MA3D9 Example Sheet 1

Without otherwise mentioned, all curves are smooth and regular.

1. A tractrix γ : (0, π) → R2 is given by t γ(t) = (sin t, cos t + ln(tan )), 2 where t is the angle that the y axis makes with the vector γ0(t). π a. Show that γ is a diﬀerentiable parametrized curve, regular except at t = 2 . b. The length of the segment of the tangent of the tractrix between the point of tangency and the y axis is constantly equal to 1.

Solution: 0 1 a.) γ (t) = (cos t, − sin t + sin t ). So it is diﬀerentiable at (0, π). It is zero if and only cos t = 0, π i.e. when t = 2 . b.) The tangent line at γ(t) is t cos t y − cos t − ln(tan ) = (x − sin t). 2 sin t

t So the intersection with y axis is (0, ln(tan 2 )). Hence the distance between the point and tangency to the intersection of tangent line with y axis is t t sin2 t + ((cos t + ln(tan )) − ln(tan ))2 = 1. 2 2

2. One often gives a plane curve in polar coordinates by r = r(θ), a < θ < b. a. Show that the arc length is Z b p 2 2 r + (rθ) dθ, a

where rθ denote the derivative relative to θ. b. Show that the curvature is 2 2 r + 2rθ − rrθθ κ(θ) = 3 . 2 2 2 (r + rθ ) c. Calculate the curvature of Archimedes’ spiral r = cθ. Solution: a.) x = r cos θ, y = r sin θ. 2 2 2 2 So γθ = (xθ, yθ) = (−r sin θ + rθ cos θ, r cos θ + rθ sin θ). So xθ + yθ = r + (rθ) . Hence the arc length is Z b q Z b 2 2 p 2 2 xθ + yθ dθ = r + (rθ) dθ. a a

b.) γθθ = (−r cos θ−2rθ sin θ+rθθ cos θ, −r sin θ+2rθ cos θ+rθθ sin θ). By our curvature formula |yttxt−xttyt| for plane curves κ = 3 , we have κ(θ) = 2 2 2 (xt +yt )

(−r sin θ + 2r cos θ + r sin θ)(−r sin θ + r cos θ) − (−r cos θ − 2r sin θ + r cos θ)(r cos θ + r sin θ) θ θθ θ θ θθ θ , 2 2 3 (r + (rθ) ) 2 which is 2 2 r + 2rθ − rrθθ 3 . 2 2 2 (r + rθ ) c.) θ2 + 2 κ(θ) = 3 c(1 + θ2) 2 3. Let h(s) be an arbitrary diﬀerentiable function on a segment (a, b). Then there is a plane curve γ for which h(s) is the signed curvature function and s is the arc length parameter. The curve is determined unique up to a translation and a rotation.

Solution: The functions x(s), y(s) and θ(s) satisfy the system of equations dx dy dθ = cos θ(s), = sin θ(s), = h(s). ds ds ds Solving this system, we get Z s Z s Z s θ(s) = θ0 + h(s)ds, x(s) = x0 + cos θ(s)ds, y(s) = y0 + sin θ(s)ds. 0 0 0 The obtained curve γ(s) = (x(s), y(s)) satisﬁes all conditions of the theorem. s is the arc length parameter since Z s Z s p(x ˙)2 + (y ˙)2ds = ds = s − a. a a Further, dθ κ (s) = = h(s). s ds

Notice that the curve γ(s) has initial point (x0, y0) and the tangent direction at this point has angle α0 with x-axis. Hence, if two curves have equal curvatures, then a rigid motion that matches their initial points and initial tangent vector also maps one curve to the other.

2 2 4. Let γ :(a, b) → R be a plane curve. Assume that there exists t0, a < t0 < b, such that the distance ||γ(t)|| from the origin to the curve will be a maximum at t0. Prove that the curvature 1 κ of γ at t0 satisﬁes |κ(t0)| ≥ . ||γ(t0)||

Solution: Without loss of generality, we could assume t is a unit speed parametrization. 2 ||γ(t)|| is maximal at t0, so 0 (γ(t) · γ(t)) |t=t0 = 0 and 00 (γ(t) · γ(t)) |t=t0 < 0. These mean

0 γ(t0) · γ (t0) = 0 (1)

0 0 γ (t0) · γ (t0) + κ(t0)γ(t0) · n(t0) < 0 (2)

Since (1), we have γ(t0) is parallel to n. So |γ(t0) · n(t0)| = ||γ(t0)||, and (2) implies |κ(t0)| ≥ 1 since γ0(t) · γ0(t) = 1. ||γ(t0)|| 5. Prove that if a curve γ(s) lies on a unit sphere, then the following equality holds:

(κ0)2 = κ2τ 2(κ2 − 1),

where κ and τ are the curvature and the torsion of the curve. (You don’t need to show: actually the inverse is also true.)

Solution: γ(s) lies on a unit sphere implies γ(s) · γ(s) = 1, (3) which implies γ(s) · t(s) = 0. Take one more derivative, we have

γ(s) · γ00(s) + γ0(s) · γ0(s) = 0,

i.e.

1 γ(s) · n = − . (4) κ(s)

Take one more derivative for (4),

1 κ0 γ(s) · (−κt + τb) + t(s) · n(s) = (− )0 = . κ κ2

3 Notice γ(s) · t(s) = 0 and t(s) · n(s) = 0, we have

κ0 1 γ(s) · b = · (5) κ2 τ

(4) and (5) imply 1 κ0 1 γ(s) = − n + · b. κ κ2 τ Plug into (3),

1 κ0 1 ( )2 + ( )2 · = 1. κ κ2 τ 2 This is (κ0)2 = κ2τ 2(κ2 − 1).