Answer:

We have \(x=\sin t\) and \(y=\sin 2t\), for \(0\leqslant t\leqslant \pi\).

First differentiate with respect to \(t\): \(\dfrac{\mathrm{d}x}{\mathrm{d}t}=\cos t\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=2\cos 2t\).

So

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}=\dfrac{2\cos 2t}{\cos t}.\)

Now differentiate again with respect to \(t\):

\(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{2\cos 2t}{\cos t}\right).\)

Using the quotient rule,

\(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{2\cos 2t}{\cos t}\right)=\dfrac{(-4\sin 2t)\cos t-(2\cos 2t)(-\sin t)}{\cos^2 t} = \dfrac{-4\sin 2t\cos t+2\cos 2t\sin t}{\cos^2 t}.\)

Then

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\dfrac{\mathrm{d}}{\mathrm{d}t}(\mathrm{d}y/\mathrm{d}x)}{\mathrm{d}x/\mathrm{d}t}=\dfrac{-4\sin 2t\cos t+2\cos 2t\sin t}{\cos^3 t}.\)

This can be simplified using \(\sin 2t=2\sin t\cos t\):

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{-8\sin t\cos^2 t+2\cos 2t\sin t}{\cos^3 t}.\)

Since \(\cos 2t=\cos^2 t-\sin^2 t\), this becomes

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{-8\sin t\cos^2 t+2\sin t(\cos^2 t-\sin^2 t)}{\cos^3 t}=\dfrac{4\sin^3 t-6\sin t}{\cos^3 t}.\)

Now for stationary points, set \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\). Since \(\cos t\neq 0\) at a stationary point of the curve, we need \(2\cos 2t=0\), so \(\cos 2t=0\).

Hence \(2t=\dfrac{\pi}{2}\) or \(\dfrac{3\pi}{2}\), giving \(t=\dfrac{\pi}{4}\) or \(\dfrac{3\pi}{4}\).

Find the coordinates:

• When \(t=\dfrac{\pi}{4}\), \(x=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\) and \(y=\sin\dfrac{\pi}{2}=1\).
• When \(t=\dfrac{3\pi}{4}\), \(x=\sin\dfrac{3\pi}{4}=\dfrac{1}{\sqrt2}\) and \(y=\sin\dfrac{3\pi}{2}=-1\).

So the stationary points are \(\left(\dfrac{1}{\sqrt2},1\right)\) and \(\left(\dfrac{1}{\sqrt2},-1\right)\).

To determine their nature, substitute into \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\):

At \(t=\dfrac{\pi}{4}\), \(\sin t=\dfrac{1}{\sqrt2}\) and \(\cos t=\dfrac{1}{\sqrt2}\), so

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{4\left(\frac{1}{2\sqrt2}\right)-6\left(\frac{1}{\sqrt2}\right)}{\left(\frac{1}{\sqrt2}\right)^3}=-8,\)

which means a maximum.

At \(t=\dfrac{3\pi}{4}\), \(\sin t=\dfrac{1}{\sqrt2}\) and \(\cos t=-\dfrac{1}{\sqrt2}\), so

\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=+8,\)

which means a minimum.