Suppose the principal curvatures at a point are one and four. A unit tangent halfway between their orthonormal directions has normal curvature 5/2. This is a weighted quadratic combination, not a linear interpolation of the tangent vectors themselves.
A curve on the surface has its own space curvature κ. Its normal acceleration relative to the surface is only one component of that curvature vector. The other component represents turning within the tangent plane, later called geodesic curvature.
Predict
Predict whether a saddle point can have a direction of zero normal curvature. Opposite principal signs allow cancellation. That does not mean every curve in that direction is a straight line in space; it can turn entirely within the tangent plane at that instant.
Predict whether normal curvature can exceed both principal values. A convex-combination formula will show that it lies between them.
Experiment
Use principal values one and four.
Angle θ from the first principal line
cos²θ, sin²θ
Normal curvature
0
1,0
1
π/4
1/2,1/2
5/2
π/2
0,1
4
Now replace the second value by minus four. Solve k_n=0 to obtain tan²θ=1/4. There are two distinct unoriented tangent lines satisfying this condition. Explain why angles differing by π describe the same line rather than two additional asymptotic directions.
Understand
Choose an orthonormal principal basis e₁,e₂. Write a unit tangent T=cos θ e₁+sin θ e₂. Since S is diagonal in that basis,
k_n=⟨ST,T⟩=k₁cos²θ+k₂sin²θ.
This is Euler’s formula. The weights are nonnegative and sum to one, so the principal values are the extreme normal curvatures. When they differ, equality occurs only on the corresponding principal lines.
For a unit-speed space curve with κ>0, let n be its Frenet principal normal and let α be the angle between n and the surface normal N. Since γ″=κn,
k_n=γ″·N=κcos α.
This is Meusnier’s relation in the stated angle convention. It compares two different normals and should not be read as κ=k_n unless they align appropriately.
At a point with K<0, relabel the principal values so k₁>0>k₂. Euler’s formula gives tan²θ=−k₁/k₂ for asymptotic directions. At a positive-curvature point the same-sign values cannot cancel; at a parabolic or planar point the count changes and must be considered separately.
Apply
Repair “Zero normal curvature means zero curve curvature.” A curve can have nonzero tangential acceleration perpendicular to its unit tangent while its normal component vanishes. On a plane every normal curvature is zero, yet a circle drawn in that plane has nonzero space curvature.
The bound |k_n|≤κ also follows from Meusnier because |cos α|≤1. It is a local necessary condition on a surface curve’s curvature, not a global existence theorem for a prescribed curve.
For optional depth, normal sections realize normal curvature by intersecting the surface with a plane containing N and the selected tangent direction, under suitable local regularity. Meusnier explains why other curves with that tangent can have larger space curvature.
Continue
Draw and label both the surface normal and the curve’s principal normal before using a curvature formula. Continue to the Gauss map to interpret the determinant K as an area-change rate of normal directions.
Apply · explain · revise
Practice and reasoning
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If k1=1,k2=5 and a unit tangent makes angle π/4 with the first principal direction, find normal curvature.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
At a point with K<0, how many unoriented asymptotic tangent lines are there?
Prove k_n(θ)=k1 cos²θ+k2 sin²θ in an orthonormal principal basis.
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For a unit-speed space curve on a surface with κ>0 and principal normal n, prove k_n=κ cos α where α is the angle between n and surface normal N.
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Derive the two asymptotic directions when k1>0>k2.
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Prove every normal curvature lies between the principal curvatures and identify equality cases when they differ.
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