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

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

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

Given the curve equation \(y = x^2 - 2x - 3\), we need to find \(\frac{dy}{dx}\).

Differentiate \(y\) with respect to \(x\):

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

We know \(\frac{dy}{dt} = 4\) and \(\frac{dx}{dt} = 6\).

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

Substitute the known values:

\(4 = (2x - 2) \cdot 6\).

Solve for \(x\):

\(4 = 12x - 12\)

\(16 = 12x\)

\(x = \frac{16}{12} = \frac{4}{3}\).

(i) The surface area \(S\) of the cuboid is given by:

\(S = 2(x \cdot 2x + 2x \cdot 4x + 4x \cdot x) = 2(2x^2 + 8x^2 + 4x^2) = 28x^2\).

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

\(V = x \cdot 2x \cdot 4x = 8x^3\).

To show \(S = 7V^{\frac{2}{3}}\), substitute \(V = 8x^3\):

\(V^{\frac{2}{3}} = (8x^3)^{\frac{2}{3}} = 4x^2\).

Thus, \(S = 7 \cdot 4x^2 = 28x^2\), which confirms the relation.

(ii) Given \(V = 1000\), find \(x\):

\(8x^3 = 1000 \Rightarrow x^3 = 125 \Rightarrow x = 5\).

\(S = 28x^2 = 28 \cdot 25 = 700\).

\(\frac{dS}{dt} = 2\), \(\frac{dS}{dx} = 56x\), \(\frac{dV}{dx} = 24x^2\).

\(\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}\).

\(\frac{dS}{dt} = \frac{dS}{dx} \cdot \frac{dx}{dt} \Rightarrow 2 = 56x \cdot \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{1}{140}\).

\(\frac{dV}{dt} = 24x^2 \cdot \frac{1}{140} = \frac{24 \cdot 25}{140} = \frac{30}{7}\).

Thus, the rate of increase of the volume is \(\frac{30}{7}\) or 4.29 cm³ s⁻¹.

Given \(\frac{dy}{dx} = x^3 - \frac{4}{x^2}\), we need to find \(\frac{dy}{dt}\) when \(x = 2\) and \(\frac{dx}{dt} = -0.05\).

First, calculate \(\frac{dy}{dx}\) at \(x = 2\):

\(\frac{dy}{dx} = 2^3 - \frac{4}{2^2} = 8 - 1 = 7\).

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

Substitute the values: \(\frac{dy}{dt} = 7 \times (-0.05) = -0.35\).

Thus, the rate of change of the \(y\)-coordinate is \(-0.35\) units/s.

(i)(a) Differentiate \(y = \frac{1}{2}(4x - 3)^{-1}\) to find \(\frac{dy}{dx} = \left[-\frac{1}{2}(4x - 3)^{-2}\right] \times 4 = -2(4x - 3)^{-2}\).

At \(x = 1\), \(m = -2\).

The gradient of the normal is \(-\frac{1}{m} = \frac{1}{2}\).

The equation of the normal is \(y - \frac{1}{2} = \frac{1}{2}(x - 1)\).

Simplifying gives \(y = \frac{1}{2}x\).

(i)(b) Substitute \(y = \frac{1}{2}x\) into the curve equation: \(\frac{1}{2}(4x - 3)^{-1} = \frac{x}{2}\).

Solving gives \(x = -\frac{1}{4}\).

(ii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

At \(x = 1\), \(\frac{dy}{dx} = -2\).

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

Given the curve \(y = 2x + \frac{5}{x}\), we need to find \(\frac{dy}{dt}\) when \(x = 1\) and \(\frac{dx}{dt} = 0.02\).

First, differentiate \(y\) with respect to \(x\):

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

Substitute \(x = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 - \frac{5}{1^2} = 2 - 5 = -3\).

Now, use the chain rule to find \(\frac{dy}{dt}\):

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

Substitute \(\frac{dy}{dx} = -3\) and \(\frac{dx}{dt} = 0.02\):

\(\frac{dy}{dt} = -3 \times 0.02 = -0.06\).

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 \(\frac{dy}{dt}\):

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

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 \(\frac{dy}{dt}\):

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

(i) To find \(\frac{dy}{dx}\), use the quotient rule. Let \(u = 12\) and \(v = 3 - 2x\). Then \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -2\).

Using the quotient rule: \(\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} = \frac{(3 - 2x) \cdot 0 - 12 \cdot (-2)}{(3 - 2x)^2} = \frac{24}{(3 - 2x)^2}\).

(ii) Given \(\frac{dy}{dt} = 0.4\) and \(\frac{dx}{dt} = 0.15\), use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{0.4}{0.15} = \frac{8}{3}\).

Equate this to the derivative found in part (i): \(\frac{24}{(3 - 2x)^2} = \frac{8}{3}\).

Cross-multiply to solve for \(x\): \(24 \cdot 3 = 8 \cdot (3 - 2x)^2\).

\(72 = 8(3 - 2x)^2\)

\(9 = (3 - 2x)^2\)

Taking the square root: \(3 - 2x = 3\) or \(3 - 2x = -3\).

Solving these gives \(x = 0\) or \(x = 3\).

Given the curve \(y = \frac{8}{x} + 2x\), we need to find \(\frac{dy}{dt}\) when \(x = 1\) and \(\frac{dx}{dt} = 0.04\).

First, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{8}{x} + 2x \right) = -\frac{8}{x^2} + 2\).

At point \(A\) where \(x = 1\), substitute \(x = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = -\frac{8}{1^2} + 2 = -8 + 2 = -6\).

Now, use the chain rule to find \(\frac{dy}{dt}\):

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

Substitute \(\frac{dy}{dx} = -6\) and \(\frac{dx}{dt} = 0.04\):

\(\frac{dy}{dt} = -6 \times 0.04 = -0.24\).

(a) To find the rate at which \(h\) is increasing, we use the chain rule. First, differentiate \(V\) with respect to \(h\):

\(\frac{dV}{dh} = \frac{4}{3} \times 3(25 + h)^2 = 4(25 + h)^2\).

At \(h = 10\), \(\frac{dV}{dh} = 4(25 + 10)^2 = 4900\).

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

Given \(\frac{dV}{dt} = 500\), we have:

\(500 = 4900 \cdot \frac{dh}{dt}\).

Solving for \(\frac{dh}{dt}\), we get \(\frac{dh}{dt} = \frac{500}{4900} = 0.102 \text{ cm/s}\).

(b) Given \(\frac{dh}{dt} = 0.075 \text{ cm/s}\), use the chain rule:

\(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 4(25 + h)^2 \cdot 0.075\).

Set \(\frac{dV}{dt} = 500\):

\(500 = 4(25 + h)^2 \cdot 0.075\).

Solving for \(h\), we have:

\((25 + h)^2 = \frac{5000}{3}\).

\(25 + h = \sqrt{\frac{5000}{3}}\).

\(h = \sqrt{\frac{5000}{3}} - 25 \approx 15.82\).

Substitute \(h\) back into the volume formula:

\(V = \frac{4}{3}(25 + 15.82)^3 - \frac{62500}{3}\).

\(V \approx 69900 \text{ cm}^3\).

The area \(A\) of a circle is given by \(A = \pi r^2\), where \(r\) is the radius.

To find the rate at which the area is increasing, we need \(\frac{dA}{dt}\).

Using the chain rule, \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\).

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

\(\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r\).

Given \(\frac{dr}{dt} = 3\) m/h and \(r = 50\) m at midday, substitute these values:

\(\frac{dA}{dt} = 2\pi (50) \times 3 = 300\pi\).

Thus, \(\frac{dA}{dt} = 300\pi\) square metres per hour, or approximately 942 square metres per hour.

(i) To find \(\frac{dy}{dx}\), we differentiate each term of \(y = 4\sqrt{x} + \frac{2}{\sqrt{x}}\).

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

Differentiate: \(\frac{dy}{dx} = 4 \cdot \frac{1}{2}x^{-\frac{1}{2}} + 2 \cdot (-\frac{1}{2})x^{-\frac{3}{2}}\).

Simplify: \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - x^{-\frac{3}{2}}\).

(ii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

Given \(\frac{dx}{dt} = 0.12\) and \(x = 4\), substitute into \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - x^{-\frac{3}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = 2(4)^{-\frac{1}{2}} - (4)^{-\frac{3}{2}} = \frac{1}{\sqrt{4}} - \frac{1}{8} = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}\).

Thus, \(\frac{dy}{dt} = \frac{3}{8} \times 0.12 = 0.105\).

Given the formula \(M = kr^3\), we need to find \(k\) when \(M = 3.2\) kg and \(r = 10\) cm.

Substitute the values into the formula:

\(3.2 = k \times 10^3\)

\(3.2 = 1000k\)

\(k = \frac{3.2}{1000} = 0.0032\)

Next, we find the rate at which the mass is increasing, \(\frac{dM}{dt}\).

Differentiate \(M = kr^3\) with respect to \(r\):

\(\frac{dM}{dr} = 3kr^2\)

Using the chain rule, \(\frac{dM}{dt} = \frac{dM}{dr} \times \frac{dr}{dt}\).

Given \(\frac{dr}{dt} = 0.1\) cm/day, substitute the values:

\(\frac{dM}{dt} = 3 \times 0.0032 \times 10^2 \times 0.1\)

\(\frac{dM}{dt} = 0.096 \text{ kg/day}\)

Given the volume of a sphere \(V = \frac{4}{3}\pi r^3\), differentiate with respect to \(r\):

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

Substitute \(r = 10\):

\(\frac{dV}{dr} = 4\pi \times 10^2 = 400\pi\).

We know \(\frac{dV}{dt} = 50\) cm³/s. Use the chain rule:

\(\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}\).

Rearrange to find \(\frac{dr}{dt}\):

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

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

(i) To find \(\frac{dx}{dt}\), we differentiate \(x = 0.7 \sqrt{(2t - 1)}\) with respect to \(t\).

Let \(u = 2t - 1\), then \(x = 0.7 u^{1/2}\).

\(\frac{dx}{du} = 0.7 \cdot \frac{1}{2} u^{-1/2} = 0.35 u^{-1/2}\).

\(\frac{du}{dt} = 2\).

By the chain rule, \(\frac{dx}{dt} = \frac{dx}{du} \cdot \frac{du}{dt} = 0.35 u^{-1/2} \cdot 2 = 0.7 (2t - 1)^{-1/2}\).

(ii) To find the rate of growth when the snake is 5 years old, substitute \(t = 5\) into \(\frac{dx}{dt}\).

\(\frac{dx}{dt} = 0.7 (2(5) - 1)^{-1/2} = 0.7 (10 - 1)^{-1/2} = 0.7 \cdot 3^{-1/2}\).

\(\frac{dx}{dt} \approx 0.23\).

(i) Differentiate \(y = \frac{12}{x^2 + 3}\) using the chain rule. Let \(u = x^2 + 3\), then \(y = \frac{12}{u}\). The derivative \(\frac{dy}{du} = -\frac{12}{u^2}\). Since \(u = x^2 + 3\), \(\frac{du}{dx} = 2x\). Therefore, \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = -\frac{12}{(x^2 + 3)^2} \times 2x\).

(ii) At \(x = 1\), the slope \(m = \frac{dy}{dx} = -\frac{3}{2}\). The slope of the normal is the negative reciprocal, \(\frac{2}{3}\). The equation of the normal is \(y - 3 = \frac{2}{3}(x - 1)\).

(iii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\). Given \(\frac{dx}{dt} = 0.012\) and \(\frac{dy}{dx} = -\frac{3}{2}\) at \(x = 1\), \(\frac{dy}{dt} = -\frac{3}{2} \times 0.012 = -0.018\).

(i) To find the gradient of the curve, we need to differentiate \(y = \frac{6}{5 - 2x}\) with respect to \(x\).

Using the chain rule, \(\frac{dy}{dx} = -6(5 - 2x)^{-2} \times (-2)\).

This simplifies to \(\frac{dy}{dx} = \frac{12}{(5 - 2x)^2}\).

Substitute \(x = 1\) into the derivative: \(\frac{dy}{dx} = \frac{12}{(5 - 2 \times 1)^2} = \frac{12}{9} = \frac{4}{3}\).

Therefore, the gradient at \(x = 1\) is \(\frac{1}{3}\).

(ii) We are given that \(\frac{dy}{dt} = 0.02\). We need to find \(\frac{dx}{dt}\) when \(x = 1\).

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

Rearrange to find \(\frac{dx}{dt} = \frac{dy}{dt} \div \frac{dy}{dx}\).

Substitute \(\frac{dy}{dt} = 0.02\) and \(\frac{dy}{dx} = \frac{1}{3}\) from part (i):

\(\frac{dx}{dt} = \frac{0.02}{\frac{1}{3}} = 0.02 \times 3 = 0.06\).

Therefore, the rate of increase of \(x\) when \(x = 1\) is \(0.015\).

(i) To find the gradient of the curve, we need to differentiate \(y = \sqrt{5x + 4}\) with respect to \(x\).

Let \(y = (5x + 4)^{1/2}\).

The derivative \(\frac{dy}{dx} = \frac{1}{2}(5x + 4)^{-1/2} \times 5\).

At \(x = 1\), \(\frac{dy}{dx} = \frac{1}{2}(5 \times 1 + 4)^{-1/2} \times 5 = \frac{5}{6}\).

(ii) We are given \(\frac{dx}{dt} = 0.03\) and need to find \(\frac{dy}{dt}\).

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

Substitute \(\frac{dy}{dx} = \frac{5}{6}\) and \(\frac{dx}{dt} = 0.03\):

\(\frac{dy}{dt} = \frac{5}{6} \times 0.03 = 0.025\).

To find the \(x\)-coordinate where the \(x\)- and \(y\)-coordinates are increasing at the same rate, we need \(\frac{dy}{dx} = 1\).

First, differentiate \(y = \frac{1}{60}(3x + 1)^2\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{60} \times 2(3x + 1) \times 3 = \frac{1}{10}(3x + 1)\).

Set \(\frac{dy}{dx} = 1\):

\(\frac{1}{10}(3x + 1) = 1\).

Solving for \(x\):

\(3x + 1 = 10\)

\(3x = 9\)

\(x = 3\)

Given the volume function \(V = 243 - \frac{1}{3}(9-x)^3\), we need to find \(\frac{dx}{dt}\) when \(x = 4\).

First, differentiate \(V\) with respect to \(x\):

\(\frac{dV}{dx} = (9-x)^2\).

Substitute \(x = 4\) into \(\frac{dV}{dx}\):

\(\frac{dV}{dx} = (9-4)^2 = 25\).

We know \(\frac{dV}{dt} = 3.6\) m³/hour.

Using the chain rule, \(\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}\).

So, \(3.6 = 25 \cdot \frac{dx}{dt}\).

Solving for \(\frac{dx}{dt}\):

\(\frac{dx}{dt} = \frac{3.6}{25}\) m/hour.

Convert \(\frac{dx}{dt}\) to cm/min:

\(\frac{dx}{dt} = \frac{3.6}{25} \times 100 \div 60 = 0.24\) cm/min.

Given the curve \(y = 18 - \frac{3}{8}x^{\frac{5}{2}}\), we need to find \(\frac{dy}{dt}\) when \(x = 4\) and \(\frac{dx}{dt} = 2\).

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

\(\frac{dy}{dx} = -\frac{5}{2} \times \frac{3}{8} x^{\frac{3}{2}} = -\frac{15}{16} x^{\frac{3}{2}}\).

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

Substitute \(x = 4\) and \(\frac{dx}{dt} = 2\):

\(\frac{dy}{dt} = -\frac{15}{16} \times 4^{\frac{3}{2}} \times 2\).

Calculate \(4^{\frac{3}{2}} = 8\), so:

\(\frac{dy}{dt} = -\frac{15}{16} \times 8 \times 2 = -15\).

Given the function \(f(x) = (4x + 2)^{-2}\), we need to find the derivative \(\frac{dy}{dx}\).

Using the chain rule, \(\frac{dy}{dx} = -2(4x + 2)^{-3} \times 4\).

This simplifies to \(\frac{dy}{dx} = -8(4x + 2)^{-3}\).

We know that the \(y\)-coordinate is decreasing at the rate of \(k\) and the \(x\)-coordinate is increasing at the rate of \(k\), so \(\frac{dy}{dx} = -1\).

Set \(-8(4x + 2)^{-3} = -1\).

Solving for \(x\), we get \((4x + 2)^{3} = 8\).

Taking the cube root, \(4x + 2 = 2\).

Solving for \(x\), \(4x = 0\) so \(x = 0\).

Substitute \(x = 0\) back into \(f(x)\) to find \(y\):

\(y = (4(0) + 2)^{-2} = 2^{-2} = \frac{1}{4}\).

Therefore, the coordinates of A are \((0, \frac{1}{4})\).

(a) Differentiate the volume equation with respect to \(r\):

\(\frac{dV}{dr} = \frac{9}{2} \left( r - \frac{1}{2} \right)^2\).

Use the chain rule to find \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{1}{\frac{dV}{dr}} \times \frac{dV}{dt}\).

Substitute \(r = 5.5\) and \(\frac{dV}{dt} = 1.5\):

\(\frac{dr}{dt} = \frac{1.5}{\frac{9}{2} (5.5 - 0.5)^2} = \frac{1.5}{112.5} = 0.0133 \text{ m/s}\).

(b) Given \(\frac{dr}{dt} = 0.1 \text{ m/s}\), use the relationship:

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

\(\frac{dV}{dr} = \frac{9}{2} \left( r - \frac{1}{2} \right)^2\).

\(0.1 = \frac{1.5}{\frac{9}{2} \left( r - \frac{1}{2} \right)^2}\).

Solve for \(r\):

\(\frac{9}{2} \left( r - \frac{1}{2} \right)^2 = 15\).

\(r = \frac{1}{2} + \sqrt{\frac{10}{3}}\).

Substitute \(r\) back into the volume equation:

\(V = \frac{3}{2} \left( \frac{1}{2} + \sqrt{\frac{10}{3}} - \frac{1}{2} \right)^3 - 1\).

\(V \approx 8.13 \text{ m}^3\).

At \(x = 1\), \(\frac{dy}{dx} = 6\).

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

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

\(\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{6}\).

Given \(\frac{dy}{dt} = 3\), substitute into the chain rule:

\(\frac{dx}{dt} = \frac{1}{6} \times 3 = \frac{1}{2}\).