Is there meaning or application to taking the cross product of the first and the second derivative of a vector valued function?

In my multivariable calculus class we have several practice problems where given a vector valued function r(t): we need to find $$r'(t) \cdot r''(t)$$

and

$$r'(t) \times r''(t)$$

For example, using the given function $r(t) = cos(t)i +sin(t)j +2tk$ which is from Larson 10e section 12.2 example 2.

The book seems to emphasize this calculation in the exercises and because of the frequency with which it appears, I wonder if this calculation has any significance. Is this just a convenient way for the authors to practice dot products, cross products, and derivatives of vector functions, or is there a meaning in math or any other field to taking the dot or cross product of a first derivative with a second derivative. I know how to solve the calculation, my question is whether there is any meaning to it.

• Look up "normal component of acceleration" in your text to see how this figures in. Nov 30 '17 at 18:30

The first of these quantities is proportional to the derivative of speed (squared): $$\frac{d}{dt}|r'|^2 = \frac{d}{dt} r'\cdot r' = r'' \cdot r' + r' \cdot r'' = 2 r' \cdot r''$$ I need to be a bit more hand-wavy about the second one. Remember that $|u \times v|$ is the area of the parallelogram spanned by $u$ and $v$. So $r'\times r''$ is a vector quantity whose length is proportional to $|r'|$, $|r''|$ and the sine of the angle between them. If the angle between $r'$ and $r''$ is large (i.e., close to a right angle), there's a lot of curving to the curve.
The curvature formula $$\kappa = \frac{|r' \times r''|}{|r'|^3}$$ comes from reparametrizing $r$ so that its speed is constant, and normalizing $r'$ to a unit tangent vector field $T= \frac{r'}{|r'|}$. If $s$ is the arc length as a coordinate, then $\kappa = |T'(s)|$. When you compute this derivative carefully, the $r'\times r''$ pops out, not for any mystical reason other than BAC-CAB.
In physics, angular momentum is defined as $L=m r(t) \times r'(t)$. Thus its derivative is $$\frac {dL}{dt} =mr'(t) \times r'(t) + m r(t)\times r''(t)=mr(t) \times r''(t).$$
Thus the angular momentum is conserved, if $mr''(t)\times r(t)=r\times F=0.$
This happens for example if the force is a central force (gravitation of the sun, electric field of a single particle), i.e. $F(r)=|F| e_r$, where $e_r$ is the unit vector in radial direction.