Step 1

\(\displaystyle{r}{\left({t}\right)}={\left(\sqrt{{{t}}}\right)}{i}+{\left(\sqrt{{{t}-{1}}}\right)}{k}\)

we know for a function to be continuous in Interval then it should be defined at every point in that interval.

we know that inside a root function there can't be a negative number so

t should greater than or equal to 0 and t - 3 should greater than or equal to 0

by taking the intersection of both t should greater than or equal to 3

But at 0 and 3 limit does not exist as the left-hand side limit is not equal to the right-hand side limit.

Step 2

Therefore, the interval is(1, co) on which the vector-valued function is continuous.

\(\displaystyle{r}{\left({t}\right)}={\left(\sqrt{{{t}}}\right)}{i}+{\left(\sqrt{{{t}-{1}}}\right)}{k}\)

we know for a function to be continuous in Interval then it should be defined at every point in that interval.

we know that inside a root function there can't be a negative number so

t should greater than or equal to 0 and t - 3 should greater than or equal to 0

by taking the intersection of both t should greater than or equal to 3

But at 0 and 3 limit does not exist as the left-hand side limit is not equal to the right-hand side limit.

Step 2

Therefore, the interval is(1, co) on which the vector-valued function is continuous.