Answer: The arc length is \(32.6\) to 3 significant figures. The mean value of \(y\) with respect to \(x\) is \(1.6\).

Differentiate the parametric equations:

\(\frac{dx}{dt}=4+3\sqrt t\), and \(\frac{dy}{dt}=4-3\sqrt t\).

Thus

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(4+3\sqrt t)^2+(4-3\sqrt t)^2=32+18t\).

The arc length is therefore

\(s=\int_0^4\sqrt{32+18t}\,dt=\frac1{27}\left(104^{3/2}-32^{3/2}\right)\).

So \(s=32.6\) to 3 significant figures.

The mean value of \(y\) with respect to \(x\) over \(0\leqslant x\leqslant32\) is

\(\frac1{32}\int_0^{32}y\,dx\).

Using the parameter \(t\),

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

Now

\(y\frac{dx}{dt}=(4t-2t^{3/2})(4+3\sqrt t)=16t+4t^{3/2}-6t^2\).

Hence

\(\int_0^4(16t+4t^{3/2}-6t^2)\,dt=\left[8t^2+\frac85t^{5/2}-2t^3\right]_0^4=\frac{256}{5}=51.2\).

Therefore the mean value is \(\frac{51.2}{32}=1.6\).