Answer: lpha=\frac{4}{3}\ln\left(1+\frac{3s}{5}\right).
For a polar curve \(r=f(\theta)\), the arc length from \(\theta=0\) to \(\theta=\alpha\) is
\(s=\int_0^{\alpha}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta\).
Here \(r=e^{\frac{3}{4}\theta}\), so
\(\frac{dr}{d\theta}=\frac{3}{4}e^{\frac{3}{4}\theta}\).
Therefore,
\(r^2=e^{\frac{3}{2}\theta}\), \qquad \(\left(\frac{dr}{d\theta}\right)^2=\frac{9}{16}e^{\frac{3}{2}\theta}\).
Hence
\(s=\int_0^{\alpha}\sqrt{e^{\frac{3}{2}\theta}+\frac{9}{16}e^{\frac{3}{2}\theta}}\,d\theta\)
\(=\frac{5}{4}\int_0^{\alpha}e^{\frac{3}{4}\theta}\,d\theta\).
Integrating,
\(s=\frac{5}{4}\left[\frac{4}{3}e^{\frac{3}{4}\theta}\right]_0^{\alpha}=\frac{5}{3}\left(e^{\frac{3}{4}\alpha}-1\right).\)
So
\(e^{\frac{3}{4}\alpha}=1+\frac{3s}{5}\).
Taking natural logarithms gives
\(\frac{3}{4}\alpha=\ln\left(1+\frac{3s}{5}\right),\)
so
\(\alpha=\frac{4}{3}\ln\left(1+\frac{3s}{5}\right).\)
