(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:

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

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

(a) The curve is a spiral starting at the pole \(\theta = 0\) and spiraling outward as \(\theta\) increases to \(2\pi\). The radius \(r\) is strictly increasing, and the gradient is correct at points of intersection with the initial line.

(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:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4} \theta} \right]_{0}^{2\pi} - \int_{0}^{2\pi} 8\theta e^{\frac{1}{4} \theta} d\theta\)

Integrate by parts again:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4} \theta} - 4(4\theta e^{\frac{1}{4} \theta} + 16)e^{\frac{1}{4} \theta} \right]_{0}^{2\pi}\)

\(= \left[ 2\theta^2 e^{\frac{1}{4} \theta} - 16\theta e^{\frac{1}{4} \theta} + 64e^{\frac{1}{4} \theta} \right]_{0}^{2\pi}\)

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

(c) Let \(y = \theta e^{\frac{1}{8} \theta} \sin \theta\). Differentiate:

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

Set \(\frac{dy}{d\theta} = 0\):

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

Check sign change:

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

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

There is a sign change between 5 and 5.05, confirming a root exists.

(a) Differentiate \(r^2 = e^{\sin \theta} \cos \theta\) with respect to \(\theta\) and set \(\frac{dr}{d\theta} = 0\). This gives:

\(-e^{\sin \theta} \sin \theta + e^{\sin \theta} \cos^2 \theta = 0\)

\(\cos^2 \theta = \sin \theta\)

\(1 - \sin^2 \theta = \sin \theta\)

\(\sin \theta = \frac{1}{2}(-1 + \sqrt{5})\)

\(\theta = 0.666\)

\(r = \sqrt{e^{\sin(0.666)} \cos(0.666)} = 1.208\)

(b) Use \(x = e^{\sin \theta} \cos \theta\) and differentiate with respect to \(\theta\):

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

\(-3\sin \theta + \cos^2 \theta = 0\)

\(-3\sin \theta + 1 - \sin^2 \theta = 0\)

\(\sin \theta = \frac{1}{2}(-3 + \sqrt{13})\)

\(\theta = 0.308\)

\(r = \sqrt{e^{\sin(0.308)} \cos(0.308)} = 1.136\)

(c) The sketch shows a closed loop passing through O and one other point on the initial line. It does not go into the 2nd and 3rd quadrants and is more in the 1st than the 4th quadrant.

(d) The area is given by:

\(\frac{1}{2} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} e^{\sin \theta} \cos \theta \, d\theta\)

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

\(= \frac{1}{3}(e - e^{-1})\)

(a) The curve is a spiral shape on \([0, 2\pi]\), starting at the pole with one other point of intersection with the initial line. The maximum distance from the pole is \(a\).

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

\(A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_0^{2\pi} \tan^2\left(\frac{1}{8}\theta\right) \, d\theta\)

Using \(\tan^2 A = \sec^2 A - 1\), we have:

\(= \frac{1}{2} a^2 \left[ 8 \tan\left(\frac{1}{8}\theta\right) - \theta \right]_0^{2\pi}\)

\(= \frac{1}{2} a^2 (8 - 2\pi) = a^2 (4 - \pi)\)

(c) Let \(y = a \tan\left(\frac{1}{8}\theta\right) \sin\theta\). The derivative is:

\(\frac{dy}{d\theta} = a \left( \tan\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta \sec^2\left(\frac{1}{8}\theta\right) \right) = 0\)

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

\(\Rightarrow \sin\left(\frac{1}{4}\theta\right) \cos\theta + \frac{1}{4} \sin\theta = 0\)

\(\Rightarrow 4 \sin\left(\frac{1}{4}\theta\right) \cos\theta + \sin\theta = 0\)

Checking values:

\(4 \sin\left(\frac{1}{4} \times 4.95\right) \cos 4.95 + \sin 4.95 \approx -0.0822\)

\(4 \sin\left(\frac{1}{4} \times 5\right) \cos 5 + \sin 5 \approx 0.118\)

There is a sign change, confirming a root between 4.95 and 5.

(a) Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation \((x^2 + y^2)^2 = 6xy\).

This gives \((r^2)^2 = 6r^2 \cos \theta \sin \theta\).

Thus, \(r^4 = 6r^2 \cos \theta \sin \theta\).

Divide both sides by \(r^2\) (assuming \(r \neq 0\)) to get \(r^2 = 6 \cos \theta \sin \theta\).

Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have \(r^2 = 3 \sin 2\theta\).

(b) The sketch shows a single loop symmetrical about \(\theta = \frac{1}{4}\pi\). The maximum distance from the pole is \(\sqrt{3}\).

(c) The area enclosed by \(C\) is given by \(\frac{1}{2} \int_0^{\frac{1}{2}\pi} r^2 \, d\theta\).

Substitute \(r^2 = 3 \sin 2\theta\) to get \(\frac{3}{2} \int_0^{\frac{1}{2}\pi} \sin 2\theta \, d\theta\).

Integrate to find \(\frac{3}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_0^{\frac{1}{2}\pi} = \frac{3}{2}\).

(d) To find the maximum distance from the initial line, set \(y = r \sin \theta = 3^{\frac{3}{4}} \sin^{\frac{1}{2}} 2\theta \sin \theta\).

Set \(\frac{dy}{d\theta} = 0\) and solve \(\sin^2 2\theta \cos \theta + \sin^{-\frac{1}{2}} 2\theta \cos 2\theta \sin \theta = 0\).

Using trigonometric identities, solve for \(\theta = \frac{1}{3}\pi\).

The maximum distance is \(\frac{3^{\frac{3}{4}}}{2^{\frac{1}{2}}} = 1.40\).

(a) The polar equation is \(r = a(\cos \theta + \sin \theta)\). Using \(x = r\cos \theta\) and \(y = r\sin \theta\), we have:

\(x = a\cos \theta (\cos \theta + \sin \theta) = a(\cos^2 \theta + \cos \theta \sin \theta)\)

\(y = a\sin \theta (\cos \theta + \sin \theta) = a(\sin \theta \cos \theta + \sin^2 \theta)\)

Adding these, \(x + y = a(\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta) = a\).

Thus, \(x + y = a\). The Cartesian equation is \(x^2 + y^2 = ax + ay\), which simplifies to \((x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = (\frac{a}{\sqrt{2}})^2\).

(b) The greatest distance from the pole is when \(\theta = \frac{\pi}{4}\), giving \(r = a\sqrt{2}\).

(c) To verify \(1.25 < \phi < 1.26\), solve \(\cos \phi + \sin \phi - \phi = 0\). Check values: \(\cos 1.25 + \sin 1.25 - 1.25 = 0.01\) and \(\cos 1.26 + \sin 1.26 - 1.26 = -0.002\), showing a sign change.

(d) The area of the smaller region is:

\(\frac{1}{2} \int_0^\phi \theta^2 \, d\theta + \frac{a^2}{2} \int_\phi^{\frac{3}{4}\pi} (\cos \theta + \sin \theta)^2 \, d\theta\)

Calculate each integral and simplify to get:

\(\frac{1}{2}a^2 \left( \frac{3}{4}\pi + \frac{1}{3}\phi^3 - \phi + \frac{1}{2}\cos 2\phi \right)\)

The area of the larger region is:

\(\frac{a^2}{2} \left( \frac{\pi}{3} + \frac{1}{3}\phi - \frac{1}{2}\cos 2\phi \right)\)

(a) Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation:

\((r^2)^2 = 6r^2 \sin \theta \cos \theta\)

\(r^4 = 6r^2 \sin \theta \cos \theta\)

\(r^2 = 3 \sin 2\theta\)

(b) The sketch shows a single loop symmetrical about \(\theta = \frac{1}{4}\pi\). The maximum distance from the pole is \(\sqrt{3}\).

(c) The area enclosed by \(C\) is given by:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_0^{\frac{1}{2}\pi} 3 \sin 2\theta \, d\theta\)

\(= \frac{3}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_0^{\frac{1}{2}\pi}\)

= \(\frac{3}{2}\)

(d) To find the maximum distance from the initial line, set \(y = 3^{\frac{1}{2}} \sin^{\frac{1}{2}} 2\theta \sin \theta\) and solve \(\frac{dy}{d\theta} = 0\).

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

\(\sin 2\theta \cos \theta + \cos 2\theta \sin \theta = 0\)

\(\tan 2\theta = -\tan \theta\)

\(\frac{2 \tan \theta}{1 - \tan^2 \theta} = -\tan \theta\)

\(\theta = \frac{1}{3}\pi\)

Maximum distance is \(\frac{3^{\frac{3}{2}}}{2^{\frac{3}{2}}} = 1.40\).

(a) The curve is sketched with the correct shape, and the section \(\frac{1}{2} \pi \leq \theta \leq \pi\) is correct. The polar coordinates of the point furthest from the pole are \((\sqrt{\pi \arctan \pi}, 0) = (1.99, 0)\).

(b) Using the substitution \(u = \pi - \theta\), the area \(A\) is given by:

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

or

\(A = \frac{1}{2} \int_0^\pi u \arctan(u) \, du\)

Integrating by parts:

\(\frac{1}{2} \left[ \frac{1}{2} (u^2) \arctan(u) \right]_0^\pi - \frac{1}{2} \int_0^\pi \frac{u^2}{1 + u^2} \, du\)

\(\int_0^\pi \frac{u^2}{1 + u^2} \, du = \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 \arctan(\pi) - \frac{1}{4} \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 - \frac{1}{4} \pi = 2.65\)

(c) Let \(y = \sqrt{\pi - \theta} \arctan(\pi - \theta) \sin \theta\).

Find the derivative:

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

Set \(\frac{dy}{d\theta} = 0\) and form the equation:

\(2(\pi - \theta) \arctan(\pi - \theta) \cot \theta - \frac{\pi - \theta}{1 + (\pi - \theta)^2} - \arctan(\pi - \theta) = 0\)

Check sign change between \(\theta = 1.2\) and \(\theta = 1.3\):

\(2(\pi - 1.2) \arctan(\pi - 1.2) \cot 1.2 - \frac{\pi - 1.2}{1 + (\pi - 1.2)^2} - \arctan(\pi - 1.2) = 0.151\)

and

\(2(\pi - 1.3) \arctan(\pi - 1.3) \cot 1.3 - \frac{\pi - 1.3}{1 + (\pi - 1.3)^2} - \arctan(\pi - 1.3) = -0.395\)

Thus, there is a sign change, confirming a root exists between \(\theta = 1.2\) and \(\theta = 1.3\).

(a) The polar coordinates of the point of C furthest from the pole are \((\sqrt{\pi \arctan \pi}, 0) = (1.99, 0)\).

(b) Using the substitution \(u = \pi - \theta\), the area \(A\) is given by:

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

or

\(A = \frac{1}{2} \int_0^\pi u \arctan(u) \, du\)

Integrating by parts:

\(\frac{1}{2} \left[ \frac{1}{2} (u^2) \arctan(u) \right]_0^\pi - \frac{1}{2} \int_0^\pi \frac{u^2}{1 + u^2} \, du\)

\(\int_0^\pi \frac{u^2}{1 + u^2} \, du = \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 \arctan \pi - \frac{1}{4} (\pi^2 + 1) \arctan \pi - \frac{1}{4} \pi = 2.65\)

(c) Let \(y = r \sin \theta\), then:

\(\frac{dy}{d\theta} = \frac{d}{d\theta} \left( \sqrt{(\pi - \theta) \arctan(\pi - \theta)} \sin \theta \right)\)

\(= \frac{(\pi - \theta) \arctan(\pi - \theta) \cos \theta}{\sqrt{(\pi - \theta) \arctan(\pi - \theta)}} + \arctan(\pi - \theta) \sin \theta\)

Setting \(\frac{dy}{d\theta} = 0\) gives:

\(2(\pi - \theta) \arctan(\pi - \theta) \cot \theta - \frac{\pi - \theta}{1 + (\pi - \theta)^2} - \arctan(\pi - \theta) = 0\)

For \(\theta = 1.2\),

\(2(\pi - 1.2) \arctan(\pi - 1.2) \cot 1.2 - \frac{\pi - 1.2}{1 + (\pi - 1.2)^2} - \arctan(\pi - 1.2) = 0.151\)

For \(\theta = 1.3\),

\(2(\pi - 1.3) \arctan(\pi - 1.3) \cot 1.3 - \frac{\pi - 1.3}{1 + (\pi - 1.3)^2} - \arctan(\pi - 1.3) = -0.395\)

There is a sign change, confirming a root exists between \(\theta = 1.2\) and \(\theta = 1.3\).

(a) The sketch of the curve shows two loops symmetric about the line \(\theta = \frac{1}{2} \pi\).

(b) Start with \(r^2 = \sin 2\theta \cos \theta\). Use \(\sin 2\theta = 2 \sin \theta \cos \theta\) and polar to Cartesian conversions: \(x = r \cos \theta\), \(y = r \sin \theta\). Thus, \(r^5 = 2x^2\). Therefore, \(\left( x^2 + y^2 \right)^{\frac{3}{2}} = 2x^2 y\).

(c) The total area enclosed by the curve is given by \(\frac{1}{2} \int_0^\pi \sin(2\theta) \cos \theta \, d\theta\). Simplifying, \(\int \sin \theta \cos^2 \theta \, d\theta = -\frac{1}{3} \cos^3 \theta\). Evaluating from 0 to \(\pi\), the area is \(\frac{2}{3}\).

(d) Differentiate \(r^2 = \sin 2\theta \cos \theta\) with respect to \(\theta\) to find the maximum distance. Solve \(6 \cos^3 \theta - 4 \cos \theta = 0\) to find \(\cos^2 \theta = \frac{2}{3}\). The greatest distance is \(r = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{9} \approx 0.877\).

(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\). Expand and rearrange to get \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\), which simplifies to \(x^2 + y^2 = x\). In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\). Substitute to get \(r^2 = r \cos \theta\), which simplifies to \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), we have \(r = \cos \theta\).

(b) Set \(\sin 2\theta = \cos \theta\). Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\), we have \(2\sin \theta \cos \theta = \cos \theta\). Assuming \(\cos \theta \neq 0\), divide by \(\cos \theta\) to get \(2\sin \theta = 1\), so \(\sin \theta = \frac{1}{2}\). Therefore, \(\theta = \frac{\pi}{6}\). The polar coordinates of \(P\) are \(\left( \frac{1}{2}, \frac{\sqrt{3}}{6} \pi \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\), marking the intersection point \(P\).

(d) The area \(R\) is given by \(\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\). Use the identities \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\) to integrate. The result is \(\frac{1}{4} (\pi - 3\sqrt{3})\).

(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.

(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\).

Expand and rearrange: \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\) which simplifies to \(x^2 + y^2 = x\).

In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\), so \(r^2 = r \cos \theta\).

Factor to get \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), \(r = \cos \theta\).

(b) Set \(r = \cos \theta = \sin 2\theta\).

Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), so \(\cos \theta = 2 \sin \theta \cos \theta\).

Thus, \(\cos \theta (1 - 2 \sin \theta) = 0\). Since \(\cos \theta \neq 0\), \(\sin \theta = \frac{1}{2}\).

Therefore, \(\theta = \frac{\pi}{6}\) and \(r = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\).

The polar coordinates of \(P\) are \(\left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\) on the same diagram, marking the point \(P\).

(d) The area \(R\) is given by:

\(\frac{1}{2} \int_0^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\).

Use identities: \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\).

Integrate to find:

\(\frac{1}{4} \left[ \theta - \frac{1}{4} \sin 4\theta \right]_0^{\frac{\pi}{6}} + \frac{1}{4} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\).

Calculate to get \(\frac{1}{4} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) + \frac{1}{4} \left( \frac{\pi}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\).

Simplify to \(\frac{1}{4} \left( \pi - \sqrt{3} \right)\).

(a) The curve is sketched as a semicircle with \(\theta = 0\) at the rightmost point. The polar coordinates of the point furthest from the pole are \((1, 0)\).

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

\(A = \frac{1}{2} \int_{0}^{\pi} \frac{1}{\theta^2 + 1} \, d\theta\)

Integrating, we have:

\(\frac{1}{2} \left[ \arctan \theta \right]_{0}^{\pi} = \frac{1}{2} \arctan \pi = 0.631\)

(c) Let \(y = \frac{\sin \theta}{\sqrt{\theta^2 + 1}}\). The derivative is:

\(\frac{dy}{d\theta} = \left( \theta^2 + 1 \right)^{-\frac{1}{2}} \cos \theta - \theta \left( \theta^2 + 1 \right)^{-\frac{3}{2}} \sin \theta = 0\)

Solving gives:

\(\theta \neq 0 \Rightarrow \left( \theta + \frac{1}{\theta} \right) \cot \theta - 1 = 0\)

Checking for a root between 1.1 and 1.2:

\(\left( 1.1 + \frac{1}{1.1} \right) \cot 1.1 - 1 = 0.02 \ldots\)

\(\left( 1.2 + \frac{1}{1.2} \right) \cot 1.2 - 1 = -0.209 \ldots\)

There is a sign change, confirming a root exists between 1.1 and 1.2.

(a) The sketch shows a polar graph with \(\theta = 0\) at the pole and the curve extending to \(\theta = \pi\). The point furthest from the pole is \((1, 0)\).

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

\(A = \frac{1}{2} \int_{0}^{\pi} \frac{1}{\theta^2 + 1} \, d\theta\)

\(= \frac{1}{2} \left[ \arctan \theta \right]_{0}^{\pi}\)

\(= \frac{1}{2} \arctan \pi = 0.631\)

(c) Let \(y = \frac{\sin \theta}{\sqrt{\theta^2 + 1}}\).

Differentiate and set \(\frac{dy}{d\theta} = 0\):

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

\(\theta \neq 0 \Rightarrow \left( \theta + \frac{1}{\theta} \right) \cot \theta - 1 = 0\)

Check for root between 1.1 and 1.2:

\(\left( 1.1 + \frac{1}{1.1} \right) \cot 1.1 - 1 = 0.02 \ldots\)

\(\left( 1.2 + \frac{1}{1.2} \right) \cot 1.2 - 1 = -0.209 \ldots\)

There is a sign change, confirming a root exists between 1.1 and 1.2.

(a) Start with the Cartesian equation \(x^2 - y^2 = a\).

In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\).

Substitute these into the equation: \((r \cos \theta)^2 - (r \sin \theta)^2 = a\).

This simplifies to \(r^2 (\cos^2 \theta - \sin^2 \theta) = a\).

Using the double angle identity, \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), we have \(r^2 \cos 2\theta = a\).

Thus, \(r^2 = a \sec 2\theta\).

(b) The sketch shows the curve starting at \(\theta = 0\) and increasing, with the minimum distance from the pole being \(\sqrt{a}\).

(a) The curve is sketched with \(r\) strictly increasing from \(\theta = 0\) to \(\theta = 2\). The maximum distance is found by evaluating \(r\) at \(\theta = 2\), giving \(r = \sqrt{\frac{1}{2}\arctan(1)} = \frac{1}{2}\sqrt{\pi}\).

(b) The area is calculated using the integral \(\frac{1}{2} \int_0^2 \arctan\left(\frac{1}{2}\theta\right) d\theta\). Integration by parts is applied:

\(\frac{1}{2}\left[ \frac{1}{2}\theta\arctan\left(\frac{1}{2}\theta\right) - \int \frac{\theta}{\theta^2 + 4} d\theta \right]_0^2\)

\(= \frac{1}{2}\left[ \frac{1}{2}\theta\arctan\left(\frac{1}{2}\theta\right) - \ln(\theta^2 + 4) \right]_0^2\)

\(= \frac{1}{4}\pi - \frac{1}{2}\ln 2\)

(c) To find the point furthest from \(\theta = \frac{1}{2}\pi\), set \(x = (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} \cos\theta\) and differentiate:

\(\frac{dx}{d\theta} = (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} \sin\theta + \cos\theta (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} (\theta^2 + 4)^{-1} = 0\)

\((\theta^2 + 4)\arctan\left(\frac{1}{2}\theta\right)\sin\theta - \cos\theta = 0\)

Check sign change between 0.6 and 0.7:

\((0.6^2 + 4)\arctan(0.3)\sin 0.6 - \cos 0.6 = -0.108\)

\((0.7^2 + 4)\arctan(0.35)\sin 0.7 - \cos 0.7 = 0.209\)

Thus, a root exists between 0.6 and 0.7.

(a) Start with the polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute into the Cartesian equation:

\(x^2 + xy + y^2 = r^2 \cos^2 \theta + r^2 \cos \theta \sin \theta + r^2 \sin^2 \theta = a\).

This simplifies to \(r^2 (\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta) = a\).

Using \(\cos^2 \theta + \sin^2 \theta = 1\) and \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

\(r^2 (1 + \frac{1}{2} \sin 2\theta) = a\).

Thus, \(r^2 = \frac{2a}{2 + \sin 2\theta}\).

(b) The sketch should show a polar graph with the curve in the correct domain \(0 \leq \theta \leq \frac{1}{4}\pi\), with \(r\) strictly decreasing.

(c) Given \(\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}\), substitute \(t = \tan \theta\), then \(\frac{d\theta}{dt} = \frac{1}{1 + t^2}\).

The integral becomes:

\(R = \frac{1}{2} a \int_{0}^{1} \frac{1}{t^2 + t + 1} \, dt\).

Complete the square: \(t^2 + t + 1 = (t + \frac{1}{2})^2 + \frac{3}{4}\).

Integrate: \(R = \frac{1}{\sqrt{3}} [\arctan(\frac{2t + 1}{\sqrt{3}})]_{0}^{1}\).

Evaluate: \(R = \frac{\pi a}{6\sqrt{3}}\).

(a) Start with the Cartesian equation \((x^2 + y^2)^2 = 36(x^2 - y^2)\).

In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), so \(x^2 + y^2 = r^2\).

Substitute: \(r^4 = 36(r^2 \cos^2 \theta - r^2 \sin^2 \theta)\).

Factor out \(r^2\): \(r^4 = 36r^2(\cos^2 \theta - \sin^2 \theta)\).

Use the identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\): \(r^2 = 36 \cos 2\theta\).

(b) The sketch is a closed curve symmetrical about the initial line. The maximum distance from the pole is 6.

(c) The area enclosed by \(C\) is given by:

\(\frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} r^2 \, d\theta = \frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} 36 \cos 2\theta \, d\theta\).

\(= 18 \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} \cos 2\theta \, d\theta\).

\(= 18 [\frac{1}{2} \sin 2\theta]_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\).

\(= 18 [\sin \frac{1}{2}\pi - \sin(-\frac{1}{2}\pi)]\).

\(= 18 [1 - (-1)] = 18 \times 2 = 18\).

(d) To find the maximum distance from the initial line, set \(y = 6 \cos^{\frac{1}{2}} 2\theta \sin \theta\).

Set \(\frac{dy}{d\theta} = 0\) and solve:

\(\cos^2 2\theta \cos \theta - \cos^{\frac{1}{2}} 2\theta \sin 2\theta \sin \theta = 0\).

Use trigonometric identities to solve for \(\theta\):

\(\theta = \pm \frac{1}{4}\pi\).

The maximum distance is \(3\sqrt{2}\).

(a) The polar equation is \(r = a \sec^2 \theta\). At \(\theta = 0\), \(r = a \sec^2(0) = a\), giving the point \((a, 0)\). At \(\theta = \frac{1}{4} \pi\), \(r = a \sec^2(\frac{1}{4} \pi) = 2a\), giving the point \((2a, \frac{1}{4} \pi)\).

(b) The maximum distance from the initial line is when \(\theta = \frac{1}{4} \pi\), where \(r = 2a\). The distance is \(y = r \sin \theta = 2a \sin(\frac{1}{4} \pi) = a\sqrt{2}\).

(c) The area \(A\) is given by \(\frac{1}{2} \int_0^{\frac{1}{4} \pi} r^2 \, d\theta\). Substituting \(r = a \sec^2 \theta\), we have:

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

\(= \frac{1}{2} a^2 \int_0^{\frac{1}{4} \pi} (\sec^2 \theta)(\tan^2 \theta + 1) \, d\theta\)

\(= \frac{1}{2} a^2 \left[ \tan \theta + \frac{1}{3} \tan^3 \theta \right]_0^{\frac{1}{4} \pi}\)

\(= \frac{3}{8} a^2\)

(d) The Cartesian form is derived by substituting \(x = r \cos \theta\) and \(y = r \sin \theta\). From \(r = a \sec^2 \theta\), we have:

\(x^2 + y^2 = r^2\)

\(r^2 \cos^2 \theta = a^2 \Rightarrow x^2 = a^2 - y^2\)

\(x^4 = a^2 (x^2 + y^2)\)

\(y = \sqrt{a^2 x^4 - x^2} = x \sqrt{a^2 x^2 - 1}\)

(a) At \(\theta = 0\), \(r = a \cot\left(\frac{1}{3}\pi\right) = 2\sqrt{3}\). Solving gives \(a = 2\).

(b) The area is given by \(\frac{1}{2} \int_{0}^{\frac{1}{6}\pi} 4\cot^2\left(\frac{1}{3}\pi - \theta\right) \, d\theta\). This simplifies to \(2 \left[ \cot\left(\frac{1}{3}\pi - \theta\right) - \theta \right]_{0}^{\frac{1}{6}\pi}\), resulting in \(\frac{4}{3}\sqrt{3} - \frac{1}{3}\sqrt{3}\pi\).

(c) Using identities for \(\cos(A-B)\) and \(\sin(A-B)\), and substituting \(r = \sqrt{x^2 + y^2}\), \(x = r \cos \theta\), \(y = r \sin \theta\), the Cartesian equation is shown as \(2(x + y\sqrt{3}) = (x\sqrt{3} - y)\sqrt{x^2 + y^2}\).

(a) The curve is approximately correct, passing through the pole, \(O\), with the correct domain. The curve \(r\) is strictly increasing over the domain \(0\) to \(\frac{\pi}{2}\). It has the correct form at \(O\) and is approximately tangential to the initial line at \(O\).

(b) To find the area, use the formula:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} \left( \frac{1}{\pi - \theta} - \frac{1}{\pi} \right)^2 \, d\theta\)

Expanding the integrand:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} \left( \frac{1}{(\pi - \theta)^2} - \frac{2}{\pi(\pi - \theta)} + \frac{1}{\pi^2} \right) \, d\theta\)

Integrating each term separately:

\(\frac{1}{2} \left[ \frac{1}{\pi - \theta} + \frac{2}{\pi} \ln(\pi - \theta) + \frac{\theta}{\pi^2} \right]_0^{\frac{1}{2}\pi}\)

Substituting the limits:

\(\frac{1}{2} \left( \frac{2}{\pi} - \frac{\pi}{2} + \frac{2}{\pi} \ln \frac{\pi}{2} + \frac{1}{2\pi} \right) = \frac{1}{2} \left( \frac{3}{2\pi} + \frac{2}{\pi} \ln \frac{1}{2} \right)\)

Simplifying the logarithmic terms:

\(= \frac{3 - 4 \ln 2}{4\pi}\)

(a) To find the point furthest from the pole, differentiate \(r\) with respect to \(\theta\):

\(\frac{dr}{d\theta} = 2(\cos \theta) - (1 + \sin \theta) \sin \theta = 0\)

Simplifying gives:

\(1 - s - 2s^2 = 0\)

Solving the quadratic in \(\sin \theta\), we find \(\sin \theta = -1\) or \(\sin \theta = \frac{1}{2}\), leading to \(\theta = \frac{1}{6} \pi\).

The polar coordinates are \(\left( \frac{3}{2} \sqrt{3}, \frac{1}{6} \pi \right)\).

(b) The sketch of C shows a correct shape, tangential to \(\theta = \frac{\pi}{2}\) at the pole, with \(r\) strictly increasing to \(\frac{\pi}{6}\) then decreasing. The curve is in the first quadrant.

(c) The area of the region bounded by C and the initial line is found using:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\frac{1}{2}\pi} 4c^2(1+s)^2 \, d\theta\)

Using double angle formulae and integrating, we obtain:

\(\left[ \frac{5}{4} \theta + \frac{1}{2} \sin 2\theta - \frac{4}{3} \cos^3 \theta - \frac{1}{16} \sin 4\theta \right]_{0}^{\frac{1}{2}\pi}\)

The exact area is \(\frac{5}{8} \pi + \frac{4}{3}\).

(a) The curve \(C\) has polar equation \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \pi\). It is a limacon, symmetric about the line \(\theta = \frac{\pi}{2}\); the completed sketch shows a single loop above the pole.

(b) The line \(l\) has polar equation \[ r\sin\theta = 2 \quad\Rightarrow\quad r = 2\csc\theta \quad (\sin\theta \ne 0). \] Points of intersection satisfy \[ 3 + 2\sin\theta \;=\; 2\csc\theta. \] Multiplying by \(\sin\theta\), \[ 3\sin\theta + 2\sin^2\theta = 2. \] Let \(\sin\theta = s\). Then \[ 2s^2 + 3s - 2 = 0 \quad\Rightarrow\quad s = \frac{-3 \pm 5}{4} \;\Rightarrow\; s = \frac12 \text{ or } s = -2. \] \(\sin\theta = -2\) is impossible, so \(\sin\theta = \frac12\), giving \[ \theta = \frac{\pi}{6},\; \frac{5\pi}{6}. \] At these points, \[ r = 3 + 2\sin\theta = 3 + 2\cdot\frac12 = 4. \] So the points of intersection are \[ \left(4,\; \frac{\pi}{6}\right),\quad \left(4,\; \frac{5\pi}{6}\right). \]

(c) The region \(R\) is enclosed by \(C\) and the line \(l\), and contains the pole. For a given \(\theta\), the boundary closest to the pole is:

• \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \frac{\pi}{6}\) and \(\frac{5\pi}{6} \le \theta \le \pi\) (the curve is nearer than the line),
• \(r = 2\csc\theta\) for \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\) (the line is nearer than the curve).

Hence the area of \(R\) is \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi/6} (3 + 2\sin\theta)^2\,d\theta + \frac12 \int_{\pi/6}^{5\pi/6} (2\csc\theta)^2\,d\theta + \frac12 \int_{5\pi/6}^{\pi} (3 + 2\sin\theta)^2\,d\theta. \] It is convenient to rewrite this as \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta \;-\; \frac12 \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta. \]

First, \[ \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta = \int_{-\pi}^{\pi} \bigl(9 + 12\sin\theta + 4\sin^2\theta\bigr)\,d\theta = 22\pi. \]

Next, \[ (3 + 2\sin\theta)^2 - (2\csc\theta)^2 = 9 + 12\sin\theta + 4\sin^2\theta - \frac{4}{\sin^2\theta}, \] so \[ \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta = 5\sqrt{3} + \frac{22\pi}{3}. \]

Therefore, \[ \text{Area}(R) = \frac12\left(22\pi - \left(5\sqrt{3} + \frac{22\pi}{3}\right)\right) = \frac12\left(\frac{44\pi}{3} - 5\sqrt{3}\right) = \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}. \]

So the exact area of \(R\) is \[ \boxed{\displaystyle \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}}. \]

Answer:

1. The polar equation is \(r=\sin 4\theta\).
2. The curve is a single loop in the first quadrant, starting and ending at the origin. Its line of symmetry is \(\theta=\frac{\pi}{8}\).
3. The area enclosed by \(C\) is \(\frac{\pi}{16}\).
4. The maximum distance from the line \(\theta=\frac{\pi}{2}\) is \(0.93\) to 2 decimal places.

1. Use \(x=r\cos\theta\), \(y=r\sin\theta\) and \(x^2+y^2=r^2\).

   The Cartesian equation is

   \((x^2+y^2)^{5/2}=4xy(x^2-y^2)\).

   Substituting gives

   \(r^5=4(r\cos\theta)(r\sin\theta)(r^2\cos^2\theta-r^2\sin^2\theta)\),

   so

   \(r^5=4r^4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)\).

   For \(r e 0\), divide by \(r^4\):

   \(r=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)\).

   Using \(2\sin\theta\cos\theta=\sin2\theta\) and \(\cos^2\theta-\sin^2\theta=\cos2\theta\),

   \(r=2\sin2\theta\cos2\theta=\sin4\theta\).

   Thus the polar equation is \(r=\sin4\theta\).

2. For \(0\leq\theta\leq\frac{\pi}{4}\), \(r=\sin4\theta\geq0\). Also \(r=0\) at \(\theta=0\) and \(\theta=\frac{\pi}{4}\), and \(r\) is greatest when \(4\theta=\frac{\pi}{2}\), i.e. \(\theta=\frac{\pi}{8}\).

   So \(C\) is one loop in the first quadrant, starting at the origin on the initial line, reaching its furthest point on \(\theta=\frac{\pi}{8}\), and returning to the origin on \(\theta=\frac{\pi}{4}\).

   The line of symmetry is \(\theta=\frac{\pi}{8}\).

3. The area enclosed by a polar curve is

   \(A=\frac12\int_\alpha^\beta r^2\,d\theta\).

   Here \(r=\sin4\theta\), with \(0\leq\theta\leq\frac{\pi}{4}\), so

   \(A=\frac12\int_0^{\pi/4}\sin^2 4\theta\,d\theta\).

   Using \(\sin^2 u=\frac12(1-\cos2u)\),

   \(A=\frac12\int_0^{\pi/4}\frac12(1-\cos8\theta)\,d\theta\)

   \(=\frac14\left[\theta-\frac18\sin8\theta\right]_0^{\pi/4}\).

   Since \(\sin2\pi=0\),

   \(A=\frac14\cdot\frac{\pi}{4}=\frac{\pi}{16}\).

4. The line \(\theta=\frac{\pi}{2}\) is the \(y\)-axis, so the distance of a point on this loop from the line is \(x\), since \(x\geq0\) on \(C\).

   Now \(x=r\cos\theta\). Using the given identity,

   \(x=\sin4\theta\cos\theta\)

   \(=(4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta)\cos\theta\)

   \(=4\sin\theta\cos^4\theta-4\sin^3\theta\cos^2\theta\).

   Differentiate and set equal to zero:

   \(\frac{dx}{d\theta}=4\cos\theta(\cos^4\theta-7\sin^2\theta\cos^2\theta+2\sin^4\theta)=0\).

   In \(0\leq\theta\leq\frac{\pi}{4}\), \(\cos\theta e 0\), so

   \(\cos^4\theta-7\sin^2\theta\cos^2\theta+2\sin^4\theta=0\).

   Dividing by \(\cos^4\theta\),

   \(1-7\tan^2\theta+2\tan^4\theta=0\).

   Let \(u=\tan^2\theta\). Then

   \(2u^2-7u+1=0\),

   so

   \(u=\frac{7\pm\sqrt{41}}{4}\).

   Since \(0\leq\tan^2\theta\leq1\),

   \(\tan^2\theta=\frac{7-\sqrt{41}}{4}\).

   Hence \(\theta\approx0.369\). Substituting into \(x=\sin4\theta\cos\theta\),

   \(x\approx0.929\).

   The endpoints give \(x=0\), so the maximum distance from \(\theta=\frac{\pi}{2}\) is \(0.93\) to 2 decimal places.

Answer:

1. (a) At \(P\), \(2\theta\tan\theta-1=0\), with a root between \(0.6\) and \(0.7\).
2. (b) \(Q\) has polar coordinates \(\left(\frac{\pi\sqrt{2}}{8},\frac{\pi}{4}\right)\).
3. (c) The sketch should show both curves from the pole, meeting again at \(Q\), with \(C_1\) initially outside \(C_2\) for \(0<\theta<\frac{\pi}{4}\).
4. (d) The required area is \(\displaystyle \frac{\pi^2}{64}-\frac{1}{8}\).

(a) The line \(\theta=\frac{\pi}{2}\) is the positive \(y\)-axis. The distance of a point from this line is its \(x\)-coordinate.

For \(C_1\), \(r=\theta\cos\theta\), so

\(x=r\cos\theta=\theta\cos^2\theta\).

To find the point furthest from the line, maximise \(x(\theta)=\theta\cos^2\theta\). Differentiate:

\(\displaystyle \frac{dx}{d\theta}=\cos^2\theta-2\theta\cos\theta\sin\theta\).

At an interior maximum, \(\frac{dx}{d\theta}=0\), so

\(\cos^2\theta-2\theta\cos\theta\sin\theta=0\).

Since \(0<\theta<\frac{\pi}{2}\), \(\cos\theta e0\), and therefore

\(\cos\theta-2\theta\sin\theta=0\).

Dividing by \(\cos\theta\) gives

\(1-2\theta\tan\theta=0\),

so \(2\theta\tan\theta-1=0\).

Let \(f(\theta)=2\theta\tan\theta-1\). Then

\(f(0.6)=2(0.6)\tan0.6-1\approx -0.179\),

and

\(f(0.7)=2(0.7)\tan0.7-1\approx 0.179\).

Since the signs are different, there is a root between \(0.6\) and \(0.7\).

(b) At the non-pole intersection,

\(\theta\cos\theta=\theta\sin\theta\).

Since \(\theta e0\), \(\cos\theta=\sin\theta\), so \(\tan\theta=1\). Hence \(\theta=\frac{\pi}{4}\).

Then

\(r=\theta\cos\theta=\frac{\pi}{4}\cdot\frac{\sqrt{2}}{2}=\frac{\pi\sqrt{2}}{8}\).

Therefore

\(Q=\left(\frac{\pi\sqrt{2}}{8},\frac{\pi}{4}\right)\).

(c) Both curves start at the pole. The curve \(C_1:r=\theta\cos\theta\) is initially outside \(C_2\) for \(0<\theta<\frac{\pi}{4}\), and the two curves meet again at \(Q\). The curve \(C_2:r=\theta\sin\theta\) continues outwards towards \(\theta=\frac{\pi}{2}\). A sketch should show this intersection and the first-quadrant polar shape.

(d) For \(0\le\theta\le\frac{\pi}{4}\), \(\cos\theta\ge\sin\theta\), so \(C_1\) is outside \(C_2\). The required area is

\(\displaystyle A=\frac{1}{2}\int_0^{\pi/4}\left[(\theta\cos\theta)^2-(\theta\sin\theta)^2\right]d\theta\).

Thus

\(\displaystyle A=\frac{1}{2}\int_0^{\pi/4}\theta^2\cos2\theta\,d\theta\).

Integrating by parts,

\(\displaystyle \int \theta^2\cos2\theta\,d\theta=\frac{\theta^2}{2}\sin2\theta+\frac{\theta}{2}\cos2\theta-\frac{1}{4}\sin2\theta+C\).

Therefore

\(\displaystyle A=\frac{1}{2}\left[\frac{\theta^2}{2}\sin2\theta+\frac{\theta}{2}\cos2\theta-\frac{1}{4}\sin2\theta\right]_0^{\pi/4}\).

At \(\theta=\frac{\pi}{4}\), \(\sin2\theta=1\) and \(\cos2\theta=0\), so

\(\displaystyle A=\frac{1}{2}\left(\frac{\pi^2}{32}-\frac{1}{4}\right)=\frac{\pi^2}{64}-\frac{1}{8}\).

Answer: (i) At \(P\), \(2\theta\tan\theta=1\), with a root between \(0.6\) and \(0.7\).

(ii) \(\theta=\dfrac{\pi}{4}\).

(iii) \(C_1\) starts at \(O\), lies outside \(C_2\), and meets \(C_2\) again at \(Q\).

(iv) The enclosed area is \(\dfrac14\ln 2+\dfrac{\pi}{8}\left(\dfrac{\pi}{4}-1\right)\).

(i) The line \(\theta=\dfrac{\pi}{2}\) is the positive \(y\)-axis. For a point with polar coordinates \((r,\theta)\), the perpendicular distance from this line is

\(x=r\cos\theta\).

On \(C_1\), \(r^2=2\theta\), so \(r=\sqrt{2\theta}\). We maximise

\(x=\sqrt{2\theta}\cos\theta\).

Differentiate:

\(\dfrac{\mathrm d}{\mathrm d\theta}\left(\sqrt{2\theta}\cos\theta\right)=\sqrt2\left(-\theta^{1/2}\sin\theta+\dfrac12\theta^{-1/2}\cos\theta\right)\).

At the furthest point this is zero, so

\(-\theta^{1/2}\sin\theta+\dfrac12\theta^{-1/2}\cos\theta=0\).

Multiplying by \(2\theta^{1/2}\) gives

\(\cos\theta=2\theta\sin\theta\),

and hence

\(2\theta\tan\theta=1\).

Let \(f(\theta)=2\theta\tan\theta-1\). Then

\(f(0.6)=2(0.6)\tan(0.6)-1\approx -0.179\),

whereas

\(f(0.7)=2(0.7)\tan(0.7)-1\approx 0.179\).

Since the signs are different, the equation has a root between \(0.6\) and \(0.7\).

(ii) At \(Q\), the two curves have the same value of \(r\), and \(\theta\neq0\). Therefore

\(2\theta=\theta\sec^2\theta\).

Since \(\theta\neq0\), divide by \(\theta\):

\(\sec^2\theta=2\).

Thus \(\cos^2\theta=\dfrac12\). In the interval \(0\leq\theta\leq\dfrac{\pi}{4}\), this gives

\(\theta=\dfrac{\pi}{4}\).

(iii) The curve \(C_1\) starts at \(O\) when \(\theta=0\). For \(0<\theta<\dfrac{\pi}{4}\),

\(2\theta>\theta\sec^2\theta\),

so \(C_1\) has the larger radius and lies outside \(C_2\). It then meets \(C_2\) at \(Q\), where \(\theta=\dfrac{\pi}{4}\). The sketch should therefore show a smooth curve from \(O\) to \(Q\), outside \(C_2\), with the second intersection at \(Q\).

(iv) The enclosed area is

\(\dfrac12\int_0^{\pi/4}\left(r_1^2-r_2^2\right)\,\mathrm d\theta\).

Using \(r_1^2=2\theta\) and \(r_2^2=\theta\sec^2\theta\),

\(\text{Area}=\dfrac12\int_0^{\pi/4}\left(2\theta-\theta\sec^2\theta\right)\,\mathrm d\theta\).

Write the integrand as \(\theta(2-\sec^2\theta)\). Integrating by parts with \(u=\theta\) and \(\mathrm dv=(2-\sec^2\theta)\,\mathrm d\theta\), we have \(v=2\theta-\tan\theta\). Hence

\(\text{Area}=\dfrac12\left[\theta(2\theta-\tan\theta)\right]_0^{\pi/4}-\dfrac12\int_0^{\pi/4}(2\theta-\tan\theta)\,\mathrm d\theta\).

Now

\(\int(2\theta-\tan\theta)\,\mathrm d\theta=\theta^2+\ln(\cos\theta)\).

Therefore

\(\text{Area}=\dfrac12\left[\theta(2\theta-\tan\theta)-\theta^2-\ln(\cos\theta)\right]_0^{\pi/4}\).

Substitute \(\theta=\dfrac{\pi}{4}\), \(\tan\dfrac{\pi}{4}=1\), and \(\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\):

\(\text{Area}=\dfrac12\left(\dfrac{\pi}{4}\left(\dfrac{\pi}{2}-1\right)-\dfrac{\pi^2}{16}+\dfrac12\ln2\right)\).

So

\(\text{Area}=\dfrac14\ln2+\dfrac{\pi}{8}\left(\dfrac{\pi}{4}-1\right)\).

Answer: (i) The curve starts at the pole when \(\theta=0\), since then \(r^2=\ln 1=0\). As \(\theta\) increases from \(0\) to \(2\pi\), \(r^2=\ln(1+\theta)\) increases steadily, so the curve moves continuously away from the pole in the direction of increasing \(\theta\), crossing the initial line again only as part of the full sweep through angles. A correct sketch is an outward spiral-like curve beginning at the pole on the initial line and expanding as \(\theta\) increases to \(2\pi\).

(ii) The area enclosed by a polar curve from \(\theta=\alpha\) to \(\theta=\beta\) is \(\frac12\int_{\alpha}^{\beta} r^2\,d\theta\). Here the region bounded by \(C\) and the initial line corresponds to \(0\le \theta\le 2\pi\), so

\(A=\frac12\int_0^{2\pi}\ln(1+\theta)\,d\theta.\)

Use \(u=1+\theta\), so that \(du=d\theta\). When \(\theta=0\), \(u=1\); when \(\theta=2\pi\), \(u=1+2\pi\). Hence

\(A=\frac12\int_1^{1+2\pi}\ln u\,du.\)

Now integrate by parts, or use \(\int \ln u\,du=u\ln u-u\):

\(A=\frac12\Big[ u\ln u-u\Big]_1^{1+2\pi}.\)

So

\(A=\frac12\Big((1+2\pi)\ln(1+2\pi)-(1+2\pi) - (1\cdot 0 -1)\Big)\)

\(=\frac12\Big((1+2\pi)\ln(1+2\pi)-2\pi\Big).\)

Therefore the area is \(\frac12\big((1+2\pi)\ln(1+2\pi)-2\pi\big)\).

(i) Since \(r^2=\ln(1+\theta)\), we have \(r=0\) when \(\theta=0\). Also, as \(\theta\) increases, \(\ln(1+\theta)\) increases, so \(r\) increases continuously. The curve therefore begins at the pole on the initial line and spirals steadily outwards as \(\theta\) goes from \(0\) to \(2\pi\).

(ii) The area in polar coordinates is

\(A=\frac12\int r^2\,d\theta.\)

Hence

\(A=\frac12\int_0^{2\pi}\ln(1+\theta)\,d\theta.\)

Let \(u=1+\theta\). Then \(du=d\theta\), and the limits change from \(\theta=0\) to \(u=1\), and from \(\theta=2\pi\) to \(u=1+2\pi\). So

\(A=\frac12\int_1^{1+2\pi}\ln u\,du.\)

Now

\(\int \ln u\,du=u\ln u-u,\)

so

\(A=\frac12\Big[u\ln u-u\Big]_1^{1+2\pi}.\)

Therefore

\(A=\frac12\left((1+2\pi)\ln(1+2\pi)-(1+2\pi)- (0-1)\right)\)

\(=\frac12\left((1+2\pi)\ln(1+2\pi)-2\pi\right).\)

So the exact area is \(\frac12\left((1+2\pi)\ln(1+2\pi)-2\pi\right)\).

Answer: (i) The curve is a single loop of \(r=\cos 2\theta\), symmetric about the \(x\)-axis, from \(\theta=-\frac{\pi}{4}\) to \(\theta=\frac{\pi}{4}\), with maximum radius \(1\) at \(\theta=0\).

(ii) The enclosed area is \(\frac{\pi}{8}\).

(iii) A Cartesian equation is \(\left(x^2+y^2\right)^{3/2}=x^2-y^2\).

(i) The curve meets the pole when

\(\cos 2\theta=0\),

so \(2\theta=\pm\frac{\pi}{2}\), and therefore \(\theta=\pm\frac{\pi}{4}\).

At \(\theta=0\), \(r=1\), so the curve reaches \((1,0)\). On \(-\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4}\), \(r\geq0\), giving one loop symmetric about the \(x\)-axis.

(ii) The polar area is

\(A=\frac{1}{2}\int_{-\pi/4}^{\pi/4} r^2\,d\theta=\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2 2\theta\,d\theta.\)

Using \(\cos^2 2\theta=\frac{1}{2}(1+\cos4\theta)\),

\(A=\frac{1}{4}\left[\theta+\frac{1}{4}\sin4\theta\right]_{-\pi/4}^{\pi/4}.\)

Since \(\sin\pi=\sin(-\pi)=0\),

\(A=\frac{1}{4}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\frac{\pi}{8}.\)

So the area enclosed by \(C\) is \(\boxed{\frac{\pi}{8}}\).

(iii) Since

\(\cos 2\theta=\cos^2\theta-\sin^2\theta=\frac{x^2}{r^2}-\frac{y^2}{r^2}=\frac{x^2-y^2}{r^2},\)

the equation \(r=\cos2\theta\) gives

\(r=\frac{x^2-y^2}{r^2}.\)

Hence

\(r^3=x^2-y^2.\)

Using \(r^2=x^2+y^2\), we get

\(\boxed{\left(x^2+y^2\right)^{3/2}=x^2-y^2}.\)

Answer: (i) The intersections satisfy both polar equations, so

\(a = 2a|\cos\theta|\), hence \(|\cos\theta| = \tfrac12\).

For \(0 \le \theta \le \pi\), this gives \(\theta = \tfrac{\pi}{3}\) or \(\tfrac{2\pi}{3}\).

So the points are

\(P_1\!:\left(a,\tfrac{\pi}{3}\right)\), \(P_2\!:\left(a,\tfrac{2\pi}{3}\right)\).

(ii) \(C_1\) is the circle \(r=a\), centred at the pole with radius \(a\). \(C_2\) is the loop of the curve \(r=2a|\cos\theta|\), which is symmetric about the line \(\theta=0\) (the polar axis). The line of symmetry is therefore the polar axis.

(iii) The required region \(R\) is inside the circle \(r=a\) but outside the curve \(r=2a|\cos\theta|\), between the two intersection angles.

By symmetry about the polar axis, the area can be found as twice the area in the upper half-plane:

\(A=2\cdot \frac12\int_{\pi/3}^{\pi/2}\big(a^2-(2a\cos\theta)^2\big)\,d\theta.\)

So

\(A=\int_{\pi/3}^{\pi/2}\left(a^2-4a^2\cos^2\theta\right)d\theta = a^2\int_{\pi/3}^{\pi/2}(1-4\cos^2\theta)\,d\theta.\)

Use \(\cos^2\theta=\tfrac12(1+\cos2\theta)\):

\(1-4\cos^2\theta=1-2(1+\cos2\theta)=-1-2\cos2\theta.\)

Hence

\(A=a^2\int_{\pi/3}^{\pi/2}(-1-2\cos2\theta)\,d\theta = a^2\left[-\theta-\sin2\theta\right]_{\pi/3}^{\pi/2}.\)

Now

\(A=a^2\left(\left(-\frac\pi2-\sin\pi\right)-\left(-\frac\pi3-\sin\frac{2\pi}{3}\right)\right) = a^2\left(-\frac\pi2+\frac\pi3+\frac{\sqrt3}{2}\right).\)

Therefore

\(A= a^2\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right) = \frac{a^2}{6}(3\sqrt3-\pi).\)

We solve the three parts in turn.

(i) At an intersection, the same point must satisfy both equations, so

\(a=2a|\cos\theta|.\)

Since \(a>0\), divide by \(a\):

\(1=2|\cos\theta|\quad\Rightarrow\quad |\cos\theta|=\frac12.\)

For \(0\le \theta\le \pi\), this happens when

\(\theta=\frac\pi3\quad\text{or}\quad \theta=\frac{2\pi}{3}.\)

Because \(r=a\) at the intersection, the points are

\(P_1=\left(a,\frac\pi3\right),\qquad P_2=\left(a,\frac{2\pi}{3}\right).\)

(ii) The curve \(C_1:r=a\) is a circle of radius \(a\) centred at the pole.

The curve \(C_2:r=2a|\cos\theta|\) has the same shape as \(r=2a\cos\theta\) for \(-\tfrac\pi2\le\theta\le\tfrac\pi2\), reflected to keep \(r\ge 0\). It is symmetric about the polar axis \((\theta=0)\).

So the sketch should show:

• the circle \(r=a\),
• the curve \(r=2a|\cos\theta|\), passing through the pole and intersecting the circle at angles \(\theta=\pi/3\) and \(2\pi/3\),
• the line of symmetry: the polar axis.

(iii) The region \(R\) is enclosed by the arc of the circle from \(P_1\) to \(P_2\) and the corresponding arc of \(C_2\) from \(P_1\) to \(P_2\), together with the pole. Since the figure is symmetric about the polar axis, we can find the area in the upper half and double it.

For \(0\le \theta\le \pi/2\), we have \(|\cos\theta|=\cos\theta\), so the relevant branch of \(C_2\) is \(r=2a\cos\theta\). The area between the curves from \(\theta=\pi/3\) to \(\theta=\pi/2\) is

\(\frac12\int_{\pi/3}^{\pi/2}\Big(a^2-(2a\cos\theta)^2\Big)\,d\theta.\)

Doubling this gives

\(A=\int_{\pi/3}^{\pi/2}\left(a^2-4a^2\cos^2\theta\right)d\theta = a^2\int_{\pi/3}^{\pi/2}(1-4\cos^2\theta)\,d\theta.\)

Now use \(\cos^2\theta=\frac12(1+\cos2\theta)\):

\(1-4\cos^2\theta=1-2(1+\cos2\theta)=-1-2\cos2\theta.\)

So

\(A=a^2\int_{\pi/3}^{\pi/2}(-1-2\cos2\theta)\,d\theta = a^2\left[-\theta-\sin2\theta\right]_{\pi/3}^{\pi/2}.\)

Evaluate the bracket:

\(A=a^2\left(\left(-\frac\pi2-\sin\pi\right)-\left(-\frac\pi3-\sin\frac{2\pi}{3}\right)\right).\)

Since \(\sin\pi=0\) and \(\sin\tfrac{2\pi}{3}=\tfrac{\sqrt3}{2}\),

\(A=a^2\left(-\frac\pi2+\frac\pi3+\frac{\sqrt3}{2}\right) = a^2\left(\frac{\sqrt3}{2}-\frac\pi6\right).\)

Hence

\(\boxed{A=\frac{a^2}{6}(3\sqrt3-\pi)}.\)

Answer: (a) One loop in the first quadrant, symmetric about \(\theta=\frac\pi6\).

(b) \(\dfrac\pi{12}\)

(c) \(\dfrac9{16}\)

(d) \((x^2+y^2)^2=y(3x^2-y^2)\)

(a)

Since \(r=0\) at \(\theta=0\) and \(\theta=\frac\pi3\), the curve forms one loop. The maximum value of \(r\) occurs when \(3\theta=\frac\pi2\), so \(\theta=\frac\pi6\). Therefore the loop is symmetric about

\(\theta=\frac\pi6.\)

(b)

The polar area is

\(\frac12\int_0^{\pi/3}r^2\,d\theta=\frac12\int_0^{\pi/3}\sin^2 3\theta\,d\theta.\)

Using \(\sin^2 3\theta=\frac12(1-\cos6\theta)\),

\(\frac12\int_0^{\pi/3}\sin^2 3\theta\,d\theta=\frac14\int_0^{\pi/3}(1-\cos6\theta)\,d\theta.\)

Thus

\(\frac14\left[\theta-\frac16\sin6\theta\right]_0^{\pi/3}=\frac14\cdot\frac\pi3=\frac\pi{12}.\)

(c)

The distance from the initial line is the \(y\)-coordinate:

\(y=r\sin\theta=\sin3\theta\sin\theta.\)

Using the identity,

\(y=(3\sin\theta-4\sin^3\theta)\sin\theta=3\sin^2\theta-4\sin^4\theta.\)

Differentiate:

\(\frac{dy}{d\theta}=6\sin\theta\cos\theta-16\sin^3\theta\cos\theta.\)

For the interior maximum, \(\sin\theta\cos\theta\ne0\), so

\(6-16\sin^2\theta=0\quad\Rightarrow\quad\sin^2\theta=\frac38.\)

Therefore

\(y=3\cdot\frac38-4\left(\frac38\right)^2=\frac98-\frac9{16}=\frac9{16}.\)

(d)

Starting with \(r=\sin3\theta\),

\(r=3\sin\theta-4\sin^3\theta.\)

Since \(y=r\sin\theta\), we have \(\sin\theta=\frac yr\), so

\(r=\frac{3y}{r}-\frac{4y^3}{r^3}.\)

Multiplying by \(r^3\),

\(r^4=3yr^2-4y^3=y(3r^2-4y^2).\)

Using \(r^2=x^2+y^2\),

\((x^2+y^2)^2=y(3(x^2+y^2)-4y^2)=y(3x^2-y^2).\)

Mark scheme

Answer: (a) Greatest distance from the pole is \(1\).

(b) \(\dfrac18\ln2\)

(c) \(x^4-2xy-y^4=0\)

(d) \(\dfrac18\sqrt2-\dfrac18\ln2\)

(a)

Since \(r^2=\tan2\theta\), at \(\theta=0\), \(r=0\), so the curve starts at the pole. As \(\theta\) increases to \(\pi/8\), \(\tan2\theta\) increases to \(\tan(\pi/4)=1\), so \(r\) increases to \(1\).

The greatest distance from the pole is therefore \(1\).

(b)

For polar area,

\(A=\frac12\int r^2\,d\theta.\)

Here \(r^2=\tan2\theta\), with \(0\le\theta\le\pi/8\), so

\(A=\frac12\int_0^{\pi/8}\tan2\theta\,d\theta.\)

Since \(\int\tan2\theta\,d\theta=-\frac12\ln(\cos2\theta)\),

\(A=\left[-\frac14\ln(\cos2\theta)\right]_0^{\pi/8}.\)

Thus

\(A=-\frac14\ln\left(\frac{\sqrt2}{2}\right)=\frac18\ln2.\)

(c)

Use

\(\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.\)

Since \(\tan\theta=\frac yx\),

\(r^2=\frac{2(y/x)}{1-y^2/x^2}=\frac{2xy}{x^2-y^2}.\)

Also \(r^2=x^2+y^2\). Therefore

\(x^2+y^2=\frac{2xy}{x^2-y^2}.\)

Multiplying by \(x^2-y^2\),

\((x^2+y^2)(x^2-y^2)=2xy.\)

Hence

\(x^4-y^4=2xy,\)

so

\(x^4-2xy-y^4=0.\)

(d)

The required area is the area of the triangle formed by the pole, the point \(r=1,\theta=\pi/8\), and its projection on the \(x\)-axis, minus the polar area from part (b).

The triangle area is

\(\frac12\cos\frac{\pi}{8}\sin\frac{\pi}{8}=\frac14\sin\frac{\pi}{4}=\frac{\sqrt2}{8}.\)

Therefore the required area is

\(\frac{\sqrt2}{8}-\frac18\ln2.\)

Mark scheme