Difficulty: Intermediate | Prerequisites: Arc length and reparametrisation (first half of Ch. 13.3), unit tangent vector T(t) from Ch. 13.2, cross product.
Arc length tells you how far along a curve you have gone. Curvature tells you how sharply the curve is bending at each point. This section defines curvature rigorously, gives you three equivalent formulas for computing it, and introduces the unit normal vector N, which points in the direction the curve is turning. Together with the unit tangent vector T from Ch. 13.2, these tools set up the TNB frame you will use in Ch. 13.4.
Curvature κ measures how fast a curve changes direction per unit distance. A straight line has curvature 0, and a circle of radius R has curvature 1/R. The unit normal vector N points perpendicular to the curve in the direction it is turning.
Curvature (κ, "kappa")
The magnitude of the rate of change of the unit tangent vector with respect to arc length: κ = |dT/ds|. In simple terms, curvature measures how much the curve bends per unit of distance travelled. High curvature means a tight turn, low curvature means a gentle sweep.
Unit tangent vector T(t)
The normalised derivative of the position vector: T(t) = r'(t) / |r'(t)|. Think of it as the direction you are facing as you walk along the curve.
Unit normal vector N(t)
The normalised derivative of the unit tangent: N(t) = T'(t) / |T'(t)|. It points perpendicular to T in the direction the curve is turning. Think of it as the direction the steering wheel is pulling you.
Osculating circle
The circle that best fits the curve at a given point. It has the same tangent, the same curvature, and its centre lies along the unit normal vector. Its radius equals 1/κ. In simple terms, if you zoomed in enough, the curve and the osculating circle would look the same.
Imagine walking along a curve at a constant speed of 1 (i.e. parametrised by arc length). Curvature measures how fast your direction changes as you move.
Formally: κ = |dT/ds|, where T is the unit tangent vector and s is arc length.
A straight line has κ = 0 everywhere (your direction never changes). A circle bends at a constant rate.
Definition form (conceptual, rarely used for computation): κ = |dT/ds|
Better for computation: κ = |T'(t)| / |r'(t)|
This avoids needing to reparametrise by arc length.
Easiest in R³ (cross product formula): κ = |r'(t) × r''(t)| / |r'(t)|³
This is the go-to formula for most exam problems in three dimensions. You only need r'(t) and r''(t), then take the cross product.
Let r(t) = ⟨R cos t, R sin t, 0⟩ (a circle of radius R in the xy-plane).
r'(t) = ⟨−R sin t, R cos t, 0⟩, so |r'(t)| = R.
r''(t) = ⟨−R cos t, −R sin t, 0⟩.
Cross product: r'(t) × r''(t) = ⟨0, 0, R² sin²t + R² cos²t⟩ = ⟨0, 0, R²⟩.
Magnitude of the cross product: |r'(t) × r''(t)| = R².
Curvature: κ = R² / R³ = 1/R.
Result: the curvature of a circle of radius R is 1/R. Smaller circles curve more sharply.
Because the curvature of a circle of radius R is 1/R, you can estimate the curvature at any point on a curve by fitting a circle to the curve at that point.
A point where the curve bends tightly corresponds to a small osculating circle (large κ). A nearly flat section corresponds to a large osculating circle (small κ).
N(t) = T'(t) / |T'(t)|.
N is orthogonal to T. This is not an accident; it follows from the fact that T is a unit vector.
Proof: T · T = 1. Differentiate both sides with the product rule: T' · T + T · T' = 0, so 2(T · T') = 0. Therefore T and T' are orthogonal, and normalising T' gives a unit vector perpendicular to T.
Students sometimes assume that the direction of T'(t) is the same as the direction of r''(t), and try to compute N as r''(t) / |r''(t)|. This is wrong.
The reason: T(t) = r'(t) / |r'(t)| is a quotient. Differentiating it with the quotient rule produces extra terms, so T'(t) and r''(t) generally point in different directions.
Always compute T(t) first, then differentiate T(t) to get T'(t), then normalise.
\kappa = \left|\frac{d\mathbf{T}}{ds}\right| = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}Osculating circle radius = 1/κ.
Curvature determines how roads and railways are banked: a tighter curve (higher κ) needs a steeper bank angle to keep vehicles safe. In computer graphics and font design, curvature continuity between joined curve segments is what makes a shape look smooth rather than kinked.
Students often confuse curvature with the second derivative r''(t). Curvature is a scalar that depends on both r' and r''. A curve can have a large r'' but low curvature if it is also moving fast.
A common error is computing N(t) as r''(t)/|r''(t)| instead of T'(t)/|T'(t)|. These are different because T(t) = r'(t)/|r'(t)| involves a quotient, and its derivative is not simply r''(t)/|r'(t)|.
Students sometimes think curvature depends on how fast you traverse the curve. It does not. Curvature is a geometric property of the curve itself, independent of parametrisation.
A circle of radius 5 has curvature 1/5 everywhere. Students sometimes write 5 instead of 1/5. Remember: bigger radius means gentler bending, so smaller curvature.
⚠️ The cross product formula κ = |r' × r''| / |r'|³ is the workhorse for exam computation. Memorise it and practise the cross product step.
⚠️ Know the curvature of a circle of radius R (κ = 1/R). This result often appears as a quick standalone question or as a sanity check.
⚠️ Computing N(t) requires differentiating T(t), not r(t). Do not skip the intermediate step of finding T(t) first.
⚠️ The proof that T and T' are orthogonal (differentiating T · T = 1) is a classic "show that" exam problem.
Fill in the blank: The curvature of a circle of radius 3 is ___. (1/3)
True or False: N(t) = r''(t) / |r''(t)|. (False, N = T'/|T'|)
True or False: Curvature is always non-negative. (True)
Fill in the blank: If T · T = 1, then differentiating gives 2(T · T') = ___, proving T and T' are ___. (0, orthogonal)
True or False: A straight line has curvature 1. (False, it has curvature 0)
Q: Find the curvature of r(t) = ⟨t, t², 0⟩ at the point where t = 0.
A: r'(t) = ⟨1, 2t, 0⟩ and r''(t) = ⟨0, 2, 0⟩. At t = 0: r'(0) = ⟨1, 0, 0⟩ and r''(0) = ⟨0, 2, 0⟩. Cross product: r' × r'' = ⟨0, 0, 2⟩, so |r' × r''| = 2. Also |r'(0)| = 1. Therefore κ = 2/1³ = 2.
Q: Show that T(t) and T'(t) are orthogonal for any smooth curve.
A: T · T = 1 for all t. Differentiate both sides: T' · T + T · T' = 0, so 2(T · T') = 0, which gives T · T' = 0.
Q: A curve has curvature κ = 4 at a point. What is the radius of the osculating circle there?
A: Radius = 1/κ = 1/4.
Q: Using the cross product formula, find the curvature of the helix r(t) = ⟨3 cos t, 3 sin t, 4t⟩.
A: r'(t) = ⟨−3 sin t, 3 cos t, 4⟩, |r'(t)| = √(9 + 16) = 5. r''(t) = ⟨−3 cos t, −3 sin t, 0⟩. The cross product r' × r'' = ⟨12 sin t, −12 cos t, 9⟩, so |r' × r''| = √(144 + 9) = √153. Then κ = √153 / 125. (Simplifying: √153 = 3√17, so κ = 3√17/125.)
Curvature and N(t) combine with the binormal vector B(t) = T(t) × N(t) to form the TNB (Frenet-Serret) frame in Ch. 13.4, which describes a curve's twisting in addition to its bending. The idea of fitting a circle to a curve at a point reappears in differential geometry and in approximation theory (osculating circles are the curvature analogue of tangent lines). In physics, κ enters the formula for the normal component of acceleration: aₙ = κ|v|².
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