(a) The curve is a spiral on \([0, 2\pi]\), starting at the pole with \(\theta = 0\) and increasing as \(\theta\) increases, intersecting the initial line at one other point.

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_0^{2\pi} \theta^2 e^{\frac{1}{4}\theta} \, d\theta\)

Integrate by parts twice:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4}\theta} \right]_0^{2\pi} - \frac{1}{2} \int_0^{2\pi} 8\theta e^{\frac{1}{4}\theta} \, d\theta\)

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4}\theta} \right]_0^{2\pi} - 4 \left[ 4\theta e^{\frac{1}{4}\theta} \right]_0^{2\pi} + 16 \int_0^{2\pi} e^{\frac{1}{4}\theta} \, d\theta\)

\(= \left[ 2\theta^2 e^{\frac{1}{4}\theta} - 16\theta e^{\frac{1}{4}\theta} + 64e^{\frac{1}{4}\theta} \right]_0^{2\pi}\)

\(= (8\pi^2 - 32\pi + 64)e^{\frac{1}{2}\pi} - 64\)

(c) Use \(y = r\sin \theta\) and differentiate:

\(\frac{dy}{d\theta} = \theta e^{\frac{1}{8}\theta} \cos \theta + \sin \theta \left( \frac{1}{8}\theta e^{\frac{1}{8}\theta} + e^{\frac{1}{8}\theta} \right)\)

\(e^{\frac{1}{8}\theta} \neq 0 \Rightarrow \theta \cos \theta + \left( \frac{1}{8}\theta + 1 \right) \sin \theta = 0\)

Check sign change:

\(5\cos 5 + \left( \frac{5}{8} + 1 \right)\sin 5 = -0.1399\)

\(5.05\cos 5.05 + \left( \frac{5.05}{8} + 1 \right)\sin 5.05 = 0.1336\)