Module 34: Vector-valued motion, arc length, and curvature

A. Prerequisites and observable goals#

B. The central idea: a changing vector describes a trajectory#

A vector-valued function $\vv r(t)=\langle x(t),y(t),z(t)\rangle$ assigns a position to each time or parameter. Limits, derivatives, and integrals act componentwise. Velocity $\vv v=\vv r'$ is a vector tangent to the trajectory; speed $v=\|\vv r'\|$ is a scalar. Acceleration $\vv a=\vv r''$ may be nonzero even at constant speed because the velocity’s direction can turn.

C. Regular curves, length, and curvature#

For a sufficiently smooth curve with $\vv r'(t)\ne\vv0$, the unit tangent is $\vv T=\vv r'/\|\vv r'\|$. Arc length from a starting parameter is $s(t)=\int_{t_0}^t\|\vv r'(u)\|\dd u$ for $t\ge t_0$, so $\dd s/\dd t=v$. Curvature measures turning per unit distance:

The cross-product formula applies to a twice differentiable regular spatial curve; a zero velocity requires separate examination. Where $\kappa>0$, the principal normal is $\vv N=\vv T'/\|\vv T'\|$. If $\kappa=0$, that formula does not supply a unique normal.

For regular motion, the tangential acceleration is $a_T=v'=\vv v\cdot\vv a/\|\vv v\|$, and the nonnegative normal component is $a_N=\kappa v^2$. Thus $\vv a=a_T\vv T+a_N\vv N$ wherever these directions are defined. The alternative magnitude relation is $a_N=\sqrt{\|\vv a\|^2-a_T^2}$. Vector curve definitions: [O3Curves].

D. Why constant speed can coexist with acceleration#

Write $\vv v=v\vv T$. Product differentiation gives $\vv a=v'\vv T+v\vv T'$. Since $\vv T\cdot\vv T=1$, differentiation gives $\vv T\cdot\vv T'=0$; turning is perpendicular to the tangent. With $\vv T'=\kappa v\vv N$, the acceleration decomposition follows. This connects a geometric turning rate to the familiar centripetal acceleration of circular motion.

On a circle of radius $R$, a direction change $\Delta\theta$ corresponds to traveled distance $R\Delta\theta$, so curvature is $1/R$. Changing traversal speed does not change the circle’s curvature, but it changes $\kappa v^2$ and therefore the required turning acceleration.

E. A calculation sequence#

Differentiate components, compute speed, and identify any zero-speed points before normalizing. Integrate speed for length, not the vector velocity itself. For curvature, choose the unit-tangent or cross-product expression that simplifies most. For a motion interpretation, compute both $a_T$ and $a_N$ rather than inferring speeding up from the magnitude of acceleration. If reparameterizing by length, solve $s=s(t)$ on a monotone branch and substitute back into the position vector.

F. Worked examples#

A helix. For $\vv r(t)=\langle2\cos t,2\sin t,3t\rangle$, velocity is $\langle-2\sin t,2\cos t,3\rangle$ and speed is $\sqrt{13}$. Acceleration is $\langle-2\cos t,-2\sin t,0\rangle$, perpendicular to velocity. Therefore $a_T=0$, $a_N=2$, and $\kappa=a_N/v^2=2/13$. One turn has length $2\pi\sqrt{13}$.

Recovering motion from acceleration. If $\vv a=\langle0,0,-g\rangle$, then integrating gives $\vv v=\vv v_0+\langle0,0,-gt\rangle$ and $\vv r=\vv r_0+t\vv v_0+\langle0,0,-gt^2/2\rangle$. The integration constants are vectors determined component by component by the initial conditions.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 34#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Retrieve M12.02 and M31.08. Explain why speed, tangential acceleration, normal acceleration, and curvature have different units and meanings. Advance when you can check the regularity needed before dividing by speed.