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

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

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