We are given the initial value problem:

$\frac{dr}{d\theta} = \sin(-3\pi \theta), \quad r\left(-\frac{1}{6}\right) = 10$

Step 1: Solve the differential equation

The equation $\frac{dr}{d\theta} = \sin(-3\pi \theta)$ is a separable differential equation. We can solve it by integrating both sides with respect to $\theta$.

First, rewrite the equation:

$\frac{dr}{d\theta} = \sin(-3\pi \theta)$

We can integrate both sides:

$r(\theta) = \int \sin(-3\pi \theta) \, d\theta$

Use the substitution $u = -3\pi \theta$, so that $du = -3\pi \, d\theta$, and the integral becomes:

$r(\theta) = \int \sin(u) \cdot \frac{du}{-3\pi}$

$r(\theta) = -\frac{1}{3\pi} \int \sin(u) \, du$

The integral of $\sin(u)$ is $-\cos(u)$, so:

$r(\theta) = -\frac{1}{3\pi} (-\cos(u)) = \frac{1}{3\pi} \cos(-3\pi \theta)$

Since $\cos(-x) = \cos(x)$, this simplifies to:

$r(\theta) = \frac{1}{3\pi} \cos(3\pi \theta) + C$

where $C$ is the constant of integration.

Step 2: Apply the initial condition

We are given that $r\left(-\frac{1}{6}\right) = 10$. Substituting this into the equation:

$10 = \frac{1}{3\pi} \cos\left(3\pi \cdot \left(-\frac{1}{6}\right)\right) + C$

Simplify the cosine term:

$\cos\left(3\pi \cdot \left(-\frac{1}{6}\right)\right) = \cos\left(-\frac{\pi}{2}\right) = 0$

So the equation becomes:

$10 = \frac{1}{3\pi} \cdot 0 + C$

$C = 10$

Step 3: Final solution

Thus, the solution to the initial value problem is:

$r(\theta) = \frac{1}{3\pi} \cos(3\pi \theta) + 10$

This is the general solution, with the constant determined from the initial condition.

Would you like further details or have any questions?

Here are five related questions to expand on the information:

1. What is the importance of the initial condition in solving differential equations?
2. How does the cosine function behave when the argument involves $\pi$?
3. Could we solve this equation using a different method, like a numerical approach?
4. What would happen if the differential equation involved $\sin(3\pi \theta)$ instead of $\sin(-3\pi \theta)$?
5. How would the solution change if the initial condition was different?

Tip: Always check if the differential equation is separable or if another method like exact equations can be applied before starting to solve.