-t} \) and \( e^{-t} \) are defined for all real values of \( t \). Exponential functions have no restrictions.

2. **Logarithmic term:**
   The term \( \ln(t - 1) \) is only defined when its argument \( t - 1 > 0 \), or equivalently:
   \[
   t > 1
   \]
   This restricts \( t \) to values greater than 1.

### Domain
The domain of \( r(t) \) is the set of \( t \)-values where all components of the function are defined. Since the logarithmic term restricts \( t \) to values greater than 1, the domain of \( r(t) \) is:
\[
(1, \infty)
\]

### Final Answer:
The domain of \( r(t) \) is \( (1, \infty) \).

Would you like further details on the logarithmic properties or any other part of the solution?

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Here are 5 related questions to expand on this:

1. What is the range of the logarithmic function \( \ln(t - 1) \)?
2. How does the behavior of \( r(t) \) change as \( t \to \infty \)?
3. What happens to the value of \( e^{-t} \) as \( t \) approaches negative infinity?
4. Can you explain how to differentiate the vector function \( r(t) \)?
5. How does the domain change if the logarithmic term was \( \ln(t + 1) \) instead?

**Tip:** Always check for restrictions in logarithmic functions by ensuring the argument is positive!

Find the domain of r(t) = 5e−t i + e−t j + ln(t − 1)k. (Enter your answer using interval notation.)

This solution provides a step-by-step approach to finding the domain of the vector function r(t) = 5e−t i + e−t j + ln(t − 1)k. Key mathematical concepts such as exponential functions, logarithmic functions, and domain restrictions are explained. Suitable for undergraduate students studying calculus.

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