A particle moves along the polar curve \( r=-2 \sin (-3 \theta) \) in such a way that the distance between the particle and the origin increases at a constant rate of 4 units per second. Find the rate at which the angle \( \theta \) changes with respect to time at the instant when the position of the particle corresponds to \( \theta=0 \), rounding to the nearest thousandth.

To solve the problem, we need to find the rate at which the angle \( \theta \) changes with respect to time, \( \frac{d\theta}{dt} \), when \( \theta = 0 \). We start with the polar equation given: \[ r = -2 \sin(-3\theta) = 2 \sin(3\theta) \] because \( \sin(-x) = -\sin(x) \). Next, we find \( \frac{dr}{dt} \) using the chain rule: \[ \frac{dr}{dt} = \frac{dr}{d\theta} \cdot \frac{d\theta}{dt} \] We first compute \( \frac{dr}{d\theta} \): \[ \frac{dr}{d\theta} = \frac{d}{d\theta}(2 \sin(3\theta)) = 2 \cdot 3 \cos(3\theta) = 6 \cos(3\theta) \] Now we know that the distance from the origin is increasing at a constant rate of 4 units per second, so \( \frac{dr}{dt} = 4 \). Substituting this in, we have: \[ 4 = 6 \cos(3\theta) \cdot \frac{d\theta}{dt} \] Next, we evaluate this at \( \theta = 0 \): \[ \cos(3 \cdot 0) = \cos(0) = 1 \] Thus, the equation simplifies to: \[ 4 = 6 \cdot 1 \cdot \frac{d\theta}{dt} \] Solving for \( \frac{d\theta}{dt} \): \[ \frac{d\theta}{dt} = \frac{4}{6} = \frac{2}{3} \] Now, rounding to the nearest thousandth, we find: \[ \frac{d\theta}{dt} \approx 0.667 \] Therefore, the rate at which the angle \( \theta \) changes with respect to time when \( \theta = 0 \) is approximately: \[ \boxed{0.667} \]