Space curves, speed and curvature

Try

A particle follows r(t)=(cos t,sin t,2t), with angle measured in radians. Its horizontal projection goes around the unit circle while its height rises steadily. At t=0 its position is (1,0,0); at t=π it is (−1,0,2π). The displacement between those positions is not the distance traveled. Sketch a cylinder and mark these positions on it, labeling the perspective drawing as schematic. The parameter t represents time only when the model explicitly assigns it that role.

Predict

Predict whether the particle speeds up as its height increases. The curve looks increasingly distant from the origin, but distance from the origin is not speed. Differentiate component by component: r′(t)=(−sin t,cos t,2). Its magnitude is √5 at every time. Thus velocity changes direction while speed stays constant. Acceleration r″(t)=(−cos t,−sin t,0) is nonzero. Explain why constant speed does not imply zero acceleration: it fixes the velocity’s magnitude, not the full vector.

Experiment

Compare a second traversal q(u)=r(3u) with the first. Compute q′=3r′(3u) and q″=9r″(3u). Make a table at corresponding points, using t=3u rather than comparing equal parameter numbers. The faster traversal has triple the speed and nine times the acceleration. Its geometric track has not changed.

For the original helix from t=0 to t=2, arc length is ∫₀²|r′(t)|dt=2√5. Try replacing the vertical component 2t by bt. The speed becomes √(1+b²), showing exactly how increased vertical advance changes length per revolution.

Understand

A regular curve has nonzero derivative. Its unit tangent is T=r′/|r′| and its arc length satisfies ds/dt=|r′|. Curvature κ is |dT/ds|: turning per unit distance, not per unit time. For a twice differentiable regular curve, κ=|r′×r″|/|r′|³. The denominator requires nonzero speed.

For the helix above, the cross product is (2sin t,−2cos t,1), with magnitude √5. Consequently κ=√5/(√5)³=1/5. A unit circle has curvature 1, so adding steady vertical motion reduces the turning per unit distance even though the horizontal projection remains circular. Curvature has inverse-length units. A circle of radius R has κ=1/R; a straight line has zero curvature wherever its parametrization is regular.

Apply

Diagnose “acceleration equals curvature.” The statement confuses dimensions and ignores speed. When speed is nonzero, acceleration separates into a tangential component controlling speed change and a normal component of magnitude κv². Here speed is √5 and κ=1/5, giving normal acceleration magnitude 1, which agrees with |r″|.

Compute a displacement by integrating velocity: ∫ₐᵇr′(t)dt=r(b)−r(a). Compute distance by integrating its magnitude instead. The order of taking a norm and integrating matters. A full circle returns to its starting point, so its displacement is zero while its distance is positive.

Continue

For optional depth, derive the cross-product curvature formula by differentiating the unit tangent and removing the acceleration component parallel to velocity. This explains why cross products appear: they retain only the perpendicular part responsible for turning.

A stopped parametrization, such as r(t)=(t³,0,0) at zero, makes the formula undefined even though the geometric track is a line. Do not assign a curvature using division by zero; discuss whether a regular alternative parametrization is available. Save a motion comparison with position, velocity, speed and curvature separately labeled. The next lesson studies surfaces through domains and level sets before differentiating functions of several independent variables.