Learning targets.
- Define curvature \(\kappa\) as a quantitative measure of turning.
- Construct the unit tangent \(T\) and show \(T\cdot T’=0\).
- Define arc length \(s\) and obtain the intrinsic formula \(\kappa = \left|\dfrac{dT}{ds}\right|\).
- Compute curvature for a line and for a circle (recovering \(\kappa=0\) and \(\kappa=1/R\)).
Board plan (five boxes).
- (B1) Regular curve \(r(t)\) and unit tangent \(T(t)\).
- (B2) Orthogonality: \(T\cdot T’=0\).
- (B3) Arc length: \(s(t)=\int_{t_0}^t |r'(u)|\,du\), hence \(ds/dt=|r'(t)|\).
- (B4) Curvature: \(\kappa=\left|dT/ds\right| = \left|T'(t)\right|/|r'(t)|\).
- (B5) Examples: line \(\kappa=0\); circle \(\kappa=1/R\).
Active learning rhythm.
- After each section: 30–60s “stop & check”.
- Do one complete proof in class (orthogonality of \(T\) and \(T’\)).
- Use reveal blocks for longer derivations.
Instructor pacing (suggested). 0–10 intuition; 10–25 unit tangent; 25–40 arc length; 40–55 curvature formula; 55–65 examples; 65–72 angle interpretation; 72–75 exit ticket.
Curvature \(\kappa\) measures how sharply a curve turns at a point.
Two benchmarks.
- Straight line: no turning, so \(\kappa=0\).
- Circle of radius \(R\): uniform turning; smaller \(R\) means sharper turning, so \(\kappa\) should scale like \(1/R\).
Setup. Let \(r:I\to\mathbb{R}^2\) or \(\mathbb{R}^3\) be a \(C^2\) curve.
Regular curve. The curve is regular if \(r'(t)\neq 0\) for all \(t\in I\). This ensures the direction of motion is well-defined.
Unit tangent. Define the unit tangent vector
\[ T(t) := \frac{r'(t)}{|r'(t)|}. \]
Then \(|T(t)|=1\) for all \(t\).
Key lemma. \(T(t)\cdot T'(t)=0\). Thus \(T'(t)\) is orthogonal to \(T(t)\).
Let \(T(t)=\dfrac{r'(t)}{|r'(t)|}\). Since \(T(t)\) is a unit vector, \(T(t)\cdot T(t)=|T(t)|^2=1\) for all \(t\).
Differentiate the identity \(T\cdot T=1\):
\[ \frac{d}{dt}(T\cdot T) = T’\cdot T + T\cdot T’ = 2\,T\cdot T’ = 0. \]
Hence \(T(t)\cdot T'(t)=0\) for all \(t\). Therefore the change in direction of \(T\) is orthogonal to \(T\) itself.
Stop & check (45s). Explain in one sentence why \(T’\perp T\) should be geometrically reasonable.
Curvature is a geometric property of the traced curve; it should not depend on how fast we traverse the curve.
Arc length function. Fix \(t_0\in I\). Define
\[ s(t) := \int_{t_0}^{t} |r'(u)|\,du. \]
Then \(\dfrac{ds}{dt} = |r'(t)|\).
Consequences. If \(r\) is regular, then \(ds/dt>0\), hence \(s(t)\) is strictly increasing and can be used as a new parameter (“arc-length parametrization”).
Definition (Curvature). For a regular \(C^2\) curve, define curvature by
\[ \kappa(t) := \left|\frac{dT}{ds}\right|. \]
Main identity.
\[ \kappa(t)=\frac{|T'(t)|}{|r'(t)|}. \]
Definition. Curvature is defined intrinsically by
\[ \kappa := \left|\frac{dT}{ds}\right|. \]
By the chain rule,
\[ \frac{dT}{ds} = \frac{dT/dt}{ds/dt} = \frac{T'(t)}{|r'(t)|}. \]
Taking norms gives
\[ \kappa(t)=\frac{|T'(t)|}{|r'(t)|}. \]
Computation-ready formulas.
For a \(C^2\) regular curve \(r(t)\) in \(\mathbb{R}^3\), one has the standard identity
\[ \kappa(t) = \frac{|r'(t)\times r”(t)|}{|r'(t)|^3}. \]
For a planar curve \(r(t)=(x(t),y(t))\), this becomes
\[ \kappa(t)=\frac{|x'(t)y”(t)-y'(t)x”(t)|}{\big(x'(t)^2+y'(t)^2\big)^{3/2}}. \]
These formulas are often the most convenient for computation.
Stop & check (60s). Why does curvature involve dividing by speed \( |r'(t)| \)?
Example (Line). Let \(r(t)=r_0 + tv\). Then \(r'(t)=v\) is constant, hence \(T\) is constant and \(T'(t)=0\). Therefore \(\kappa(t)=0\).
Example (Circle). Let \(r(t)=(R\cos t, R\sin t)\). Then
\[ r'(t)=(-R\sin t, R\cos t), \quad |r'(t)|=R. \]
Hence \(T(t)=(-\sin t,\cos t)\) and \(T'(t)=(-\cos t,-\sin t)\), so \(|T'(t)|=1\). Thus
\[ \kappa(t)=\frac{|T'(t)|}{|r'(t)|} = \frac{1}{R}. \]
In the plane, curvature is literally a “turning rate.”
Interpretation. If \(\varphi\) is the tangent angle, then curvature is the rate of change of that angle per unit arc length.
Assume \(r(t)\subset\mathbb{R}^2\). Let \(\varphi(s)\) be the angle the unit tangent makes with the positive \(x\)-axis. Then
\[ T(s)=(\cos\varphi(s),\,\sin\varphi(s)). \]
Differentiating with respect to \(s\) gives
\[ \frac{dT}{ds} = \varphi'(s)\,(-\sin\varphi(s),\,\cos\varphi(s)). \]
The vector \((-\sin\varphi,\cos\varphi)\) is a unit normal. Therefore
\[ \kappa(s)=\left|\frac{dT}{ds}\right| = |\varphi'(s)|. \]
With an orientation choice, one may define a signed curvature \(\kappa(s)=\varphi'(s)\).
Practical computation. For curves given parametrically, the cross-product formula is typically the most robust.
(* Mathematica: curvature for r(t) in R^3 *)
r[t_] := {a Cos[t], a Sin[t], b t};
kappa[t_] := Norm[Cross[D[r[t], t], D[r[t], {t, 2}]]]/Norm[D[r[t], t]]^3;
Plot[kappa[t], {t, -10, 10}]
(* Mathematica: curvature for planar r(t) = {x(t), y(t)} *)
r2[t_] := {t, t^2};
kappa2[t_] := Abs[Det[{D[r2[t], t], D[r2[t], {t, 2}]}]]/Norm[D[r2[t], t]]^3;
Plot[kappa2[t], {t, -2, 2}]Embedded lecture video. (VideoPress)
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Answer quickly.
- State the definition \(\kappa = |dT/ds|\).
- Why do we assume \(r'(t)\neq 0\)?
- What is \(\kappa\) for a circle of radius \(R\)?