[Cambridge/2006]

1. Differentiate the equation of the curve implicitly with respect to $x$.

$\frac{d}{dx}(3x^2 + xy + y^2) = \frac{d}{dx}(33)$

$6x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$

2. Apply the condition for the tangent to be parallel to the $x$-axis and find the relationship between $x$ and $y$.

For the tangent to be parallel to the $x$-axis, $\frac{dy}{dx} = 0$.

Substituting $\frac{dy}{dx} = 0$ into the differentiated equation:
$6x + y + x(0) + 2y(0) = 0$
$y = -6x$

3. Substitute $y = -6x$ into the original curve equation to solve for $x$.

$3x^2 + x(-6x) + (-6x)^2 = 33$
$3x^2 - 6x^2 + 36x^2 = 33$
$33x^2 = 33$
$x^2 = 1$
$x = 1 \text{ or } x = -1$

4. Find the corresponding $y$-coordinates and state the final points.

When $x = 1$, $y = -6(1) = -6$.
When $x = -1$, $y = -6(-1) = 6$.

The coordinates of the points are $(1, -6)$ and $(-1, 6)$.

1. Calculate the remaining volume of the liquid 3 minutes after the start.

$t = 3 \text{ minutes} = 180 \text{ s}$
$\text{Volume lost} = 180 \times 2 = 360 \text{ cm}^3$
$\text{Remaining volume, } V = 390 - 360 = 30 \text{ cm}^3$

2. Express the volume $V$ entirely in terms of the depth $h$.

$\text{Let } r \text{ be the radius of the liquid surface.}$
$\text{Using the semi-vertical angle of } 45^\circ: \frac{r}{h} = \tan 45^\circ = 1 \implies r = h$
$\text{Volume of cone, } V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h^3$

3. Relate the rate of change of volume to the rate of change of depth using the chain rule.

$\frac{dV}{dh} = \pi h^2$
$\text{By the chain rule: } \frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$
$\frac{dV}{dt} = \pi h^2 \frac{dh}{dt}$

4. Find the depth $h$ when the volume is 30 cm³.

$\text{When } V = 30: \quad \frac{1}{3}\pi h^3 = 30$
$h^3 = \frac{90}{\pi} \implies h = \left(\frac{90}{\pi}\right)^{\frac{1}{3}} \approx 3.0598 \text{ cm}$

5. Substitute the known values to calculate the rate at which the depth is decreasing.

$\text{Substitute } \frac{dV}{dt} = -2 \text{ and } h = \left(\frac{90}{\pi}\right)^{\frac{1}{3}}:$
$-2 = \pi \left(\left(\frac{90}{\pi}\right)^{\frac{1}{3}}\right)^2 \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{-2}{\pi^{\frac{1}{3}} 90^{\frac{2}{3}}} \approx -0.06800 \text{ cm s}^{-1}$
$\text{The rate of decrease of the depth is } 0.0680 \text{ cm s}^{-1} \text{ (to 3 s.f.).}$