Worked problems

In the following video, we work out the velocity, acceleration, unit tangent vector, principal normal vector, arc length and curvature of the helix $${\bf r}(t) = \langle a \cos(t), a \sin(t), bt \rangle.$$ For reference, the calculations are repeated, in text form, below the video.

The velocity is $$\displaystyle{{\bf r}'(t) = \langle -a \sin(t), a \cos(t), b \rangle.}$$

The speed is $$\displaystyle{ \|{\bf r}'(t)\| = \sqrt{a^2 \sin^2(t) + a^2 \cos^2(t) + b^2} = \sqrt{a^2+b^2}}.$$

Since the speed is constant, the arc length is $$s = \sqrt{a^2+b^2}\; t.$$

The unit tangent vector is $${\bf T} = \frac{{\bf r}'}{\|{\bf r}'\|} = \frac{1}{\sqrt{a^2+b^2}} \langle -a \sin(t), a \cos(t), b \rangle.$$

Since $\displaystyle{\frac{d {\bf T}}{dt} = \frac{1}{\sqrt{a^2+b^2}} \langle -a \cos(t), -a \sin(t) ,0 \rangle}$, the principal normal vector is $$ {\bf N} = \frac{{\bf T}'}{\|{\bf T}'\|} = \langle -\cos(t), -\sin(t), 0 \rangle.$$

The acceleration is $$\displaystyle{{\bf r}'' = \langle - a \cos(t), -a\sin(t), 0 \rangle = a {\bf N}}.$$ Usually, the acceleration has a component in the direction of motion, related to changes in speed, and a component in the principal normal direction, related to changes in direction. In this case the speed is constant, so the acceleration is entirely in the normal direction.

We compute the curvature from the definition: $$\kappa = \left \| \frac{d {\bf T}}{ds} \right \| = \frac{\| {\bf T}' \|}{ds/dt} = \frac{a/\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = \frac{a}{a^2+b^2}.$$This makes the radius of curvature $$R = \frac{1}{\kappa} = \frac{a^2+b^2}{a}.$$

Notice that $a = \kappa v^2 = v^2/R$, where $v$ is the speed, so that the acceleration is $${\bf r}'' = a {\bf N} = \frac{v^2}{R} {\bf N} = \kappa v^2 {\bf N}.$$

We could have computed the curvature a different way, using the formula $$\kappa = \frac{\|{\bf r}' \times {\bf r}''\|}{\|{\bf r}'\|^3},$$but working from the definition is easier.