Given the volume of water \(V = \pi \left( \frac{1}{2}h^2 + h \right)\), differentiate with respect to \(h\) to find \(\frac{dV}{dh}\).

\(\frac{dV}{dh} = \pi (h + 1)\).

We know \(\frac{dV}{dt} = 2\) cm\(^3\) s\(^{-1}\). We need to find \(\frac{dh}{dt}\) when \(h = 3\) cm.

Using the chain rule: \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\).

Substitute \(\frac{dV}{dh} = \pi (h + 1)\) and \(\frac{dV}{dt} = 2\):

\(2 = \pi (h + 1) \cdot \frac{dh}{dt}\).

When \(h = 3\), \(\pi (3 + 1) = 4\pi\).

\(2 = 4\pi \cdot \frac{dh}{dt}\).

\(\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}\) cm s\(^{-1}\).

The volume of a cube is given by the formula:

\(V = x^3\)

To find the rate of change of the volume with respect to time, we use the chain rule:

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

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

\(\frac{dV}{dx} = 3x^2\)

We are given \(\frac{dx}{dt} = 0.01\) cm/min.

Substitute \(x = 20\) into the expression:

\(\frac{dV}{dt} = 3 \times 20^2 \times 0.01\)

Calculate:

\(\frac{dV}{dt} = 3 \times 400 \times 0.01 = 12\)

Therefore, the rate of increase of \(V\) when \(x = 20\) is 12 cm³/min.

(i) To find where the curve crosses the x-axis, set \(y = 0\):

\(0 = 3 + \frac{12}{2-x}\)

\(-3 = \frac{12}{2-x}\)

\(-3(2-x) = 12\)

\(-6 + 3x = 12\)

\(3x = 18\)

\(x = 6\)

The curve crosses the x-axis at \((6, 0)\).

Find \(\frac{dy}{dx}\):

\(y = 3 + \frac{12}{2-x}\)

\(\frac{dy}{dx} = 0 + 12(2-x)^{-2}(-1)\)

\(\frac{dy}{dx} = \frac{12}{(2-x)^2}\)

At \(x = 6\), \(\frac{dy}{dx} = \frac{12}{(2-6)^2} = \frac{12}{16} = \frac{3}{4}\)

The equation of the tangent is \(y = \frac{1}{4}(x - 6)\) or \(4y = 3x - 18\).

(ii) Given \(\frac{dx}{dt} = 0.04\) and \(x = 4\), find \(\frac{dy}{dt}\):

\(\frac{dy}{dx} = 3\) when \(x = 4\).

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = 3 \times 0.04 = 0.12\)

(i) The base of the pyramid is a square with side \(\frac{1}{2}h\), so the area \(A\) of the base is \(\left(\frac{1}{2}h\right)^2 = \frac{1}{4}h^2\).

The volume \(V\) of the pyramid is given by \(V = \frac{1}{3}Ah\).

Substituting for \(A\), we have \(V = \frac{1}{3} \times \frac{1}{4}h^2 \times h = \frac{1}{12}h^3\).

(ii) We know \(\frac{dV}{dt} = 20 \text{ cm}^3/\text{min}\).

From (i), \(V = \frac{1}{12}h^3\), so \(\frac{dV}{dh} = \frac{1}{4}h^2\).

Using the chain rule, \(\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} = \frac{4}{h^2} \times 20\).

When \(h = 10\), \(\frac{dh}{dt} = \frac{4}{10^2} \times 20 = 0.8 \text{ cm/min}\).

(i) Differentiate \(y = 2 + \frac{3}{2x - 1}\) using the chain rule:

Let \(u = 2x - 1\), then \(\frac{du}{dx} = 2\).

\(y = 2 + \frac{3}{u}\), so \(\frac{dy}{du} = -\frac{3}{u^2}\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{3}{(2x-1)^2} \times 2\).

(ii) The curve has no stationary points because \(\frac{dy}{dx} = 0\) is impossible as the derivative is never zero.

(iii) At \(x = 2\), \(\frac{dy}{dx} = \frac{-6}{9} = -\frac{2}{3}\) and \(y = 3\).

The slope of the normal is the negative reciprocal: \(m = \frac{9}{6} = \frac{3}{2}\).

The equation of the normal is \(y - 3 = \frac{3}{2}(x - 2)\).

Substitute \(x = 0\) to find \(y = 0\), showing it passes through the origin.

(iv) Given \(\frac{dx}{dt} = -0.06\), find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = -\frac{2}{3} \times -0.06 = 0.04\).