Question #41271

Let $\vec{u}(t) = \hat{i} + t\hat{j} + \frac{t^2}{2}\hat{k}$, $\vec{v}(t) = \hat{i} + \sin(t)\hat{j} + (1-\cos t)\hat{k}$, and $\vec{w}(t) = \hat{i} + \tan(t)\hat{j} + \ln\left(1+\frac{t^2}{2}\right)\hat{k}$ be three vectors in $\mathbb{R}^3$ for $t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Let $V(t)$ denote the volume of the parallelepiped determined by $\vec{u}(t)$, $\vec{v}(t)$, and $\vec{w}(t)$. Let $\vec{n}(t) = \vec{u}(t) \times \vec{v}(t) = n_x(t)\hat{i} + n_y(t)\hat{j} + n_z(t)\hat{k}$ be the normal vector to the plane containing $\vec{u}(t)$ and $\vec{v}(t)$. If $m(t) = \frac{n_y(t)}{n_x(t)}$ represents the slope of the projection of $\vec{n}(t)$ onto the $xy$-plane, evaluate the limit: $\lim_{t \to 0} \left( m(t) \cdot \frac{V(t)}{t^6} \right)$