Answer: (i) \(\frac{dy}{dx}=\frac{6t^{1/2}}{9-t}\), and \(\frac{d^2y}{dx^2}=\frac{3(9+t)}{2t^{1/2}(9-t)^3}\).

(ii) The mean value is \(\frac{3}{70}\).

(iii) The surface area is \(\frac{169984}{35}\pi\), approximately \(15300\).

Since \(x=18t-t^2\), we have \(\frac{dx}{dt}=18-2t=2(9-t)\). Also, since \(y=8t^{3/2}\), \(\frac{dy}{dt}=12t^{1/2}\).

Therefore \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{12t^{1/2}}{18-2t}=\frac{6t^{1/2}}{9-t}\).

Differentiate \(\frac{dy}{dx}\) with respect to \(t\):

\(\frac{d}{dt}\left(\frac{6t^{1/2}}{9-t}\right)=\frac{3t^{-1/2}(9-t)+6t^{1/2}}{(9-t)^2}\).

Using \(\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}\) and \(\frac{dt}{dx}=\frac{1}{18-2t}=\frac{1}{2(9-t)}\),

\(\frac{d^2y}{dx^2}=\frac{3t^{-1/2}(9-t)+6t^{1/2}}{2(9-t)^3}=\frac{3t^{-1/2}(9+t)}{2(9-t)^3}=\frac{3(9+t)}{2t^{1/2}(9-t)^3}\).

For the mean value over \(0\le x\le56\), use \(x=18t-t^2\). The endpoints correspond to \(t=0\) and \(t=4\). Hence

\(\text{mean}=\frac{1}{56}\int_0^{56}\frac{d^2y}{dx^2}\,dx=\frac{1}{56}\left[\frac{dy}{dx}\right]_0^{56}\).

In terms of \(t\), this is

\(\frac{1}{56}\left[\frac{6t^{1/2}}{9-t}\right]_0^4=\frac{1}{56}\left(\frac{12}{5}-0\right)=\frac{3}{70}\).

For the surface area about the \(x\)-axis, use \(S=2\pi\int y\,ds\), where \(\frac{ds}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\).

Now \(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(18-2t)^2+144t=4(t+9)^2\), so \(\frac{ds}{dt}=2(t+9)\).

Thus

\(S=2\pi\int_0^4 8t^{3/2}\cdot2(t+9)\,dt=32\pi\int_0^4\left(t^{5/2}+9t^{3/2}\right)dt\).

So

\(S=32\pi\left[\frac{2}{7}t^{7/2}+\frac{18}{5}t^{5/2}\right]_0^4=32\pi\left(\frac{256}{7}+\frac{576}{5}\right)=\frac{169984}{35}\pi\).