(a) Differentiate the volume equation with respect to \(r\):

\(\frac{dV}{dr} = \frac{9}{2} \left( r - \frac{1}{2} \right)^2\).

Use the chain rule to find \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{1}{\frac{dV}{dr}} \times \frac{dV}{dt}\).

Substitute \(r = 5.5\) and \(\frac{dV}{dt} = 1.5\):

\(\frac{dr}{dt} = \frac{1.5}{\frac{9}{2} (5.5 - 0.5)^2} = \frac{1.5}{112.5} = 0.0133 \text{ m/s}\).

(b) Given \(\frac{dr}{dt} = 0.1 \text{ m/s}\), use the relationship:

\(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\).

\(\frac{dV}{dr} = \frac{9}{2} \left( r - \frac{1}{2} \right)^2\).

\(0.1 = \frac{1.5}{\frac{9}{2} \left( r - \frac{1}{2} \right)^2}\).

Solve for \(r\):

\(\frac{9}{2} \left( r - \frac{1}{2} \right)^2 = 15\).

\(r = \frac{1}{2} + \sqrt{\frac{10}{3}}\).

Substitute \(r\) back into the volume equation:

\(V = \frac{3}{2} \left( \frac{1}{2} + \sqrt{\frac{10}{3}} - \frac{1}{2} \right)^3 - 1\).

\(V \approx 8.13 \text{ m}^3\).

At \(x = 1\), \(\frac{dy}{dx} = 6\).

Using the chain rule, \(\frac{dx}{dt} = \left( \frac{dx}{dy} \times \frac{dy}{dt} \right)\).

First, find \(\frac{dx}{dy}\):

\(\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{6}\).

Given \(\frac{dy}{dt} = 3\), substitute into the chain rule:

\(\frac{dx}{dt} = \frac{1}{6} \times 3 = \frac{1}{2}\).