Signed curvature and turning angle

Try

For γ(t)=(t,t²), the tangent at zero is horizontal and the curve bends upward. Its signed curvature there is two when the plane has its usual counterclockwise orientation. Traversing the same parabola in reverse changes the sign but not the magnitude.

Curvature measures change of direction per unit distance, not acceleration per unit time. A vehicle speeding up on a straight line has acceleration but zero geometric curvature. The tangential part of acceleration must be separated from the bending part.

Predict

Predict the curvature of a circle of radius R. A full circuit turns the tangent through 2π while traveling 2πR, suggesting 1/R. Smaller circles bend more sharply.

Predict whether an inflection point can occur on a regular curve. Yes: velocity can stay nonzero while the tangent’s rate of turning vanishes. Regularity and nonzero curvature are different conditions.

Experiment

Compute exact curvature for the parabola.

Parameter t	Speed	Signed curvature
0	1	2
1/2	√2	1/√2
1	√5	2/(5√5)

Use γ′=(1,2t) and γ″=(0,2). The vertical second derivative stays two, yet curvature decreases because the tangent direction changes more slowly per unit distance. Explain why using only y″ would miss the speed correction.

Understand

For a unit-speed plane curve let T be its unit tangent and let J rotate vectors counterclockwise by π/2. Since T·T=1, differentiation gives T′·T=0. Therefore T′=kJT for a signed scalar k. Locally write T=(cos θ,sin θ); differentiation gives k=θ′, where the derivative is with respect to arc length.

For a general regular parameter, write v=||γ′|| and γ′=vT. Differentiating gives γ″=v′T+v²kJT, because ds/dt=v. Taking the oriented determinant of γ′ and γ″ yields v³k. Thus k=det(γ′,γ″)/||γ′||³. The unsigned curvature is |k|.

For γ(t)=(R cos t,R sin t), the determinant is R² and speed is R, giving k=1/R. Reversing traversal reverses the determinant sign but not speed. If k vanishes identically on a connected interval, then the unit tangent is constant and integration places the curve on a straight line.

Apply

The osculating circle at a point with k≠0 has signed center γ+(1/k)JT and radius 1/|k|. It matches position, tangent and second-order bending under arc length. This is a local approximation, not a claim that a finite segment of the curve lies on that circle.

Repair “Curvature zero at one point proves the curve is a line.” The theorem requires zero curvature throughout an interval. The graph y=x³ has zero curvature at zero and bends elsewhere.

For optional depth, a tangent-angle lift exists locally because the circle of unit directions has local angle coordinates. On a full closed curve, the angle may return shifted by an integer multiple of 2π; that global winding will be studied separately.

A global tangent lift distinguishes closure from one traversal

Let γ be a regular C² closed plane curve with matching endpoint tangents. On its compact parameter interval the unit tangent T is continuous. Subdivide the interval finely enough that on each piece its image lies in an open semicircle. Such a subdivision exists by uniform continuity. Each semicircle has a continuous real angle coordinate. At a shared endpoint, add an integer multiple of 2π to the next coordinate so it agrees with the previous one. This constructs a continuous angle lift θ on the whole interval; since T is C¹, the locally defined lift is C¹ as well.

Because T has equal endpoint values, θ(b)−θ(a)=2πm for an integer m. Locally dθ/ds=k, so integrating gives total signed curvature ∫k ds=2πm. Changing the initial lift adds a constant multiple of 2π and does not change m. This proves integrality for a regular closed curve, not that m must be 1 or −1. The latter assertion needs the separate simple-closed-curve turning theorem.

Worked retracing case. For γ(t)=(cos(2t),sin(2t)), 0≤t≤2π, speed is 2, signed curvature is 1 and tangent angle may be chosen as 2t+π/2. The total signed curvature is 4π and m=2. The image is a circle, but the parametrized curve traverses it twice. Looking only at the set of plotted points loses that multiplicity.

Changed written task. Traverse a radius-3 circle clockwise once. Compute speed, signed curvature, length and both curvature integrals. Feedback: using γ(t)=(3cos t,−3sin t), speed is 3, signed curvature −1/3, length 6π, signed total −2π and absolute total 2π. More generally, |∫k ds|≤∫|k|ds gives a lower bound 2π|m|. It does not supply the full total-curvature theorem when m=0. Corners require additional exterior-angle contributions and are outside this C² proof.

Predict, change the case, and repair the claim

Case Two traversalsReverse traversal

γ(t)=(cos(2t),sin(2t)), 0≤t≤2π.

What is the signed total curvature?

2π 4π 0

Check this claim Reset practice

Before checking, write the hypothesis or calculation that decides your answer. Change cases and explain which assumption changed. These fixed cases are public guided practice, not a saved score or independent proof assessment.

Continue

In a curvature report state the plane orientation, traversal direction and whether the value is signed. Continue to space curves and Frenet frames to distinguish bending within a plane from twisting out of it.