Vector Functions — Question 1

Question 1 – Solution

Strategy. A vector function is defined only where all three component functions are defined. We therefore intersect their domains.

Step 1: Square-root condition Real values require $$t+2\ge 0\quad\Longrightarrow\quad t\ge-2.$$

Step 2: Logarithm condition The logarithm requires a strictly positive argument: $$5-t>0\quad\Longrightarrow\quad t<5.$$

Step 3: Denominator condition We must have $$t^2-1\ne 0\quad\Longrightarrow\quad t\ne-1,1.$$ Intersecting the three conditions gives $$\boxed{\operatorname{Dom}(\mathbf r)=[-2,-1)\cup(-1,1)\cup(1,5)}.$$

Step 4: Evaluation $$\boxed{\mathbf r(2)=\left\langle 2,\ln 3,\tfrac 13\right\rangle}.$$

Verification. At $t=-2$ the square root is zero, so that endpoint is included. At $t=5$ the logarithm sees zero; below $-2$ its first component is nonreal; and at $t=\pm 1$ the third component divides by zero.