Answer: (i) The curve starts at the pole when \(\theta=0\), since \(r=a(1-e^0)=0\). As \(\theta\) increases, \(r\) increases monotonically and tends to \(a\) as \(\theta\to\infty\), so the curve spirals out from the pole and approaches the circle \(r=a\). Near the pole, the initial tangent is along the initial line \(\theta=0\).

(ii) The area of a region in polar coordinates is \(\frac12\int r^2\,d\theta\). Hence

\(A=\frac12\int_{\ln 2}^{\ln 4} a^2\left(1-e^{-\theta}\right)^2\,d\theta\)

\(=\frac{a^2}{2}\int_{\ln 2}^{\ln 4}\left(1-2e^{-\theta}+e^{-2\theta}\right)d\theta\).

Integrating term by term gives

\(A=\frac{a^2}{2}\left[\theta+2e^{-\theta}-\frac12 e^{-2\theta}\right]_{\ln 2}^{\ln 4}.\)

Now evaluate at the limits:

At \(\theta=\ln 4\), \(e^{-\theta}=\frac14\) and \(e^{-2\theta}=\frac1{16}\), so

\(\ln 4+2\cdot\frac14-\frac12\cdot\frac1{16}=\ln 4+\frac12-\frac1{32}.\)

At \(\theta=\ln 2\), \(e^{-\theta}=\frac12\) and \(e^{-2\theta}=\frac14\), so

\(\ln 2+2\cdot\frac12-\frac12\cdot\frac14=\ln 2+1-\frac18.\)

Therefore

\(A=\frac{a^2}{2}\left(\ln 4-\ln 2+\frac12-\frac1{32}-1+\frac18\right).\)

Since \(\ln 4-\ln 2=\ln 2\), this simplifies to

\(A=\frac{a^2}{2}\left(\ln 2-\frac{13}{32}\right).\)

(i) For \(r=a(1-e^{-\theta})\), when \(\theta=0\) we have \(r=0\), so the curve passes through the pole at the start of the graph. For increasing \(\theta\), \(e^{-\theta}\) decreases, so \(r\) increases and tends to \(a\) as \(\theta\) becomes large. Hence the curve spirals out from the pole and approaches the circle \(r=a\). The initial direction at the pole is along \(\theta=0\), so the curve is tangent to the initial line there.

(ii) The area enclosed between \(\theta=\ln 2\) and \(\theta=\ln 4\) is

\(A=\frac12\int_{\ln 2}^{\ln 4} r^2\,d\theta=\frac12\int_{\ln 2}^{\ln 4} a^2(1-e^{-\theta})^2\,d\theta.\)

Expanding the square gives

\(A=\frac{a^2}{2}\int_{\ln 2}^{\ln 4}(1-2e^{-\theta}+e^{-2\theta})\,d\theta.\)

Integrate term by term:

\(\int (1-2e^{-\theta}+e^{-2\theta})\,d\theta=\theta+2e^{-\theta}-\frac12 e^{-2\theta}.\)

So

\(A=\frac{a^2}{2}\left[\theta+2e^{-\theta}-\frac12 e^{-2\theta}\right]_{\ln 2}^{\ln 4}.\)

Now substitute the limits:

\(A=\frac{a^2}{2}\left(\ln 4+2\cdot\frac14-\frac12\cdot\frac1{16}-\ln 2-2\cdot\frac12+\frac12\cdot\frac14\right).\)

Since \(\ln 4=2\ln 2\), this becomes

\(A=\frac{a^2}{2}\left(\ln 2+\frac12-\frac1{32}-1+\frac18\right)=\frac{a^2}{2}\left(\ln 2-\frac{13}{32}\right).\)