Answer: \(\displaystyle \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(1+9t^2)^2\).

Arc length: \(\displaystyle \frac{2}{\sqrt3}\).

Surface area: \(\displaystyle \frac{\pi}{3}\).

A cartesian equation is \(x^3-x^2+3y^2=0\), and the polar equation is \(r=\sec\theta(1-3\tan^2\theta)\), with \(0\leqslant\theta\leqslant\frac{\pi}{6}\).

The enclosed area is \(\displaystyle \frac{4\sqrt3}{45}\).

For the parametric equations \(x=1-3t^2\) and \(y=t(1-3t^2)\), we have

\(\dfrac{dx}{dt}=-6t\) and \(\dfrac{dy}{dt}=1-9t^2\).

So

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=36t^2+(1-9t^2)^2=1+18t^2+81t^4=(1+9t^2)^2.\)

Hence the speed is \(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=1+9t^2\), since this is positive on the given domain.

(i) Arc length of \(C\)

\(L=\int_0^{1/\sqrt{3}} (1+9t^2)\,dt = \left[t+3t^3\right]_0^{1/\sqrt{3}}.\)

Now \(\left(\frac{1}{\sqrt{3}}\right)^3=\frac{1}{3\sqrt{3}}\), so

\(L=\frac{1}{\sqrt{3}}+3\cdot\frac{1}{3\sqrt{3}}=\frac{2}{\sqrt{3}}.\)

(ii) Surface area when rotated about the \(x\)-axis

For rotation about the \(x\)-axis,

\(S=2\pi\int_0^{1/\sqrt{3}} y\,ds = 2\pi\int_0^{1/\sqrt{3}} y(1+9t^2)\,dt.\)

Since \(y=t(1-3t^2)\),

\(S=2\pi\int_0^{1/\sqrt{3}} t(1-3t^2)(1+9t^2)\,dt.\)

Expanding,

\((1-3t^2)(1+9t^2)=1+6t^2-27t^4,\)

so

\(S=2\pi\int_0^{1/\sqrt{3}} (t+6t^3-27t^5)\,dt.\)

Thus

\(S=2\pi\left[\frac{t^2}{2} + \frac{6t^4}{4} - \frac{27t^6}{6}\right]_0^{1/\sqrt{3}} = 2\pi\left[\frac{t^2}{2} + \frac{3t^4}{2} - \frac{9t^6}{2}\right]_0^{1/\sqrt{3}}.\)

With \(t^2=\frac13\), \(t^4=\frac19\), \(t^6=\frac1{27}\), this gives

\(S=2\pi\left(\frac{1}{6}+\frac{1}{6}-\frac{1}{6}\right)=\frac{\pi}{3}.\)

Cartesian equation

Using \(t=\dfrac{y}{x}\), substitute into \(x=1-3t^2\):

\(x=1-3\left(\frac{y}{x}\right)^2.\)

Multiplying by \(x^2\),

\(x^3=x^2-3y^2,\)

so a cartesian equation is

\(x^3-x^2+3y^2=0.\)

Polar equation

With \(x=r\cos\theta\) and \(y=r\sin\theta\), we have \(t=\dfrac{y}{x}=\tan\theta\). Then

\(r\cos\theta = 1-3\tan^2\theta,\)

so

\(r=\sec\theta\,(1-3\tan^2\theta).\)

Since \(0\le t\le \frac{1}{\sqrt3}\), we have \(0\le \tan\theta\le \frac{1}{\sqrt3}\), hence

\(0\le \theta\le \frac{\pi}{6}.\)

Area enclosed between \(C\) and the initial line

The region is above the initial line and under the curve, so

\(A=\int y\,dx.\)

Using the parameter,

\(A=\int_0^{1/\sqrt3} y\left(-\frac{dx}{dt}\right)dt = \int_0^{1/\sqrt3} t(1-3t^2)(6t)\,dt.\)

Hence

\(A=6\int_0^{1/\sqrt3} (t^2-3t^4)\,dt = 6\left[\frac{t^3}{3}-\frac{3t^5}{5}\right]_0^{1/\sqrt3}.\)

Now \(t^3=\frac{1}{3\sqrt3}\) and \(t^5=\frac{1}{9\sqrt3}\), so

\(A=6\left(\frac{1}{9\sqrt3}-\frac{1}{15\sqrt3}\right)=\frac{4}{15\sqrt3}=\frac{4\sqrt3}{45}.\)