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 \(0.015\).