Tangent and normal vectors

So far we have described parametrized curves via their positions, velocities and accelerations. However, these can depend on the parametrization. Two cars driving down the same road may have very different velocities and acceleration, even though they both trace the same curve. In this module we will develop ways to describe the curve itself, and answer four basic questions:

1. What direction are we going? This is given by a tangent vector $\bf T$, which gives the direction of the velocity vector.
2. Which direction are we curving? Are we turning left, right, up, down, or a combination? This is given by a vector called the principal normal ${\bf N}$.
3. How fast are we turning? By ``fast'' we mean a derivative with respect to distance, not with respect to time. How much will our direction change in the next inch, or foot, or mile? This is described by the curvature $\kappa$, and by the radius of curvature $R = 1/\kappa$. The bigger the curvature, the tighter the turn and the smaller the radius of curvature.
4. What plane are we currently moving in? The binormal vector ${\bf B} = {\bf T} \times {\bf N}$ is perpendicular to the instantaneous plane of motion.

For a space curve given parametrically by ${\bf r}(t)$, the tangent and normal vectors at the point ${\bf r}(t)$ are the unit vectors defined respectively by $${\bf T}(t) \ = \ \frac{{\bf r}'(t)}{\| {\bf r}'(t)\|}\,, \qquad {\bf N}(t) \ = \ \frac{{\bf T}'(t) }{\|{\bf T}'(t)\| }\,.$$ Frequently, ${\bf N}(t)$ is called the principal normal and the cross product ${\bf B}(t) = {\bf T}(t)\times {\bf N}(t)$ is called the binormal.

Notice that the tangent and principal normal vectors are perpendicular in the sense that ${\bf T}(t) \, {\Large\cdot}\, {\bf N}(t) = 0$ since $$\|{\bf T}(t)\|^2 \,=\, {\bf T}(t)\, {\Large\cdot} \,{\bf T}(t) \,=\, 1 \quad \Longrightarrow \quad \frac{d}{dt}\big( {\bf T}(t)\, {\Large\cdot} \,{\bf T}(t) \big) \,=\, 2 {\bf T}'(t)\, {\Large\cdot} \,{\bf T}(t) \,=\, 0\,.$$ Thus $\big\{ {\bf T}(t), \, {\bf N}(t),\, {\bf B}(t)\big\}$ forms a right-handed system of unit vectors at each point ${\bf r}(t)$ on the curve. For the case of the helix these are easy to calculate.

Example 4: for the position function $${\bf r} (t) \ = \ \cos t\, {\bf i} + \sin t \, {\bf j} + t\, {\bf k}\,,$$ whose path is a helix we see that the velocity is $$ {\bf v} (t) = {\bf r}'(t) =\ -\sin t\, {\bf i} + \cos t \, {\bf j} + {\bf k}\,,$$ so $\|{\bf r}'(t)\| \, = \, \sqrt{2}$, while $${\bf T} (t) \,=\, \frac{1}{\sqrt{2}} \big (-\sin t\, {\bf i} + \cos t \, {\bf j} + {\bf k} \big). $$

But then $${\bf T}' (t) =- \frac{1}{\sqrt{2}}\big (\cos t\, {\bf i} +\sin t \, {\bf j} \big)\,,$$ so $${\bf N} (t)= \frac{{\bf T}'(t)}{\|{\bf T}'(t)\|}= -\big (\cos t\, {\bf i} +\sin t \, {\bf j}\big)\,,$$ while $${\bf B} (t)= {\bf T}(t)\times {\bf N}(t) = \frac{1}{\sqrt{2}}\big (\sin t\, {\bf i} - \cos t \, {\bf j} + {\bf k}\big) \,.$$