Difficulty: Intermediate to Advanced Prerequisites: Arc length for space curves (Section 13.3 first half), chain rule, cross product, unit vectors.
Curvature measures how sharply a curve bends at a given point. A straight line has zero curvature everywhere; a tight circle has high curvature. This concept links the geometry of a curve to its calculus: you take derivatives of the position vector, build the unit tangent vector, and measure how fast that tangent direction is turning relative to distance travelled along the curve. The TNB frame (unit tangent, unit normal, binormal) then gives you a moving coordinate system that rides along the curve, which is essential for describing motion in physics.
Curvature kappa = |T'(t)| / |r'(t)|, or equivalently |r'(t) x r''(t)| / |r'(t)|^3. For a circle of radius r, kappa is always 1/r. The TNB frame consists of the unit tangent, the unit normal (direction the curve is turning), and the binormal (perpendicular to both).
Curvature (kappa, κ)
A scalar that measures how quickly the direction of a curve is changing per unit of arc length. In simple terms, it tells you how "bendy" the curve is at a particular point. A large kappa means a tight turn; kappa = 0 means the curve is locally straight.
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 heading at parameter value t, stripped of any information about speed.
Unit normal vector N(t)
The normalised derivative of T: N(t) = T'(t) / |T'(t)|. It points in the direction the curve is turning. In simple terms, if you are driving along the curve, N points toward the centre of the turn.
Binormal vector B(t)
B(t) = T x N. It is perpendicular to both the tangent and the normal, completing a right-handed coordinate frame that moves along the curve.
TNB frame (Frenet-Serret frame)
The set of three mutually perpendicular unit vectors {T, N, B} attached to a point on the curve. This moving frame is the natural coordinate system for describing motion along a space curve.
Radius of curvature
The reciprocal of curvature: R = 1/kappa. It equals the radius of the circle that best approximates the curve at that point. For a circle of radius r, the radius of curvature is simply r.
Curvature is defined as kappa = |dT/ds|, the magnitude of the rate of change of the unit tangent vector with respect to arc length. The problem is that curves are usually parametrised by t, not by s. The derivation below re-expresses kappa in terms of t.
From arc length, we know ds/dt = |r'(t)|. By the chain rule:
dT/dt = (dT/ds)(ds/dt)
Rearranging: dT/ds = (dT/dt) / (ds/dt) = T'(t) / |r'(t)|
Taking magnitudes: kappa = |dT/ds| = |T'(t)| / |r'(t)|
This is the first working formula for curvature. To use it, you need to find T(t) first (normalise r'), then differentiate T, then divide by |r'|.
The T'/|r'| formula can be awkward because normalising r' and then differentiating often produces messy quotients. An alternative derivation avoids T entirely:
Write r' = (ds/dt) T, then differentiate with the product rule: r'' = (d^2s/dt^2) T + (ds/dt) T'
Compute r' x r'' and use the fact that T x T = 0 and T is perpendicular to T'
After simplification: |r' x r''| = |r'|^2 |T'|
Therefore: kappa = |T'|/|r'| = |r' x r''| / |r'|^3
This second formula is often faster because you only need r' and r'', with no normalisation step.
For r(t) = <3 cos(t), 3 sin(t)>:
r'(t) = <-3 sin(t), 3 cos(t)>, so |r'(t)| = 3
T(t) = <-sin(t), cos(t)>
T'(t) = <-cos(t), -sin(t)>, so |T'(t)| = 1
kappa = 1/3
This confirms the general result: for any circle of radius r, kappa = 1/r. The curvature is constant and equals the reciprocal of the radius. A tighter circle (smaller r) has larger curvature.
Using the cross-product formula:
r'(t) = <1, 2t, 3t^2>, so r'(0) = <1, 0, 0>, |r'(0)| = 1
r''(t) = <0, 2, 6t>, so r''(0) = <0, 2, 0>
r'(0) x r''(0) = <0, 0, 2>, so |r'(0) x r''(0)| = 2
kappa = 2 / 1^3 = 2
Important: you must finish all differentiation before substituting t = 0. Plugging in t = 0 too early kills terms you still need.
The three vectors are defined as:
T(t) = r'(t) / |r'(t)| (unit tangent, direction of travel)
N(t) = T'(t) / |T'(t)| (unit normal, direction the curve is turning)
B(t) = T x N (binormal, perpendicular to the plane of T and N)
These three vectors form an orthonormal basis that rides along the curve. T points forward, N points toward the centre of curvature, and B points out of the plane in which the curve is bending.
This is a standard helix. The class notes ask you to compute T, N and B for this curve. The method is: differentiate to get r', normalise to get T, differentiate T to get T', normalise to get N, then cross T and N to get B.
\kappa = \left|\frac{d\vec{T}}{ds}\right| \quad \text{(definition)}\kappa = \frac{|\vec{T}'(t)|}{|\vec{r}'(t)|} \quad \text{(first working formula)}\kappa = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3} \quad \text{(cross-product formula, often faster)}\kappa = \frac{1}{r} \quad \text{(circle of radius } r \text{)}\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} \qquad \vec{N}(t) = \frac{\vec{T}'(t)}{|\vec{T}'(t)|} \qquad \vec{B}(t) = \vec{T} \times \vec{N}The cross-product formula is the one to reach for first on exams. It avoids the normalisation step that makes the |T'|/|r'| route messy.
Curvature determines how much a road must bank at a bend to keep vehicles from skidding. In roller-coaster design, engineers use curvature to control the g-forces passengers experience. In optics, the curvature of a lens surface determines its focal length.
Students sometimes confuse curvature with the second derivative f''(x). They are related for plane curves, but kappa involves a normalisation that f'' does not.
A common error is substituting a specific value of t (like t = 0) before completing all the differentiation steps. Derivatives must be finished first, then you plug in.
Students often think the cross-product formula only works in 3D. For a 2D curve, embed it in 3D with z = 0 and the formula still applies (the cross product just gives a vector in the z-direction).
Forgetting that T x T = 0 (the cross product of any vector with itself is the zero vector) is a frequent error in the derivation.
Expect at least one curvature computation on the final. Know both formulas and choose the cleaner one for the given problem.
The circle result kappa = 1/r is a useful sanity check: if your answer for a circle does not come out to the reciprocal of the radius, you have made an error.
"Find the curvature at a specific point" means finish all calculus, then substitute. This is a favourite exam trap.
Understanding how kappa = |dT/ds| is rewritten in terms of t demonstrates mastery of the chain rule and the Fundamental Theorem of Calculus, both of which appear elsewhere on the exam.
True or false: A straight line has curvature kappa = 1.
False. A straight line has kappa = 0.
Fill in the blank: For a circle of radius 5, kappa = _______.
1/5
True or false: T and T' are always perpendicular to each other.
True. Because T is a unit vector, its derivative is perpendicular to it.
Fill in the blank: The cross-product curvature formula is kappa = |r' x r''| / _______.
|r'|^3
True or false: You can substitute a specific t-value before differentiating when computing curvature.
False. All derivatives must be completed first.
Q: Find the curvature of a circle of radius 7.
A: kappa = 1/7. For any circle, kappa = 1/r.
Q: For r(t) =
A: r'(0) = <1, 0, 0>, r''(0) = <0, 2, 0>. The cross product is <0, 0, 2>, with magnitude 2. |r'(0)|^3 = 1. So kappa = 2.
Q: What is the unit tangent vector for r(t) = <3 cos(t), 3 sin(t)>?
A: r'(t) = <-3 sin(t), 3 cos(t)>, |r'(t)| = 3. So T(t) = <-sin(t), cos(t)>.
Q: Why does the cross-product formula avoid computing T(t) directly?
A: Because it works with r' and r'' only, skipping the normalisation step (dividing by |r'|) and its subsequent differentiation, which often produces complicated quotient-rule expressions.
Q: State the three vectors in the TNB frame and how each is computed.
A: T = r'/|r'| (normalise the velocity). N = T'/|T'| (normalise the derivative of T). B = T x N (cross product of the first two).
Curvature depends directly on the arc length material from the first half of Section 13.3, specifically the relationship ds/dt = |r'(t)|. The TNB frame reappears in Section 13.4 when decomposing acceleration into tangential and normal components (a_T and a_N). Curvature also connects to the second-derivative test for plane curves and to surface curvature in multivariable calculus (Sections 14-16).
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