Answer
$$
x=t+\sqrt{t} , \quad y=t-\sqrt{t},\quad 0 \leq t \leq 1
$$
the length of the given curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{1} \sqrt{2+\frac{1}{2 t}} d t \\
&=\lim _{t \rightarrow 0^{+}} \int_{t}^{1} \sqrt{2+\frac{1}{2 t}} d t\\
& \approx 2.0915
\end{aligned}
$$
Work Step by Step
$$
x=t+\sqrt{t} , \quad y=t-\sqrt{t},\quad 0 \leq t \leq 1
$$
then
$$
d x / d t=1+\frac{1}{2 \sqrt{t}} , \quad d y / d t=1-\frac{1}{2 \sqrt{t}},
$$
so
$$
\begin{aligned}
(d x / d t)^{2}+(d y / d t)^{2}& =\left(1+\frac{1}{2 \sqrt{t}}\right)^{2}+\left(1-\frac{1}{2 \sqrt{t}}\right)^{2} \\
&=1+\frac{1}{\sqrt{t}}+\frac{1}{4 t}+1-\frac{1}{\sqrt{t}}+\frac{1}{4 t} \\
&=2+\frac{1}{2 t}
\end{aligned}
$$
Thus the length of the given curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{1} \sqrt{2+\frac{1}{2 t}} d t \\
&=\lim _{t \rightarrow 0^{+}} \int_{t}^{1} \sqrt{2+\frac{1}{2 t}} d t\\
& \approx 2.0915
\end{aligned}
$$
