(a) The curve is sketched with the initial line drawn. The shape is correct with \(r\) strictly decreasing. The greatest distance from the pole is \(1 - e^{-\frac{1}{2}\pi}\).

(b) The area is calculated using the integral:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} \left( e^{-\theta} - e^{-\frac{1}{2}\pi} \right)^2 \, d\theta\)

Expanding and integrating:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} \left( e^{-2\theta} - 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi} \right) \, d\theta\)

\(\frac{1}{2} \left[ -\frac{1}{2}e^{-2\theta} + 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi}\theta \right]_{0}^{\frac{1}{2}\pi}\)

\(\frac{1}{2} \left( -\frac{1}{2}e^{-\pi} + 2e^{-\pi} + \frac{1}{2}\pi e^{-\pi} - 2e^{-\frac{1}{2}\pi} \right) = \frac{3}{4}e^{-\pi} + \frac{1}{4}\pi e^{-\pi} - e^{-\frac{1}{2}\pi} + \frac{1}{4}\)

(c) Using \(y = (e^{-\theta} - e^{-\frac{1}{2}\pi}) \sin \theta\), differentiate:

\(\frac{dy}{d\theta} = (e^{-\theta} - e^{-\frac{1}{2}\pi}) \cos \theta + \sin \theta (-e^{-\theta}) = 0\)

\(\left[ \theta = \frac{1}{2}\pi \Rightarrow 1 + \frac{-e^{-\theta}}{e^{-\theta} - e^{-\frac{1}{2}\pi}} \right] \tan \theta = 0 \Rightarrow 1 - e^{\theta - \frac{1}{2}\pi} - \tan \theta = 0\)

Checking values:

\(1 - e^{0.56 - \frac{1}{2}\pi} - \tan 0.56 = 0.00912\)

\(1 - e^{0.57 - \frac{1}{2}\pi} - \tan 0.57 = -0.00856\)

There is a sign change, confirming a root between 0.56 and 0.57.