Given a normal distribution on $\mathbb{R}^3$
$$p(\mathbf{x})\propto\exp(-\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu}))$$
and a circular trajectory through $\mathbb{R}^3$
$$\mathbf{x}(\theta)=r(\mathbf{\hat{i}}\cos\theta+\mathbf{\hat{j}}\sin\theta)$$
where
- $r$ is the radius of curvature
- $\mathbf{\mu}$ is the distribution's mean
- $\Sigma$ is the distribution's variance matrix (which is symmetric)
Can you suggest how to solve (or alternatively approximate) $\theta$ maximizing $p(\mathbf{x}(\theta))$?
Note that $x \mapsto \exp(- \frac 12 x)$ is a decreasing function. Thus, it suffices to find the $\mathbf x(\theta)$ that minimizes $$ f(\mathbf x) = (\mathbf x - \boldsymbol \mu)^T\Sigma^{-1}(\mathbf x - \boldsymbol \mu). $$ One approach is to simply plug in. Denote $\mathbf x = (x_1,x_2,x_3)$, $\boldsymbol \mu = (\mu_1,\mu_2,\mu_3)$, $M = \Sigma^{-1}$ and the entries of $M$ are $m_{ij}$. We have $$ f(\mathbf x) = \sum_{i,j = 1}^3 m_{ij}(x_i - \mu_i)(x_j - \mu_j). $$ Plugging in $\mathbf x(\theta)$ yields $$ f(\mathbf x(\theta)) = m_{11}(\cos \theta - \mu_1)^2 + 2m_{12}(\cos \theta - \mu_1)(\sin \theta - \mu_2) + m_{22}(\sin \theta - \mu_2)^2. $$ This function can be maximized in the usual fashion; just set $f'(\theta) = 0$ and solve.