Final Answer:
The domain of the vector function r(t) = 9 - t², e^(-3t), ln(t¹) is 0 < t < ∞ (excluding zero).
Step-by-step explanation:
In a vector function r(t) comprising three component functions, each component's domain must be considered separately.
The first component, 9 - t², doesn't involve any restrictions as it's a polynomial. Therefore, its domain covers all real numbers.
The second component, e^(-3t), is a part of the exponential function, where the base (e) is positive and defined for all real numbers t. Thus, its domain extends to -∞ < t < ∞.
The third component, ln(t¹), involves the natural logarithm function. For ln(t¹) to be defined, the argument t¹ must be positive. Therefore, t should be greater than zero. Hence, the domain of this component is t > 0.
Combining the domains of the three components, we find the overall domain of r(t). Considering all constraints, the domain of the vector function r(t) is 0 < t < ∞ (excluding zero) to satisfy the conditions of each component function within the vector.