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 γ 0 (t).
a. Show that γ is a differentiable parametrized curve, regular except at t = π2 .
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:
a.) γ 0 (t) = (cos t, − sin t +
i.e. when t = π2 .
1
).
sin t
So it is differentiable at (0, π). It is zero if and only cos t = 0,
b.) The tangent line at γ(t) is
t
cos t
y − cos t − ln(tan ) =
(x − sin t).
2
sin t
So the intersection with y axis is (0, ln(tan 2t )). 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 bp
r2 + (rθ )2 dθ,
a
where rθ denote the derivative relative to θ.
b. Show that the curvature is
κ(θ) =
r2 + 2rθ2 − rrθθ
3
(r2 + rθ2 ) 2
c. Calculate the curvature of Archimedes’ spiral r = cθ.
.
Solution:
a.) x = r cos θ, y = r sin θ.
So γθ = (xθ , yθ ) = (−r sin θ + rθ cos θ, r cos θ + rθ sin θ). So x2θ + yθ2 = r2 + (rθ )2 . Hence the arc
length is
Z bp
Z bq
2
2
xθ + yθ dθ =
r2 + (rθ )2 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
for plane curves κ = |ytt x2t −x2tt3yt | , we have κ(θ) =
(xt +yt ) 2
(−r sin θ + 2rθ cos θ + rθθ sin θ)(−r sin θ + rθ cos θ) − (−r cos θ − 2rθ sin θ + rθθ cos θ)(r cos θ + rθ sin θ)
3
(r2 + (rθ )2 ) 2
which is
r2 + 2rθ2 − rrθθ
3
(r2 + rθ2 ) 2
.
c.)
κ(θ) =
θ2 + 2
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 sp
(ẋ)2 + (ẏ)2 ds =
ds = s − a.
a
a
Further,
κs (s) =
dθ
= h(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
,
4. Let γ : (a, b) → R2 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
κ of γ at t0 satisfies |κ(t0 )| ≥ ||γ(t10 )|| .
Solution: Without loss of generality, we could assume t is a unit speed parametrization.
||γ(t)||2 is maximal at t0 , so
(γ(t) · γ(t))0 |t=t0 = 0
and
(γ(t) · γ(t))00 |t=t0 < 0.
These mean
γ(t0 ) · γ 0 (t0 ) = 0
(1)
γ 0 (t0 ) · γ 0 (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 γ 0 (t) · γ 0 (t) = 1.
||γ(t0 )||
5. Prove that if a curve γ(s) lies on a unit sphere, then the following equality holds:
(κ0 )2 = κ2 τ 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) · γ 00 (s) + γ 0 (s) · γ 0 (s) = 0,
i.e.
γ(s) · n = −
1
.
κ(s)
Take one more derivative for (4),
κ0
1 0
γ(s) · (−κt + τ b) + t(s) · n(s) = (− ) = 2 .
κ
κ
3
(4)
Notice γ(s) · t(s) = 0 and t(s) · n(s) = 0, we have
γ(s) · b =
κ0 1
·
κ2 τ
(4) and (5) imply
κ0 1
1
γ(s) = − n + 2 · b.
κ
κ τ
Plug into (3),
κ0 2 1
1 2
( ) + ( 2 ) · 2 = 1.
κ
κ
τ
This is
(κ0 )2 = κ2 τ 2 (κ2 − 1).
4
(5)