First, find \(\frac{dy}{dx}\) for the curve \(y = x^2 - \frac{10}{3}x^{3/2} + 5x\).

\(\frac{dy}{dx} = 2x - 5x^{1/2} + 5\).

Given that \(\frac{dy}{dt} = 2\), we equate \(\frac{dy}{dx}\) to 2:

\(2x - 5x^{1/2} + 5 = 2\).

Simplify to get:

\(2x - 5x^{1/2} + 3 = 0\).

Factor the quadratic equation:

\((2x^{1/2} - 3)(x^{1/2} - 1) = 0\).

Solving gives:

\(x^{1/2} = \frac{3}{2}\) or \(x^{1/2} = 1\).

Thus, \(x = \frac{9}{4}\) and \(x = 1\).

Given \(\frac{dy}{dx} = 2 - 8(3x + 4)^{-\frac{1}{2}}\).

When \(x = 0\), \(\frac{dy}{dx} = 2 - 8(3(0) + 4)^{-\frac{1}{2}} = 2 - 8(4)^{-\frac{1}{2}} = 2 - 8 \times \frac{1}{2} = 2 - 4 = -2\).

The rate of change of the \(x\)-coordinate is \(\frac{dx}{dt} = 0.3\).

The rate of change of the \(y\)-coordinate is given by:

\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = -2 \times 0.3 = -0.6\).

Therefore, the rate of change of the \(y\)-coordinate is 0.6 units per second.

(i) To find the volume \(V\) of water in the tank, we first determine the area of the triangle \(ABE\). Since \(\angle ABE = \angle BAE = 30^\circ\), the triangle is isosceles with \(\angle AEB = 120^\circ\). Using the tangent of \(60^\circ\), we have:

\(\tan 60 = \frac{x}{h} \Rightarrow x = h \tan 60\)

\(\tan 60 = \sqrt{3}\), so \(x = h\sqrt{3}\).

The area \(A\) of triangle \(ABE\) is:

\(A = \frac{1}{2} \times x \times h = \frac{1}{2} \times h\sqrt{3} \times h = \frac{h^2\sqrt{3}}{2}\)

The volume \(V\) of the prism is:

\(V = \text{Area of } ABE \times AD = \frac{h^2\sqrt{3}}{2} \times 40 = 20h^2\sqrt{3}\)

Thus, \(V = (40\sqrt{3})h^2\).

(ii) Given \(\frac{dV}{dt} = 200\) cm\(^3\) s\(^{-1}\), we find \(\frac{dh}{dt}\) when \(h = 5\).

Differentiate \(V = (40\sqrt{3})h^2\) with respect to \(h\):

\(\frac{dV}{dh} = 80h\sqrt{3}\)

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

\(200 = 80h\sqrt{3} \cdot \frac{dh}{dt}\)

When \(h = 5\):

\(200 = 80 \times 5 \times \sqrt{3} \cdot \frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{200}{400\sqrt{3}} = \frac{1}{2\sqrt{3}}\)

Thus, \(\frac{dh}{dt} = \frac{1}{2\sqrt{3}}\) or approximately 0.289.

(i) The base of triangle \(XPQ\) is \(2 + p\) and the height is \(2p^2\) (since \(Q\) is on the curve \(y = 2x^2\) at \(x = p\), so \(y = 2p^2\)).

The area \(A\) of triangle \(XPQ\) is given by:

\(A = \frac{1}{2} \times (2 + p) \times 2p^2 = \frac{1}{2} \times (2p^2 + p^3)\)

\(A = p^3 + 2p^2\)

(ii) To find the rate of change of \(A\) with respect to time \(t\), we use the chain rule:

\(\frac{dA}{dt} = \frac{dA}{dp} \times \frac{dp}{dt}\)

First, find \(\frac{dA}{dp}\):

\(\frac{dA}{dp} = 3p^2 + 4p\)

Given \(\frac{dp}{dt} = 0.02\), substitute \(p = 2\):

\(\frac{dA}{dt} = (3(2)^2 + 4(2)) \times 0.02\)

\(\frac{dA}{dt} = (12 + 8) \times 0.02 = 20 \times 0.02 = 0.4\)

First, find \(\frac{dy}{dx}\) for the curve \(y = (7x^2 + 1)^{\frac{1}{3}}\).

Using the chain rule, \(\frac{dy}{dx} = \left[ \frac{1}{3} (7x^2 + 1)^{-\frac{2}{3}} \right] \times [14x]\).

When \(x = 3\), \(\frac{dy}{dx} = \frac{1}{3} \times (64)^{-\frac{2}{3}} \times 42\).

Calculate \((64)^{-\frac{2}{3}} = \left( \frac{1}{64} \right)^{\frac{2}{3}} = \left( \frac{1}{4} \right)^2 = \frac{1}{16}\).

Thus, \(\frac{dy}{dx} = \frac{1}{3} \times \frac{1}{16} \times 42 = \frac{7}{8}\).

Given \(\frac{dx}{dt} = 8\), use the chain rule to find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = \frac{7}{8} \times 8 = 7\).