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

To find the domain of the vector-valued function \( \mathbf{r}(t)=\sqrt{t+3} i+\sqrt{t-3} j \), we need to ensure that the expressions under the square root functions are non-negative. 1. For \( \sqrt{t+3} \), we require: \[ t + 3 \geq 0 \implies t \geq -3 \] 2. For \( \sqrt{t-3} \), we require: \[ t - 3 \geq 0 \implies t \geq 3 \] Combining these conditions, the most restrictive one is \( t \geq 3 \). Therefore, the domain of the function is \( \{t: t \geq 3\} \). A. \( \{t: t \geq 3\} \) --- For a bit of fun, did you know that the concept of a function's domain goes way back to the origins of calculus in the 17th century? Mathematicians like Newton and Leibniz were already grappling with the idea of limits and the conditions under which functions are defined long before calculators were a twinkle in anyone's eye! If you're curious about functions and calculus, a great read is "Calculus Made Easy" by Silvanus P. Thompson. It's a classic that demystifies the concepts of calculus in a way that’s engaging and accessible, perfect for anyone looking to strengthen their math skills without diving into overly complex jargon!