(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \sqrt{4x + 1}\).

\(y = \int \sqrt{4x + 1} \, dx = \frac{2}{3}(4x+1)^{\frac{3}{2}} + C\)

Using the point \((2, 5)\), substitute to find \(C\):

\(5 = \frac{2}{3}(4(2)+1)^{\frac{3}{2}} + C\)

\(5 = \frac{2}{3}(9)^{\frac{3}{2}} + C\)

\(5 = 12 + C\)

\(C = \frac{1}{2}\)

Thus, the equation is \(y = \frac{2}{3}(4x+1)^{\frac{3}{2}} + \frac{1}{2}\).

(ii) Given \(\frac{dy}{dt} = 0.06\), use the chain rule:

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\)

\(0.06 = \sqrt{4(2) + 1} \cdot \frac{dx}{dt}\)

\(0.06 = 3 \cdot \frac{dx}{dt}\)

\(\frac{dx}{dt} = 0.02\)

(iii) Differentiate \(\frac{dy}{dx} = \sqrt{4x + 1}\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{1}{2}(4x+1)^{-\frac{1}{2}} \cdot 4\)

\(\frac{d^2y}{dx^2} = \frac{2}{\sqrt{4x+1}}\)

Then, \(\frac{d^2y}{dx^2} \times \frac{dy}{dx} = \frac{2}{\sqrt{4x+1}} \times \sqrt{4x+1} = 2\)