r(t) = < ln(13t), sqrt(t+19), 1/sqrt(7t) >
2 Answers

Consider the domain of each component of r(t). Then the domain of r(t) is the intersection of the domains of the components.
The domain of ln(13t) (assuming we’re working with the real numbers) is t > 0. In interval notation, (0,∞).
The domain of √(t + 19) is t ≥ 19 (the argument must be nonnegative). In interval notation, [19, ∞).
The last component is a bit tricky. The domain of √(7 – x) is the set of numbers for which 7 – x ≥ 0, that is, for x ≤ 7. But we cannot allow t = 7, since that produces division by zero, which is not defined. Hence, the domain of 1/√(7 – t) is t < 7. In interval notation, (∞, 7).
So now we need to find (0,∞) ∩ [19, ∞) ∩ (∞, 7).
First, do (0,∞) ∩ [19, ∞). This is clearly (0, ∞) since it is wholly contained in [19, ∞)
Now, consider (0,∞) ∩ (∞, 7). It might be easiest to solve this by drawing a number line and graphing each interval. You will find that the intersection is (0, 7).
Thus, the domain of the vector function r(t) is (0, 7).

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