About  Differential Geometry 

Math

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250 Homework #

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This homework assignment covers the notes 1-5. (Review of Linear Algebra, Arclength an

d

Rectifiability, Framed Curves and the Frenet Frame, More on Curvature and Torsion, and The
Bishop Frame). Until we are used to distinguishing between vector and scalar-valued functions,
we’re going to write vectors with an arrow over them (~α) and leave scalar variables alone (s)

.

1. (Derivatives and inverse functions.) Suppose that f(x) is a function, and g(x) is its inverse

function; that is, that
f(g(x)) = x,

Find a relationship between f ′ and g′ by differentiating both sides of the equation above.

2. (Reparametrizing by arclength.) Suppose that ~α(t) : R → R

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is a space curve. Let s(t) be the
arclength function

s(t) =

∫ t
0

|~α′(x)| dx.

a. Compute the derivative d
dt
s(t) using the Fundamental Theorem of Calculus.

b. Suppose that i(t) is the inverse function of s(t), so that i(s(t)) = t. Compute d
dt
i(t) using

the above (and the first problem).

c. Suppose that we let β(t) = α(i(t)). Compute
∣∣ d
dt
β(t)

∣∣ and conclude that β(t) is parametrized
by arclength.

3. (Hyperbolic trig functions) We are used to the usual trigonometric functions sin(x) and cos(x).
We’re now going to introduce two new functions: sinh(x) and cosh(x), defined by

coshx =
ex + e−x

2
, sinhx =

ex −e−x

2
.

a. Prove that cosh(0) = 1 and sinh(0) = 0.

b. Prove that d
dx

sinh(x) = cosh(x) and d
dx

cosh(x) = + sinh(x). (Notice that this is different
from the regular trig functions.)

c. Prove that cosh2(x)− sinh2(x) = 1 for all x.

d. Use Mathematica or Desmos to graph coshx and sinhx for x between 0 and 4.

e. Use Mathematica1 or Desmos2 to plot the parametrized curves

~α(t) = (cost,sint), ~β(t) = (cosht,sinht)

for −10 ≤ t ≤ 10. Can you recognize these curves? Why are cosht and sinht called
hyperbolic trigonometric functions?

1Use ParametricPlot.
2Any expression in the form (x(t),y(t)) will be plotted as a parametric curve in Desmos, but only if t is the

variable!.
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4. (Derivatives of Inverse Trig and Hyperbolic Trig Functions) Recall that the inverses of the trig
(and, as it turns out, the hyperbolic trig) functions are preceded by “arc”3. This means that

sin(arcsinx) = x, cos(arccosx) = x, sinh(arcsinhx) = x, cosh(arccoshx) = x.

We can compute the derivatives of the inverse trig functions and inverse hyperbolic trig func-
tions by using the inverse function/chain rule trick from Problem 1. For example:

d

dx
sin(arcsinx) =

d

dx
x

cos(arcsinx)
d

dx
arcsinx = 1

d

dx
arcsinx =

1

cos(arcsinx

)
.

At this point, we remember that cos2 x + sin2 x = 1, so cosx =
√
1− sin2 x and so we can

continue:

d
dx
arcsinx =

1√
1− sin2(arcsinx)

d
dx
arcsinx =

1
√
1−x2

.

a. For practice, repeat the above derivation to figure out d
dx

arccosx.

b. Make a similar argument to find out d
dx

arcsinhx and d
dx

arccoshx.

c. Now integrate both sides to express these derivatives as integrals. Hang on to them; you’ll
need them in a problem or two!

5. (Autonomous differential equations) Recall that a first order differential equation for an un-
known function u(t) is an equation in the form

u′(t) = F(u(t), t)

(where F is some function of u′(t) and t) and that the equation is autonomous if the right hand
side can be written only in terms of u(t) as

(?)

u′(t) = F(u(t))

3Why? Well arcsinx is the angle θ for which sinθ = x. And if you measure θ in radians, the angle θ is the angle
covered by an arc of length θ on the unit circle. So arcsinx literally means “the arc whose sine is x”.

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As you learned in MATH 2

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00, every autonomous first-order ODE may be solved as follows

u′(t) = F(u(t))

u′(t)

F(u(t))
= 1∫

u′(t)

F(u(t))
dt =

∫
1dt∫

1

F(u)
du = t + k

G(u) = t + k

where G(u) is any antiderivative of 1
F(u)

. If we can find an inverse function G−1(u) for G(u),
we can solve explicitly for u as

(♠) u(t) = G−1(t + k)
This is the most general solution to equation (?). If we know some value u(t0) = u0 we can
plug it in to (♠) to solve for k.
a. Use the method above to find the most general solution to

u′(t) = u(t)2.

b. Suppose u(t0) = u0. Plug it in to the results of (a) to write down the general solution in
terms of t0 and u0.

c. Suppose t0 = 0 and u0 = 2. Plug in to the results of (b) to write down the particular solution
with u(0) = 2.

6. (The Caternary Curve) Suppose we parametrize the graph of a function f(x) by ~α(t) = (t,f(t)).
a. Prove that the arclength formula is given by

s(t) =
∫ t
0

√
1 + (f ′(x))2 dx.

b. Show that if f(x) = coshx then s(t) = sinht.

c. Reparametrize the caternary curve by arclength, following the model you established in
Problem 2. Try to simplify the results as much as possible using hyperbolic trig identities.

7. (The planar operator ⊥) Suppose that ~v = (v1,v2) is vector in R2. We define the operator ⊥ by
~v⊥ = (−v2,v1).
a. Show that

⟨
~v,~v⊥

⟩
= 0, and hence that ~v and ~v⊥ are orthogonal to one another.

b. Show that
⟨
~v, ~w⊥

⟩
= |~v| |~w|sinθ, where θ is the angle between ~v and ~w.

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8. (The square-wheeled car) Consider the figure below, where a square with sidelength 2 is rolling
along a bumpy road.

We will parametrize the progress of the square by the arclength s it has rolled along the road
since starting in the horizontal position with center C(0) = O. Assume that the square rolls
smoothly, so that as a curve,

~C(s) = (c1(s),c2(s)) = (c1(s),0)

Assume that ~α(s) is an arclength parametrization of the road, so that ~P(s) = ~α(s) = (x(s),y(s)),
and assume that ~α(0) = (0,−1). Further, assume that ~Q(s) is always the midpoint of that edge
of the square. Since the square has rolled without slipping along the portion of the edge between
~Q(s) and ~P(s), we know that

∣∣∣~Q(s)− ~P(s)∣∣∣ = s, and that the edge of the square is tangent to
α(s) at the point of contact ~P(s). Assume that ~O(s) = ~0 is the origin for all t.

Note: The parameter s is an arclength parameter for ~α(s) = ~P(s) (only). That is, ~O(s),
~Q(s), and ~C(s) certainly trace out parametrized curves in the plane as the square wheel turns.
But we don’t have any reason to believe that those curves are arclength parametrized.

a. Write down a formula for ~P(s)− ~O(s) in terms of x(s), y(s), and s.

b. Write down a formula for ~Q(s)− ~P(s) in terms of x(s), y(s), and s.

c. Write down a formula for ~C(s)− ~Q(s) in terms of x(s), y(s), and s.

d. Add the results of (a)-(c) to find a formula for ~C(s)− ~O(s) = (c1(s),c2(s)) in terms of x(s),
y(s) and s. Compute the derivative c′2(s) = 0 to get a relationship between s, x

′′(s), and
y′′(s).

e. Remember that 1 = |~α′(s)|2 = x(s)2 + y(s)2. Differentiate both sides with respect to s to
get a relationship between x′(s), x′′(s), y′(s) and y′′(s).

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f. Combine the results of (d) and (e) to get a relationship between s, x′(s), and y′(s).

g. Now there is some function f so that y(s) = f(x(s)). Differentiating both sides with respect
to s), we know that

y′(s) = f ′(x(s))x′(s)

Use this and the results of (f) to find a formula for f ′(x) in terms of s.

h. Note that we can reparametrize ~α(s) as ~β(t) = (t,f(t)) as we did in (4). Further, remember
that s is the arclength along ~β(t) from 0 to x. Thus

s(x) =

∫ x
0

∣∣∣~β′(t)∣∣∣ dt
and so

s′(x) =
∣∣∣~β(x)∣∣∣

i. Differentiate the results of (g) by x (on both sides) to get an expression for f ′′(x) in terms of
s′(x), and use the results of (h) to write f ′′(x) = F(f ′(x)) for some function F . Solve this
autonomous differential equation for f ′(x) using the technique you reviewed in and the fact
that you know f ′(0). Remember Problem 4.

j. Integrate your formula for f ′(x) to get f(x). Use the fact that you know f(0) to eliminate
the constant of integration.

9. (Frenet Apparatus) Find the Frenet frame ~T(s), ~N(s), ~B(s), κ(s) and τ(s) for the arclength-
parametrized curve

~α(s) =

(
1

3
(1 + s)3/2,

1

3
(1−s)3/2,

1
√
2
s

)
.

10. (The Darboux Vector) If γ(s) is an arclength-parametrized curve with nonzero curvature, find
a vector ω(s), expressed as a linear combination of T , N, and B so that

T ′(s) = ω(s)×T(s)
N ′(s) = ω(s)×N(s)
B′(s) = ω(s)×B(s)

This vector is called the Darboux vector. Find a formula for the length of the Darboux vector
in terms of the curvature κ(s) and torsion τ(s) of the curve.

11. (The curvature of the graph of a function) Prove that a plane curve ~α(t) = (t,f(t),0) has
curvature

κ(t) =
|f ′′(t)|

(1 + f ′(t)2)3/2

You’ll need the formula

(♣) κ(t) =
|~α′(t)×~α′′(t)|
|~α′(t)|3

from the notes “More on Curvature and Torsion”.
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12. (The torsion of non arclength-parametrized curve) Reread the proof of (♣) in the notes “More
on Curvature and Torsion”. Now make the same kind of argument to show that

τ(t) =
〈~α′(t),~α′′(t)×~α′′′(t)〉
|~α′(t)×~α′′(t)|

for any space curve ~α(t), regardless of whether or not ~α(t) is parametrized by arclength.

1. CHALLENGE PROBLEMS

1. (The Shortest Curve Between Points) Let α: I → R3 be a a differentiable parametrized curve.
Suppose [a,b] ∈ I and α(a) = p while α(b) = q.

a. Show that for any constant vector v with |v| = 1,

〈q −p,v〉 =
∫ b
a

〈α′(t),v〉 dt ≤
∫ b
a

|α′(t)|dt.

b. Let

v =
q −p
|q −p|

and show that

|α(b)−α(a)| ≤
∫ b
a

|α′(t)|dt.

That is, the curve of shortest length joining two points is the straight line!

2. (Constant Breadth) A closed planar curve ~α(s) is said to have constant breadth if the distance
between parallel tangent line of ~α(s) is always µ. A circle is an example of such a curve, but
it’s not the only one:

We’ll assume that ~α(s) : [0,L] → R2 is an arclength parametrization of ~α(s) with ~α(0) = ~α(L)
(and all derivatives of ~α(s) equal at 0 and L as well. We’ll also assume that κ(s) 6= 0.

a. Let’s call two points with parallel tangent lines opposite points. Suppose that ~β(s) is always
the opposite point to ~α(s). (Note that just because s is an arclength parameter for ~α(s)
doesn’t mean that s is an arclength parameter for ~β(s); it’s probably not.) Since the tangent
and normal vectors ~T(s), ~N(s) at ~α(s) are a basis for the plane, there are some coefficients
c1(s) and c2(s) so that

~β(s)−~α(s) = c1(s)~T(s) + c2(s) ~N(s).

Prove that c2(s) = µ and then prove that c1(s) = 0. Conclude that the chord joining opposite
points is normal to the curve at both ends.

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b. Let κα(s) be the curvature of ~α at s, and κβ(s) be the curvature of ~β at s. Prove that
1

κα(s)
+

1

κβ(s)
= µ

Hint: Let ~Tβ(s) and ~Nβ(s) denote the tangent and normal vectors to ~β(s). How are they
related to ~T(s) and ~N(s)? Can you use (♣) to compute κβ(s)?

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• 1. Challenge Problems