Answer: a) \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2t-2\sin t}{1+\cos t}\)

b) \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{2t\sin t}{(1+\cos t)^3}\)

c) \(\frac{\mathrm{d}y}{\mathrm{d}x}\) increases with \(x\) over \(-\pi\lt t\lt \pi\).

Differentiate the parametric equations with respect to \(t\):

\(\frac{\mathrm{d}x}{\mathrm{d}t}=1+\cos t\), so \(\frac{\mathrm{d}x}{\mathrm{d}t}\gt 0\) for \(-\pi\lt t\lt \pi\) except at the endpoints where \(\cos t=-1\).

Also \(\frac{\mathrm{d}y}{\mathrm{d}t}=2t-2\sin t\).

Hence

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}=\frac{2t-2\sin t}{1+\cos t}.\)

To find the second derivative, differentiate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(t\), then divide by \(\frac{\mathrm{d}x}{\mathrm{d}t}\):

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2t-2\sin t}{1+\cos t}\right)=\frac{(2-2\cos t)(1+\cos t)-(2t-2\sin t)(-\sin t)}{(1+\cos t)^2}.\)

This simplifies to

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{(2-2\cos t)(1+\cos t)+\sin t(2t-2\sin t)}{(1+\cos t)^2}.\)

Therefore

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\frac{\mathrm{d}}{\mathrm{d}t}(\mathrm{d}y/\mathrm{d}x)}{\mathrm{d}x/\mathrm{d}t}=\frac{(2-2\cos t)(1+\cos t)+\sin t(2t-2\sin t)}{(1+\cos t)^3}.\)

Now simplify the numerator:

\((2-2\cos t)(1+\cos t)=2-2\cos^2 t=2\sin^2 t,\)

so the numerator becomes

\(2\sin^2 t+2t\sin t-2\sin^2 t=2t\sin t.\)

Thus

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{2t\sin t}{(1+\cos t)^3}.\)

For monotonicity, consider the sign of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\). For \(-\pi\lt t\lt 0\), both \(t\) and \(\sin t\) are negative, so \(t\sin t\gt 0\). For \(0\lt t\lt \pi\), both are positive, so again \(t\sin t\gt 0\). Also \(1+\cos t\gt 0\) throughout the interval \(-\pi\lt t\lt \pi\). Hence \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\gt 0\) for all non-zero \(t\) in the interval.

So \(\frac{\mathrm{d}y}{\mathrm{d}x}\) increases with \(x\) over the given interval.