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