水泥以 $36\pi\ \text{cm}^3/\text{s}$ 的恒定速率注入一个空的正圆锥形模具中。圆锥的高度始终是其半径的两倍。

1. Calculate the total volume of cement poured after 1.5 minutes

Time $t = 1.5 \times 60 = 90\ \text{s}$
Volume $V = 36\pi \times 90 = 3240\pi\ \text{cm}^3$

2. Express volume in terms of radius $r$ and solve for $r$

Given $h = 2r$,
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3$
Equating the volumes:
$\frac{2}{3}\pi r^3 = 3240\pi$
$r^3 = 3240 \times \frac{3}{2} = 4860$
$r = \sqrt[3]{4860} \approx 16.938\ \text{cm}$
$r \approx 16.94\ \text{cm}$ (shown)

1. Find the value of the constant $k$.

Given $I = \frac{k}{x^2}$.
When $x = 4$, $I = 12$:
$12 = \frac{k}{4^2}$
$k = 12 \times 16 = 192$

2. Differentiate $I$ with respect to $x$ to find the rate of change formula.

$I = 192x^{-2}$
$\frac{dI}{dx} = 192(-2)x^{-3}$
$\frac{dI}{dx} = -\frac{384}{x^3}$

3. Evaluate the rate of change when $x = 8$.

When $x = 8$:
$\frac{dI}{dx} = -\frac{384}{8^3}$
$\frac{dI}{dx} = -\frac{384}{512} = -0.75$