已知 $y = (x - 2)(2x - 5)^3$。

1. Differentiate the function using the product rule and chain rule.

Let $y = (x - 2)(2x - 5)^3$.
Using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$:
$\frac{dy}{dx} = (x - 2) \cdot \frac{d}{dx}[(2x - 5)^3] + (2x - 5)^3 \cdot \frac{d}{dx}(x - 2)$
$\frac{dy}{dx} = (x - 2)[3(2x - 5)^2(2)] + (2x - 5)^3(1)$
$\frac{dy}{dx} = 6(x - 2)(2x - 5)^2 + (2x - 5)^3$

2. Factorise the expression to obtain the required form.

Factor out the common term $(2x - 5)^2$:
$\frac{dy}{dx} = (2x - 5)^2 [6(x - 2) + (2x - 5)]$
$\frac{dy}{dx} = (2x - 5)^2 (6x - 12 + 2x - 5)$
$\frac{dy}{dx} = (2x - 5)^2 (8x - 17)$
$\frac{dy}{dx} = (8x - 17)(2x - 5)^2$

1. State the condition for a decreasing function and use the derivative from part (a)

From part (a), $\frac{dy}{dx} = (8x - 17)(2x - 5)^2$.
For $y$ to be a decreasing function, $\frac{dy}{dx} < 0$.
$(8x - 17)(2x - 5)^2 < 0$

2. Solve the inequality

Since $(2x - 5)^2 \ge 0$ for all real $x$, the inequality holds when:
$8x - 17 < 0$
$8x < 17$
$x < \frac{17}{8}$

1. Find the value of $\frac{dy}{dx}$ when $x = 3$ using the expression from part (a)

From part (a), $\frac{dy}{dx} = (8x - 17)(2x - 5)^2$.
When $x = 3$,
$\frac{dy}{dx} = (8(3) - 17)(2(3) - 5)^2$
$\frac{dy}{dx} = (24 - 17)(6 - 5)^2$
$\frac{dy}{dx} = 7(1)^2 = 7$

2. Use the chain rule to relate the rates of change

By the chain rule for connected rates of change:
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

3. Substitute known values to find $\frac{dx}{dt}$

Given $\frac{dy}{dt} = 0.35$ when $x = 3$:
$0.35 = 7 \times \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{0.35}{7} = 0.05$

1. Differentiate $z$ with respect to $t$ using the chain rule

Given $z = y^2$, differentiating with respect to $t$ gives:
$\frac{dz}{dt} = 2y \frac{dy}{dt}$

2. Evaluate $y$ when $x = 3$ and substitute to find the relationship between the rates

When $x = 3$,
$y = (3 - 2)(2(3) - 5)^3 = (1)(1)^3 = 1$
Substitute $y = 1$ into the rate equation:
$\frac{dz}{dt} = 2(1) \frac{dy}{dt}$
$\frac{dz}{dt} = 2 \frac{dy}{dt}$
Thus, when $x = 3$, $z$ is increasing at twice the rate of $y$.