Curvature of Plane Curves What is arc length parametrization? Let 2 γ :[a, b] ⊆ R → R , t 7→ γ(s) = (x(s), y(s)) be a nice curve. We define its arc length from t = a to t = b to be

Z b ||γ0(t)||dt. a We say that γ(s) is an arc length parametrization provided ||γ0(s)|| ≡ 1. We also call such a parametrized curve a unit speed curve. Let c : R → R2, t 7→ c(t) be a curve We can always (assuming that ||c0(t)|| never equals 0) construct an arc length parametrization for a curve. Define a function Z τ=t 0 s : R → R, t 7→ s, t 7→ ||c(τ) ||dτ. τ=a Observe that ds = ||c0(t)||. ( ) dt ♦ Define 2 γ : R → R , s → γ(s) by γ ◦ s = c. We have the diagram

R A A c s AA AA A  2 R γ / R Using the chain rule we have ds dc γ0(s) · = . dt dt dc ds 0 From equation ♦ we have that || dt || = dt . Hence ||γ || ≡ 1. We say that a curve γ(s) that is an arclength parametrization is a unit speed curve. We have that the length of the curve from γ(a) to γ(b) is b − a for any a, b > a. Definition of Curvature We start with an arclength parametrized curve γ, so that ||γ0(s)|| ≡ 1. Hence we can write γ0(s) = (cos(θ(s)), sin(θ(s))) = t(s). The curvature is how fast the direction of the tangent vector changes, that is, the curvature is

κ(s) = θ0(s).

Notice that γ00(s) = θ0(s)(− sin(θ(s), cos(θ(s)) = θ0(s)n(s).

1 and that n(s) = (− sin(θ(s), cos(θ(s)) is a normal to the curve so that the two unit vectors

(n(s), t(s)) form an orietnted basis of R2. From this we can see the difference between curvature that is positive and curvature that is negative. See the figure. Observe that the curvature is the area of the rectangle spanned by γ0(t) and γ00(t). Hence  γ0(s)  κ(s) = det . gamma00(s)

A formula for the curvature of an arbitrarily (not arclength) parmetrized curve. Given c(t) we can theoretically construct a corresponding arc length parametrized curve γ(s). But only theoretically. How do we explicitly compute the curvature of c(t) in terms 0 00 ds of c (t), c (t), dt ? We use the diagram

RO A A c s AA AA A  2 R γ / R The chain rule tells how to compute derivatives of c in terms of the derivatives of s and dγ dγ ds dc ds . We solve this relation to obtain ds in terms of dt and dt . We apply this idea a second d2γ d2c dc ds time to obtain ds2 in terms of dt2 , dt , dt . Applying the chain rule to γ ◦ s = c we obtain dγ ds dc = . (?) ds dt dt Differentiating this with respect to t, we obtain

ds 2 d2s d2c γ00(s) + γ0(s) = . dt dt2 dt2 Using the relation replace γ0(s) with a multiple of c0(t) we find we have

ds 2 dc d2c γ00(s) + α = . dt dt dt2 Solve for γ00(s) gives d2c dc 2 − β γ00(s) = dt dt ds 2 ( dt ) for an appropriate constant β.

2 We use the determinant formula for curvature. This gives

 c0(t)   ds   0     −3  0  γ (s)  dt  ds c (t) κ(t) = 00 = det   = det 00 . γ (s)  00 0  dt c (t) c (t) − βc (t)  ds 2  ( ) dt If c(t) = (x(t), y(t), this gives x0y00 − x00y0 κ = . (x02 + y02)3/2 Specializing more to the case of a curve y = f(x) which we parametrized by (x(t) = t, y(t)) we get y00 κ = , (1 + y02)3/2

00 d2y . but y = dx2 .

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