(a) Differentiate \(y = 2x^{\frac{1}{2}} - 1\) to find the gradient of the tangent: \(\frac{dy}{dx} = x^{-\frac{1}{2}}\). At \(x = 4\), \(\frac{dy}{dx} = \frac{1}{2}\). The gradient of the normal is \(-\frac{1}{\frac{1}{2}} = -2\). The equation of the normal through \((4, 3)\) is \(y - 3 = -2(x - 4)\), which simplifies to \(y = -2x + 11\).

(b) Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). At \(A\), \(\frac{dy}{dx} = \frac{1}{2}\) and \(\frac{dx}{dt} = 3\). Thus, \(\frac{dy}{dt} = \frac{1}{2} \times 3 = \frac{3}{2}\).

(c) The required gradient is \(-\frac{dy}{dx} = -2\). Using the chain rule, \(\frac{dx}{dt} = \frac{1}{\text{normal gradient}} \times (-5) = \frac{1}{-2} \times (-5) = \frac{5}{2}\).

First, find \(f'(4)\) by substituting \(x = 4\) into \(f'(x) = 6x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}\).

\(f'(4) = 6(4)^{-\frac{1}{2}} - 4(4)^{-\frac{3}{2}}\)

\(= 6 \times \frac{1}{2} - 4 \times \frac{1}{8}\)

\(= 3 - 0.5\)

\(= \frac{5}{2}\)

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

Given \(\frac{dx}{dt} = 0.12\), substitute \(\frac{dy}{dx} = f'(4) = \frac{5}{2}\).

\(\frac{dy}{dt} = \frac{5}{2} \times 0.12\)

\(= 0.3\)

The volume \(V\) of a sphere is given by \(V = \frac{4}{3}\pi r^3\).

Differentiate with respect to time \(t\):

\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\).

We know \(\frac{dV}{dt} = 50\) cm³s^-1 and \(r = 10\) cm.

Substitute these values into the differentiated equation:

\(50 = 4\pi (10)^2 \frac{dr}{dt}\).

Simplify:

\(50 = 400\pi \frac{dr}{dt}\).

Solve for \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{50}{400\pi} = \frac{1}{8\pi}\).

Thus, the rate at which the radius is increasing is \(\frac{1}{8\pi}\) cm/s or approximately 0.0398 cm/s.

(a) Differentiate y with respect to x: \(\frac{dy}{dx} = \frac{1}{2}(5x-1)^{-1/2} \times 5\).

Use the chain rule to find \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Given \(\frac{dx}{dt} = 2\), substitute to find \(\frac{dy}{dt} = 2 \times \frac{1}{2}(5 \times 1 - 1)^{-1/2} \times 5\).

Calculate \(\frac{dy}{dt} = 2 \times \frac{5}{2} = \frac{5}{2}\).

(b) Set \(2 \times \frac{1}{2}(5x-1)^{-1/2} \times 5 = \frac{5}{8}\).

Simplify to find \((5x-1)^{1/2} = 8\).

Square both sides: \(5x - 1 = 64\).

Solve for x: \(x = 13\).

(a) The volume of the balloon after 30 seconds is given by:

\(V = 600 \times 30 = 18000 \text{ cm}^3\)

The formula for the volume of a sphere is:

\(V = \frac{4}{3} \pi r^3\)

Equating the two expressions for volume:

\(\frac{4}{3} \pi r^3 = 18000\)

Solving for \(r\):

\(r^3 = \frac{18000 \times 3}{4 \pi}\)

\(r = \sqrt[3]{\frac{54000}{4 \pi}}\)

\(r \approx 16.3 \text{ cm}\)

(b) The rate of change of volume with respect to radius is:

\(\frac{dV}{dr} = 4 \pi r^2\)

Using the chain rule, the rate of change of radius with respect to time is:

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

Substituting the known values:

\(\frac{dr}{dt} = \frac{1}{4 \pi r^2} \times 600\)

At \(r = 16.3 \text{ cm}\):

\(\frac{dr}{dt} = \frac{600}{4 \pi (16.3)^2}\)

\(\frac{dr}{dt} \approx 0.181 \text{ cm per second}\)