Answer
$$
x=t \sin t, \quad y=t \cos t, \quad 0 \leq t \leq 1 $$
the exact length of the curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{1} \sqrt{t^{2}+1} d t \\
&=\frac{1}{2}[\sqrt{2}+\ln (1+\sqrt{2})] .
\end{aligned}
$$
Work Step by Step
$$
x=t \sin t, \quad y=t \cos t, \quad 0 \leq t \leq 1
$$
then
$$
d x / d t=t \cos t+\sin t , \quad d y / d t=-t \sin t+\cos t,
$$
so
$$
(d x / d t)^{2}+(d y / d t)^{2} =t^{2} \cos ^{2} t+2 t \sin t \cos t+\sin ^{2} t+t^{2} \sin ^{2} t-2 t \sin t \cos t+\cos ^{2} t=t^{2}+1
$$
Thus the exact length of the curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{1} \sqrt{t^{2}+1} d t \\
& \quad \quad \quad \left[\text {Let } t=\tan \theta , \text {then } d t =\sec^{2} \theta d \theta \right]\\
&=\int_{0}^{\pi / 4} \sec \theta \sec ^{2} \theta d \theta \\
&=\int_{0}^{\pi / 4} \sec ^{3} \theta d \theta\\
&=\frac{1}{2}[\sec \theta \tan \theta+\ln |\sec \theta+\tan \theta|]_{0}^{\pi / 4} \\
& =\frac{1}{2}[\sqrt{2} \cdot 1+\ln (1+\sqrt{2})-0-\ln (1+0)] \\
&=\frac{1}{2}[\sqrt{2}+\ln (1+\sqrt{2})] .
\end{aligned}
$$
