Answer
$\sqrt{5}e^{\pi/4}-\sqrt{5}$
Work Step by Step
Recall: The length of the polar curve $r=f(\theta)$ for $a\leq \theta\leq b$ is formulated by $L=\int_a^b\sqrt{r^2+(dr/d\theta)^2}d\theta$.
Given: $f(\theta)=e^{\theta/2}$ for $0\leq \theta\leq \pi/2$
Find $dr/\theta$:
$dr/d\theta=\frac{d}{d\theta}(e^{\theta/2})=\frac{1}{2}\cdot e^{\theta/2}=\frac{e^{\theta/2}}{2}$
Evaluate the length of the curve:
$L=\int_0^{\pi/2}\sqrt{(e^{\theta/2})^2+(\frac{e^{\theta/2}}{2})^2}d\theta$
$L=\int_0^{\pi/2}\sqrt{e^\theta+\frac{e^{\theta}}{4}}d\theta$
$L=\int_0^{\pi/2}\sqrt{\frac{5e^{\theta}}{4}}d\theta$
$L=\int_0^{\pi/2}\frac{\sqrt{5}}{2}e^{\theta/2}d\theta$
$L=[\sqrt{5}e^{\theta/2}]_0^{\pi/2}$
$L=\sqrt{5}e^{(\pi/2)/2}-\sqrt{5}e^{0/2}$
$L=\sqrt{5}e^{\pi/4}-\sqrt{5}e^0$
$L=\sqrt{5}e^{\pi/4}-\sqrt{5}$
Thus, the length is $\sqrt{5}e^{\pi/4}-\sqrt{5}$.
