Answer: (i) \(\frac{d^2y}{dx^2}=\frac{4-3t}{16t^3(2-t)^{3/2}}.\)

(ii) The mean value is

\(\frac{8\sqrt2}{15}.\)

(iii) The \(y\)-coordinate of the centroid is

\(\frac{5\sqrt2}{16}.\)

First differentiate the parametric equations:

\(\frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=-\frac12(2-t)^{-1/2}.\)

Hence

\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac{1}{4t(2-t)^{1/2}}.\)

Differentiate this with respect to \(t\):

\(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac14t^{-2}(2-t)^{-1/2}-\frac18t^{-1}(2-t)^{-3/2}.\)

Therefore

\(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}=\frac{4-3t}{16t^3(2-t)^{3/2}}.\)

For the mean value, use \(dx=2t\,dt\):

\(\int_0^4 y\,dx=\int_0^2 2t(2-t)^{1/2}\,dt.\)

Integrating by parts gives

\(\int_0^4 y\,dx=\frac{32\sqrt2}{15}.\)

So the mean value over \(0\leq x\leq4\) is

\(\frac14\cdot\frac{32\sqrt2}{15}=\frac{8\sqrt2}{15}.\)

For the centroid,

\(\bar y=\frac{\frac12\int_0^4y^2\,dx}{\int_0^4y\,dx}.\)

Now \(y^2=2-t\), so

\(\frac12\int_0^4y^2\,dx=\frac12\int_0^2 2t(2-t)\,dt=\left[t^2-\frac13t^3\right]_0^2=\frac43.\)

Thus

\(\bar y=\frac{4/3}{32\sqrt2/15}=\frac{5}{8\sqrt2}=\frac{5\sqrt2}{16}.\)