(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\).

\(y = \int (x^{\frac{1}{2}} - x^{-\frac{1}{2}}) \, dx = \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} + c\).

Substitute the point \((4, \frac{2}{3})\) into the equation:

\(\frac{2}{3} = \frac{2}{3}(4)^{\frac{3}{2}} - 2(4)^{\frac{1}{2}} + c\).

\(\frac{2}{3} = \frac{16}{3} - 4 + c\).

\(c = -\frac{2}{3}\).

Thus, the equation of the curve is \(y = \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} - \frac{2}{3}\).

(ii) Differentiate \(\frac{dy}{dx} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\) to find \(\frac{d^2y}{dx^2}\).

\(\frac{d^2y}{dx^2} = \frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2}x^{-\frac{3}{2}}\).

(iii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(x^{\frac{1}{2}} - x^{-\frac{1}{2}} = 0 \Rightarrow \frac{x-1}{\sqrt{x}} = 0\).

\(x = 1\).

Substitute \(x = 1\) into the equation of the curve:

\(y = \frac{2}{3}(1)^{\frac{3}{2}} - 2(1)^{\frac{1}{2}} - \frac{2}{3} = -2\).

Thus, the stationary point is \((1, -2)\).

Check the nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} \bigg|_{x=1} = \frac{1}{2}(1)^{-\frac{1}{2}} + \frac{1}{2}(1)^{-\frac{3}{2}} = 1 > 0\).

Hence, the stationary point is a minimum.

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = x + \frac{4}{x^2}\).

\(y = \int \left( x + \frac{4}{x^2} \right) \, dx = \int x \, dx + \int \frac{4}{x^2} \, dx\)

\(y = \frac{x^2}{2} - \frac{4}{x} + c\)

Using the point \(P(4, 8)\), substitute \(x = 4\) and \(y = 8\):

\(8 = \frac{4^2}{2} - \frac{4}{4} + c\)

\(8 = 8 - 1 + c\)

\(c = 1\)

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

(ii) To find when the gradient has a minimum value, find the second derivative:

\(\frac{d^2y}{dx^2} = \frac{d}{dx} \left( x + \frac{4}{x^2} \right) = 1 - \frac{8}{x^3}\)

Set \(\frac{d^2y}{dx^2} = 0\) to find critical points:

\(1 - \frac{8}{x^3} = 0\)

\(\frac{8}{x^3} = 1\)

\(x^3 = 8\)

\(x = 2\)

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 + \frac{4}{2^2} = 2 + 1 = 3\)

Check the nature of the critical point using the third derivative or sign change:

\(\frac{d}{dx} \left( 1 - \frac{8}{x^3} \right) = \frac{24}{x^4}\)

Since \(\frac{24}{x^4} > 0\) for \(x > 0\), the gradient has a minimum value at \(x = 2\).

Thus, the minimum gradient is 3 when \(x = 2\).

(a) To find the value of \(a\), set \(\frac{dy}{dx} = 0\) at \(x = a\):

\(2(a + 3)^{\frac{1}{2}} - a = 0\)

\(2(a + 3)^{\frac{1}{2}} = a\)

Square both sides:

\(4(a + 3) = a^2\)

\(a^2 - 4a - 12 = 0\)

Factorize:

\((a - 6)(a + 2) = 0\)

\(a = 6\) (since \(a\) is positive)

(b) To determine the nature of the stationary point, find the second derivative:

\(\frac{d^2y}{dx^2} = (x + 3)^{-rac{1}{2}} - 1\)

Substitute \(x = a = 6\):

\(\frac{d^2y}{dx^2} = (6 + 3)^{-rac{1}{2}} - 1 = \frac{1}{3} - 1 = -\frac{2}{3}\)

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

(c) To find the equation of the curve, integrate \(\frac{dy}{dx} = 2(x + 3)^{\frac{1}{2}} - x\):

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

\(y = \frac{4}{3}(x + 3)^{\frac{3}{2}} - \frac{1}{2}x^2 + c\)

Substitute \(x = 6\) and \(y = 14\):

\(14 = \frac{4}{3}(9)^{\frac{3}{2}} - 18 + c\)

\(14 = 36 - 18 + c\)

\(c = -4\)

Thus, the equation of the curve is:

\(y = \frac{4}{3}(x+3)^{\frac{3}{2}} - \frac{1}{2}x^2 - 4\)

(i) Given \(\frac{d^2y}{dx^2} = -4x\), integrate to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int -4x \, dx = -2x^2 + c\)

Since the curve has a maximum at \(x = 2\), \(\frac{dy}{dx} = 0\) when \(x = 2\):

\(0 = -2(2)^2 + c \Rightarrow c = 8\)

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (-2x^2 + 8) \, dx = -\frac{2x^3}{3} + 8x + C\)

Use the point (2, 12) to find \(C\):

\(12 = -\frac{2(2)^3}{3} + 8(2) + C\)

\(12 = -\frac{16}{3} + 16 + C\)

\(C = \frac{4}{3}\)

Thus, the equation of the curve is \(y = -\frac{2x^3}{3} + 8x + \frac{4}{3}\).

(ii) To find \(\frac{dy}{dt}\), use the chain rule:

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

From part (i), \(\frac{dy}{dx} = -2x^2 + 8\).

At \(x = 3\), \(\frac{dy}{dx} = -2(3)^2 + 8 = -18 + 8 = -10\).

Given \(\frac{dx}{dt} = 0.05\),

\(\frac{dy}{dt} = -10 \times 0.05 = -0.5\).

Thus, the \(y\)-coordinate is decreasing at 0.5 units per second.

(i) Since \(P(3, -10)\) is a stationary point, \(f'(3) = 0\). Substituting into \(f'(x) = 2x^2 + kx - 12\), we get:

\(2(3)^2 + 3k - 12 = 0\)

\(18 + 3k - 12 = 0\)

\(3k = -6\)

\(k = -2\)

To find the other stationary point, set \(f'(x) = 0\):

\(2x^2 - 2x - 12 = 0\)

\(x^2 - x - 6 = 0\)

\((x - 3)(x + 2) = 0\)

\(x = 3\) or \(x = -2\)

Thus, the \(x\)-coordinate of \(Q\) is \(-2\).

(ii) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = 4x - 2\)

At \(x = 3\), \(f''(3) = 4(3) - 2 = 10 > 0\), so \(P\) is a minimum.

At \(x = -2\), \(f''(-2) = 4(-2) - 2 = -10 < 0\), so \(Q\) is a maximum.

(iii) Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int (2x^2 - 2x - 12) \, dx\)

\(f(x) = \frac{2}{3}x^3 - x^2 - 12x + c\)

Using \(P(3, -10)\), substitute to find \(c\):

\(-10 = \frac{2}{3}(3)^3 - (3)^2 - 12(3) + c\)

\(-10 = 18 - 9 - 36 + c\)

\(c = 17\)

Thus, \(f(x) = \frac{2}{3}x^3 - x^2 - 12x + 17\).