The diagram shows part of the curve $y = 3x^{2}$. The points P and Q lie on the curve, and have the same y-coordinates. The points P and Q move along the curve such that their y-coordinates are changing at a rate of 3 units per second.

1. Determine the length of the base PQ and the perpendicular height of the triangle

Since P and Q lie on the curve $y = 3x^2$ and are symmetric about the y-axis, their coordinates are $(-x, 3x^2)$ and $(x, 3x^2)$.
Length of base $PQ = x - (-x) = 2x$.
The perpendicular height of the triangle from the origin $O(0,0)$ to the line $PQ$ is the y-coordinate, which is $3x^2$.

2. Calculate the area of the triangle OPQ

$A = \frac{1}{2} \times \text{base} \times \text{height}$
$A = \frac{1}{2} \times (2x) \times (3x^2)$
$A = 3x^3 \text{ (shown)}$

1. Express the length $PQ$ in terms of $x$.

Since $P$ is at $(-x, 3x^2)$ and $Q$ is at $(x, 3x^2)$ with $x > 0$, the length $PQ = x - (-x) = 2x$.

2. Find the relationship between $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Given $y = 3x^2$, differentiating with respect to $t$ gives:
$\frac{dy}{dt} = 6x \frac{dx}{dt}$

3. Calculate $\frac{dx}{dt}$ when $x = 4$.

We are given $\frac{dy}{dt} = 3$.
When $x = 4$:
$3 = 6(4) \frac{dx}{dt}$
$3 = 24 \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{3}{24} = \frac{1}{8}$

4. Calculate the rate of change of $PQ$.

Let $L = PQ = 2x$.
$\frac{dL}{dt} = 2 \frac{dx}{dt}$
$\frac{dL}{dt} = 2 \left(\frac{1}{8}\right) = \frac{1}{4}$

1. Find the rate of change of $x$ at $x = 4$.

Given $y = 3x^2$, we have $\frac{dy}{dx} = 6x$.
At $x = 4$, $\frac{dy}{dx} = 6(4) = 24$.
Using the chain rule $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$:
$3 = 24 \times \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{3}{24} = \frac{1}{8}$

2. Find the derivative of the area $A$ with respect to $x$.

Given $A = 3x^3$, we have $\frac{dA}{dx} = 9x^2$.
At $x = 4$, $\frac{dA}{dx} = 9(4)^2 = 144$.

3. Calculate the rate of change of $A$ using the chain rule.

$\frac{dA}{dt} = \frac{dA}{dx} \times \frac{dx}{dt}$
$\frac{dA}{dt} = 144 \times \frac{1}{8} = 18$