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Find the domain of the following vector-valued function. \( r(t)=\cos 3 t i+e^{2 \sqrt{t}} j+\frac{15}{t} k \) Select the correct choice below and fill in any answer box(es) to complete your choice. A. \( \{t: t \leq \square\} \) B. \( \{t: t \leq \square \) or \( t \geq \square\} \) C. \( \{t: \square \leq t \leq \square\} \) D. \( \{t: t>\square\} \) E. \( \{t: t \geq \square\} \) F. \( \{t: \square\square\} \)

Ask by Boyd Watkins. in the United States
Feb 03,2025

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Answer

The domain of the function is all real numbers greater than 0. So, the correct choice is D. \( \{t: t > 0\} \).

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The function \( r(t) = \cos(3t)i + e^{2\sqrt{t}}j + \frac{15}{t}k \) has specific constraints for its domain. The term \( e^{2\sqrt{t}} \) is defined for \( t \geq 0 \), while \( \frac{15}{t} \) requires that \( t \neq 0 \) (as division by zero is not defined). Therefore, taking these into consideration, the domain of the function is \( t > 0 \). The result leads us to conclude that the answer is: D. \( \{t: t > 0\} \). When you explore vector-valued functions like this, you're often combining multiple mathematical concepts. This particular function showcases neat features such as oscillating behavior (from the cosine component) and exponential growth (from the \( e^{2\sqrt{t}} \)). It's a compact little mixture that emphasizes the beauty of mathematics! For real-world applications, vector-valued functions can be used to describe various phenomena, like the motion of a particle in space. Think of a roller coaster: as it moves through the air, its position can be defined using a vector for each point in time, making these mathematical tools quite handy in physics and engineering!

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